Important Questions

Balbharti Maharashtra State Board 11th Physics Important Questions Chapter 1 Units and Measurements Important Questions and Answers.

Maharashtra State Board 11th Physics Important Questions Chapter 1 Units and Measurements

Question 1.
What is a measurement? How is measured quantity expressed?
Answer:

1. A measurement is a comparison with internationally accepted standard measuring unit.
2. The measured quantity (M) is expressed in terms of a number (n) followed by a corresponding unit (u) i.e., M = nu.

Example:
Length of a wire when expressed as 2 m, it means value of length is 2 in the unit of m (metre).

Question 2.
State true or false. If false correct the statement and rewrite. Different quantities are measured in different units.
Answer: True.
[Note: Choice of unit depends upon its suitability for measuring the magnitude of a physical quantity under consideration. Hence, we choose different scales for same physical quantity.]

Question 3.
Describe briefly different types of systems of units.
Answer:
System of units are classified mainly into four types:

1. C.G.S. system:
   It stands for Centimetre-Gram-Second system. In this system, length, mass and time are measured in centimetre, gram and second respectively.
2. M.K.S. system:
   It stands for Metre-Kilogram-Second system. In this system, length, mass and time are measured in metre, kilogram and second respectively.
3. F.P.S. system:
   It stands for Foot-Pound-Second system. In this system, length, mass and time are measured in foot, pound and second respectively.
4. S.I. system:
   It stands for System International. This system has replaced all other systems mentioned above. It has been internationally accepted and is being used all over world. As the SI units use decimal system, conversion within the system is very simple and convenient.

Question 4.
What are fundamental quantities? State two examples of fundamental quantities. Write their S.J. and C.G.S. units.
Answer:
Fundamental quantities:
The physical quantities which do not depend on any other physical quantity for their measurements i.e., they can be directly measured are called fundamental quantities.
Examples: mass, length etc.

Question 5.
What are fundamental units? State the S.l. units of seven fundamental quantities.
Answer:
Fundamental units:
The units used to measure fundamental quantities are called fundamental units.
S.I. Units of fundamental quantities:

Question 6.
State and describe the two supplementary units.
Answer:
The two supplementary units are:
i) Plane angle (dθ):
a. The ratio of kngth of arc (ds) of an circle to the radius (r) of the circle is called as Plane angle (dθ)
i.e., dθ = \(\frac{\mathrm{ds}}{\mathrm{r}}\)

b. Thus, dθ is angle subtended by the arc at the centre of the circle.
c. Unit: radian (rad)
d. Denoted as θ^c
e. Length of arc of circle = Circumference of circle = 2πr.
∴ plane angle subtended by entire circle at its centre is θ = \(\frac{2 \pi \mathrm{r}}{\mathrm{r}}\) = 2π^c

ii) Solid angle (dΩ):
a. solid angle is 3-dimensional analogue of plane angle.
b. Solid angle is defined as area of a portion of surface of a sphere to the square of radius of the sphere.
i.e., dΩ = \(\frac{\mathrm{d} \mathrm{A}}{\mathrm{r}^{2}}\)

c. Unit: Steradian (sr)
d. Denoted as (Ω)
e. Surface area of sphere = 4πr²
∴ solid angle subtended by entire sphere at its centre is Ω = \(\frac{4 \pi r^{2}}{r^{2}}\) = 4π sr

Question 7.
Derive the relation between radian and degree. Also find out 1” and 1’ in terms of their respective values in radian. (Take π = 3.1416)
Answer:
We know that, 2 π^c = 360°
∴ π^c = 180°

Question 8.
What are derived quantities and derived units? State two examples. State the corresponding S.L. and C.G.S. units of the examples.
Answer:

1. Derived quantities: Physical qUantities other than fundamental quantities which depend on one or more fundamental quantities for their measurements are called derived quantities.
2. Derived units: The units of derived quantities which are expressed in terms of fundamental units for their measurements are called derived units.
3. Examples and units:
   

Question 9.
Classify the following quantities into fundamental and derived quantities: Length, Velocity, Area, Electric current, Acceleration, Time, Force, Momentum, Energy, Temperature, Mass, Pressure, Magnetic induction, Density.
Answer:
Fundamental Quantities: Length, Electric current, Time, Temperature, Mass.

Derived Quantities: Velocity, Area, Acceleration, Force, Momentum, Energy. Pressure, Magnetic induction, Density

Question 10.
Classify the following units into fundamental, supplementary and derived units:
newton, metre, candela, radian, hertz. square metre, tesla, ampere, kelvin, volt, mol, coulomb, farad, steradian.
Answer:

Question 11.
List the conventions followed while using SI units.
Answer:
Following conventions should be followed while writing S.I. units of physical quantities:

1. Unit of every physical quantity should be represented by its symbol.
2. Full name of a unit always starts with smaller letter even if it is named after a person, eg.: 1 newton, 1 joule, etc. But symbol for unit named after a person should be in capital letter, eg.: N after scientist Newton, J after scientist Joule, etc.
3. Symbols for units do not take plural form.
4. Symbols for units do not contain any full stops at the end of recommended letter.
5. The units of physical quantities in numerator and denominator should be written as one ratio. For example the SI unit of acceleration is m/s² or m s^-2 but not m/s/s.
6. Use of combination of units and symbols for units is avoided when physical quantity is expressed by combination of two. For example, The unit J/kg K is correct while joule/kg K is not correct.
7. A prefix symbol is used before the symbol of the unit.
   • a. Prefix symbol and symbol of unit constitute a new symbol for the unit which can be raised to a positive or negative power of 10.
     For example,
     1 ms = 1 millisecond = 10^-3 s
     1 μs = 1 microsecond = 10^-6 s
     1 ns = 1 nanosecond = 10^-9 s
   • b. Use of double prefixes is avoided when single prefix is available
     10^-6 s = 1 μs and not 1 mms
     10^-9 s = 1 ns and not 1 mμs
8. Space or hyphen must be introduced while indicating multiplication of two units e.g., m/s should be written as m s^-1 or m-s^-1.

Solved Examples

Question 12.
What is the solid angle subtended by the moon at any point of the Earth, given the diameter of the moon is 3474 km and its distance from the Earth 3.84 × 10⁸ m?
Solution:
Given: Diameter (D) = 3474 km
∴ Radius of moon (R) = 1737 km
= 1.737 × 10⁶ m
Distance from Earth r = 3.84 × 10⁸ m
To find: Solid angle (dΩ)
Formula: dΩ = \(\frac{\mathrm{d} \mathrm{A}}{\mathrm{r}^{2}}\)

Calculation:
From formula,
dΩ = \(\frac{\pi \mathrm{R}^{2}}{\mathrm{r}^{2}}\) ……..( cross-sectional area of disc of moon = πR²)
dΩ = \(\frac{\pi \times\left(1.737 \times 10^{5}\right)^{2}}{\left(3.84 \times 10^{8}\right)^{2}}\)
= \(\frac{3.412 \times(1.737)^{2} \times 10^{10}}{(3.84)^{2} \times 10^{16}}\)
= antilog{log(3.142) + 2log(1.737) – 2log(3.84)} × 10^-6
= antilog {0.4972 + 2(0.2397) – 2(0.5843)} × 10^-6
= antilog{0.4972 + 0.4794 – 1.1686} × 10^-6
= antilog{\(\overline{1}\) .8080} × 10^-6
= 6.428 × 10^-1 × 10^-6
= 6.43 × 10^-5 sr
Solid angle subtended by moon at Earth is 6.43 × 10^-5 sr
[Note: Above answer is obtained substituting value of r as 3.142]

Question 13.
Pluto has mean diameter of 2,300 km and very eccentric orbit (oval shaped) around the Sun, with a perihelion (nearest) distance of 4.4 × 10⁹ km and an aphelion (farthest) distance of 7.3 × 10⁹ km. What are the respective solid angles subtended by Pluto from Earth’s perspective? Assume that distance from the Sun can be neglected.
Solution:
Given: Radius of Pluto. R = \(\frac{2300}{2}\) km
= 1150km
Perihelion distance r_p = 4.4 × 10⁹ km
Aphelion distance r_a = 7.3 × 10⁹ km
To find: Solid angles (dΩ_p and dΩ_a)

Solid angle at perihelion distance is 2.146 × 10^-13 sr and at aphelion distance is 7.798 × 10^-14 sr.

Question 14.
Define a metre.
Answer:
The metre is the length of the path travelled by light in vacuum during a time interval of 1/299, 792, 458 of a second.
Answer:

Question 15.
What ¡s parallax?
Answer:

1. Parallax is defined as the apparent change in position of an object due to a change in position of an observer.
2. Explanation: When a pencil is held in front of our eyes and we look at it once with our left eye closed and then with our right eye closed, pencil appears to move against the background. This effect is called parallax effect.

Question 16.
What is parallax angle?
Answer:
i) Angle between the two directions along which a star or planet is viewed at the two points of observation is called parallax angle (parallactic angle).

ii) It is given by θ = \(\frac{b}{D}\)
where, b = Separation between two points of observation.
D = Distance of source from any point of observation.

Question 17.
Explain the method to determine distance of a planet from the Earth.
Answer:

1. Parallax method is used to determine distance of different planets from the Earth.
2. To measure the distance ‘D’ of a far distant planet S, select two different observatories (E₁ and E₂).
3. The planet should be visible from E₁ and E₂ observatories simultaneously i.e. at the same time.
4. E₁ and E₂ are separated by distance ‘b’ shown in figure.
   ∴ E₁E₃ = b
   
5. The angle between the two directions along which the planet is viewed, can be measured. It is parallax angle, which in this case is L ∠E₁E₂ = θ
6. The planet is far away from the (Earth) observers, hence
   b < <D
   ∴ \(\frac{b}{D}\) < < 1 and ‘θ’ is also very small.
   Hence, E₁E₂ can be considered as arc of length b of circle with S as centre and D as radius.
   :. E₁S = E₂S = D
   ∴ θ = \(\frac{b}{D}\) . . . .(θ is taken in radian)
   ∴ D = \(\frac{b}{\theta}\)
   Thus, the distance ‘D’ of a far away planet ‘S’ can be determined using the parallax method.

Question 18.
Explain how parallax method is used to measure distance of a star from Earth.
Answer:

1. The parallax measured from two farthest distance points on Earth for stars will be too small and hence cannot be measured.
2. Instead, parallax between two farthest points (i.e., 2 ΔU apart) along the orbit of Earth around the Sun (s) is measured.
   

Question 19.
Explain how size of a planet or star is measured.
Answer:

1. To determine the diameter (d) of a planet or star, two diametrically opposite points of the planet are viewed from the same observatory.
2. If d is diameter of planet or star, angle subtended by it at any single point on the Earth is called angular diameter of planet.
3. Let angle α be angle between these two directions.
4. If distance between the Earth and planet or star (D) is known, α = \(\frac{\mathrm{d}}{\mathrm{D}}\)
5. This relation gives, d = α D
   Thus, diameter (d) of planet or star can be determined.

Question 20.
Name the devices used to measure very small distances such as atomic size.
Answer:
Devices used are:
Electron microscope, tunnelling electron microscope.

Question 21.
Just as large distances are measured in AU, parsec or light year, atomic or nuclear distances are measured with the help of microscopic units. Match the units given in column A with their corresponding SI unit given in column B.

Answer:
i. – (b)
ii. – (a)

Solved Examples

Question 22.
A star is 5.5 light years away from the Earth. How much parallax in arcsec will it subtend when viewed from two opposite points along the orbit of the Earth?

Solution:
Two opposite points-A and B along the orbit of the Earth are 2 AU apart. The angle subtended by AB at the position of the star is

= antilog{log(2.992) – log(5.5) – log(9.46)} × 10^-4
= antilog {0.4761 – 0.7404 – 0.9759} × 10^-4
= antilog {\(\overline{2}\).7598} × 10^-4
= 5.751 × 10^-2 × 10^-4
= 5.75 × 10^-6
= 5.75 × 10^-6 rad
= 5.75 × 10^-6 × 57.297 × 60 × 60 arcsec
…. (converting radian into arcsecond)
= 1.186 arcsec
Parallax is 1.186 arcsec

Question 23.
The moon is at a distance of 3.84 × 10⁸ m from the Earth. If viewed from two diametrically opposite points on the Earth, the angle subtended at the moon is 1° 54′. What is the diameter of the Earth?
Solution:
Given
Distance (D) = 3.84 × 10⁸ m
Subtended angle (α)
= 1° 54′ = (60’+ 54′)= 114′
= 114 × 2.91 × 10^-4 rad
= 3.317 × 10^-2 rad
To find: Diameter of Earth (d)
Formula: d = αD
Calculation: From formula,
d = 3.317 × 10^-2 × 3.84 × 10⁸
= 1.274 × 10⁷ m
Diameter of Earth is 1.274 × 10⁷ m.

Question 24.
Explain the method to measure mass.
Answer:
Method for measurement of mass:

1. Mass, until recently, was measured with a standard mass of the international prototype of the kilogram (a platinum-iridium alloy cylinder) kept at international Bureau of Weights and Measures, at Serves, near Paris, France.
2. As platinum – iridium piece was seen to pick up microparticles and found to be affected by atmosphere, its mass could no longer be treated as constant.
3. Hence, a new definition of mass was introduced in terms of electric current on 20th May 2019.
4. Now, one kilogram mass is described in terms of amount of current which has to be passed through electromagnet to pull one side of extremely sensitive balance to balance the other side which holds one standard kg mass.
5. To measure mass of small entities such as atoms and nucleus, atomic mass unit (amu) is used.
   It is defined as (\(\frac{1}{12}\))^th mass of an unexcited atom of carbon -12(C¹²).
   1 amu ≈ 1.66 × 10^-27 kg.

Question 25.
That can he the reason for choosing Carbon-12 to define atomic mass unit?
Answer:

1. Unlike oxygen and hydrogen, which exhibit various isotopes in higher proportions, carbon- 12 is the single most abundant (98% of available carbon) isotope of carbon.
2. it is also very stable.
   Hence, it makes more accurate unit of measuring mass and is used to define atomic mass unit.

Question 26.
Define mean solar day. Explain the method for measurement of time.
Answer:

1. A mean solar day is the average time interval from one noon to the next noon.
   Method for measurement of time:
2. The unit of time, the second, was considered to be \(\frac{1}{86400}\) of the mean solar day, where a mean solar day = 24 hours
   = 24 × 60 × 60
   = 86400 s
3. However, this definition proved to be unsatisfactory to define the unit of time precisely because solar day varies gradually due to gradual slowing down of the Earth’s rotation. Hence, the definition of second was replaced by one based on atomic standard of time.
4. Atomic standard of time is now used for the measurement of time. In atomic standard of time, periodic vibrations of caesium atom is used.
5. One second is time required for 9,192.631,770 vibrations of the radiation corresponding to transition between two hyperfine energy states of caesium-133 (Cs- 133) atom.

Question 27.
Define dimensions and dimensional formula of physical quantities. Give few examples of dimensional formula.
Answer:

1. Dimensions:
   The dimensions of a physical quantity are the powers to which the fundamental units must be raised in order to obtain the unit of a given physical quantity.
2. Dimensional formula:
   When any derived quantity is represented with appropriate powers of symbols of the fundamental quantities, SUCh an expression is called dimensional formula.
   It is expressed by square bracket with no comma in between the symbols.
3. Examples of dimensional formula:
   a. Speed = \(\frac{\text { Distance }}{\text { time }}\)
   ∴ Dimensions of speed = \(\frac{[\mathrm{L}]}{[\mathrm{T}]}\) = [L¹M⁰T^-1]
   [Note: As power of M is zero, it can be omitted from dimensional formula. Therefore, dimensions of speed can be written as [L¹T¹]
   

Question 28.
A book with many printing errors contains four different formulae for the displacement y of a particle undergoing a certain periodic function:
i) y = a sin \(\frac{2 \pi t}{T}\)
ii) y = a sin v t
iii) y = \(\frac{\mathrm{a}}{\mathbf{T}} \sin \frac{\mathrm{t}}{\mathrm{a}}\)
iv) y = \(\frac{a}{\sqrt{2}}\left[\sin \frac{2 \pi t}{T}+\cos \frac{2 \pi t}{T}\right]\)
Here, a is maximum displacement of particle, y ¡s speed of particle, T is time period of motion. Rule out the wrong formulae on dimensional grounds.
Answer:
The argument of trigonometrical function, i.e., angle is dimensionless. Now,
i) The argument, \(\left[\frac{2 \pi \mathrm{t}}{\mathrm{T}}\right]=\frac{[\mathrm{T}]}{[\mathrm{T}]}\) = 1 = [L⁰M⁰T⁰]
which is a dimensionless quantity.
Hence, formula (i) is correct.

ii) The argument,
[vt] = [LT^-1] [T] = [L] = [L¹M⁰T⁰]
which is not a dimensionless quantity.
Hence, formula (ii) is incorrect.

iii) The argument,
\(\left[\frac{\mathrm{t}}{\mathrm{a}}\right]=\frac{[\mathrm{T}]}{[\mathrm{L}]}\) = [L^-1M⁰T¹]
which is not a dimensionless quantity.
Hence, formula (iii) is incorrect.

iv) The argument,
\(\left[\frac{2 \pi \mathrm{t}}{\mathrm{T}}\right]=\frac{[\mathrm{T}]}{[\mathrm{T}]}\) = 1 = [L⁰M⁰T⁰]
which is a dimensionless quantity.
Hence, formula (iv) is correct.

Question 29.
State principle of homogeneity of dimensions.
Answer:
Principle of homogeneity of dimensions: The dimensions of all the terms on the two sides of a physical equation relating different physical quantities must be same.

Question 30.
State the uses of dimensional analysis.
Answer:
Uses of dimensional analysis:

To check the correctness of a physical equation.
Correctness of a physical equation by dimensional analysis:

1. A physical equation is correct only if the dimensions of all the terms on both sides of that equations are the same.
2. For example, consider the equation of motion.
   v = u + at ……………. (1)
3. Writing the dimensional formula of every term, we get
   Dimensions of LH.S. [v] [L¹M⁰T^-1],
   Dimensions of R.H.S. = [u] + [at]
   = [L¹M⁰T^-1] + [L¹M⁰T^-2] [L¹M⁰T^-1]
   = [L¹M⁰T^-1] + [L¹M⁰T^-1]
   ⇒ [L.HS.] = [R.H.S.]
4. As dimensions of both side of equation is same, physical equation is dimensionally correct.

To derive the relationship between related physical quantities.
Expression for time period of a simple pendulum by dimensional analysis:

1. Time period (T) of a simple pendulum depends upon length (l) and acceleration due to gravity (g) as follows:
   T ∝ l^a g^b
   i.e., T = k l^a g^b ………… (1)
   where, k = proportionality constant, which is dimensionless.
2. The dimensions of T = [L⁰M⁰T¹)
   The dimensions of l = [L¹M⁰T⁰]
   The dimensions of g = [L¹M⁰T²]
   Taking dimensions on both sides of equation (1),
   [L⁰M⁰T¹] = [L¹M⁰T⁰]^a [L¹M⁰T^-2]^b
   [L⁰M⁰T¹] = [L^a+bM⁰T^-2b]
3. Equating corresponding power of L, M and T on both sides, we get
   a + b = 0 …………. (2)
   and -2b = 1
   ∴ b = –\(\frac{1}{2}\)
4. Substituting ‘b’ in equation (2), we get
   a = \(\frac{1}{2}\)
5. Substituting values of a and b in equation (1),
   we have,
   
6. Experimentally, it ¡s found that k = 2π
   ∴ T = 2π \(\sqrt{\frac{l}{\mathrm{~g}}}\)
   This is the required expression for time period of a simple pendulum.

To find the conversion factor between the units of the same physical quantity in two different systems of units.
Conversion factor between units of same physical quantity:

1. let ‘n’ be the conversion factor between the units of work.
   ∴ 1 J = n erg ………….. (1)
2. Dimensions of work in S.l. system are \(\left[\mathrm{L}_{1}^{2} \mathrm{M}_{1}^{\prime} \mathrm{T}_{1}^{-2}\right]\) and in CGS system are \(\left[\mathrm{L}_{2}^{2} \mathrm{M}_{2}^{1} \mathrm{~T}_{2}^{-2}\right]\)
3. From (1),
   \(1\left[\mathrm{~L}_{1}^{2} \mathrm{M}_{1}^{1} \mathrm{~T}_{1}^{-2}\right]=\mathrm{n}\left[\mathrm{L}_{2}^{2} \mathrm{M}_{2}^{1} \mathrm{~T}_{2}^{-2}\right]\)
   
   n= 10⁴ × 10³ × 1 = 10⁷
   Hence, the conversion factor, n = 10⁷
   There fore, from equation (1), we have,
   ∴ 1 J = 10⁷ erg.

Question 31.
Explain the use of dimensional analysis to check the correctness of a physical equation.
Answer:
Correctness of a physical equation by dimensional analysis:

1. A physical equation is correct only if the dimensions of all the terms on both sides of that equations are the same.
2. For example, consider the equation of motion.
   v = u + at ……………. (1)
3. Writing the dimensional formula of every term, we get
   Dimensions of LH.S. [v] [L¹M⁰T^-1],
   Dimensions of R.H.S. = [u] + [at]
   = [L¹M⁰T^-1] + [L¹M⁰T^-2] [L¹M⁰T^-1]
   = [L¹M⁰T^-1] + [L¹M⁰T^-1]
   ⇒ [L.HS.] = [R.H.S.]
4. As dimensions of both side of equation is same, physical equation is dimensionally correct.

Question 32.
Time period of a simple pendulum depends upon the length of pendulum (l) and acceleration due to gravity (g). Using dimensional analysis, obtain an expression for time period of a simple pendulum.
Answer:
Expression for time period of a simple pendulum by dimensional analysis:
i) Time period (T) of a simple pendulum depends upon length (l) and acceleration due to gravity (g) as follows:
T ∝ l^a g^b
i.e., T = k l^a g^b ………… (1)
where, k = proportionality constant, which is dimensionless.

ii) The dimensions of T = [L⁰M⁰T¹)
The dimensions of l = [L¹M⁰T⁰]
The dimensions of g = [L¹M⁰T²]
Taking dimensions on both sides of equation (1),
[L⁰M⁰T¹] = [L¹M⁰T⁰]^a [L¹M⁰T^-2]^b
[L⁰M⁰T¹] = [L^a+bM⁰T^-2b]

iii) Equating corresponding power of L, M and T
on both sides, we get
a + b = 0 …………. (2)
and -2b = 1
∴ b = –\(\frac{1}{2}\)

iv) Substituting ‘b’ in equation (2), we get
a = \(\frac{1}{2}\)

v) Substituting values of a and b in equation (1),
we have,

vi) Experimentally, it ¡s found that k = 2π
∴ T = 2π \(\sqrt{\frac{l}{\mathrm{~g}}}\)
This is the required expression for time period of a simple pendulum.

Question 33.
Find the conversion factor between the S.I. and the C.OES. units of work using dimensional analysis.
Answer:
Conversion factor between units of same physical quantity:

1. let ‘n’ be the conversion factor between the units of work.
   ∴ 1 J = n erg ………….. (1)
2. Dimensions of work in S.l. system are \(\left[\mathrm{L}_{1}^{2} \mathrm{M}_{1}^{\prime} \mathrm{T}_{1}^{-2}\right]\) and in CGS system are \(\left[\mathrm{L}_{2}^{2} \mathrm{M}_{2}^{1} \mathrm{~T}_{2}^{-2}\right]\)
3. From (1),
   \(1\left[\mathrm{~L}_{1}^{2} \mathrm{M}_{1}^{1} \mathrm{~T}_{1}^{-2}\right]=\mathrm{n}\left[\mathrm{L}_{2}^{2} \mathrm{M}_{2}^{1} \mathrm{~T}_{2}^{-2}\right]\)
   
   n= 10⁴ × 10³ × 1 = 10⁷
   Hence, the conversion factor, n = 10⁷
   There fore, from equation (1), we have,
   ∴ 1 J = 10⁷ erg.

Question 34.
State the limitations of dimensional analysis.
Answer:
Limitations of dimensional analysis:

• The value of dimensionless constant can be obtained with the help of experiments only.
• Dimensional analysis cannot be used to derive relations involving trigonometric (sin θ, cos θ, etc.), exponential (e^x, e^x2, etc.), and logarithmic functions (log x, log x³, etc) as these quantities are dimensionless.
• This method is not useful if constant of proportionality is not a dimensionless quantity.
• If the correct equation contains some more terms of the same dimension, it is not possible to know about their presence using dimensional equation.

Question 35.
If two quantities have same dimensions, do they always represent the same physical content?
Answer:
When dimensions of two quantities are same, they do not always represent the same physical content.
Example:
Force and momentum both have same dimensions but they represent different physical content.

Question 36.
A dimensionally correct equation need not actually be a correct equation but dimensionally incorrect equation is necessarily wrong. Justify.
Answer:
i) To justify a dimensionally correct equation need not be actually a correct equation, consider equation, v² = 2as
Dimensions of L.H.S. = [v²] = [L²M⁰T²]
Dimensions of R.H.S. = [as]= [L²M⁰T²]
⇒ [L.H.S.] = [R.H.S.]
This implies equation v² = 2as is dimensionally correct.
But actual equation is, v² = u² + 2as
This confirms a dimensionally correct equation need not be actually a correct equation.

ii) To justify dimensionally incorrect equation is necessarily wrong, consider the formula,
\(\frac{1}{2}\) mv = mgh
Dimensions of L.H.S. = [mv] = [L¹M¹T^-1]
Dimensions of R.H.S. = [mgh] = [L²M¹T^-2]
Since the dimensions of R.H.S. and L.H.S. are not equal, the formula given by equation must be incorrect.
This confirms dimensionally incorrect equation is necessarily wrong.

Question 37.
State, whether all constants are dimensionless or unitless.
Answer:
All constants need not be dimensionless or unitless.
Planck’s constant, gravitational constant etc., possess dimensions and units. They are dimensional constants.

Solved Examples

Question 38.
If length ‘L’, force ‘F’ and time ‘T’ are taken as fundamental quantities, what would be the dimensional equation of mass and density?
Solution:
i) Force = Mass × Acceleration Force

∴ Dimensional equation of mass

= [F¹L^-4T²]

i) The dimensional equation of mass is [F¹L^-1T²].
ii) The dimensional equation of density is [F¹L^-4T²].

Question 39.
A calorie is a unit of heat and it equals 4.2 J, where 1 J = kg m² s^-2. A distant civilisation employs a system of units in which the units of mass, length and time are α kg, β m and δ s. Also J’ is their unit of energy. What will be the magnitude of calorie in their units?
Solution:
1 cal = 4.2 kg m² s^-2

New unit of energy is J’
Dimensional formula of energy is [L²M¹T^-2] According to the question,

Question 40.
Assume that the speed (v) of sound in air depends upon the pressure (P) and density (ρ) of air, then use dimensional analysis to obtain an expression for the speed of sound.
Solution:
It is given that speed (v) of sound in air depends upon the pressure (P) and density (ρ) of the air.
Hence, we can write, v = k P^a ρ^b ……….. (1)
where, k is a dimensionless constant and a and b are powers to be determined.
Dimensions of y = [L¹M⁰T^-1]
Dimensions of P = [L^-1M¹T^-2]
Dimensions of ρ = [L^-3M¹T⁰]
Substituting the dimensions of the quantities on both sides of equation (1),
∴ [L¹M⁰T^-1] = [L^-1M¹T^-2]^a [L^-3M¹T⁰]^b
∴ [L¹M⁰T^-1] = [L^-aM^aT^-2a] [L^-3bM^bT⁰]
∴ [L¹M⁰T^-1] = [L^-a-3bM^a+bT^-2a]
Comparing the powers of L, M and T on both sides, we get,
-2a = -1
∴ a = \(\frac{1}{2}\)
Also, a + b = O
∴ \(\frac{1}{2}\) + b = 0 b = – \(\frac{1}{2}\)
Substituting values of a and b in equation (1), we get
y = k P^{\(\frac{1}{2}\)} ρ^{–\(\frac{1}{2}\)}
∴ v = k \(\sqrt{\frac{\mathrm{p}}{\rho}}\)

Question 41.
Density of oil is 0.8 g cm³ in C.G.S. unit. Find its value in S.I. units.
Solution:
Dimensions of density is [L^-3M¹T⁰]

= 0.8 [10^-3] [10^-2]^-3
= 0.8 [10^-3] [10]⁶
n = 0.8 × 10³
Substituting the value of ‘n’ in equation (1).
we get, 0.8 g cm3 = 0.8 × 10³ kg m^-3.
Density of oil in S.l unit is 0.8 × 10³ kg m^-3.

Question 42.
The value of G in C.G.S system is 6.67 × 10^-8 dyne cm² g^-2. Calculate its value in S.l. system.
Solution:
Dimensional formula of gravitational constant
[L³M^-1T^-2]

n = 6.67 × 10^-8 × 10^-6 × 10³
n = 6.67 × 10^-11
From equation (1),
6.67 × 10^-8 dyne cm² g^-2
= 6.67 × 10^-11 N-m² kg^-2
Value ofG in S.l. system is 6.67 × 10^-11 N-m² kg^-2.

Question 43.
What is accuracy?
Answer:
Accuracy is how close a measurement is to the actual value of that quantity.

Question 44.
What is precision?
Answer:
Precision is a measure of how consistently a device records nearly identical values i.e., reproducible results.

Question 45.
A scale in a lab measures the mass of object consistently more by 500 g than their actual mass. How would you describe the scale in terms of accuracy and precision?
Answer:
The scale is precise but not accurate.
Explanation: Precision measures how consistently a device records the same answer; even though it displays the wrong value. Hence, the scale is precise.

Accuracy is how well a device measures something against its accepted value. As scale in the lab is always off by 500 g, it is not accurate.
[Note: The goal of the observer should be to get accurate as well as precise measurements.]

Question 46.
List reasons that may introduce possible uncertainties in an observation.
Answer:
Possible uncertainties in an observation may arise due to following reasons:

1. Quality of instrument used,
2. Skill of the person doing the experiment,
3. The method used for measurement,
4. External or internal factors affecting the result of the experiment.

Question 47.
What is systematic error? Classify errors into different categories.
Answer:

1. Systematic errors are errors that are not determined by chance but are introduced by an inaccuracy (involving either the observation or measurement process) inherent to the system.
2. Classification of errors:
   Errors are classified into following two groups:
3. Systematic errors:
   • Instrumental error (constant error),
   • Error due to imperfection in experimental technique,
   • Personal error (human error).
4. Random error (accidental error)

Question 48.
What is instrumental (constant) error?
Answer:
Instrumental error:

1. It arises due to defective calibration of an instrument.
2. Example: If a thermometer is not graduated properly, i.e., one degree on the thermometer actually corresponds to 0.99°, the temperature measured by such a thermometer will differ from its value by a constant amount.

Question 49.
What is error due to imperfection in experimental technique?
Answer:
Error due to imperfection in experimental technique:

• The errors which occur due to defective setting of an instrument is called error due to imperfection in experimental technique.
• For example the measured volume of a liquid in a graduated tube will be inaccurate if the tube is not held vertical.

Question 50.
What is personal error?
Answer:
Personal error (Human error):

• The errors introduced due to fault of an observer taking readings are called personal errors.
• For example, while measuring the length of an object with a ruler, it is necessary to look at the ruler from directly above. If the observer looks at it from an angle, the measured length will be wrong due to parallax.

Question 51.
What is random error (accidental)?
Answer:

1. Random error (accidental):
   The errors which are caused due to minute change in experimental conditions like temperature, pressure change in gas or fluctuation in voltage, while the experiment is being performed are called random errors.
2. They can be positive or negative.
3. Random error cannot be eliminated completely but can be minimized by taking multiple observations and calculating their mean.

Question 52.
State general methods to minimise effect of systematic errors.
Answer:
Methods to minimise effect of systematic errors:

1. By using correct instrument.
2. Following proper experimental procedure.
3. Removing personal error.

Question 53.
Define the term:
Arithmetic mean
Answer:
Arithmetic mean:
a. The most probable value of a large number of readings of a quantity is called the arithmetic mean value of the quantity. This value can be considered to be true value of the quantity.

b. If a₁, a₂, a₃, …………… a_n are ‘n’ number of readings taken for measurement of a quantity, then their mean value is given by,
a_mean = \(\frac{a_{1}+a_{2}+\ldots \ldots .+a_{n}}{n}\)
∴ a_mean = \(\frac{1}{n} \sum_{i=1}^{n} a_{i}\)

Question 54.
What does a = a_mean ± ∆a_mean signify?
Answer:
a = a_mean ± ∆ a_mean signifies that the actual value of a lies between (a_mean – ∆ a_mean) and (a_mean + ∆ a_mean).

Question 55.
What is meant by the term combination of errors?
Answer:
Derived quantities may get errors due to individual errors of fundamental quantities, such type of errors are called as combined errors.

Question 56.
Explain errors in sum and in difference of measured quantity.
Answer:
Errors in sum and in difference:
i) Suppose two physical quantities A and B have measured values A ± ∆A and B ± ∆B. respectively, where ∆A and ∆B are their mean absolute errors.

ii) Then, the absolute error ∆Z in their sum.
Z = A + B
Z ± ∆Z = (A ± ∆A) + (B ± ∆B)
= (A + B) ± ∆A ± ∆B
∴ ± ∆Z = ± ∆A ± ∆B.

iii) For difference. i.e.. if Z = A – B.
Z ± ∆Z = A ± ∆A) – (B ± ∆B)
= (A – B) ± ∆A ∓ ∆B
∴ ± ∆Z = ± ∆A ∓ ∆B,

iv) There are four possible values for ∆Z. namely (+∆A – ∆B), (+∆A + ∆B), (-∆A -∆B), (-∆A + ∆B). Hence, maximLim value of absolute error is ∆Z = (∆A+ ∆B) in both the cases.

v) Thus. when two quantities are added or subtracted, the absolute error in the final result is the sum of the absolute errors in the individual quantities.

Question 57.
Explain errors in product of measured quantity.
Answer:
Errors in product:
i) Suppose Z = AB and measured values of A and B are (A ± ∆A) and (B ± ∆B) then,
Z ± ∆Z = (A ± ∆A) (B ± ∆B)
= AB ± A∆B ± B∆A ± ∆A∆B
Dividing L.H.S by Z and R.H.S. by AB we get
\(\left(1 \pm \frac{\Delta Z}{Z}\right)=\left[1 \pm \frac{\Delta B}{B} \pm \frac{\Delta A}{A} \pm\left(\frac{\Delta A}{A}\right)\left(\frac{\Delta B}{B}\right)\right]\)
Since ∆A/A and ∆B/B are very small, product is neglected. Hence, maximum relative error in Z is \(\frac{\Delta Z}{Z}=\frac{\Delta A}{A}+\frac{\Delta B}{B}\)

ii) Thus, when two quantities are multiplied, the maximum relative error in the result is the sum of relative errors in each quantity.

Question 58.
Explain errors due to power (index) of measured quantity.
Answer:
Errors due to the power (index) of measured quantity:

1. Suppose
   Z = A³ = A × A × A
   then, \(\)
2. Hence the relative error in Z = A³ is three times the relative error in A.
3. This means if Z = Aⁿ
   
4. This implies, the quantity in the formula which has large power is responsible for maximum error.

Question 59.
The radius of a sphere measured repeatedly yields values 5.63 m, 5.54 m, 5.44 m, 5.40 m and 5.35 m. Determine the most probable value of radius and the mean absolute, relative and percentage errors.
Solution:
Given: a₁ = 5.63 m, a₂ = 5.54 m, a₃ = 5.44 m
a₄ = 5.40 m, a₅ = 5.35 m,
To find:
i) Most probable value (Mean value)
ii) Mean absolute error
iii) Relative error
iv) Percentage error

From formula (ii),
Absolute errors:
∆a₁ = |a_mean – a₁| = |5.472 – 5.63| = 0.158
∆a₂ = |a_mean – a₂| = |5.472 – 5.54| = 0.068
∆a₃ = |a_mean – a₃| = |5.472 – 5.44| = 0.032
∆a₄ = |a_mean – a₄| = |5.472 – 5.40| = 0.072
∆a₅ = |a_mean – a₅| = |5.472 – 5.35| = 0.122

From formula (ii),
∆a_mean = \(\frac{0.158+0.068+0.032+0.072+0.122}{5}\)
= \(\frac{0.452}{5}\)
= 0.0904 m
From formula (iii),
Relative error = \(\frac{0.0904}{5.472}\)
= 1.652 × 10^-2
(after rounding off to correct significant digits)
= 1.66 × 10^-2
= 0.0166
∴ Percentage error = 1.66 × 10^-2 × 100 = 1.66%
i) The mean value is 5.472 m.
ii) The mean absolute error is 0.0904 m.
iii) The relative error is 0.0166.
iv) The percentage error is 1.66%
[Note: Answer to relative error is rounded off using rules of significant figures and of rounding off]

Question 60.
Lin an experiment to determine the volume of an object, mass and density are recorded as m = (5 ± 0.15) kg and p = (5 ± 0.2) kg m³ respectively. Calculate percentage error in the measurement of volume.
Solulion:
Given: M = 5kg, ∆M = 0.15 kg, ρ = 5 kg/m³,
∆ρ = 0.2 kg/m³
To find: Percentage error in volume (V)

The percentage error in the determination of volume is 7%.

Question 61.
The acceleration due to gravity is determined by using a simple pendulum of length l = (100 ± 0.1) cm. If its time period is T = (2 ± 0.01) s, find the maximum percentage error in the measurement of g.
Solution:
Given: ∆l = 0.1 cm, l = 100 cm, ∆T = 0.01 s,
T = 2s
To find: Percentage error

Percentage error in measurement of g is 1.1 %.

Question 62.
Find the number of significant figures in the following numbers,
i. 25.42
ii. 0.004567
iii. 35.320
iv. 91.000
Answer:

Solved Examples

Question 63.
Add 7.21, 12.141 and 0.0028 and express the result to an appropriate number of significant figures.
Solution:

In the given problem, minimum number of digits after decimal is 2.
∴ Result will be rounded off upto two places of decimal.
Corrected rounded off sum is 19.35.

Question 64.
The mass of a box measured by a grocer’s balance is 2.3 kg. Two gold pieces of masses 20.15 g and 20.17 g are added to the box. What is (i) the total mass of the box? (ii) the difference in the masses of the pieces to correct significant figures?
Solution:
i) Total mass of the box
= (2.3+ 0.02017 + 0.02015) kg
= 2.34032 kg
Since, the last number of significant figure is 2, therefore, the total mass of the box = 2.3 kg

ii) Difference of mass = (20.17 – 20.15) = 0.02g Since, there are two significant figures so the difference in masses to the correct significant figures is 0.02 g.

i) The total mass of the box to correct significant figures is 2.3 kg.
ii) The difference in the masses to correct significant figures is 0.02 g.

Apply Your Knowledge

Question 65.
Write the dimensions of a and b in the relation
E = \(\frac{b-x^{2}}{a}\)
Where E is energy, x ¡s distance and t is time.
Answer:
The given relation is E = \(\frac{b-x^{2}}{a}\)
As x is subtracted from b,
∴ dimensions of b are x²;
i.e., b = [L²]
∴ We can write equation as E = \(\frac{\mathrm{L}^{2}}{\mathrm{a}}\)
Or a = \(\frac{\mathrm{L}^{2}}{\mathrm{E}}=\frac{\mathrm{L}^{2}}{\left[\mathrm{~L}^{2} \mathrm{MT}^{-2}\right]}\) = [L⁰M^-1T²]

Question 66.
What is the difference between 6.0 and 6.00? which Is more accurate?
Answer:
6.0 indicates the measurement is correct up to first decimal place, whereas 6.00 indicates that the measurement is correct up to second decimal place. Thus, 6.00 is a more accurate value than 6.0.

Question 67.
A child walking on a footpath notices that the width of the footpath is uneven. He reported this to his school principal and the complaint was forwarded to the municipal officer.
i. What is the possible error encountered?
ii. What is the relative error in width of footpath if width of footpath in 10 m length are noted as 5 m, 5.5 m, 5 m, 6 m and 4.5 m?
Answer:
i) The error encountered is personal error.

ii) Mean value of widths

The relative error in width of footpath is 0.084.

Question 68.
A factory owner kept five identical spheres between two wooden blocks on a ruler as shown in figure. He called all his workers and told them to take reading, to check their efficiency and knowledge.

i. What is the area of central sphere?
ii. What is the absolute error in reading of diameter of second sphere?
Answer:
i) From above diagram radius of central sphere is
r = 1 cm
∴ Area = πr² = 3.142 × (1)²= 3.142 cm²
The area of central sphere is 3.142 cm².

ii) Mean value of all reading of diameters
d_mean = \(\frac{\mathrm{d}_{1}+\mathrm{d}_{2}+\mathrm{d}_{3}+\mathrm{d}_{4}+\mathrm{d}_{5}}{5}=\frac{2+2+2+2+2}{5}\)
= \(\frac{10}{5}\) = 2 cm
Absolute error in reading of second sphere.
∆d₂ = |d_mean – d₂| = 2 – 2 = 0
The absolute error in reading of diameter of second sphere is zero.

Question 69.
A potential difference of V = 100 ± 2 volt, when applied across a resistance R gives a current of 10 ± 0.5 ampere. Calculate percentage error in R given by R V/I.
Answer:
Here. V = 100 ± 2 volt and I = 10 ± 0.5 ampere
Expressing limits of error as percentage error,
We have
V = 100 volt ± \(\frac{2}{100}\) × 100% = 10 volt ± 2%
and I = 10 ampere ± \(\frac{0.5}{10}\) × 100%
= 10 ampere ± 5%
∴ R = \(\frac{V}{I}\)
∴ %error in R = %error in V + %error in I
= 2% + 5% = 7%

Quick Review

Multiple Choice Questions

Question 1.
A physical quantity may be defined as
(A) the one having dimension.
(B) that which is immeasurable.
(C) that which has weight.
(D) that which has mass.
Answer:
(A) the one having dimension.

Question 2.
Which of the following is the fundamental unit?
(A) Length, force, time
(B) Length, mass, time
(C) Mass, volume, height
(D) Mass, velocity, pressure
Answer:
(B) Length, mass, time

Question 3.
Which of the following is NOT a fundamental quantity?
(A) Temperature
(B) Electric charge
(C) Mass
(D) Electric current
Answer:
(B) Electric charge

Question 4.
The distance of the planet from the earth is measured by __________.
(A) direct method
(B) directly by metre scale
(C) spherometer method
(D) parallax method
Answer:
(D) parallax method

Question 5.
The two stars S₁ and S₂ are located at distances d₁ and d₂ respectively. Also if d₁ > d2₂ then following statement is true.
(A) The parallax of S₁ and S₂ are same.
(B) The parallax of S₁ is twice as that of S₂
(C) The parallax of S₁ is greater than parallax of S₂
(D) The parallax of S₂ is greater than parallax of S₁
Answer:
(D) The parallax of S₂ is greater than parallax of S₁

Question 6.
Which of the following is NOT a unit of time?
(A) Hour
(B) Nano second
(C) Microsecond
(D) parsec
Answer:
(D) parsec

Question 7.
An atomic clock makes use of _________.
(A) cesium-133 atom
(B) cesium-132 atom
(C) cesium-123 atom
(D) cesium-131 atom
Answer:
(A) cesium-133 atom

Question 8.
S.I. unit of energy is joule and it is equivalent to
(A) 10⁶ erg
(B) 10^-7 erg
(C) 10⁷ erg
(D) 10⁵ erg
Answer:
(C) 10⁷ erg

Question 9.
[L¹M¹T^-1] is an expression for __________.
(A) force
(B) energy
(C) pressure
(D) momentum
Answer:
(D) momentum

Question 10.
Dimensions of sin θ is
(A) [L²]
(B) [M]
(C) [ML]
(D) [M⁰L⁰T⁰]
Answer:
(D) [M⁰L⁰T⁰]

Question 11.
Accuracy of measurement is determined by
(A) absolute error
(B) percentage error
(C) human error
(D) personal error
Answer:
(B) percentage error

Question 12.
Zero error of an instrument introduces .
(A) systematic error
(B) random error
(C) personal error
(D) decimal error
Answer:
(A) systematic error

Question 13.
The diameter of the paper pin is measured accurately by using ________.
(A) Vernier callipers
(B) micrometer screw gauge
(C) metre scale
(D) a measuring tape
Answer:
(B) micrometer screw gauge

Question 14.
The number of significant figures in 11.118 × 10^-6 is
(A) 3
(B) 4
(C) 5
(D) 6
Answer:
(C) 5

Question 15.
0.00849 contains ___________ significant figures.
(A) 6
(B) 5
(C) 3
(D) 2
Answer:
(C) 3

Question 16.
3.310 × 10² has ___________ significant figures.
(A) 6
(B) 4
(C) 2
(D) 1
Answer:
(B) 4

Question 17.
The Earth’s radius is 6371 km. The order of magnitude of the Earth’s radius is
(A) 10³ m
(B) 10⁹ m
(C) 10⁷ m
(D) 10² m
Answer:
(C) 10⁷ m

Question 18.
__________ is the smallest measurement that can be made using the given instrument
(A) Significant number
(B) Least count
(C) Order of magnitude
(D) Relative error
Answer:
(B) Least count

Competitive Corner

Question 1.
In an experiment, the percentage of error occurred in the measurement of physical quantities A, B, C and D are 1%, 2%, 3% and 4% respectively. Then the maximum percentage of error in the measurement X,
where X = \(\frac{A^{2} \frac{1}{B^{2}}}{C^{\frac{1}{3}} D^{3}}\), will be:
(A) -10 %
(B) 10 %
(C) \(\left(\frac{3}{13}\right) \%\)
(D) 16 %
Answer:
(D) 16 %
Hint:


∴ Percentage error in x is given as,
\(\frac{\Delta x}{x}\) × 100 – (error contributed by A) – (error contributed by B) + (error contributed by C) + (error contributed by D)
= 2% + 1% + 1% + 12%
= 16%

Question 2.
The main scale of a vernier callipers has n divisions/cm. n divisions of the vernier scale coincide with (n – 1) divisions of main scale. The least count of the vernier callipers is,
(A) \(\frac{1}{n(n+1)}\) cm
(B) \(\frac{1}{(n+1)(n-1)}\) cm
(C) \(\frac{1}{n}\) cm
(D) \(\frac{1}{n^{2}}\) cm
Answer:
(D) \(\frac{1}{n^{2}}\) cm
Hint:
1 V.S.D. = \(\frac{(n-1)}{n}\) M.S.D.
LC. = 1 M.S.D. – 1 V.S.D.
= 1 M.S.D. – \(\frac{(n-1)}{n}\) M.S.D.
= \(\frac{1}{n}\) M.S.D.
= \(\frac{1}{n}\) × \(\frac{1}{n}\) cm
∴ L.C. = \(\frac{1}{n^{2}}\) cm

Question 3.
A student measures time for 20 oscillations of a simple pendulum as 30 s. 32 s, 35 s and 31 s. 1f the minimum division in the measuring clock is I s, then correct mean time in second is
(A) 32 ± 3
(B) 32 ± 1
(C) 32 ± 2
(D) 32 ± 5
Answer:
(C) 32 ± 2
Hint:

Hence rounding off,
∆t = ± 2 s
∴ t ± ∆t = 32 ± 2 s

Question 4.
A student measured the diameter of a small steel ball using a screw gauge of least count 0.001 cm. The main scale reading is 5 mm and zero of circular scale division coincides with 25 divisions above the reference leveL If screw gauge has a zero error of— 0.004 cm, the correct diameter of the ball is
(A) 0.521 cm
(B) 0.525 cm
(C) 0.053 cm
(D) 0.29 cm
Answer:
(D) 0.29 cm

Hint:
Least count of screw gauge = 0.001 cm = 0.01mm
Main scale reading = 5 mm.
Zero error = – 0.004 cm = -0.04 mm
Zero correction = +0.04 mm
Observed reading = Mainscale reading + (Division × least count)
Observed reading = 5 + (25 × 0.01) = 5.25 mm
Corrected reading = Observed reading + Zero correction
Corrected reading = 5.25 + 0.04
= 5.29 mm = 0.529 cm

Question 5.
The density of the material in the shape of a cube is determined by measuring three sides of the cube and its mass. 1f the relative errors in measuring the mass and length are respectively 1.5% and 1%, the maximum error in determining the density is:
(A) 4.5%
(B) 6%
(C) 2.5°
(D) 3.5%
Answer:
(A) 4.5%

Question 6.
Let x = \(\left[\frac{a^{2} b^{2}}{c}\right]\) be the physical quantity. If the percentage error in the measurement of physical quantities a, b and c is 2, 3 and 4 percent respectively then percentage en-or in the measurement of x is
(A) 7%
(B) 14%
(C) 21%
(D) 28%
Answer:
(B) 14%
Hint:
Given: x = \(\frac{a^{2} b^{2}}{c}\)
Percentage error is given by.
\(\frac{\Delta x}{x}=\frac{2 \Delta a}{a}+\frac{2 \Delta b}{b}+\frac{\Delta c}{c}\)
= (2 × 2) + (2 × 3) + 4
= 4 + 6 + 4 = 14
∴ \(\frac{\Delta \mathrm{x}}{\mathrm{x}} \%\) = 14%

Question 7.
A physical quantity of the dimensions of length that can be formed out of c, G and \(\frac{\mathrm{e}^{2}}{4 \pi \varepsilon_{0}}\) is [c is velocity of light, G is universal constant of gravitation and e is charge]:

Answer:
(A) \(\frac{1}{\mathrm{c}^{2}}\left[\mathrm{G} \frac{\mathrm{e}^{2}}{4 \pi \varepsilon_{0}}\right]^{1 / 2}\)

Hint:
Let the physical quantity formed of the dimensions of length be given as.
[L] = [c]^x [G]^y \(\left[\frac{\mathrm{e}^{2}}{4 \pi \varepsilon_{0}}\right]^{z}\) …………….. (i)
Now,
Dimensions of velocity of light [c]^x = [LT^-1]^x
Dimensions of universal gravitational constant
[G]^y = [L³T²M^-1]^y
Dimensions of \(\left[\frac{\mathrm{e}^{2}}{4 \pi \varepsilon_{0}}\right]^{z}\) = [ML³T^-2]^z
Substitrning these in equation (i)
[L] [LT^-1]^x [M^-1L³T^-2]^y [ML³T^-2]^z
= L^x+3y+3z M^-y+z T^-x-2y-2z
Solving for x, y, z
x + 3y + 3z = 1
-y + z = 0
x + 2y + 2z = O
Solving the above equation,
x = -2, y = \(\frac{1}{2}\), z = \(\frac{1}{2}\)
∴ L = \(\frac{1}{c^{2}}\left[\mathrm{G} \frac{\mathrm{e}^{2}}{4 \pi \varepsilon_{0}}\right]^{1 / 2}\)

Question 8.
The following observations were taken for determining surface tension T of water by capillary method:
diameter of capillary, D = 1.25 × 10^-2 m
rise of water, h = 1.45 × 10^-2 m
Using g = 9.80 m/s² and the simplified relation
T = \(\frac{\mathrm{rhg}}{2}\) × 10³ N/m, the possible error in surface tension is closest to:
(A) 0.15%
(B) 1.5%
(C) 2.4%
(D) 10%
Answer:
(B) 1.5%
Hint:
D = 1.25 × 10^-2 m; h = 1.45 × 10^-2 m
The maximum permissible error in D
= ∆D = 0.01 × 10^-2 m
The maximum permissible error in h
= ∆h = 0.01 × 10^-2 m
g is given as a constant and is errorless.

Balbharti Maharashtra State Board 11th Biology Important Questions Chapter 1 Living world Important Questions and Answers.

Maharashtra State Board 11th Biology Important Questions Chapter 1 Living World

Question 1.
What are the basic principles of life?
Answer:
The basic principles of life are as follows:

1. Metabolism: Metabolism is breaking of molecules (catabolism) and making of new molecules (anabolism). An organism performs metabolism in order to obtain energy and various chemical molecules essential for survival.
2. Growth and development: Organisms tend to grow and develop in a well-orchestrated process from birth onwards.
3. Ageing: It is the process during which molecules, organs and systems begin to lose their effective working and become old.
4. Reproduction: For continuity of race (species), organisms reproduce (asexually or sexually) to produce young ones like themselves. However, mules and worker bees do not reproduce, yet are living.
5. Death: As the body loses its capacity to perform metabolism, an organism dies.
6. Responsiveness: Living organisms respond to thermal, chemical or biological changes in their surroundings.

Question 2.
Enlist the characters of living organisms.
Answer:
The basic principles of life are as follows:

1. Metabolism: Metabolism is breaking of molecules (catabolism) and making of new molecules (anabolism). An organism performs metabolism in order to obtain energy and various chemical molecules essential for survival.
2. Growth and development: Organisms tend to grow and develop in a well-orchestrated process from birth onwards.
3. Ageing: It is the process during which molecules, organs and systems begin to lose their effective working and become old.
4. Reproduction: For continuity of race (species), organisms reproduce (asexually or sexually) to produce young ones like themselves. However, mules and worker bees do not reproduce, yet are living.
5. Death: As the body loses its capacity to perform metabolism, an organism dies.
6. Responsiveness: Living organisms respond to thermal, chemical or biological changes in their surroundings.

Question 3.
What are taxonomical aids? Give examples.
Answer:
Taxonomical aids are used to study biodiversity, e.g. Herbaria, botanical gardens, museums, biodiversity parks, etc.

Question 4.
What is a herbarium?
Answer:
Herbarium is a dried plant specimen that is pressed, treated and mounted on a standard size sheet in order to preserve it.
[Note: Herbarium is a collection of dried, pressed and labelled plant specimens arranged by a classification system.]

Question 5.
What information is mentioned in the label of a plant specimen preserved in herbarium?
Answer:
It is also essential to record the date, place of collection along with detailed classification and highlighting with its ecological peculiarities, characters of the plant on a sheet. Local names of plant specimens and name of the collector may be added. This information is given at lower right comer of sheet and is called ‘label’.

Question 6.
What are botanical gardens?
Answer:
Botanical gardens are places where plants of different varieties collected from different parts of the world are grown in vivo in a scientific and systematic manner.

Question 7.
Define biodiversity.
Answer:
Biodiversity is the degree of variation of life forms in an ecosystem.

Question 8.
Define conservation.
Answer:
Conservation involves attempting to slow down, stop or even reverse the loss in the natural habitat of an organism.

Question 9.
What is a museum? What are the various specimens found in a museum?
Answer:
1. Museums are places where collections of preserved plant and animal specimens are kept.
2. The different types of specimens found in a museum include;
(a) Plant and animal specimens preserved in formalin (10% to 40% formaldehyde) in transparent jars.
(b) Larger animals like birds and mammals, usually stuffed and preserved.
(c) Certain specimens in dried forms are also kept in a museum.
(d) Systematic collections of shells, skeletons of animals and insect boxes are also found in museums.

Question 10.
What is taxidermy?
Answer:
Taxidermy is a science in which larger animals like birds and mammals are usually stuffed and preserved.

Question 11.
Write a note on zoological park.
Answer:

1. Zoological park (zoo) is a place where wild animals are kept in captivity.
2. Wild animals are kept in a protected environment and care is taken to provide conditions similar to their natural habitat.
3. It is a form of ex situ conservation of species i.e. away from their natural habitat.
4. A naturalist can study the food habits and behaviour of animals in a zoological park.

Question 12.
Mention some tools of maintaining biodiversity records.
Answer:
Flora, manuals, monographs and catalogues are some tools of maintaining biodiversity records.

Question 13.
Explain the different tools used for maintaining biodiversity records.
Answer:
The different tools used for maintaining biodiversity records are as follows:

• Flora: It is the plant life occurring in a particular area at a particular time.
• Monograph: It describes any one selected biological group.
• Manual: It provides information and keys about identification of species found in a particular area.

Question 14.
Define biodiversity park.
Answer:
Biodiversity park is an ecological assemblage of species that form self-sustaining communities on degraded/ barren landscape, e.g. Uttamrao Patil Biodiversity Park, Gureghar, Mahabaleshwar.

Question 15.
Write a note on ‘key’ used as a taxonomical aid.
Answer:

1. Key is a taxonomical aid used in the classification of plants and animals.
2. Keys are based on contrasting characters. One of the contrasting characters gets accepted and the other gets rejected.
3. The statement in a key is called a lead.
4. Normally keys are analytical in nature.

Question 16.
Name the following.

1. A collection of dried plant specimen that are pressed, treated and mounted on a standard size sheet in order to preserve it.
2. Places where collections of preserved plant and animal specimens are kept.
3. Taxonomical aid used for classification of plants and animals which is based on contrasting characters.

Answer:

1. Herbarium
2. Museum
3. Key

Question 17.
Fill in the blanks:

1. The extent of complexity and density of ________ can be regarded as a measure of health of an ecosystem.
2. In a museum, plant and animal specimens are preserved in _________ in transparent jars.
3. A naturalist can study food habits and behaviour of animals in a ___________.
4. Study of _________ is a must, to understand interrelations between organisms and maintain harmony on planet earth.
5. The statement in a key is called a _________.

Answer:

1. biodiversity
2. formalin
3. zoo/ zoological park
4. biodiversity
5. lead

Question 18.
Rakesh went for a study tour to the nearest national park. There he found some different plant species. He was not aware about their names and family. He wanted to bring that plants to his college and keep them for longer period of time, so that he can study them thoroughly. What should he do in such a situation?
Answer:
1. Rakesh can press and mount the plant specimen on the herbarium sheet and can preserve the dried plant material.
2. He can also write any information he knows about the plant on herbarium sheet, which can be used for further studies.

Question 19.
While doing his Ph.D. in Plant Taxonomy your friend has come across a plant, which he feels is a new species. How can he confirm the same?
Answer:
1. The newly discovered plant can be identified with the help of taxonomic keys, monographs, floras, herbaria and preserved plant specimens.
2. A separate taxonomic key is available for each taxonomic category.
3. The individual would have to study the morphological and anatomical features of the plant and compare it with the existing information available in the scientific literature.
Conservation of Biodiversity

Question 20.
Quick Review:
Answer:

Question 21.
Multiple Choice Questions:

Question 1.
Which one of the following aspects is an inclusive characteristic of living things?
(A) Isolated metabolic reactions occurring in vitro
(B) Reproduction
(C) Irritability
(D) Increase in mass by accumulation of material on surface
Answer:
(C) Irritability

Question 2.
Which of the following property is shown by both living and non-living things?
(A) Growth
(B) Consciousness
(C) Ageing
(D) Metabolism
Answer:
(A) Growth

Question 3.
Herbarium is
(A) a collection of living plants which are medicinally important
(B) a place where plants collected from different parts of the world are grown
(C) a garden where herbs are cultivated
(D) a collection of dried and preserved plants
Answer:
(D) a collection of dried and preserved plants

Question 4.
A zoological park does not
(A) have wild animals in captivity under human care.
(B) provide conditions similar to their natural habitat of animals.
(C) have a systematic collection of shells and skeletons of animals
(D) enable naturalists to study the food habits and behaviour of wild animals.
Answer:
(C) have a systematic collection of shells and skeletons of animals

Question 5.
A naturalist can study food habits and behaviour of animals in a
(A) museum
(B) zoological park
(C) botanical garden
(D) herbarium
Answer:
(B) zoological park

Question 6.
Which of the following is NOT a tool of maintaining biodiversity records?
(A) Flora
(B) Monograph
(C) Fauna
(D) Manual
Answer:
(C) Fauna

Question 7.
Which of the following tools provides information for identification of names of species found in a particular area?
(A) Catalogues
(B) Manuals
(C) Flora
(D) Monographs
Answer:
(B) Manuals

Question 8.
Keys are taxonomical aids that
(A) are used to identify plants and animals based on similarities and dissimilarities.
(B) contains the account of habitat and distribution of plants in a given area.
(C) provides an index to the plant species found in a particular area.
(D) provide information for identification of species found in an area.
Answer:
(A) are used to identify plants and animals based on similarities and dissimilarities.

Question 22.
Competitive Corner:

Question 1.
Match the items given in Column I with those in Column II and select the correct option given below: [NEET (UG) 2018]
Answer:

Column I	Column II
1. Herbarium	(a) It is a place having a collection of preserved plants and animals
2. Key	(b) A list that enumerates methodically all the species found in an area with brief description aiding identification
3. Museum	(c) It is a place where dried and pressed plant specimens mounted on sheets are kept
4. Catalogue	(d) A booklet containing a list of characters and their alternates which are helpful in identification of various taxa.

(A) i-b, ii-d, iii-c, iv-a
(B) i-c, ii-b, iii-a, iv-d
(C) i-a, ii-d, iii-c, iv-b
(D) i-c, ii-d, iii-a, iv-b
Answer:
(D) i-c, ii-d, iii-a, iv-b

Question 2.
The label of a herbarium sheet does not carry information on
(A) height of the plant
(B) date of collection
(C) name of collector
(D) local names
Answer:
(A) height of the plant

Question 3.
Which one of the following is NOT a correct statement? [NEET 2013]
(A) Herbarium houses dried, pressed and preserved plant specimens.
(B) Botanical gardens have collection of living plants for reference.
(C) A museum has collection of photographs of plants and animals.
(D) Key is a taxonomic aid for identification of specimens.
Answer:
(C) A museum has collection of photographs of plants and animals.

Question 23.
Basic Principles of Life

Question 1.
Define metabolism.
Answer:
Metabolism: Metabolism is breaking of molecules (catabolism) and making of new molecules (anabolism). An organism performs metabolism in order to obtain energy and various chemical molecules essential for survival.

Question 2.
Enlist the basic principles of life.
Answer:
The basic principles of life are as follows:
(i) Metabolism: Metabolism is breaking of molecules (catabolism) and making of new molecules (anabolism). An organism performs metabolism in order to obtain energy and various chemical molecules essential for survival.
(ii) Growth and development: Organisms tend to grow and develop in a well-orchestrated process from birth onwards.
(iii) Ageing: It is the process during which molecules, organs and systems begin to lose their effective working and become old.
(iv) Reproduction: For continuity of race (species), organisms reproduce (asexually or sexually) to produce young ones like themselves. However, mules and worker bees do not reproduce, yet are living.
(v) Death: As the body loses its capacity to perform metabolism, an organism dies.
(vi) Responsiveness: Living organisms respond to thermal, chemical or biological changes in their surroundings.

Question 3.
Reproduction is not an inclusive character of life. Explain.
Answer:
No, we cannot call reproduction as an inclusive character of life. Certain organisms like mules and worker bees do not reproduce and are still living. Thus, reproduction cannot be considered as an all inclusive defining characteristic of living organisms.

Question 4.
Define taxonomical aids and give two examples
Answer:
Taxonomical aids are used to study biodiversity, e.g. Herbaria, botanical gardens, museums, biodiversity parks, etc.

Question 5.
1. Define herbarium.
2. Mention any four essentials of a good herbarium.
Answer:
1. Herbarium is a dried plant specimen that is pressed, treated and mounted on a standard size sheet in order to preserve it.
[Note: Herbarium is a collection of dried, pressed and labelled plant specimens arranged by a classification system.]

2. The essentials of a good herbarium are as follows:
(i) It is essential to identify and label the collected specimen correctly.
(ii) Specimens should be stored in a dry place.
(iii) The plants are usually pressed and mounted on the sheet of paper known as herbarium sheets. Some plants are not suitable for pressing or mounting, like succulents, seeds, cones, etc. They need to be preserved in suitable liquid like formaldehyde, acetic alcohol, etc.
(iv) In order to preserve the specimen for longer durations, acid-free paper, special glues and inks must be used to mount the specimen so that the specimen does not deteriorate.

Question 6.
Shanaya found a unfamiliar plant on her visit to Tamil Nadu. She wants to study the plant thoroughly in her laboratory? How can she do so?
Answer:
1. Riya can press and mount the plant specimen on a herbarium sheet and preserve the dried plant material, until she returns back from her visit.
2. She can also write any available information regarding the collected specimen on the herbarium sheet, which can be useful for further studies with her biology teacher.
3. Various taxonomical aids can be useful to get information about this peculiar plant.
[Note: In order to conserve the local flora, Riya can collect photographs ofplant and describe it’s structure to her teacher.]

Question 7.
Manas wants to prepare a herbarium of plants.
1. What is a herbarium?
2. What are the essentials he should keep in mind to prepare a good herbarium?
3. What information should be added on the label of a herbarium?
Answer:
1. Herbarium is a dried plant specimen that is pressed, treated and mounted on a standard size sheet in order to preserve it.
[Note: Herbarium is a collection of dried, pressed and labelled plant specimens arranged by a classification system.]

2. (i) It is essential to identify and label the collected specimen correctly.
(ii) Specimens should be stored in a dry place.
(iii) The plants are usually pressed and mounted on the sheet of paper known as herbarium sheets. Some plants are not suitable for pressing or mounting, like succulents, seeds, cones, etc. They need to be preserved in suitable liquid like formaldehyde, acetic alcohol, etc.
(iv) In order to preserve the specimen for longer durations, acid-free paper, special glues and inks must be used to mount the specimen so that the specimen does not deteriorate.
(v) The specimens should be dried well before preparing a herbarium in order to prevent rotting of specimen.

3. It is also essential to record the date, place of collection along with detailed classification and highlighting with its ecological peculiarities, characters of the plant on a sheet. Local names of plant specimens and name of the collector may be added. This information is given at lower right comer of sheet and is called ‘label’.

Question 8.
Can humans help in conservation of biodiversity? Explain your answer.
Answer:

1. Due to rapid increase in human population and industrialization, humans have over utilized natural resources; leading to degradation of the environment and hence only humans can help conserve the ecosystem.
2. Humans are capable of conserving and improving the quality of nature and thus, can play a major role in biodiversity conservation.
3. In order to conserve biodiversity and its environmental resources, humans must use the resources rationally and avoid excessive degradation of environment.
4. Human beings are stakeholders of the environment and need to come together to overcome pollution and improve the environment quality in order to conserve biodiversity.

E.g. Ban or limit on use of harmful products (plastic, chemicals, etc.) that are toxic to various birds, animals, etc. Human beings also play a role in conservation of biodiversity by establishment of various sites for in situ (national parks, wildlife sanctuaries and biosphere reserves) and ex situ (botanical gardens, culture collections and zoological parks) conservation.

Question 9.
Write a note on botanical gardens.
Answer:
Botanical gardens are places where plants of different varieties collected from different parts of the world are grown in vivo in a scientific and systematic manner.
The importance of botanical gardens is as follows:

1. It is a place where there is an assemblage of living plants maintained for botanical teaching and research purpose.
2. Botanical gardens are important for their records of local flora.
3. Botanical gardens provide facilities for the collection of living plant materials for botanical studies.
4. Botanical gardens also supply seeds and material for botanical investigations.
5. The development of botanical gardens in any country is associated with its history of civilization, culture, heritage, science, art, literature and various other social and religious expressions.
6. Botanical gardens besides possessing an outdoor garden may contain herbaria, research laboratory, greenhouses and library.
7. Botanical gardens are not only important for botanical studies, but also to develop tourism in the country.

Question 10.
Botanical gardens are important in botanical studies. Justify.
Answer:
Metabolism can be considered as an all-inclusive (defining) feature of life since it is exhibited by all living organisms and does not take place in non-living things. Another all-inclusive characteristic of life is responsiveness or irritability. This is a unique property of living beings since all living beings are conscious of their surroundings.

Question 11.
Suggest any three measures you can take to prevent loss of biodiversity.
Answer:
The essentials of a good herbarium are as follows:

1. It is essential to identify and label the collected specimen correctly.
2. Specimens should be stored in a dry place.
3. The plants are usually pressed and mounted on the sheet of paper known as herbarium sheets. Some plants are not suitable for pressing or mounting, like succulents, seeds, cones, etc. They need to be preserved in suitable liquid like formaldehyde, acetic alcohol, etc.
4. In order to preserve the specimen for longer durations, acid-free paper, special glues and inks must be used to mount the specimen so that the specimen does not deteriorate.
5. The specimens should be dried well before preparing a herbarium in order to prevent rotting of specimen.
6. It is also essential to record the date, place of collection along with detailed classification and highlighting with its ecological peculiarities, characters of the plant on a sheet.

Local names of plant specimens and name of the collector may be added. This information is given at lower right comer of sheet and is called ‘label’.

Question 12.
1. Define biodiversity.
2. How does loss of biodiversity affect the ecosystem?
Answer:
1. Biodiversity is the degree of variation of life forms in an ecosystem.
2. (i) The loss of biodiversity is an moral and ethical issue.
(ii) Biodiversity helps to maintain stability in an ecosystem.
(iii) Humans share the environment with various other organisms and harm to these species can result in loss of biodiversity.
(iv) The loss of even one variety of organisms can affect the entire ecosystem.
Hence, due to all these reasons, loss of biodiversity matters.

Question 13.
Define botanical garden and write a note on importance of greenhouses in botanical gardens.
Answer:
Botanical gardens are places where plants of different varieties collected from different parts of the world are grown in vivo in a scientific and systematic manner.

1. Greenhouse is a structure with suitable walls and a roof in which plants are grown under regulated climatic conditions.
2. Most botanical gardens exhibit ornamental plants which require stringent/ optimum climatic conditions for their growth and/or flowering.
3. The greenhouse associated with botanical gardens are also used to grow and propagate those plants that may not survive seasonal changes.

Question 14.
Which science is used to animals at museums? preserve larger
Answer:
Taxidermy is a science in which larger animals like birds and mammals are usually stuffed and preserved.

Question 15.
What is a museum?
Answer:
Museums are places where collections of preserved plant and animal specimens are kept.

Question 16.
What chemical is used to preserve plant and animal specimens in transparent jars at museums?
Answer:
Plant and animal specimens preserved in formalin (10% to 40% formaldehyde) in transparent jars.

Question 17.
Define the following terms:
1. Flora
2. Monograph
3. Manual
Answer:
The different tools used for maintaining biodiversity records are as follows:
1. Flora: It is the plant life occurring in a particular area at a particular time.
2. Monograph: It describes any one selected biological group.
3. Manual: It provides information and keys about identification of species found in a particular area.

Question 18.
Define the following terms:

1. Botanical garden
2. Zoological parks
3. Biodiversity parks
4. Museum
5. Herbarium

Answer:

1. Botanical gardens are places where plants of different varieties collected from different parts of the world are grown in vivo in a scientific and systematic manner.
2. Zoological park (zoo) is a place where wild animals are kept in captivity.
3. Biodiversity park is an ecological assemblage of species that form self-sustaining communities on degraded/ barren landscape, e.g. Uttamrao Patil Biodiversity Park, Gureghar, Mahabaleshwar.
4. Museums are places where collections of preserved plant and animal specimens are kept.
5. Herbarium is a dried plant specimen that is pressed, treated and mounted on a standard size sheet in order to preserve it.

[Note: Herbarium is a collection of dried, pressed and labelled plant specimens arranged by a classification system.]

Question 19.
On what characters is the ‘key’ based on? taxonomical aid
Answer:

1. Key is a taxonomical aid used in the classification of plants and animals.
2. Keys are based on contrasting characters. One of the contrasting characters gets accepted and the other gets rejected.
3. The statement in a key is called a lead.
4. Normally keys are analytical in nature.

Balbharti Maharashtra State Board 11th Commerce Book Keeping & Accountancy Important Questions Chapter 1 Introduction to Book Keeping and Accountancy Important Questions and Answers.

Maharashtra State Board 11th Commerce BK Important Questions Chapter 1 Introduction to Book Keeping and Accountancy

1. Answer in One Sentence:

Question 1.
What is Transaction?
Answer:
The exchange of goods or services for money or money’s worth is called a transaction.

Question 2.
What is Cash Discount?
Answer:
The amount which is deducted by the seller from the amount due at the time of the receipt is called cash discount.

Question 3.
What is the entity concept?
Answer:
The business entity concept states that a business has a separate entity and has an independent legal existence distinct from the person who owns it.

Question 4.
What is the money measurement concept?
Answer:
The money measurement concept implies that every business transaction must be recorded in a common unit of measurement i.e. in terms of money only.

Question 5.
What is the consistency concept?
Answer:
The consistency concept implies that any policy adopted for accounting should be continuous or consistent throughout the business and it need not be changed generally unless and until circumstances demand.

Question 6.
What is conservatism?
Answer:
The concept of conservatism states that while deciding the policy of the enterprise the businessman has to anticipate no profit but provide for all possible losses.

Question 7.
What is accounting?
Answer:
Accounting is a process of recording, classifying, summarising, analyzing, and interpreting financial transactions and communicating the result thereof to the users of such information.

Question 8.
Name the financial statements prepared by a business.
Answer:
A businessman prepares the financial statements such as income statements i.e. Trading Account and Profit and Loss Account and Position Statement i.e. Balance Sheet.

Question 9.
Explain the term ‘Profit’.
Answer:
Excess of revenue income over revenue expenditure is termed as profit.

Question 10.
Explain the term ‘Cost Concept’.
Answer:
The concept according to which assets are recorded in the books of accounts at the price at which they are acquired or purchased is called the cost concept.

2. Give the word term or phrase which can substitute each of the following statements:

Question 1.
Dealings between two persons.
Answer:
Transaction

Question 2.
Business transaction in which cash is not paid or received immediately.
Answer:
Credit Transaction

Question 3.
An allowance is given by the receiver of the cash to the giver of cash at the time of payment.
Answer:
Cash Discount

Question 4.
Liability depends on the happening or not happening a certain event.
Answer:
Contingent Liability

Question 5.
Expenditure on fixed assets increases the earning capacity of the business.
Answer:
Capital Expenditure

Question 6.
System in which entry is recorded for cash as well as credit transactions.
Answer:
Accrual System

Question 7.
The concept under which comparison of one accounting period with the other period is possible.
Answer:
Consistency Concept

Question 8.
A science and art of correctly recording in the books of accounts, all those business transactions that result in the transfer of money or money’s worth.
Answer:
Bookkeeping

Question 9.
Exchange of goods and services either for cash or any other goods or services.
Answer:
Transaction

Question 10.
Sale or purchase of goods or services for immediate cash payment.
Answer:
Cash Transaction

Question 11.
Sale or purchase of goods or services for a certain value to be receivable or payable in the future.
Answer:
Credit Transaction

Question 12.
Commodity purchased or produced for sale.
Answer:
Goods

Question 13.
Excess of assets over liabilities.
Answer:
Capital

Question 14.
Excess of profit over normal profit.
Answer:
Super Profit

Question 15.
The total amount of goods and services are withdrawn by the proprietor for self-use.
Answer:
Drawings

Question 16.
Amount of discount deducted from the invoice price.
Answer:
Trade discount

Question 17.
The reputation of a business is expressed in terms of money.
Answer:
Goodwill

Question 18.
The concept states that assets when purchased should be recorded at cost price.
Answer:
Cost concept

Question 19.
The concept states that business operations will continue forever.
Answer:
Concept of Going Concern

Question 20.
A concept on which a double-entry bookkeeping system is based.
Answer:
Dual Aspect Concept

Question 21.
Rules of conduct are accepted universally to record business transactions.
Answer:
Accounting Principles

Question 22.
General ideas which convey certain meanings.
Answer:
Concepts

Question 23.
The general notion or abstract ideas on which accounting is based.
Answer:
Accounting Concept

Question 24.
The accounting concept suggests that a business has a separate identity from its owner.
Answer:
Business Entity Concept

Question 25.
The accounting concept states that monetary transactions are only recorded in the books of accounts.
Answer:
Money Measurement Concept

Question 26.
The money value for which assets are acquired or manufactured.
Answer:
Cost

Question 27.
Proprietor’s contribution to the business.
Answer:
Capital

Question 28.
Valuable things are owned by the business.
Answer:
Assets

Question 29.
Period of time for which accounts of the business are prepared.
Answer:
Financial year/Accounting year

Question 30.
Amount received after selling of goods or services.
Answer:
Revenue

Question 31.
Excess of revenue over its cost.
Answer:
Profit

Question 32.
Name the accounting concept on the basis of which the income statement is prepared.
Answer:
Accrual Accounting Concept

3. Select the most appropriate alternatives from those given below and rewrite the statements.

Question 1.
Money value of the reputation of business is known as ____________
(a) Copyright
(b) Goodwill
(c) Patents
(d) Trademark
Answer:
(b) Goodwill

Question 2.
Heavy advertising expenditure for launching a new product is called as ____________
(a) Capital expenditure
(b) Revenue Expenditure
(c) Deferred Revenue Expenditure
(d) None of these
Answer:
(c) Deferred Revenue Expenditure

Question 3.
Expenditure incurred on purchase of fixed asset is ____________
(a) Revenue Expenditure
(b) Capital Expenditure
(c) Deferred Revenue Expenditure
(d) None of these
Answer:
(b) Capital Expenditure

Question 4.
Concept which provides a line between present and future is known as ____________
(a) Going Concern Concept
(b) Cost Concept
(c) Accrual Concept
(d) Entity Concept
Answer:
(a) Going Concern Concept

Question 5.
Totalling of Journal or Ledger is called as ____________
(a) Posting
(b) Folio
(c) Casting
(d) Journalising
Answer:
(c) Casting

Question 6.
In cash transaction, goods or services are exchanged for ____________
(a) other goods
(b) other services
(c) immediate cash
(d) grains
Answer:
(c) immediate cash

Question 7.
The work of book-keeping is of ____________ nature.
(a) competitive
(b) primary/basic
(c) secondary
(d) none of these
Answer:
(b) primary/basic

Question 8.
Capital is ____________ of the business.
(a) asset
(b) liability
(c) property
(d) goodwill
Answer:
(b) liability

Question 9.
The work of accounting depends upon ____________
(a) book-keeping
(b) cash book
(c) subsidiary books
(d) ledger
Answer:
(a) book-keeping

Question 10.
____________ is the amount invested by the owner of a business.
(a) Cash
(b) Money
(c) Asset
(d) Capital
Answer:
(d) Capital

Question 11.
Revenue expenditure is ____________ in nature.
(a) capital
(b) outstanding
(c) recurring
(d) contingent
Answer:
(c) recurring

Question 12.
Amount withdrawn by the owner for his personal expenses is called ____________
(a) Drawings
(b) Personal expenses
(c) Cash
(d) Assets
Answer:
(a) Drawings

Question 13.
Financial statements are a part of ____________
(a) Book keeping
(b) Planning
(c) Accounting
(d) None of these
Answer:
(c) Accounting

Question 14.
A ____________ liability is an uncertain liability.
(a) Long term
(b) Contingent
(c) Current
(d) Fixed
Answer:
(b) Contingent

Question 15.
A Derson who is unable to nay his debts, is called ____________
(a) insolvent
(b) solvent
(c) well to do
(d) poor
Answer:
(a) insolvent

Question 16.
According to ____________ concept business shall go on for a long time.
(a) going concerned
(b) consistency
(c) materiality
(d) dual aspects
Answer:
(a) going concerned

Question 17.
According to ____________ concept, every business transaction has two aspects.
(a) going concerned
(b) materiality
(c) business entity
(d) dual aspects
Answer:
(d) dual aspects

Question 18.
According to ____________ concept revenue is recognised when it is earned.
(a) Realisation
(b) Accounting period
(c) Accrual
(d) Matching cost
Answer:
(c) Accrual

Question 19.
According to ____________ convention, while preparing planning anticipate losses.
(a) Materiality
(b) Consistency
(c) Conservatism
(d) disclosure
Answer:
(c) Conservatism

Question 20.
Customs and traditions which guide the accountants to prepare accounting statements are called ____________
(a) Conventions
(b) Principles
(c) Concepts
(d) Procedure
Answer:
(a) Conventions

Question 21.
According to ____________ concept, assets are recorded at a price paid to acquire them.
(a) Cost
(b) Money measurement
(c) Entity
(d) Dual aspect
Answer:
(a) Cost

4. State whether the following statements are true or false with reasons:

Question 1.
The trading concern is established for rendering services to society.
Answer:
This statement is False.
The main objective of trading concern is to earn profit. It is for traders’ livelihood, so the trading concern is not established for rendering services to the society.

Question 2.
Book-keeping is an art as well as a science.
Answer:
This statement is True.
Book-keeping is also considered as an art of recording business transactions because the writing of accounts in a specific style and format requires education, knowledge, training, skill, and experience. From another point of view, Book-keeping is a continuous process of collecting, analyzing, classifying, summarising, and recording the different types of business transactions. In brief, bookkeeping may be defined as “A science as well as an art of collecting, analyzing, classifying, summarising and recording all types of business transactions in a significant manner and in terms of money in a separate set of books.”

Question 3.
In Book-keeping & Accountancy non-monetary transactions are also recorded.
Answer:
This statement is False.
Only monetary transactions are recorded in Book-keeping & Accountancy. It is done to find financial results for the purpose of analyzing and interpreting.

Question 4.
Perpetual succession is explained by the concept of the entity.
Answer:
This statement is False.
The concept of entity is different from the perpetual succession. Entity means separate existence of business from the owner whereas perpetual succession means long life and continuation.

Question 5.
Accounting is useful only to the owner.
Answer:
This statement is False.
Accounting is not only useful for the owner but also to the suppliers, government, customers, lenders, etc.

Question 6.
Writing of account does not require specific skill and knowledge.
Answer:
This statement is False.
Writing of account requires specific skill and knowledge as book-keeping is an art and science of correctly recording business transactions.

Question 7.
The main objective of Bookkeeping is to keep permanent records of business transactions.
Answer:
This statement is True.
The record of business transactions is prepared for a specific period of time i.e. one year and it is preserved for a long period of time.

Question 8.
In credit transactions, goods or services are purchased for cash only.
Answer:
This statement is False.
In credit transactions, goods and services are purchased and sold for credit only. Credit means post-pone payments for a future date.

Question 9.
In barter transactions, goods or services are purchased for other goods or services.
Answer:
This statement is True.
Barter means exchange of goods or services for goods or services. No money is involved in barter transactions.

Question 10.
In cash transactions, goods or services are purchased for a certain value to be paid in the future.
Answer:
This statement is False.
In a cash transaction, goods or services are purchase or sold for spot payments or receipts. No credit is allowed.

Question 11.
Good is a commodity that is purchased for self-use.
Answer:
This statement is False.
Good is a commodity that is purchased for resale purposes. It is an article or a commodity in which businessman deals.

Question 12.
Amount due from other person is known as debt.
Answer:
This statement is True.
Debt is a total sum of money due from a person with whom the business has dealings. Accordingly, a person from whom such debt is due to a business is called the debtor.

Question 13.
Liabilities represent the debts or obligations that a business must receive in terms of money.
Answer:
This statement is False.
Liabilities refer to the total amount of obligations that a business has to pay in the future. Liability means the total amount owed by the business to other persons.

Question 14.
Capital = Liabilities – Assets.
Answer:
This statement is False.
Excess of business assets over business liabilities is called capital.
Capital = Assets – liabilities.

Question 15.
Drawings made by the businessman increase his capital.
Answer:
This statement is False.
Drawing means cash or goods withdrawn by the proprietor for his personal or family use. Drawings made by the businessman decreases his capital.

Question 16.
A person whose assets are equal to or greater than liabilities is known as insolvent.
Answer:
This statement is False.
Insolvent means a person whose liabilities are greater than assets is known as ‘Insolvent’.

Question 17.
A person whose assets are less than business liabilities is known as insolvent.
Answer:
This statement is True.
When a person is not able to pay liabilities is called insolvent. Liabilities are greater than assets.

Question 18.
The cost or sacrifice which is incurred to run the business is known as an income.
Answer:
This statement is False.
The cost or sacrifice which is incurred to run the business is known as an expenditure.

Question 19.
Benefits which are given by giver to the receiver is known as discount.
Answer:
This statement is False.
Discount is a concession given by the seller to the buyer. There are two types of discount,

• Cash discount
• Trade discount

Question 20.
Posting implies recording a transaction in a journal.
Answer:
This statement is False.
Posting means transferring or recording journal entries from the journal to the respective ledger accounts. Ledger is the main book of accounting.

Question 21.
Contingent liabilities can also be called doubtful liabilities.
Answer:
This statement is True.
The contingent liabilities are the liabilities whose occurrence depends upon the happening of a certain event that may or may not take place. So a contingent liabilities can also be called doubtful liabilities.

Question 22.
A sports club is a trading concern.
Answer:
This statement is False.
The sports club is a nontrading concern as they do not work for profit-making. They work for the promotion of sports.

Question 23.
A cash discount is an incentive allowed for the speedy recovery of income.
Answer:
This statement is True.
A cash discount is given by the creditor or seller to the debtor or buyer to induce him to make prompt payment. It is given at the time of cash payment.

Question 24.
Excess of revenue over expenses is called loss.
Answer:
This statement is False.
Excess of revenue over expenses is called profit. Your income is less and expenditures are less so you get profit.
Income ₹ 50,000 – Expenses ₹ 35,000 = Profit is ₹ 15,000.

Question 25.
Excess of liabilities over assets represents the solvency of a business.
Answer:
This statement is False.
Excess of liabilities over assets represents insolvency of business. A trader cannot pay his debts as liabilities are greater than his assets.
Liabilities ₹ 1,50,000 Assets ₹ 80,000.
₹ 1,50,000 – ₹ 80,000 = ₹ 70,000 deficiency.

Question 26.
A business can not run without bookkeeping.
Answer:
This statement is False.
Book-keeping is a recording of business financial transactions which is useful to prepare ledger, trial balance, financial statements. All the above are useful to take decisions, planning, liabilities, and assets so a businessman cannot run his business without book-keeping.

Question 27.
Bookkeeping is necessary only for organizing with profit objectives.
Answer:
This statement is False.
Books of accounts are maintained by trading and non-trading concerns. Non-trading concerns do not earn any profits so book-keeping is not necessary only with the objective of profit.

Question 28.
The figures of profit and net worth are disclosed by books of accounts.
Answer:
This statement is True.
The purpose of preparing books of accounts is to know financial details of organization from journal ledger, trial balance, final accounts proprietor to get an idea of his financial status. The purpose of financial books of accounts is to get profit and net worth etc.

Question 29.
Financial statements are an effective weapon in the hands of management.
Answer:
This statement is True.
With help of financial statements, management does planning, decision-making for the future. The past record gives ideas to the management to improve financial decisions like more profit better financial conditions etc. so financial statements are effective weapons in the hands of management.

Question 30.
Irregularities, frauds, and misappropriations can be detected only because of books accounts.
Answer:
This statement is False.
Irregularities frauds and misappropriations can be detected by auditing. Continuous and periodical auditing will detect the irregularities frauds and misappropriations which is done by a chartered accountant.

5. Do you agree or disagree with the following statements:

Question 1.
Goodwill is a tangible asset.
Answer:
Disagree

Question 2.
For income tax purposes an accounting year starts on 1st January and ends on 31st March.
Answer:
Disagree

Question 3.
Earning profits is an aim of Not for Profit Concern.
Answer:
Disagree

Question 4.
Trade discount is recorded in the books of accounts.
Answer:
Disagree

Question 5.
Wages paid on installation of machinery are revenue expenditures.
Answer:
Disagree

Question 6.
An amount that is not recoverable from debtors is called bad debt.
Answer:
Agree

Question 7.
The double-entry book-keeping system is invented in India.
Answer:
Disagree

Question 8.
Book-keeping does not have any rules and regulations for recording business transactions.
Answer:
Disagree

Question 9.
The owner of the business does not have any utility of book-keeping.
Answer:
Disagree

Question 10.
Bookkeeping cannot be used as financial evidence in a court of law.
Answer:
Disagree

Question 11.
It is not possible for management to plan and take decisions in business with help of book-keeping and accountancy.
Answer:
Disagree

Question 12.
Cash-withdrawn by the proprietor from his business for personal use is called capital.
Answer:
Disagree

Question 13.
A person who has to pay the business for getting goods and services on credit is known as the debtor.
Answer:
Agree

Question 14.
Trade discount is not recorded in the books of accounts.
Answer:
Agree

Question 15.
Purchase of goods is revenue expenditure.
Answer:
Agree

Question 16.
Heavy expenditure on advertising is deferred revenue expenditure.
Answer:
Agree

Question 17.
A solvent person’s assets are more than his liabilities.
Answer:
Agree

Question 18.
Net Worth = Total Assets – Outsiders Liabilities
Answer:
Agree

Question 19.
Conservatism’s concept means to play safe.
Answer:
Agree

Question 20.
The dual aspect means every transaction has two effects i.e. debit and credit.
Answer:
Agree

6. Complete the following sentences:

Question 1.
In barter transaction goods are exchange for ____________
Answer:
Goods

Question 2.
Asset is ____________ of the business.
Answer:
property

Question 3.
Capital expenditures are ____________ in nature.
Answer:
Non Recurring

Question 4.
A ____________ liability is an uncertain liability.
Answer:
Contingent

Question 5.
A person whose liabilities are more than his assets is known as ____________
Answer:
Insolvent

Question 6.
An article or a commodity which is purchase for resale in business is called ____________
Answer:
Goods

Question 7.
Customs and traditions which guide the accountants to prepare accounting systems are called ____________
Answer:
Conventions

Question 8.
Discount given by the creditor to the debtor on payment of cash is called ____________
Answer:
Cash discount

Question 9.
For tax purpose an year starts on 1st April and ends on 31st March is called ____________
Answer:
Accounting year

Question 10.
An institution which provides standard of accounting in India called ____________
Answer:
Institute of Chartered Accountants of India

Question 11.
Which accounting standard is followed for depreciation on fixed assets ____________
Answer:
AS-6

Question 12.
Brief explanation of an entry is called as ____________
Answer:
Narration

Question 13.
Businessman open ____________ type of bank account.
Answer:
Current

Question 14.
The main objective of business concern ____________
Answer:
making profits

Question 15.
Assets which are held in the business for a period of less than one year ____________
Answer:
Current Assets

Question 16.
10 years loan taken from bank by organisation is ____________ term loan.
Answer:
long

Question 17.
Heavy expenses paid on formation of an organisation is known as ____________ expenditure.
Answer:
Deferred Revenue

Question 18.
Dividend received on shares is ____________ income.
Answer:
Revenue

Question 19.
Loss on sale of old machine is ____________ loss.
Answer:
Capital

Balbharti Maharashtra State Board 12th Commerce Book Keeping & Accountancy Important Questions Chapter 2 Accounts of ‘Not for Profit’ Concerns Important Questions and Answers.

Maharashtra State Board 12th Commerce BK Important Questions Chapter 2 Accounts of ‘Not for Profit’ Concerns

1. Objective Type Questions:

A. Select the most appropriate alternatives from those given below.

Question 1.
Purchases of stationery is a ______________ expenditure.
(a) capital
(b) revenue
(c) long-term
(d) deferred revenue
Answer:
(b) revenue

Question 2.
Usually ______________ is a major source of revenue income for ‘Not for Profit’ concerns.
(a) subscription
(b) donations
(c) legacies
(d) entrance fees
Answer:
(a) subscription

Question 3.
An Income and Expenditure Account and a Balance Sheet are prepared as final accounts by a ______________
(a) ‘Not for Profit’ concern
(b) Trading concern
(c) commercial organisation
(d) Public Limited Company
Answer:
(a) ‘Not for Profit’ concern

Question 4.
Non-cash items are not recorded in ______________
(a) Income and Expenditure Account
(b) Receipts and Payments Account
(c) Balance Sheet
(d) Profit and Loss Account
Answer:
(b) Receipts and Payments Account

Question 5.
The excess of assets over liabilities is termed as ______________
(a) surplus
(b) deficit
(c) capital fund
(d) loan
Answer:
(c) capital fund

Question 6.
For a library, expenditure on the purchase of books is a ______________ Expenditure.
(a) Capital
(b) Revenue
(c) General
(d) Recurring
Answer:
(a) Capital

Question 7.
______________ Account starts with the opening Cash Balance.
(a) Income and Expenditure
(b) Receipts and Payments
(c) Capital Fund
(d) Subscriptions
Answer:
(b) Receipts and Payments

Question 8.
Only ______________ incomes and expenses are shown in the Income and Expenditure Account.
(a) revenue
(b) capital
(c) business
(d) non-recurring
Answer:
(a) revenue

Question 9.
A debit balance on the Income and Expenditure Account denotes ______________
(a) deficit
(b) surplus
(c) profit
(d) excess of income over expenditure
Answer:
(a) deficit

Question 10.
Non-trading organisation writes summary of all cash transactions in the ______________ Account.
(a) Cash
(b) Receipts and Payments
(c) Income and Expenditure
(d) Profit and Loss
Answer:
(b) Receipts and Payments

Question 11.
Both capitalised receipts and capital expenditure are shown in the ______________
(a) Profit and Loss A/c
(b) Balance Sheet
(c) Trading A/c
(d) Income and Expenditure A/c
Answer:
(b) Balance Sheet

Question 12.
‘Not for Profit’ organisation prepares ______________ to find out its financial position.
(a) Receipts and Payments A/c
(b) Balance Sheet
(c) Trading A/c
(d) Income and Expenditure A/c
Answer:
(b) Balance Sheet

Question 13.
Subscriptions received from the members is considered as ______________ receipts.
(a) capital
(b) revenue
(c) non-recurring
(d) commercial
Answer:
(b) revenue

Question 14.
Fund which provides permanent source of income to non-trading organisation is called ______________ fund.
(a) endowment
(b) general
(c) specific
(d) capital
Answer:
(a) endowment

Question 15.
For a public hospital, expenditure on the purchase of medicines is a ______________ Expenditure.
(a) General
(b) Non-recurring
(c) Capital
(d) Revenue
Answer:
(d) Revenue

Question 16.
Which of the following items will not appear in the Balance Sheet of a club?
(a) Subscriptions received in advance
(b) Special donation received during the year
(c) Subscriptions due for the year
(d) Entrance fees paid by new members
Answer:
(d) Entrance fees paid by new members

Question 17.
The main purpose of incurring a ______________ expenditure is to earn an income or to increase the earning capacity of the business.
(a) recurring
(b) capital
(c) revenue
(d) business
Answer:
(b) capital

Question 18.
Excess of Expenditure over Income is termed as ______________
(a) Surplus
(b) Deficit
(c) Capital Fund
(d) Profit
Answer:
(b) Deficit

Question 19.
Receipts and Payments Account is a ______________
(a) Personal Account
(b) Real Account
(c) Nominal Account
(d) Profit and Loss Account
Answer:
(b) Real Account

B. Write the Word/Term/Phrase which can substitute each of the following statements.

Question 1.
Such concerns, which are formed for rendering some useful services to its members without having profit motive.
Answer:
‘Not for Profit’ concern

Question 2.
Excess of expenditure over income of ‘Not for Profit’ concerns.
Answer:
Deficit

Question 3.
A statement showing the financial position of a concern on a particular date.
Answer:
Balance Sheet

Question 4.
The debit balance of an Income and Expenditure Account.
Answer:
Deficit

Question 5.
Fees received from the member only once at the time of his entry into the ‘Not for Profit’ concern.
Answer:
Life membership fees

Question 6.
Specific amount paid by the members annually to non-trading organisation to get certain services or benefits.
Answer:
Subscription

Question 7.
Donation or gift received from the members or outsiders for specific purpose.
Answer:
Specific donation

Question 8.
Donation received for general purpose like welfare of the members or society.
Answer:
General donation

Question 9.
The gifts received from legal representatives as per the will of a deceased person.
Answer:
Legacies

Question 10.
An expenditure which is incurred for carrying the day-to-day business activities.
Answer:
Revenue Expenditure

Question 11.
Capital resources which are owned and possessed by the ‘Not for Profit’ concern.
Answer:
Capital Fund

Question 12.
An account which is prepared by the ‘ Not for Profit’ concern to record summary of all types of receipts and payments.
Answer:
Receipts and Payments Account

Question 13.
Closing debit balance of Receipts and Payments Account.
Answer:
Cash in Hand and or Cash at Bank

Question 14.
A payment made by the non-trading organisation periodically for consecutive issue of magazines, newspapers, etc.
Answer:
Subscriptions paid

Question 15.
The major source of revenue to a ‘Not for Profit’ concern, from its members.
Answer:
Subscriptions

Question 16.
Fees paid by persons to become members of a ‘Not for Profit’ concern.
Answer:
Entrance Fees or Admission Fees

Question 17.
The concerns which prepare Income and Expenditure Account instead of Profit and Loss Account.
Answer:
‘Not for Profit’ concern

Question 18.
The expenditure which is recurring in nature.
Answer:
Revenue Expenditure

Question 19.
The expenditure which is non-recurring in nature.
Answer:
Capital Expenditure

Question 20.
An account opened by non-trading concerns, to find out surplus/deficit during the particular financial year.
Answer:
Income and Expenditure Account

C. State whether the following statements are True or False with reasons.

Question 1.
All receipts are the items of revenue income.
Answer:
This statement is False.
In ‘Not for Profit’ concern receipts includes revenue receipts as well as capital receipts of current year or of previous year or of next year, so we can say that all receipts are not the items of revenue income.

Question 2.
In the Income and Expenditure Account, all incomes received during the year irrespective of the year for which they are received, are to be recorded.
Answer:
This statement is False.
In the Income and Expenditure Account, all revenue incomes, received during the current year, are to be recorded only.

Question 3.
Receipts and Payments Account shows the amount of profit earned or loss suffered during a year.
Answer:
This statement is False.
Receipts and Payments Account shows the amount of receipts and payments (in cash or through bank) of any year in the current year and do not shows any profit earned or loss suffered during a year.

Question 4.
Every year ‘Not for Profit’ concerns, prepares Income and Expenditure Account.
Answer:
This statement is True.
To get the idea of sufficient income, other expenditure, for smooth run of concern, every year ‘Not for Profit’ concern prepares Income and Expenditure Account.

Question 5.
‘Deficit’ means excess of income over expenditure in the Income and Expenditure Account.
Answer:
This statement is False.
In the Income and Expenditure Account ‘Deficit’ means excess of expenditure over incomes.

Question 6.
‘Revenue receipts’ means receipts which are not recurring in nature.
Answer:
This statement is False.
Revenue Receipts means receipts which frequently takes place and which are recurring in nature.

D. Fill in the blanks.

Question 1.
An Income and Expenditure Account and a Balance Sheet are prepared as final account by a ______________
Answer:
‘Not for Profit’ concern

Question 2.
______________ is a major source of revenue income for ‘Not for Profit’ concern.
Answer:
Subscription

Question 3.
Non-cash items are not recorded in ______________
Answer:
Receipts and Payments Account

Question 4.
______________ concerns have profit motive.
Answer:
Trading

Question 5.
In a Trading concerns ______________ is prepared to know the financial position of the business.
Answer:
Balance Sheet

Question 6.
Excess of Receipts over Payments means ______________
Answer:
Cash or Bank balance

Question 7.
Legacy is received by ‘Not for Profit’ concerns on a ______________ basis.
Answer:
permanent

Question 8.
Incomes which are to be capitalised are to be added to ______________
Answer:
Capital Fund

Question 9.
Subscription received in advance, in current year, is to be ______________ from subscription amount.
Answer:
subtracted

Question 10.
Subscription received in advance, in previous year is to be ______________ to subscription amount.
Answer:
added

Question 11.
Locker’s rent is ______________ for ‘Not for Profit’ concern.
Answer:
revenue income

Question 12.
Life membership fees, Legacy, Surplus, etc. are to be ______________ to Capital Fund.
Answer:
added

Question 13.
All revenue incomes and revenue expenses are to be recorded in ______________
Answer:
Income and Expenditure Account

Question 14.
All Receipts and Payments are recorded in the Receipts and Payment Account ______________ of the year.
Answer:
irrespective

Question 15.
Incomes or Expenses having recurring nature are known as ______________ incomes or expenses.
Answer:
revenue

E. Answer in one sentence only.

Question 1.
What is deferred revenue expenditure?
Answer:
Expenditure which is basically revenue expenditure but benefits of which accured to the organisation for more than one year is called deferred revenue expenditure.

Question 2.
What is Entrance Fees?
Answer:
The fees which is paid by the persons who wish to become a member of the organisation are called Entrance Fees.

Question 3.
What is Deficit?
Answer:
Excess of expenditure over income shown by Income and Expenditure Account is called Deficit for the financial year.

Question 4.
State the meaning of Revenue Expenditure.
Answer:
An expenditure which is incurred for carrying day-to-day business activities and maintaining fixed assets in working condition is called Revenue Expenditure.

Question 5.
What do you mean by Capital Expenditure?
Answer:
An expenditure which is non recurring is nature and incurred to purchase new fixed assets, to increase earning capacity, efficiency and working life of the existing fixed assets and to achieve economy of operation an existing fixed assets is called Capital Expenditure.

Question 6.
Which final accounts do the ‘Not for Profit’ concern prepare?
Answer:
‘Not for Profit’ concern prepare Income and Expenditure Account and Balance Sheet as their final accounts.

Question 7.
Give the examples for ‘Not for Profit’ concerns.
Answer:
Examples of ‘Not for Profit’ concerns are: sports club, charitable hospitals, schools, colleges, universities, welfare association, chamber of commerce, etc.

Question 8.
What do you mean by Recurring Expenses?
Answer:
Recurring expenses are those expenses, benefit of which lasts for a maximum period of one year and is increased on purchase of goods or services, in order to carry out the main activity of the business.

Question 9.
Why Receipts and Payments Account is prepared?
Answer:
To record all cash and Bank transactions taken place in the organisation, Receipt and Payment Account is prepared.

Question 10.
In Income and Expenditure Account, which kind of incomes and expenses are to be recorded?
Answer:
In Income and Expenditure Account, ‘Revenue’ incomes and expenses are to be recorded.

Solved Problems

Question 1.
With the information given below regarding ‘Subscription’ give accounting effects of it in the Final Accounts of a ‘Not for Profit’ concern:

Additional Information:
1. Subscriptions received during the year includes:
Subscriptions received for 2018-2019 ₹ 8,750 and for 2020-21 ₹ 7,500.
2. There are 500 members of the concern and each member pays ₹ 500 as annual subscription.
3. During the year 2018-19 subscription received for the year 2019-20 was ₹ 12,500.
Solution:
In the books of _____________________

Balance Sheet as on 31st March, 2020

Working Notes:
Amount of subscriptions outstanding for current year 2019-20 is calculated as follows:
Subscriptions outstanding (receivable) = (Subscriptions due from all members) – (Subscriptions received)
= (500 members × ₹ 500 per member) – (Subscriptions received during current year + Subscription received during previous year)
= (500 × 500) – (2,27,500 + 12,500)
= 2,50,000 – 2,40,000
= ₹ 10,000.

Question 2.
Show the following items in the Income and Expenditure Account for the year ended 31st March, 2018:

Adjustments:
1. Outstanding salaries for 2016-17 is ₹ 11,250 and of 2017 – 18 is ₹ 8,125.
2. Opening stock of stationery is ₹ 5,000 and Closing stock of stationery is ₹ 2,500.
3. There are 70 members paying annual subscription of ₹ 250/- each.
4. Insurance is paid for year ended 30th June, 2018.
Solution:
In the books of _____________________

Working Notes:
1. Outstanding subscriptions for the current year 2017-18 are calculated as follows:
Outstanding subscriptions = Subscriptions due or receivable – Subscriptions received
= 70 × 250 – 15,250
= 17,500 – 15,250
= ₹ 2,250.
Subscriptions for 2016 – 17 and 2018 – 19 will not appear in the Income and Expenditure Account prepared for 2017 – 18.

2. Prepaid insurance is calculated as follows:
Insurance is paid in advance for 3 months i.e. from 1st April, 2018 to 30th June, 2018.
Prepaid insurance = \(\frac{3}{12}\) × Insurance premium paid
= \(\frac{3}{12}\) × 10,000
= ₹ 2,500.
Prepaid insurance deducted from Insurance on the Debit side of Income and Expenditure Account.

Question 3.
The following is the Receipts and Payments Account for the year ended on 31st March, 2019:

Adjustments:
1. Outstanding picnic receipts ₹ 2,550.
2. Furniture was purchased on 01 – 10 – 2018 and it is to be depreciated @ 10% p.a.
3. Outstanding subscriptions for current year ₹ 4,920.
4. Stock of Stationery on 1st April 2018 was ₹ 390 and on 31st March 2019 was ₹ 690.
5. Entire amount of legacies and 50 % of donations are to be capitalized.
With the above information, you are required to prepare Income and Expenditure Account for the year ending on 31st March 2019.
Solution:
In the books of _____________________

Question 4.
From the information given below of Sarthi Education, you are required to prepare Income and Expenditure Account and Balance Sheet for the year ended on 31st March, 2019:
Balance Sheet as on 1st April, 2018


Adjustments:
1. Tuition fees outstanding ₹ 6,750.
2. Outstanding interest on loan ₹ 30,000.
3. Entire admission Fees are to be capitalized.
4. Depreciation is to be written off as under:
Library Books ₹ 25,000, Furniture ₹ 15,000, Laboratory Equipment ₹ 10,000, Building ₹ 15,000.
Solution:
In the books of Sarthi Education

Balance Sheet as on 31st March, 2019

Working Notes:
1. Tuition fees outstanding ₹ 6,750 are first added to Tuition Fees on the credit side of Income and Expenditure A/c and then such outstanding tuition fees are shown on the Assets side of the Balance Sheet.
2. Outstanding interest on Loan ₹ 30,000 is first debited to Income and Expenditure A/c and it is added to Loan on the Liabilities side of Balance Sheet.
3. Government grant (Revenue income) ₹ 1,75,000 is recorded on the credit side of the Income and Expenditure Account.
4. Debit balance of Income and Expenditure Account ₹ 55,850 indicates a deficit. It is deducted from the Capital fund on the Liabilities side of the Balance Sheet.

Balbharti Maharashtra State Board Class 11 Political Science Important Questions Chapter 5 Concept of Representation Important Questions and Answers.

Maharashtra State Board 11th Political Science Important Questions Chapter 5 Concept of Representation

1A. Choose the correct alternative and complete the following statements.

Question 1.
Today, in most countries, the form of government is ___________ democracy. (direct, indirect, concurrent, national)
Answer:
indirect

Question 2.
After the Civil War (1640’s) UK become a ___________ (Republic, Constitutional Monarchy, Absolute Monarchy, Federation)
Answer:
Constitutional Monarchy

Question 3.
The first general elections was held in the year ___________ in India. (1947-48, 1950, 1951-52, 1935)
Answer:
1951-52

Question 4.
In the plurality method, ___________ constituency is required. (single member, multi member, transferable, non-official)
Answer:
single member

Question 5.
___________ System is used in India for presidential elections. (List, FPTP, Majority, Single Transferable vote)
Answer:
Majority

Question 6.
In India ___________ classifies parties as ‘State’ or ‘National’ and allots symbols to them. (President, Parliament, Election Commission, Judiciary)
Answer:
Election Commission

1B. Identify the incorrect pair in every set, correct it and rewrite.

Question 1.
BMS – Trade Union.
ABVP – Peasant’s Union.
FICCI – Business Group
BKU – Agricultural Unions
Answer:
ABVP – Student Union

Question 2.
Government of India Act – 1935
Queen’s Proclamation – 1858
French Revolution – 776
Morley – Minto Reforms – 1909
Answer:
French Revolution – 1789

1C. State the appropriate concept for the given statements.

Question 1.
The idea in the middle ages in Europe was that the king was God’s representative on earth.
Answer:
Divine Rights of Kings

Question 2.
Distinct geographical areas from which representatives are elected.
Answer:
Constituencies

Question 3.
The electoral system in which the candidate who secures the maximum number of votes is declared as elected.
Answer:
Plurality System

Question 4.
Views, objectives of a political party taken together.
Answer:
Ideology

Question 5.
All Indian parties have at least 11 seats in the Lok Sabha from at least three states.
Answer:
National Party

1D. Answer in one sentence.

Question 1.
What is a single-member constituency?
Answer:
A single-member constituency is one from which only a single member can be declared elected.

Question 2.
What is a multi-member constituency?
Answer:
A constituency from which several members can be elected is called a multi-member constituency.

Question 3.
What is the First Part of the Post System?
Answer:
It is a system wherein the candidate with the maximum number of votes is declared elected.

Question 4.
What is Single Transferable Vote System?
Answer:
It is a type of Proportional Representation where voters rank candidates in order of preference.

Question 5.
What is meant by the ideology of a political party?
Answer:
The reflection of the overall views, objectives, and policies of a political party is called its ideology.

Question 6.
Name some state parties in Maharashtra.
Answer:
Shiv Sena, Maharashtra Navnirman Sena, Vanchit Bahujan Aghadi, Rashtriya Samaj Paksha.

1E. Complete the following sentence using the appropriate reason.

Question 1.
Representative democracy is referred to as responsible Government because
(a) representatives are ultimately responsible to the people.
(b) people are responsible for electing the government.
(c) direct democracy is not possible today.
Answer:
(a) representatives are ultimately responsible to the people.

Question 2.
In many European countries, a struggle between the representative assemblies and monarchs arose because
(a) monarchs believed in the Divine Rights of Kings.
(b) it was a period of national awakening.
(c) representative assemblies started insisting on a share in the decision-making process.
Answer:
(c) representative assemblies started insisting on a share in the decision-making process.

Question 3.
Elections to the Lok Sabha is called the ‘First Past the Post’ method because
(a) it is a single-member constituency.
(b) candidates are ranked in order of preference.
(c) the candidate who gets a maximum number of votes is declared as elected.
Answer:
(c) the candidate who gets a maximum number of votes is declared as elected.

1F. Find the odd word in the given set.

Question 1.
Indian National Congress, Bharatiya Janata Party, Shiv Sena, Communist Party of India (Marxist).
Answer:
Shiv Sena (it is a regional party)

Question 2.
Nationalist Congress Party, All India Anna Dravida Munnetra Kazhagam, Telugu Desam Party, Akali Dal.
Answer:
Nationalist Congress Party (it is a national party)

Question 3.
National Students Union of India, Hind Mazdoor Sangh, Akhil Bharatiya Vidyarthi Parishad, Student Federation of India.
Answer:
Hind Mazdoor Sangh (it is a labour pressure group)

2A. State whether the following statements are true or false with reasons.

Question 1.
Today, most countries have an indirect or representative democracy.
Answer:
This statement is True.

• Today, most countries have large territories and populations. Hence, direct democracy is not possible. The form of democracy today is indirect democracy or representative democracy.
• People elect representatives from among themselves to govern the country for e.g., in India, members of Parliament (MP’s), Members of State Legislative Assemblies/Councils (MLA’s, MLC’s), of corporations, etc. are all our representatives.

Question 2.
First Past the Post system is followed for Lok Sabha elections.
Answer:
This statement is True.

• Lok Sabha (general) elections are held all over the country.
• The candidate who gets the maximum number of votes is declared as elected from that constituency. He / She does not need a majority of votes.

Question 3.
Proportional Representation has limited scope.
Answer:
This statement is True.

• In Proportional Representation the number of candidates of a political party to be elected depends on the proportion of votes that it receives.
• It is a lengthy process and needs a multi-member constituency. Hence it is unsuitable for large-scale elections such as elections to the Lok Sabha.

Question 4.
Political parties are important channels for political representation.
Answer:
This statement is True.

• Political parties serve as the primary channels of political representation. In democracies, parties seek to obtain power through elections. Members of various parties contest elections as candidates of their respective parties.
• During the election, the parties present before the voters a programme based on their ideology and promise them that this programme would be implemented if elected to power. Thus, the aspirations and wishes of the voters are represented in the decision-making process through the channel or the medium of a given political party.

Question 5.
Telugu Desam (TDP) is a national party.
Answer:
This statement is False.

• A national party must have a political presence in at least four states.
• TDP is significant only in Andhra Pradesh and to some extent in Telangana.

Question 6.
The Communist Party of India (CPI) can be described as the first political party in the country.
Answer:
This statement is False.

• CPI was formed in 1925 by people who were influenced by the Bolshevik Revolution in Russia (1917) and the communist ideology.
• In 1885, the Indian National Congress was formed as a united front against British Rule. It is considered the first political party in India.

2B. Complete the concept maps.

Question 1.

Answer:

3. Explain the correlation between the following.

Question 1.
Political Parties and Pressure Groups.
Answer:
Political parties are the most important channels for political representation. They are organized groups comprising of persons who hold similar views on a variety of issues or have similar objectives. They seek to obtain political power, generally, through the process of elections. The views of a party taken together are called the party’s ideology. At election times, political parties issue ‘Manifestos’ i.e., what policies/programmes they would implement if voted to power. Every party puts up its candidates who contest election.

Interest and Pressure groups are informal channels that seek to represent the people. A pressure group is an interest group that is organized to influence public opinion and government policy towards the fulfillment of its objectives and without active participation in the electoral process. This includes interest groups in the fields of business such as the Federation of Indian Chambers of Commerce and Industry (FICCI), for labour e.g., Indian National Trade Union Congress (INTUC), Bharatiya Kamgar Sena (BKS), for peasants such as Shetkari Sanghatana, for students such as Akhil Bharatiya Vidyarthi Parishad (ABVP), National Students Union of India (NSUI). In the USA, pressure groups are also called Lobby Groups.

Many pressure groups today are closely affiliated with political parties e.g., ABVP is the student wing of BJP and BKS is the Shiv Sena’s Trade Union. Both the pressure group and the political party will then support each other in times of elections and decision-making in aspects like finance, manpower, and publicity.

4. Answer the following questions.

Question 1.
Explain the Divine Rights Theory.
Answer:
The Divine Rights of Kings Theory was propagated in Europe by Kings like James I (England). It explained that the Monarchs were God’s representatives on earth to whom, the people had to render habitual obedience. The King was infallible and unquestionable as he derived his power from God. Disobedience to the King was akin to sinning against God. This theory was used to strengthen the Absolute Monarchy in Europe.

Question 2.
Explain Representative Assemblies in Europe.
Answer:
Representative Democracy has its origins in medieval Europe. Till that time, Absolute Monarchies existed in most countries. The Divine Rights of Kings Theory was in application. As time went by, monarchs in many countries like England started having Representative Assemblies that represented the population. Soon, these assemblies asked for a share in the decision-making process of the country leading to conflicts between the monarchs and the Assemblies for e.g. French Revolution. Most conflicts ended with reduced/power to the monarchs. The Representative assemblies, now become Political Representatives as they dealt with all government activities.

Question 3.
What are the three methods of representation?
Answer:
The three methods of representation are

• Electoral Method: Persons are directly or indirectly elected by the citizens to govern them as members of representative assemblies e.g., General elections to Lok Sabha and Legislative Assemblies.
• Non-electoral Method: Representatives occupy their position through nomination or appointment for e.g., the President of India appoints 12 members to the Rajya Sabha.
• Non-official Method: Civil society represents the people through various pressure groups like trade unions, student groups, etc.

Question 4.
Explain the Proportional System of Representation.
Answer:
Proportional Representation is generally used in multi-member constituencies. In this system, the number of candidates of a given political party to be elected depends upon the proportion of votes that it receives. For instance, if a political party receives 40% of the votes in a five-member constituency, then two of its candidates will be elected from that constituency. This system is not used in India. There is a sub-type of the proportional system which is known as the Single-Transferable Vote (STV) system. Here the voters have to rank the candidates in order of preferences. This system is used in elections to the Rajya Sabha and to the State Legislative Councils in India.

Question 5.
What is a National Party?
Answer:
Political parties can be classified as National or State parties. The Election commission has decided that a political party shall be eligible to be recognized as a National party if-

• It secures at least six percent (6%) of the valid votes polled in any four or more states, at a general election to the
• House of the People or, to the State Legislative Assembly; and
• In addition, it wins at least four seats in the House of the People from any State or state.

OR

• It wins at least two percent (2%) seats in the House of the People (i.e., 11 seats in the existing House having 543 members), and these members are elected from at least three different States.
• In India, some National Parties include I.N.C., B.J.P., G.P.M., N.C.P., etc.

Question 6.
Name six regional (state) parties.
Answer:

• Telugu Desam Party (TDP) – Andhra Pradesh,
• Telangana Rashtriya Samiti – Telangana,
• Dravida Munnetra Kazhagam (DMK) and All India Dravida Munnetra Kazhagam (AIADMK) – Tamil Nadu,
• National Conference – Jammu & Kashmir
• Assam Gana Parishad (AGP) – Assam.
• Shiv Sena – Maharashtra.

Question 7.
Name 4 trade unions and the political parties they are affiliated to.
Answer:

Trade Unions	Political Parties affiliated to
Indian National Trade Union Congress (INTUC)	Indian National Congress
All India Trade Union Congress (AITU)	Communist Party of India
Bharatiya Kamgar Sena (BKS)	Shiv Sena
The Bharatiya Mazdoor Sangh (BMS)	Bharatiya Janata Party (B.J.P)
Centre for Indian Trade Unions (CITU)	Communist Party of India (Marxist) (CPM)

Question 8.
How do pressure groups differ from social movements?
Answer:
Pressure groups are different from social movements. The pressure groups usually have a more formalized structure. This is why sometimes interest groups are described as representing ‘organized interest’. Social movements usually do not have a formal structure or organisation. They take up a cause and pursue it. (Example: Chipko Movement)

Question 9.
Write two examples of NGOs in India in the following field.

1. Child Welfare
2. Animal Welfare
3. Aged Persons
4. Disabled Persons
5. Environment
6. Women’s Welfare

Answer:

1. Child welfare – Child Rights and You (CRY), Akansha
2. Animal Welfare – PETA, People for Animals (PFA)
3. Aged Persons – Help Age, Dignity Foundation.
4. (iv) Disabled Persons – National Association for the Blind (NAB), GCCI, ADAPT.
5. Environment – BNHS, BEAG.
6. Women’s Welfare – SEWA, SNEHA, WIT

5. Observe the given image and writes in brief about it.

Answer:
This is a heart-warming and motivating photograph.
We can observe the following about it.

• In India, women have participated in the election process since 1950 when they were given voting rights.
• It shows political awareness and participation of women, dressed in traditional attire, braving the hot sun standing in the queue to vote.
• They are proudly holding up their identity proof which shows how motivated and proud they are to have the political right to vote.

Balbharti Maharashtra State Board Class 11 Political Science Important Questions Chapter 4 Constitutional Government Important Questions and Answers.

Maharashtra State Board 11th Political Science Important Questions Chapter 4 Constitutional Government

1A. Choose the correct alternative and complete the following statements.

Question 1.
Constitution of the ____________ was made by the Constitutional Convention. (USA, UK, India, France)
Answer:
USA

Question 2.
The Magna Carta has it’s origin in ____________ (USA, England, France, Cuba)
Answer:
England

Question 3.
Till, recently the doctrine of absolute Parliamentary Sovereignty existed in ____________ (USA, Mexico, Argentina, UK)
Answer:
UK

Question 4.
____________ is an example of a ‘Holding Together’ federation. (USA, India, UK, Portugal)
Answer:
UK

Question 5.
____________ is an example of a ‘Coming Together’ federation. (USA, India, UK, Portugal)
Answer:
USA

Question 6.
Protection of rights is entrusted to the ____________ (Legislature, Executive, Civil Services, Judiciary)
Answer:
Judiciary

Question 7.
A ____________ system functions on ‘Separation of Powers’ theory. (dictatorship, parliamentary, presidential, federation)
Answer:
Presidential

1B. Identify the incorrect pair in every set, correct it and rewrite.

Question 1.
(a) England – Republic
(b) USA – Federation
(c) Portugal – Unitary System
Answer:
(a) England – Constitutional Monarchy

Question 2.
(a) Union List – Defence
(b) State List – Atomic Energy
(c) Concurrent List – Education
Answer:
(b) Union list – Atomic energy or State list – Public health and sanitation

Question 3.
(a) Senate – USA
(b) Rajya Sabha – India
(c) House of Lords – Brazil
Answer:
(c) Houses of Lords – England

1C. State the appropriate concept for the given statement.

Question 1.
The idea that there should be limitations on powers of the government.
Answer:
Constitutionalism

Question 2.
First ten amendments to the American constitution.
Answer:
Bill of Rights

Question 3.
Type of government in which Head of State assumes his/her position on a hereditary basis.
Answer:
Monarchy

Question 4.
Process of bringing out changes in some provisions of the constitution.
Answer:
Amendment

Question 5.
The manner in which those who hold power are expected to behave.
Answer:
Constitutional Morality

Question 6.
Executive in a parliamentary system in whose name all powers are exercised.
Answer:
Nominal Executive

1D. Answer in one sentence.

Question 1.
What is the modern view of constitutionalism?
Answer:
The modern view of constitutionalism is the idea of restricting the powers of the government as a whole.

Question 2.
Explain the doctrine of Parliamentary Sovereignty.
Answer:
The doctrine of Parliamentary Sovereignty means that the Parliament which represents the citizens has the power to make laws with no restrictions on it’s jurisdiction.

Question 3.
What was the ruling in the Kesavananda Bharati case?
Answer:
The ruling in the Kesavananda Bharati case was that ‘the basic structure of the constitution could not be altered by any amendments carried out by the legislature.

Question 4.
What is Constitutional Morality?
Constitutional Morality refers to the values which are the foundation of the constitution and the manner in which those in political power are expected to behave.

Question 5.
In a Parliamentary system who constitutes the real executive?
Answer:
The Prime Minister and the Council of Ministers i.e., the Ministry constitute the real executive in a Parliamentary System.

Question 6.
Name the two kinds of executive in a parliamentary system of government.
Answer:
The two kinds of executive in a parliamentary system of Government are nominal executive and real executive.

Question 7.
Name the two houses of legislature in the following:

1. Indian
2. England
3. USA

Answer:

1. India – Lok Sabha, Rajya Sabha
2. England – House of Commons, House of Lords
3. USA – Senate, House of Representatives

Question 8.
What is the ‘veto power’ of US President The US President has the right to reject a law passed by the legislature. This is the ‘veto power’.

Question 9.
What is the unitary system of Government?
Answer:
Countries with small territory usually have a single government at the centre which is called unitary government.

Question 10.
What is the significance of the Seventh Schedule?
Answer:
The Seventh Schedule consists of the Union, State and the Concurrent lists on the basis of which government powers are distributed in India.

1E. Find the odd word out in the given set.

Question 1.
USA, UK, India, Australia.
Answer:
UK (not a federation)

Question 2.
USA, Brazil, Argentina, Japan.
Answer:
Japan (not a presidential system)

Question 3.
USA, Canada, Australia, India.
Answer:
India (not a ‘coming together’ federation)

2A. State whether the following statements are true or false with reasons.

Question 1.
Indian Constitution is enacted.
Answer:
This statement is True.

• The Indian Constitution was framed by the Constituent Assembly of India which functioned from December 1946 till November 1949.
• It is a product of detailed discussions, debates and deliberations. The constitution came into force on 26th January 1950.

Question 2.
Today the doctrine of Parliamentary Sovereignty no longer exists in it’s absolute form in the United Kingdom.
Answer:
This statement is True.

• According to the doctrine of Parliamentary Sovereignty, the Parliament has the authority to make any law and the only control mechanism is a vigilant public opinion.
• Today, the United Kingdom is a member of various international organizations and signatory to many international agreements which guarantee individual rights and restrict parliamentary powers.

Question 3.
In a parliamentary system, the Head of State is powerful.
Answer:
This statement is False.

• A parliamentary system makes a distinction between the Head of State and Head of government. The Head of State is the nominal executive. While the Prime Minister and his/her council are the head of government.
• All decisions and administration is conducted in the name of the nominal executive (President) by the real executive (Prime Minister and his/her Council of Ministers.)

2B. Complete the concept maps.

Question 1.

Answer:

Question 2.

Answer:

Question 3.

Answer:

3. Express your opinion of the following.

Question 1.
The Presidential System may lead to a deadlock in government functioning.
Answer:
The President is both Head of State and Head of government. He/she is directly elected by the people for a fixed tenure for e.g., US President is elected for a 4 year tenure. There is only one executive. The legislature (Congress in the USA) is also directly elected by the citizens.

There exists a separation of legislative and executive powers as well as a system of ‘checks and balances’ for e.g., Legislature can impeach the President, while the President can exercise the ‘Veto Power’ to reject any law passed by the Legislature.

Thus, there can be an impasse in government functioning for e.g., since President Trump assumed office- there have been many cases of a standoff between the office of the President and the US Congress, especially the Democrats. In 2019, the Congress voted to overturn President Trump’s emergency declaration to build a border wall with Mexico. In turn, the President ‘vetoed’ this vote.

4. Answer the following questions.

Question 1.
Explain the nature of Indian Federation.
Answer:
In India, at the time of independence, there were Princely States and areas under British administration. The States were created after independence on the basis of language i.e., linguistic reorganisation of States. The Union Government created the States. The journey of Indian Federalism has been mixed. After independence, the States had been granted additional powers. However, later economic and technological changes had led to the enhancement of the powers of the Central government.

The Indian Federation differs greatly from the US federation. India has been described as ‘quasi-federation’ or a ‘federation with an unitary spirit’ as the division of powers favors the central government for e.g., it has full control over the Union list and Residuary subjects and it’s laws have precedence over state legislations even in case of the subjects in Concurrent list.

Question 2.
Explain the components of a Constitution.
Answer:
The constitution is the highest law of the country. It reflects the objectives of the state and the rights and aspiration of its citizens. It establishes the rule of law and sets limits on government authority. A constitution is a living document that indicates the way in which a country is governed. The primary function of the constitution is to lay out the basic structure of the government according to which the people are to be governed.

A constitution has three distinct but interrelated components.

• Set of Rules – A constitution is a set of rules that describes the structure, powers and functions of the three organs of government to ensure that each organ functions without its jurisdiction. It lays down the limitations on what the government can do or cannot do.
• Set of Rights – A constitution lists the rights of the citizens, means for protection of this rights and the duties of citizens. It also lists the means of protecting the rights e.g., in India, the judiciary is entrusted with protecting the rights. The rights guaranteed by the constitution are not unlimited i.e. they are subject to reasonable limitations.
• Set of Objectives and Values – A constitution enumerates the values and objectives that it seeks to fulfill. For e.g., Indian Constitution seeks to ensure the values of justice, liberty and equality.

Question 3.
Explain Parliamentary system.
Answer:
The two main types of democratic governments are Parliamentary System (as seen in the United Kingdom, India, Canada, Australia, Japan, etc.) and Presidential System (which exists in the United State of America, Argentina, Mexico, Brazil, etc.). This distinction is mainly based on the nature of Legislature-Executive relationship.

Parliamentary System – It makes a distinction between Head of State (President of India) and Head of Government (Prime Minister and his Council of Ministers).

The main features of the parliamentary system are:

• There is a fusion of legislature and executives powers. The executive i.e., the ministry is drawn from the legislature and is subordinate to it. Ministers are also members of Parliament.
• There are two executives i.e., nominal (President of India or Monarch in England) and real (ministry). All powers are exercised by the real executive although it is conducted in the name of the nominal executive.
• It is a responsible government- The Prime Minister and the Council of Ministers stay in power only as long as they have the required majority in the Parliament. In case, the Ministry loses majority support, the Prime Minister along with his Council of Ministers has to resign.
• It may exist either as Republics or as Constitutional Monarchies depending on the nature of the nominal executive. In a Republic, the nominal executive is elected while in a Monarchy, he/she assumes position on the basis of heredity.
• Most Parliamentary systems have a Bicameral Parliament for e.g., in England, Parliament consists of House of Commons (lower house-directly elected)] and House of Lords (Upper house hereditary basis)

Question 4.
Explain Presidential system.
Answer:
The main features of Presidential system are:

• The President who is directly elected by the citizens for a fixed tenure is both, Head of State and Head of Government.
• Thus, there is only one executive.
• The Legislature is also directly elected. Members of the executive are not permitted to belong to the legislature.
• There exists a separation of legislative and executive powers as well as a system of checks and balances for e.g.
• The legislature can impeach the President while the President, can exercise the ‘Veto Power’ to reject any law passed by the
• Legislature.
• The President can continue in office irrespective of whether or not he/she enjoys majority support in the Legislature.

Question 5.
What are the two processes of forming a Federation?
Answer:
Federation may be performed by two processes-

• Small political units ‘come together to establish a single, large political unit for e.g., thirteen colonies came together to fight for independence from British rule and the US federation came into being. This is called centripetal process.
• States are created by the union government for e.g., in India, States were reorganized on the basis of language and other regional aspirations. This is the centrifugal process.

Question 6.
Explain Unitary System of Government.
Answer:
Countries that are small in size prefer to have a single, central, government. This is called the Unitary System. It is seen in Cuba, France, Bolivia, Israel, Portugal, Sri Lanka, etc, Some hitherto unitary systems change to a quasi-unitary form, through establishment of provinces and distribution of political power to somewhat autonomous units, for e.g., UK has an unitary system. However, it’s regions i.e. Scotland, Wales, Northern Ireland have their own assemblies with some degree of autonomy. These are known as ‘Holding Together’ federations.

Question 7.
Write about the Seventh Schedule of Indian Constitution.
Answer:
The Seventh Schedule of the constitution contains three lists i.e., the Union List, the State List and the Concurrent List. Each list contains subjects over which the Central Government, The State Governments (as far as their respective states are concerned), or both the governments can take decisions and make laws respectively. In cases where both the Central and State governments have made laws about subjects falling in the Concurrent List, then the decision of the former prevails. Furthermore, the State Governments can also ask the Central Government to make laws on subjects included in the State List, if such a need arises.

Question 8.
Explain Basic Structure Doctrine.
Answer:
The Supreme Court of India, in the celebrated Keshavananda Bharati (1973) laid down the restrictions on the power of the Government to amend the Constitution. It ruled that the Constitution of India possessed a basic structure which could not be altered in any manner, and that other than this there were no restrictions on the power of parliament to amend the Constitution. This is known as Basic Structure Doctrine.

5. Answer the following in detail with reference to the given points.

Question 1.
Explain Federation.
(i) What is a federation?
(ii) Features of a federation.
(iii) Processes of forming a federation.
(iv) Quasi-federal nature of Indian Federation.
Answer:
(i) A federation refers to a political structure in which there are two sets of governments i.e. one for the whole country and governments in each of the federal units (called Provinces or States). There is a distribution of powers between the Federal government (also known as Union or Central Government) and the State Governments. Federal governments are preferred in countries having large size and heterogeneous population.

(ii) The main features of a federation are

• Dual set of governments i.e., Union government and State governments.
• Division of power between the two sets of governments for e.g., in India, jurisdiction is distributed between the Union (Centre) and States on the basis of Union, State and Concurrent list (as stated in Seventh Schedule of the Constitution)
• A written constitution to enable a clear distribution of government powers.
• Independent judiciary to resolve center-state or state-state disputes.

(iii) Federation may be performed by two processes

• Small political units ‘come together to establish a single unit for e. g., thirteen colonies came together to fight for independence from British rule and the US federation came into being. This is called centripetal process.
• States are created by the union government for e.g., in India, states were reorganized on the basis of language and other regional aspirations. This is the centrifugal process.

(iv) India has been described as ‘quasi-federation’ or a ‘federation with an unitary spirit’ as the division of powers favors the central government for e.g. it has full control over the Union list as well as over residuary subjects. It’s laws have precedence even in case of the subjects in Concurrent list.

Balbharti Maharashtra State Board 12th Biology Important Questions Chapter 8 Respiration and Circulation Important Questions and Answers.

Maharashtra State Board 12th Biology Important Questions Chapter 8 Respiration and Circulation

Multiple-choice Questions

Question 1.
The nasal cavity is divided into right and left nasal chambers by a …………………..
(a) sphenoid
(b) palatine
(c) mesethmoid
(d) zygomatic
Answer:
(c) mesethmoid

Question 2.
The right lung is divided into …………………..
(a) 3 lobes
(b) 2 lobes
(c) 4 lobes
(d) 6 lobes
Answer:
(a) 3 lobes

Question 3.
Carbon dioxide is carried in the blood mainly as …………………..
(a) sodium carbonate
(b) sodium bicarbonate
(c) carbaminohaemoglobin
(d) carbonic acid
Answer:
(b) sodium bicarbonate

Question 4.
Transport of oxygen is carried out by …………………..
(a) plasma
(b) lungs
(c) RBCs
(d) nostrils
Answer:
(c) RBCs

Question 5.
Respiration taking place at the alveoli of lungs is called …………………..
(a) internal respiration
(b) external respiration
(c) cellular respiration
(d) tissue respiration
Answer:
(b) external respiration

Question 6.
The volume of air inspired or expired during normal breathing is …………………..
(a) ERV
(b) IRV
(c) TV
(d) VC
Answer:
(c) TV

Question 7.
What is the partial pressure of oxygen and carbon dioxide respectively in the atmospheric air?
(a) PPO₂ 159 mmHg, PPCO₂2 0.3 mmHg
(b) PPO₂ 104 mmHg, PPCO₂ 40 mmHg
(c) PPO₂ 40 mmHg, PPCO₂ 45 mmHg
(d) PPO₂ 95 mmHg, PPCO₂ 40 mmHg
Answer:
(b) PPO₂ 104 mmHg, PPCO₂ 40 mmHg

Question 8.
The vital capacity of human lung is equal to …………………..
(a) 3500 ml
(b) 4600 ml
(c) 500 ml
(d) 1200 ml
Answer:
(b) 4600 ml

Question 9.
The exchange of gases between alveolar air and alveolar capillaries occurs by …………………..
(a) osmosis
(b) active transport
(c) absorption
(d) diffusion
Answer:
(d) diffusion

Question 10.
Oxygen dissociation curve will shift to right on the decrease of …………………..
(a) acidity
(b) carbon dioxide concentration
(c) temperature
(d) pH
Answer:
(d) pH

Question 11.
Respiratory organs in scorpion are …………………..
(a) gills
(b) book lungs
(c) skin
(d) book gills
Answer:
(b) book lungs

Question 12.
Breakdown of alveoli of lungs resulting in reducing surface area for gas exchange is known as …………………..
(a) emphysema
(b) sneezing
(c) pneumonia
(d) tuberculosis
Answer:
(a) emphysema

Question 13.
During inspiration, the diaphragm …………………..
(a) relaxes
(b) contracts
(c) expands
(d) shows no change
Answer:
(b) contracts

Question 14.
Over inflation of the lungs is prevented due to …………………..
(a) Bohr’s effect
(b) Conditioned reflex
(c) Hering-Breuer reflex
(d) Haldane effect
Answer:
(c) Hering-Breuer reflex

Question 15.
Which of the following prevents collapsing of trachea?
(a) Muscles
(b) Diaphragm
(c) Ribs
(d) Cartilaginous rings
Answer:
(d) cartilaginous rings

Question 16.
Which one of the following produces antibodies ?
(a) Monocytes
(b) Erythrocytes
(c) Lymphocytes
(d) Monocytes
Answer:
(c) Lymphocytes

Question 17.
Plasma protein which initiate blood coagulation is …………………..
(a) prothrombin
(b) fibrinogen
(c) thrombin
(d) fibrin
Answer:
(a) prothrombin

Question 18.
The covering of heart is …………………..
(a) perichondrium
(b) pericardium
(c) periosteum
(d) peritoneum
Answer:
(b) pericardium

Question 19.
Left atrioventricular aperture is guarded by …………………..
(a) tricuspid valve
(b) Eustachian valve
(c) bicuspid valve
(d) semilunar valve
Answer:
(c) bicuspid valve

Question 20.
The pulmonary trunk and systemic aorta are joined by …………………..
(a) chordae tendinae
(b) columnae carnae
(c) ligamentum arteriosum
(d) Purkinje fibres
Answer:
(c) ligamentum arteriosum

Question 21.
Atrioventricular node is located in …………………..
(a) left atrium
(b) right atrium
(c) left ventricle
(d) right ventricle
Answer:
(b) right atrium

Question 22.
…………………. is most commonly used to feel pulse.
(a) Radial vein
(b) Brachial artery
(c) Brachial vein
(d) Radial artery
Answer:
(d) Radial artery

Question 23.
QRS is related to …………………..
(a) atrial contraction
(b) ventricular contraction
(c) atrial relaxation
(d) ventricular relaxation
Answer:
(b) ventricular contraction

Question 24.
Blood is a fluid connective tissue derived from …………………..
(a) ectoderm
(b) mesoderm
(c) endoderm
(d) epithelium
Answer:
(b) mesoderm

Question 25.
What is the increase in number of RBCs called?
(a) Erythropoiesis
(b) Polycythaemia
(c) Erythrocytopenia
(d) Erythroblastosis
Answer:
(b) Polycythaemia

Question 26.
What is the increase in the number of WBCs called?
(a) Leucopoiesis
(b) Leukopenia
(c) Leucocytosis
(d) Leukaemia
Answer:
(c) Leucocytosis

Question 27.
In which of the following diseases there is uncontrolled increase in number of WBCs ?
(a) Leucopoiesis
(b) Leukopenia
(c) Leucocytosis
(d) Leukaemia
Answer:
(d) Leukaemia

Question 28.
What is the decrease in the number of WBCs called?
(a) Leucopoiesis
(b) Leukopenia
(c) Leucocytosis
(d) Leukaemia
Answer:
(b) Leukopenia

Question 29.
Which is the correct arrangement of types of WBCs with respect to their number in blood?
(Consider Neutrophil = N, Eosinophil = E, Basophil = B, Monocyte = M and Lymphocyte = L)
(a) NLMEB
(b) BEMLN
(c) NEBLM
(d) MEBLN
Answer:
(a) NLMEB

Question 30.
Which is the correct order in which the proteins participate in clotting of blood?
(a) Prothrombinase → Prothrombin → Thromboplastin → Thrombin
(b) Thromboplastin → Prothrombinase → Prothrombin → Thrombin
(c) Prothrombin → Thromboplastin → Thrombin → Prothrombinase
(d) Thrombin → Prothrombin → Thromboplastin → Prothrombinase
Answer:
(b) Thromboplastin → Prothrombinase → Prothrombin → Thrombin

Question 31.
Decrease in platelet count is called …………………..
(a) thrombocytopenia
(b) thrombocytosis
(c) thrombokinase
(d) thromboplastin
Answer:
(a) thrombocytopenia

Question 32.
Atrioventricular groove is also called a …………………..
(a) foramen ovale
(b) ligamentum arteriosum
(c) coronary sulcus
(d) ductus arteriosus
Answer:
(c) coronary sulcus

Question 33.
The coronary sinus opens into the …………………..
(a) left atrium
(b) right atrium
(c) left ventricle
(d) right ventricle
Answer:
(b) right atrium

Question 34.
Name the valve from the following that guards the opening of inferior vena cava.
(a) Tricuspid valve
(b) Semilunar valve
(c) Eustachian valve
(d) Thebesian valve
Answer:
(c) Eustachian valve

Question 35.
Name the valve from the following guarding the opening of coronary sinus …………………..
(a) Thebesian valve
(b) Eustachian valve
(c) Tricuspid valve
(d) Semilunar valve
Answer:
(a) Thebesian valve

Question 36.
What is an oval aperture in the interatrial septum of the foetus called?
(a) Fossa ovalis
(b) Foramen ovalis
(c) Ligamentum arteriosum
(d) Ductus arteriosus
Answer:
(b) Foramen ovalis

Question 37.
What is the meaning of stroke volume?
(a) Amount of blood in the body
(b) Pressure of contraction of heart
(c) Amount of blood put out of the ventricles in one minute
(d) Amount of blood put out of the ventricles in one beat
Answer:
(d) Amount of blood put out of the ventricles in one beat

Question 38.
How much amount of blood is put out of the heart during one minute?
(a) Equal to cardiac output
(b) Equal to stroke volume
(c) Equal to half of blood volume
(d) Equal to quarter of blood volume
Answer:
(a) Equal to cardiac output

Question 39.
What is the time taken for one cardiac cycle of normal human being?
(a) 0.1 second
(b) 0.3 second
(c) 0.4 second
(d) 0.8 second
Answer:
(d) 0 .8 second

Question 40.
Deposition of fatty substances in the lining of arteries results in …………………..
(a) arteriosclerosis
(b) atherosclerosis
(c) hyperglycemia
(d) hypotension
Answer:
(b) atherosclerosis

Question 41.
Largest number of white blood corpuscles are …………………..
(a) eosinophils
(b) basophils
(c) neutrophils
(d) monocytes
Answer:
(c) neutrophils

Question 42.
Which of the following animal have open circulatory system?
(a) Earthworm
(b) Cockroach
(c) Frog
(d) Rabbit
Answer:
(b) Cockroach

Question 43.
Which of the following leucocytes have unlobed nucleus?
(a) lymphocyte
(b) eosinophils
(c) neutrophils
(d) basophils
Answer:
(a) lymphocyte

Question 44.
Carbonic anhydrase is found in …………………..
(a) WBC
(b) RBCs
(c) thrombocytes
(d) blood plasma
Answer:
(b) RBCs

Question 45.
The typical Lubb – Dup sounds heard in the heart of a healthy person are due to …………………..
(a) closing of cuspid valves followed by the closing of the semilunar valves
(b) closing of semilunar valves
(c) closing of tricuspid valves
(d) closing of bicuspid valves
Answer:
(a) closing of cuspid valves followed by the closing of the semilunar valves

Match the columns

Question 1.

Animal	Respiratory organ
(1) Fishes	(a) Trachea
(2) Birds/Reptiles	(b) Moist cuticle
(3) Insects	(c) Gills
(4) Earthworm	(d) Lungs

Answer:

Animal	Respiratory organ
(1) Fishes	(c) Gills
(2) Birds/Reptiles	(d) Lungs
(3) Insects	(a) Trachea
(4) Earthworm	(b) Moist cuticle

Question 2.

Respiratory organs	Alternative name
(1) Larynx	(a) Lid of larynx
(2) Trachea	(b) Air sacs
(3) Alveoli	(c) Sound box
(4) Epiglottis	(d) Windpipe

Answer:

Respiratory organs	Alternative name
(1) Larynx	(c) Sound box
(2) Trachea	(d) Windpipe
(3) Alveoli	(b) Air sacs
(4) Epiglottis	(a) Lid of larynx

Question 3.

Respiratory capacities	Respiratory volumes
(1) Residual volume (RV)	(a) 500 ml
(2) Vital capacity (VC)	(b) 2000 – 3000 ml
(3) Tidal volume (TV)	(c) 1100 – 1200 ml
(4) Inspiratory reserve volume (IRV)	(d) 4100 – 4600 ml

Answer:

Respiratory capacities	Respiratory volumes
(1) Residual volume (RV)	(c) 1100 – 1200 ml
(2) Vital capacity (VC)	(d) 4100 – 4600 ml
(3) Tidal volume (TV)	(a) 500 ml
(4) Inspiratory reserve volume (IRV)	(b) 2000 – 3000 ml

Question 4.

Disease	Symptoms
(1) Asthma	(a) Fully blown out alveoli
(2) Bronchitis	(b) Inflammation of lungs with cough and fever
(3) Emphysema	(c) Spasm of Bronchial muscles
(4) Pneumonia	(d) Inflammation of bronchi

Answer:

Disease	Symptoms
(1) Asthma	(c) Spasm of Bronchial muscles
(2) Bronchitis	(d) Inflammation of bronchi
(3) Emphysema	(a) Fully blown out alveoli
(4) Pneumonia	(b) Inflammation of lungs with cough and fever

Question 5.

Valves in heart	Location
(1) Bicuspid/Mitral valve	(a) Opening of inferior vena cava
(2) Tricuspid valve	(b) Opening of coronary sinus
(3) Eustachian valve	(c) Left atrioventricular aperture
(4) Thebesian valve	(d) Right atrioventricular aperture

Answer:

Valves in heart	Location
(1) Bicuspid/Mitral valve	(c) Left atrioventricular aperture
(2) Tricuspid valve	(d) Right atrioventricular aperture
(3) Eustachian valve	(a) Opening of inferior vena cava
(4) Thebesian valve	(b) Opening of coronary sinus

Question 6.

Blood vessel	Functions
(1) Pulmonary aorta	(a) Carries oxygenated blood to left atrium
(2) Superior vena cava	(b) Carries oxygenated blood to all body parts
(3) Pulmonary vein	(c) Carries deoxygenated blood from upper parts of body to right atrium
(4) Aorta	(d) Carries deoxygenated blood to lungs

Answer:

Blood vessel	Functions
(1) Pulmonary aorta	(d) Carries deoxygenated blood to lungs
(2) Superior vena cava	(c) Carries deoxygenated blood from upper parts of body to right atrium
(3) Pulmonary vein	(a) Carries oxygenated blood to left atrium
(4) Aorta	(b) Carries oxygenated blood to all body parts

Question 7.

Cells	Functions
(1) T-lymphocytes	(a) Phagocytic in function
(2) Neutrophils	(b) Responsible for Humoral immunity
(3) Eosinophils/Acidophils	(c) Responsible for cell-medicated immunity
(4) B-lymphocytes	(d) Anti-allergic [Antihistamine] in function

Answer:

Cells	Functions
(1) T-lymphocytes	(c) Responsible for cell-medicated immunity
(2) Neutrophils	(a) Phagocytic in function
(3) Eosinophils/Acidophils	(d) Anti-allergic [Antihistamine] in function
(4) B-lymphocytes	(b) Responsible for Humoral immunity

Question 8.

Waves recorded in ECG	Heart activity
(1) P wave	(a) Ventricular repolarization
(2) QRS complex	(b) Atrial depolarization
(3) T wave	(c) Isoelectric segment
(4) ST segment	(d) Ventricular depolarization

Answer:

Waves recorded in ECG	Heart activity
(1) P wave	(b) Atrial depolarization
(2) QRS complex	(d) Ventricular depolarization
(3) T wave	(a) Ventricular repolarization
(4) ST segment	(c) Isoelectric segment

Question 9.

Events in cardiac cycle	Time duration
(1) Atrial systole	(a) 0.3 second
(2) Atrial diastole	(b) 0.5 second
(3) Ventricular systole	(c) 0.1 second
(4) Ventricular diastole	(d) 0.7 second

Answer:

Events in cardiac cycle	Time duration
(1) Atrial systole	(c) 0.1 second
(2) Atrial diastole	(d) 0.7 second
(3) Ventricular systole	(a) 0.3 second
(4) Ventricular diastole	(b) 0.5 second

Classify the following to form Column B as per the category given in Column A

Question 1.
Classify the following composition of blood plasma given below as per Column ‘A’ and complete Column ‘B’. Select from the given options
(i) Serum albumin
(ii) Bicarbonates
(iii) Urea
(iv) Sulphates of sodium
(v) Fibrinogen
(vi) Uric acid

Column A	Column B
(1) Plasma proteins	————
(2) Nitrogenous waste	————
(3) Inorganic salts	————

Answer:

Column A	Column B
(1) Plasma proteins	Serum albumin Fibrinogen
(2) Nitrogenous waste	Urea, Uric acid
(3) Inorganic salts	Bicarbonates, Sulphates of sodium

Question 2.
Classify the following animals having different respiratory organs given below as per Column ‘A’ and complete Column ‘B’.
Select from the given options:
(i) Scorpion
(ii) Reptiles
(iii) Amphibian tadpoles of frog
(iv) Spiders
(vi) Salamanders
(v) Birds

Column A	Column B
(1) External gills	————
(2) Book lungs	————
(3) Lungs	————

Answer:

Column A	Column B
(1) External gills	Amphibian tadpoles of frog, Salamanders
(2) Book lungs	Scorpion, Spiders
(3) Lungs	Reptiles, Birds

Question 3.
Classify the following disorders of respiratory system given below as per Column A and complete Column ‘B’. Select from the given options:
(i) Pneumonia
(ii) Asbestosis
(iii) Emphysema
(iv) Laryngitis
(v) Chronic bronchitis
(vi) Silicosis

Column A	Column B
(1) Occupational disorders	————
(2) Disorders due to smoking and air pollution	————
(3) Disorders due to viruses and bacteria	————

Answer:

Column A	Column B
(1) Occupational disorders	Asbestosis, Silicosis
(2) Disorders due to smoking and air pollution	Emphysema, Chronic bronchitis
(3) Disorders due to viruses and bacteria	Pneumonia, Laryngitis

Question 4.
Classify the following white blood corpuscles given below as per Column A and complete Column ‘B’. Select from the given options:
(i) Eosinophils
(ii) T-lymphocytes
(iii) Neutrophils
(iv) Basophils
(v) B-lymphocytes
(vi) Monocytes

Column A	Column B
(1) Phagocytic cells	————
(2) Cells involved in giving immune response	————
(3) Cells that increase during allergic and anti-allergic responses	————

Answer:

Column A	Column B
(1) Phagocytic cells	Neutrophils Monocytes
(2) Cells involved in giving immune response	T-lymphocytes-B-lymphocytes
(3) Cells that increase during allergic and anti-allergic responses	Eosinophils, Basophils

Very Short Answer Questions

Question 1.
How many molecules of ATP are formed when one molecule of glucose is oxidized?
Answer:
38 molecules of ATP are formed when one molecule of glucose is oxidized.

Question 2.
What are the three regions of nasal chamber?
Answer:
Vestibule, respiratory part and sensory part are the three regions of nasal chamber.

Question 3.
What is meant by respiratory cycle?
Answer:
Alternate inspiration and expiration together make one respiratory cycle.

Question 4.
Why is it dangerous to sleep in a garage where automobiles have running engines?
Answer:
It is dangerous to sleep in a garage where automobiles have running engines because it may cause carbon monoxide poisoning.

Question 5.
In which form major part of CO₂ is transported in the blood?
Answer:
CO₂ is transported in the blood in the form of sodium and potassium bicarbonates.

Question 6.
Which are the parts of plant that help in the process of gaseous exchange?
Answer:
The parts of plants that help in the process of gaseous exchange are stomata, lenticels, etc.

Question 7.
Which respiratory membranes help in gaseous exchange between the alveolar air and the blood?
Answer:
The layer of squamous epithelium lining the alveolus, basement membrane and a layer of squamous epithelium lining the capillary wall help in gaseous exchange between the alveolar air and the blood.

Question 8.
When will the oxygen dissociation curve shift towards the right?
Answer:
The oxygen dissociation curve will shift towards the right due to increase in H⁺ concentration, increase in _PPCO₂ rise in temperature and rise in DPG (2, 3 diphosphoglycerate), formed in RBCs during glycolysis.

Question 9.
What is the action of carbonic anhydrase in the RBCs of blood?
Answer:
In the RBCs, CO₂ combines with water in the presence of a Zn containing enzyme, carbonic anhydrase to form carbonic acid. In the presence of carbonic anhydrase carbonic acid immediately dissociates into HCO₃^– and H⁺ ions leading to large accumulation of HCO₃^– inside the RBCs.

Question 10.
How much energy is required for the formation of single molecule of ATP ?
Answer:
For the formation of a single molecule of ATP about 7.3 Kcal of energy is required.

Question 11.
What is Hamburger’s phenomena?
Answer:
The diffusion of Chloride ions into the RBCs to main the ionic balance between RBCs and the plasma is called Hamburger’s phenomena or chloride shift.

Question 12.
What is the role of Hering-Breuer reflex in respiration?
Answer:
The Hering-Breuer reflex controls the depth and rhythm of respiration. It also prevents the lungs from inflating to the point of bursting.

Question 13.
How much blood is present in the human body and from which embryonic germ layer is it derived?
Answer:
An average adult has about 4 to 6 litres of blood, which is red coloured fluid connective tissue derived from embryonic mesoderm.

Question 14.
What is the percentage of plasma in the blood and how much water does it contain?
Answer:
There is 55% of plasma in the blood and it contains 90 to 92% water.

Question 15.
What is the average life span of RBCs?
Answer:
RBCs have a life span of about 120 days.

Question 16.
What is normal RBC count and total WBC count?
Answer:
Average RBC count in adult human is 5.1 to 5.8 million per cubic mm and average total WBC count in adult human is 5000 to 9000 per cubic mm.

Question 17.
What is erythropoiesis?
Answer:
The process of formation of Red Blood Cells is called erythropoiesis.

Question 18.
What is increase in the RBC number called?
Answer:
The increase in the number of RBCs is called polycythemia.

Question 19.
What is leucopenia and erythrocytopenia ?
Answer:
The decrease in the number of white blood cells is called leucopenia whereas decrease in the number of red blood cells is called erythrocytopenia.

Question 20.
Where are Eustachian valve and Thebesian valve located?
Answer:
Eustachian valve is present at the opening of inferior vena cava while Thebesian valve is present near the opening of coronary sinus.

Question 21.
What is foramen ovale and how is it related to fossa ovalis?
Answer:
Foramen ovale is an oval opening in the interatrial septum of the foetal heart representing the fossa ovalis which lies as a depression on the right side of interatrial septum.

Question 22.
When is a person described as having hypertension?
Answer:
When the blood pressure values Eire more than 140 mm Hg systolic pressure and more than 90 mm Hg diastolic pressure, then the person is described as having hypertension.

Question 23.
What are the effects of excessive hypertension?
Answer:
Excessive hypertension of values about 220/120 mm Hg can cause blindness, nephritis, stroke or paralysis.

Question 24.
What is the difference between anemia and leukemia?
Answer:
Anemia is disorder caused due to the deficiency of heaemoglobin while leukemia is blood cancer in which there is abnormal increase in the number of white blood cells.

Question 25.
What is the difference between tachycardia and bradycardia?
Answer:
The faster heart rate over 100 beats per minute is called tachycardia, while the slower heart rate below 60 beats per minute is called bradycardia.

Question 26.
What is the difference between chordae tendinae and columnae carnae?
Answer:
Chordae tendinae are chords that connect bicuspid and tricuspid valves with the papillary muscles in ventricles while columnae carnae are series of irregular muscular ridges present on the inner surface of the ventricles.

Question 27.
Which valves prevent the backward flow of blood at the time of ventricular systole?
Answer:
Semilunar valves located at the base of pulmonary artery and systemic aorta prevent the backward flow of blood at the time of ventricular systole.

Question 28.
What are the time intervals for atrial systole, ventricular systole and joint diastole?
Answer:
Atrial systole is for 0.1 second, ventricular systole is for 0.3 second and joint diastole is for 0.4 second.

Question 29.
In the electrocardiogram shown below, which wave represents ventricular diastole?

Answer:
‘T’ wave represents ventricular diastole.

Question 30.
Mention the role of pacemaker in human heart.
Answer:
Pacemaker can generate wave of contraction or cardiac impulse for rhythmic contraction of heart.

Question 31.
Which structure in the heart is called pacemaker?
Answer:
Sinuatrial node [S. A. node] in the heart wall is called a pacemaker.

Question 32.
What is electrocardiograph?
Answer:
The instrument which is used to record action potentials generated in the heart muscles is called an electrocardiograph or ECG machine.

Question 33.
What is angina pectoris?
Answer:
Angina pectoris is the pain in the chest caused due to reduction in blood supply to cardiac muscle caused due to narrowed and hardened coronary arteries.

Question 34.
What is pulse pressure?
Answer:
Difference between systolic and diastolic pressure is called pulse pressure which is normally 40 mm Hg.

Question 38.
What would happen if respiration takes place in one single step?
Answer:
If respiration takes place in one single step, then the chemical energy released at once during that step might result in a brief blast of light and heat and may lead to death of the cell. Hence respiration is a step-wise process.

Question 39.
Why do the veins have valves?
Answer:
The veins have valves at regular intervals to prevent backflow of blood as blood flows through veins with low pressure.

Question 40.
What is Bohr effect?
Answer:
Bohr effect is the shift of oxyhaemoglobin dissociation curve due to change in partial pressure of CO in blood.

Question 41.
What is Haldane effect?
Answer:
Decrease of pH of blood, due to increase in the number of H⁺ ions, HCO₃ changes into H₂O and CO₂ by the presence of oxyhaemoglobin is called Haldane effect.

Give definitions of the following

Question 1.
Respiration
Answer:
It is a biochemical process of oxidation of organic compounds in an orderly manner for the liberation of chemical energy in the form of ATP.

Question 2.
Breathing
Answer:
It is a physical process by which gaseous exchange takes place between the atmosphere and the lungs. It involves inspiration and expiration.

Question 3.
Tidal Volume (TV)
Answer:
It is the volume of un¬ inspired or expired during normal breathing. It is 500 ml.

Question 4.
Inspiratory reserve volume (IRV)
Answer:
The maximum or the extra volume of air that is inspired during forced breathing in addition to TV (2000 to 3000 ml).

Question 5.
Expiratory reserve volume (ERV)
Answer:
The maximum volume of air that is expired during forced breathing after normal expiration. (1000 to 1100 ml).

Question 6.
Dead space (DS)
Answer:
The volume of air that is present in the respiratory tract (from nose to the terminal bronchioles), but not involved in gaseous exchange (150 ml).

Question 7.
Residual volume (RV)
Answer:
The volume of air that remains in the lungs and the dead space even after maximum expiration (1100 to 1200 ml).

Question 8.
Total lung capacity
Answer:
The maximum amount of air that the lungs can hold after a maximum forceful inspiration (5200 to 5900 ml).

Question 9.
Vital capacity (VC)
Answer:
The maximum amount of air that can be breathed out after of maximum inspiration. It is the sum total of TV, IRV and ERV and is 4100 to 4600 ml.

Question 10.
Oxygen dissociation curve
Answer:
The relationship between HbO₂ saturation and oxygen tension (PPO₂) is called oxygen dissociation curve.

Question 11.
Phosphorylation
Answer:
The process that involves trapping the heat energy in the form of high energy bond of ATP molecule is called phosphorylation.

Question 12.
Artificial ventilation
Answer:
It is the method of inducing breathing in a person when natural respiration has ceased or is faltering.

Question 13.
Ventilator
Answer:
A ventilator is a machine that supports breathing and is used during surgery, treatment for serious lung diseases or other conditions when normal breathing fails.

Question 14.
Cyclosis
Answer:
Cyclosis is the streaming movement of the cytoplasm shown by almost all living organisms. E.g. Paramoecium, Amoeba, etc.

Question 15.
Single circulation
Answer:
The movement of blood once through the heart during each circulation cycle is called single circulation.

Question 16.
Double circulation
Answer:
The movement of blood twice through the heart during one circulation cycle is called double circulation.

Question 17.
Erythropoiesis
Answer:
The process of formation of RBCs is called erythropoiesis.

Question 18.
Polycythemia
Answer:
The increase in the number of RBCs is called polycythemia.

Question 19.
Erythrocytopenia
Answer:
The decrease in the number of RBCs is called Erythrocytopenia.

Question 20.
Hematocrit
Answer:
The hematocrit is ratio of the volume of RBCs to total blood volume of blood.

Question 21.
Diapedesis
Answer:
Leucocytes perform amoeboid movement. Due to this kind of movement they can move out of the capillary walls. This is called diapedesis.

Question 22.
Leucocytosis
Answer:
Increase in the number of leucocytes or WBCs is called leucocytosis.

Question 23.
Leucopenia
Answer:
The decrease in the number of white blood cells is called leucopenia.

Question 24.
Leukaemia
Answer:
Pathological Increase in the number WBCs is called leukaemia or blood cancer.

Question 25.
Thrombocytopenia
Answer:
Decrease in the number of blood platelets is called thrombocytopenia.

Question 26.
Blood Coagulation
Answer:
Conversion of liquid blood into semisolid jelly is called blood coagulation or blood clotting.

Question 27.
Pericardium
Answer:
Double layered peritoneum that covers the heart from outside is called pericardium.

Question 28.
Pacemaker
Answer:
Pacemaker is the region that has power of generation of wave of contraction. In heart, sinoatrial node is called pacemaker.

Question 29.
Heartbeat
Answer:
The rhythmic contraction and relaxation of the heart is called heartbeat.

Question 30.
Pulse
Answer:
A pressure wave that travels through the arteries after each ventricular systole is called pulse.

Question 31.
Heart rate
Answer:
The rate with which the heart beats per minute is called the heart rate.

Question 32.
Stroke volume
Answer:
The amount of blood thrown out of the ventricles during one systole is called the stroke volume.

Question 33.
Cardiac output
Answer:
The amount of blood thrown out of the ventricles during one minute is called cardiac output.

Question 34.
Tachycardia
Answer:
Higher heart rate over 100 beats per minute is called tachycardia.

Question 35.
Bradycardia
Answer:
Lower heart rate which is lesser than 60 per minute is called bradycardia.

Question 36.
Myogenic
Answer:
When the initiation and further regulation of heartbeats take place in the muscles then such a heart is called myogenic.

Question 37.
Cardiac cycle
Answer:
Consecutive systole and diastole constitutes a single heartbeat or cardiac cycle.

Question 38.
Arterial blood pressure
Answer:
The pressure exerted by blood on the wall of artery is called arterial blood pressure.

Question 39.
Angiology
Answer:
Study of blood vessels is called angiology.

Question 40.
Angiography
Answer:
X-ray or imaging of the cardiac blood vessels to locate the position of blockages is called angiography.

Question 41.
Heart transplant
Answer:
Replacement of severely damaged heart by normal heart from brain- dead or recently dead donor is called heart transplant.

Question 42.
Silent Heart Attack
Answer:
Silent heart attack, also known as silent myocardial infarction, is a type of heart attack that lacks the general symptoms of classic heart attack like extreme chest pain, hypertension, shortness of breath, sweating and dizziness.

Question 43.
Electrocardiogram
Answer:
Graphical recording of electrical variations detected at the surface of body during their propagation through the wall of heart is electrocardiogram (ECG).

Question 44.
Lymph
Answer:
It is a fluid connective tissue with almost similar composition to the blood except RBCs, platelets and some proteins.

Give functions of the following

Question 1.
Epiglottis.
Answer:
The epiglottis prevents the entry of food into the trachea by closing the glottis temporarily.

Question 2.
Carbonic anhydrase.
Answer:
Carbonic anhydrase enzyme is found inside the RBCs only to accelerate the rate of formation of carbonic acid from CO₂ and H₂O.

Question 3.
Ventilators.
Answer:
Ventilators used in hospitals are part of life supporting system, which help in breathing by

1. Pushing oxygen into the lungs
2. Removing carbon dioxide from the lungs

Question 4.
Erythrocytes.
Answer:
Erythrocytes carry oxygen to all cells of the body from the lungs and bringing carbon dioxide from all the cells back to lungs.

Question 5.
Neutrophils.
Answer:
Neutrophils are responsible for destroying pathogens by the process of phagocytosis.

Question 6.
Thrombocytes/Platelets.
Answer:
Platelets secrete platelet factors which are essential in blood clotting. They also seal v the ruptured blood vessels by formation of platelet plug/thrombus. They secrete serotonin, a local vasoconstrictor.

Question 7.
Pericardial fluid.
Answer:
Pericardial fluid acts as a shock absorber and protects the heart from mechanical injuries. It also keeps the heart moist and acts as lubricant.

Question 8.
Heart walls.
Answer:
The epicardium and endocardium are protective in function whereas myocardium is responsible for contraction and relaxation of heart.

Question 9.
Valves in heart.
Answer:
Valves in the heart prevent the backflow of the blood at the time of systole and help in maintaining a unidirectional flow of blood.

Question 10.
Chordae tendinae.
Answer:
Chordae tendinae attach the bicuspid and tricuspid valves to the ventricular wall (papillary muscles) and regulate their opening and closing.

Question 11.
Semilunar valves.
Answer:
Semilunar valves prevent the backward flow of blood from pulmonary aorta and the aorta into the respective ventricles.

Question 12.
Sinoatrial node [SA] or Pacemaker.
Answer:
SA node acts as pacemaker of heart because it has the power of generating a new wave of contraction and making the pace of contraction.

Question 13.
Electrocardiogram (ECG).
Answer:
ECG helps to diagnose the abnormality in conducting pathway, enlargement of heart chambers, damage to cardiac muscles, reduced blood supply to cardiac muscles and causes of chest pain.

Question 14.
Blood.
Answer:
Functions of blood:

1. Transport of oxygen and carbon dioxide
2. Transport of food
3. Transport of waste product
4. Transport of hormones
5. Maintenance of pH
6. Water balance
7. Transport of heat
8. Defence against infection
9. Temperature regulation
10. Blood clotting/coagulation
11. Helps in healing

Name the following

Question 1.
Name two animals in which moist skin acts as a respiratory surface.
Answer:
Earthworm, Frog

Question 2.
Name the respiratory organs in insects and fish.
Answer:
Insects – Tracheal tubes and spiracles
Fish – Internal gills

Question 3.
Name any two disorders of respiratory system.
Answer:
Asthma and pneumonia are the two disorders of respiratory system.

Question 4.
Name the structural and functional unit of lungs.
Answer:
Alveolus is the structural and functional unit of lungs.

Question 5.
Name the energy currency of cell.
Answer:
ATP is the energy currency of cell.

Question 6.
Name the site where actual exchange of O₂ and CO₂ takes place between air and blood in the body of man.
Answer:
Alveolus of lung.

Question 7.
Name any two respiratory centres required for regulation of breathing.
Answer:
Inspiratory centre, Expiratory centre, Pneumotaxic centre and Apneustic centre.

Question 8.
Name the muscles which move ribs up and down.
Answer:
External intercostal muscles.

Question 9.
Name two phyla where haemocoel is present.
Answer:
Phylum-Arthropoda and Phylum-Mollusca.

Question 10.
Name the animal-group which show single circulation.
Answer:
Fishes

Question 11.
Name the cells which produce thrombocytes.
Answer:
Megakaryocytes produce thrombocytes.

Question 12.
Name the process of formation of red blood corpuscles.
Answer:
Erythropoiesis

Question 13.
Name the space in which human heart is located.
Answer:
Mediastinum is the space in which human heart is located.

Question 14.
Name the types of lymphocytes depending upon functions.
Answer:
B-lymphocytes and T-lymphocytes.

Question 15.
Name the layers of peritoneum that surrounds the heart sequentially from outside to inside.
Answer:
Fibrous pericardium, parietal layer of serous pericardium and visceral layer of serous pericardium.

Question 16.
Name the connection between the pulmonary trunk and systemic aorta.
Answer:
Ligamentum arteriosum that represents remnant of ductus arteriosus of foetus.

Question 17.
Name the valve between left atrium and left ventricle and give its significance.
Answer:
Between left atrium and left ventricle is mitral or bicuspid valve which maintains the unidirectional flow of blood by preventing hs backflow.

Question 18.
Name the walls of an artery.
Answer:
Outer tunica externa, middle tunica media and inner tunica interna.

Question 19.
Name the instrument used to measure blood pressure.
Answer:
Sphygmomanometer is used to measure blood pressure.

Question 20.
Name the plasma proteins involved in the process of blood clotting.
Answer:
Prothrombin and fibrinogen.

Question 21.
Name the various components of conducting system of the heart.
Answer:
Conducting system of the heart consists of SA node, AV node, bundle of His and Purkinje fibers.

Question 22.
Name the neurotransmitters that decrease and increase the heart rate in human beings respectively.
Answer:
Acetylcholine decreases heart rate and adrenaline or epinephrine increases the heart rate in human.

Question 23.
Who discovered ECG?
Answer:
Willem Einthoven discovered ECG.

Distinguish between the following

Question 1.
Pharynx and Larynx.
Answer:

Pharynx	Laryix
1. Pharynx is a short, vertical tube.	1. Larynx is a sound producing organ located at the end of pharynx.
2. Mouth leads to the pharynx.	2. Larynx leads to the oesophagus.
3. Vocal cords are absent.	3. Vocal cords are present.
4. Pharynx does not increase in size at the time of puberty.	4. Larynx increases in size at the time of puberty.
5. Pharynx does not show Adam’s apple.	5. Larynx shows Adam’s apple in adult males.

Question 2.
Inspiration and Expiration.
Answer:

Inspiration	Expiration
1. Inspiration is an active process.	1. Expiration is a passive process.
2. During inspiration diaphragm contracts and becomes flattened.	2. During expiration diaphragm relaxes and becomes dome shaped.
3. During inspiration intercostal muscles contract.	3. During expiration intercostal muscles relax.
4. During inspiration ribs are pulled outwards and sternum is raised.	4. During expiration ribs are pulled inwards and sternum is lowered.
5. During inspiration the space in the thoracic cavity increases.	5. During expiration the space in the thoracic cavity decreases.
6. During inspiration pressure in the lungs decreases.	6. During expiration pressure in the lungs increases.
7. During inspiration the volume of the lungs increase.	7. During expiration the volume of the lungs decreases.
8. During inspiration air comes inside the body.	8. During expiration air goes out of the body.

Question 3.
External respiration and Internal respiration.
Answer:

External respiration	Internal respiration
1. The respiratory processes occurring in lungs is called external respiration.	1. The respiratory processes that occur in tissues is called internal respiration.
2. During external respiration O2 from the lungs enters into the lung capillaries by diffusion.	2. During internal respiration O2 from the blood enters the tissue cells.
3. During external respiration CO2 from the lung capillaries diffuse into the lungs.	3. During internal respiration CO2 from the tissues enters into the blood.
4. During external respiration exchange of gases takes place between the air and the lungs.	4. During internal respiration exchange of gases take place between the blood and the tissue.
5. Formation of oxyhaemoglobin takes place during external respiration.	5. Oxyhaemoglobin dissociates into oxygen and haemoglobin during internal respiration.
6. During external respiration CO2 is released.	6. During internal respiration carbamino haemoglobin is formed which is carried to the lungs.

Question 4.
Transport of O₂ and Transport of CO₂.
Answer:

Transport of O2	Transport of CO2
1. Transport of O2 takes place from lungs to the tissues and cells.	1. Transport of CO2 takes place from tissues and cells to the lungs.
2. Oxygen is carried as oxyhaemoglobin to the tissues with the help of RBCs.	2. Carbon dioxide is carried as carbaminohaemoglobin from the tissues with the help of plasma and RBCs.
3. Oxygen does not form oxides or other products during its transport.	3. CO2 forms bicarbonates with sodium and potassium during its transport.
4. O2 does not form acids during its transport.	4. CO2 dissolves in water to form carbonic acid.

Question 5.
Vital Capacity of Lung and Total Lung Capacity.
Answer:

Vital Capacity of Lung	Total Lung Capacity
1. It is the maximum amount of air a person can expire and inspire to their maximum extent.	1. It is the maximum amount of air that the lungs can hold after a maximum forceful inspiration.
2. It is the sum total of inspiratory reserve volume, tidal volume and expiratory reserve volume.	2. It is the sum total of vital capacity and residual volume.
3. It ranges from 4100 to 4600 ml.	3. It ranges from 5200 to 5800 ml.

Question 6.
Inspiratory Reserve Volume (IRV) and Expiratory Reserve Volume (ERV).
Answer:

Inspiratory Reserve Volume (IRV)	Expiratory Reserve Volume (ERV).
1. It is the maximum volume of air, or the extra volume of air, that is inspired during forced breathing.	1. It is the maximum volume of air that is expired during forced breathing.
2. Its value is 2000/3000 ml.	2. Its value is 1000/1100 ml.

Question 7.
T. S. of artery and T.S. of vein.
Answer:

T. S. of artery	T.S. of vein
1. Histologically in transverse section of artery there are three walls, tunica externa, tunica media and tunica interna or endothelium.	1. Histologically in transverse section of vein there are three walls, tunica externa, tunica media and tunica interna or endothelium.
2. Tunica media is thick and muscular.	2. Tunica media is thinner as compared to artery.
3. Lumen of artery is narrow.	3. Lumen of vein is broad.

Question 8.
Erythrocytes and Leucocytes.
Answer:

Erythrocytes	Leucocytes
1. Erythrocytes have a definite shape which is elliptical or oval.	1. Leucocytes do not have definite shape as they are amoeboid.
2. They are enucleated.	2. They are nucleated.
3. Erythrocytes contain haemoglobin and hence appear red.	3. Leucocytes are devoid of any respiratory pigment and hence appear colourless.
4. The normal erythrocyte count is 4.3 to 5.8 million per cubic mm of blood.	4. The normal leucocyte count is 4000 to 11000 per cubic mm of blood.
5. The life span of erythrocytes is 100 to 120 days.	5. The life span of leucocytes is 3 to 4 days.
6. The diameter of erythrocytes is 7.2 m and thickness is about 2 to 2.2 m.	6. The size of leucocytes varies with its subtypes and is of average size of 8 to 15 m.
7. Erythrocytes are formed by the process of erythropoiesis in red bone marrow.	7. Leucocytes are formed by the process of leucopoiesis in bone marrow, tonsils, lymph nodes, spleen, thymus, etc.
8. Erythrocytes transport the respiratory gases.	8. Leucocytes help in the formation of antibodies besides fighting against foreign antigens by phagocytic activity.

Question 9.
Eosinophils and Basophils.
Answer:

Eosinophils	Basophils
1. Cytoplasmic granules present in eosinophils are stained with acidic stains.	1. Cytoplasmic granules present in basophils are stained with basic stains.
2. Nucleus is bilobed.	2. Nucleus is twisted.
3. Eosinophils constitute 3% of total WBCs.	3. Basophils constitute 0.5% of total WBCs.

Question 10.
Neutrophils and Eosionophils.
Answer:

Neutrophils	Eosinophils
1. Cytoplasmic granules present in neutrophils are stained with neutral stains.	1. Cytoplasmic granules present in eosinophils are stained with acidic stains.
2. Nucleus is three to five lobes showing polymorphic form.	2. Nucleus is bilobed.
3. Neutrophils constitute 62% of total WBCs.	3. Eosinophils constitute 3% of total WBCs.

Question 11.
Lymphocytes and Monocytes.
Answer:

Lymphocytes	Monocytes
1. Large round nucleus but size of the cell is smaller.	1. Large kidney shaped nucleus and largest size among WBCs.
2. Lymphocytes form 25-33% of WBCs.	2. Monocytes form 3-9% of WBCs.

Question 12.
Granulocytes and Agranulocytes.
Answer:

Granulocytes	Agranulocytes
1. WBCs with granular cytoplasm are called granulocytes. Thus, cytoplasmic granules are present.	1. WBCs with agranular cytoplasm are called agranulocytes. Thus, cytoplasmic granules are absent.
2. Nuclei of granulocytes are variously lobed.	2. Nuclei of agranulocytes are not lobed.

Question 13.
Single circulation and Double circulation.
Answer:

Single circulation	Double circulation
1. Blood flows only once through the heart in a complete cycle.	1. Blood flows twice through the heart during one complete circulation. Systemic – to and fro ‘ from heart to body and pulmonary – to and fro from heart to lungs.
2. Heart pumps deoxygenated blood only.	2. Heart pumps both deoxygenated and oxygenated blood to lungs and body respectively.
3. Blood is oxygenated in gills.	3. Blood is oxygenated in lungs.
4. Occurs only in fishes.	4. Occurs in amphibians, reptiles, birds and mammals.

Question 14.
Systolic blood circulation and Diastolic blood circulation.
Answer:

Systolic blood circulation	Diastolic blood circulation
1. Blood is passed from right ventricle to lungs by pulmonary artery during systolic circulation. Similarly, from left ventricle the oxygenated blood is given to the entire body through systemic aorta during systolic circulation.	1. Blood is passed to left atrium from lungs by pulmonary veins during diastolic circulation. Similarly, deoxygenated blood from entire body is brought back to heart through vena cava during diastolic circulation.
2. Systolic blood circulation is under maximum pressure as heart is forcing the blood to come out of heart.	2. Diastolic blood circulation is under minimum blood pressure as heart is relaxed during diastole.

Question 15.
Atria and Ventricles.
Answer:

Atria	Ventricles
1. Atria are upper chambers of the heart.	1. Ventricles are lower chambers of the heart.
2. Atria are thin walled.	2. Ventricles are thick walled.
3. Atria are receiving chambers.	3. Ventricles are distributing chambers.
4. Interatrial septum divides the two auricles (atria).	4. Interventricular septum divides the two ventricles.
5. Right atrium is larger in size than left atrium.	5. Left ventricle is larger in size than the right ventricle.

Question 16.
S.A. Node and A.V. Node.
Answer:

S.A. Node	A.V. Node
1. Sinoatrial node is present in the right ventricle near the opening near the opening of the superior vena cava.	1. Atrioventricular node is present in the right ventricle near the opening of the coronary sinus.
2. S.A. node is the pacemaker of the heart and it starts atrial systole.	2. A.V. node starts ventricular systole through bundles of His and Purkinje’s fibre system.

Question 17.
Pulmonary circulation and Systemic circulation.
Answer:

Pulmonary circulation	Systemic circulation
1. The course of blood from the right ventricle to the left atrium of the heart through the lungs is called pulmonary circulation.	1. The course of blood from the left ventricle to the right atrium of the heart through the body is called systemic circulation.
2. Pulmonary circulation is mainly for sending the blood for oxygenation in the lungs from the heart and bringing it back to the heart after oxygenation.	2. Systemic circulation is for sending the deoxygenated blood from the body to the heart and sending oxygenated blood from the heart to the body.

Question 18.
Atrio ventricular valves and Semilunar valves.
Answer:

Atrio ventricular valves	Semilunar valves
1. Atrio ventricular valves Eire present between the atria and ventricles. On the right side there is tricuspid valve whereas on the left side there is bicuspid valve.	1. Semilunar valves are present at the opening of pulmonary artery and systemic aorta.
2. Atrio ventricular valves prevent the back flow of blood from ventricles to atria at the time of systole.	2. Semilunar valves prevent the back flow of blood from pulmonary artery and systemic aorta back to the heart.

Question 19.
Hypertension and Hypotension.
Answer:

Hypertension	Hypotension
1. Blood pressure values more than 140 mm Hg SP and 90 mm HG DP is called hypertension.	1. Blood pressure values less them 120 mm Hg SP and 70 mm HG DP is called hypotension.
2. Excessive hypertension can result into lethal complications such as stroke or paralysis.	2. Hypotension may not be lethal if immediate measures are taken to raise the blood pressure.

Give reasons

Question 1.
ATP is called energy currency of the cell.
Answer:

1. During cellular respiration, the oxidation of food (glucose) takes place.
2. This happens in the mitochondria using the oxygen present in the blood.
3. ATP molecules are formed during this oxidation.
4. ATP is used for various vital body processes and also for maintaining the body temperature to constancy.
5. Since energy is stored in the form of ATP it is called an energy currency of the cell.

Question 2.
The vestibule of nasal chamber has fine hair.
Answer:

1. Vestibule is the anterior most part of the nasal chamber.
2. The hairs present in this region trap the dust particles and prevent them from entering into the interiors of the respiratory passage.
3. Therefore, the vestibule of nasal chambers has fine hair.

Question 3.
Glottis is guarded by a flap called epiglottis.
Answer:

1. The oesophagus and trachea lie side by side.
2. There is possibility that food particles may enter respiratory passage at the time of gulping.
3. However, the epiglottis prevents the entry of food into the respiratory passage by closing it temporarily.
4. Thus, for preventing the entry of food particles into respiratory passage, the glottis is guarded by a flap called epiglottis.

Question 4.
Alveoli are very flexible.
Answer:

1. Alveoli are made up of collagen and elastin fibres.
2. They are very thin (0.0001 mm) and lined by non-ciliated squamous epithelium.
3. All the above structural components make the alveoli very flexible.

Question 5.
Expiration is called a passive process.
Answer:

1. During expiration, intercostal muscles relax. This results in pulling the ribs inwards.
2. Diaphragm also relaxes and returns to its normal dome shape.
3. The collective contraction of ribs and diaphragm results in the reduction of
   thoracic cavity and hence automatically air is pushed out of the lungs.
4. Since the pressure on the lungs increase rushing the air to outside, expiration is called a passive process.

Question 6.
Pericardium acts as a defence wall for the heart.
Answer:

1. Pericardium protects the heart. It is double layered peritoneum, having outer fibrous and inner serous pericardium layers.
2. Fibrous pericardium being tough gives protection to the heart.
3. Serous pericardium has two layers, parietal and visceral layer or epicardium.
4. In between these two layers, there is pericardial fluid, which helps to absorb shocks and provide nourishment.
5. In this way pericardium acts as a defence wall.

Question 7.
Valves are present in veins.
Answer:

1. Veins carry blood to the heart.
2. At that time the backward flow of the blood should be prevented.
3. Therefore, valves are present in veins.

Question 8.
Atria are thin walled than ventricles.
Answer:

1. Atria are receiving chambers, while ventricles are distributing chambers.
2. The blood is driven out from ventricles.
3. Ventricles are therefore, strong and with thicker walls.
4. Atria are thin walled as compared to ventricles.

Question 9.
Heart is called a pump.
Answer:

1. The heart acts as a pumping organ. It shows continuous pumping action.
2. The rhythmic contraction or systole and relaxation or diastole of heart forms one heartbeat.
3. Such heartbeats occur about 72 times per minute.
4. The heart efficiently pumps about 5 litres of blood per minute. Therefore, the heart is called a pump.

Question 10.
In normal human heart, there is no mixing of oxygenated and deoxygenated blood.
Answer:

1. In normal human heart, there is completely formed atrioventricular septum.
2. This septum keeps the deoxygenated and oxygenated blood separate.
3. Hence there is no mixing of the two types of blood.

Question 11.
Blood pressure is inversely related to the elasticity of the blood vessels.
Answer:

1. When the blood gushes through the blood vessels, the walls of blood vessels -can expand a little due to their elasticity.
2. But as the age advances, the elasticity is reduced and then the blood vessels do not expand.
3. Hence the flowing blood gets more resistance and the blood pressure can rise.
4. Lesser the elasticity more will be the blood pressure, whereas more the elasticity of the vessel wall, then the pressure will not rise.
5. In this way, the blood pressure is inversely related to the elasticity of the blood vessels.

Write short notes

Question 1.
Chloride shift or Hamburger’s phenomenon.
Answer:

1. About 70% of CO₂ is transported in the form of sodium bicarbonates/potassium bicarbonates from tissue cells to lungs.
2. In the RBCs, CO₂ combines with water in the presence of a Zn containing enzyme, carbonic anhydrase to form carbonic acid. This action is rapid in RBCs as compared to that in the plasma.
3. Carbonic acid being unstable, immediately dissociates into HCO₃^– and H⁺in the presence of same enzyme, leading to large accumulation of HCO₃^– inside the RBCs. It thus moves out of RBCs. This can bring about imbalance of the charge inside the RBCs.
4. To maintain the ionic balance between the RBCs and the plasma, Cl^– diffuses into the RBCs. This movement of chloride ions is known as chloride shift or Hamburger’s phenomenon.
5. HCO₃^– that comes in the plasma joins to Na⁺/K⁺ forming NaHCO₃/KHCO₃ which can maintain pH of blood. The remaining H⁺ ions in the RBCs are buffered by haemoglobin by the formation of oxyhaemoglobin.
6. At the level of lungs, due to the low partial pressure of carbon dioxide of the alveolar air, hydrogen ion and bicarbonate ions combine to form carbonic acid and under the influence of carbonic anhydrase again yields carbon dioxide and water.

Question 2.
Regulation of breathing.
Answer:
(1) Respiration is under dual control, i.e. nervous and chemical. Normal breathing is an involuntary process. Steady state of respiration is controlled by neurons located in the pons and medulla and are known as the respiratory centres. They regulate the rate and depth of breathing.

(2) These centres are divided into three groups : dorsal group of neurons in the medulla (inspiratory centre), ventro-lateral group of neurons in medulla (inspiratory and expiratory centre) and pneumotaxic centre located in the pons and apneustic centre which is antagonistic in action to pneumotaxic centre.

(3) During inspiration, when the lungs expand to a critical point, the stretch receptors are stimulated and impulses are sent along the vagus nerves to the expiratory centre. It then sends out inhibitory impulses to the inspiratory centre.

(4) The inspiratory muscles relax and expiration follows. As the air leaves but, the lungs are deflated and the stretch receptors are no longer stimulated. Thus, the inspiratory centre is no longer inhibited and a new respiration begins. These events are called the Hering – Breuer reflex. The Hering – Breuer reflex controls the depth and rhythm of respiration. It also prevents the lungs from inflating to the point of bursting.

(5) The respiratory centre has connections with the cerebral cortex that means we can voluntarily change our pattern of breathing. Voluntary control is protective because it enables us to prevent water or irritating gases from entering the lungs.

Question 3.
Carbon monoxide poisoning.
Answer:

1. Carbon monoxide poisoning is caused when carbon monoxide is combined with haemoglobin.
2. Haemoglobin is said to have 250 times more affinity for carbon monoxide than that for the oxygen.
3. Therefore, haemoglobin with carbon monoxide forms a stable compound, the carboxyhemoglobin.
4. Due to the formation of carboxyhaemoglobin, the haemoglobin no longer carries oxygen to the cells and tissues. Tissues then suffer from oxygen starvation. This leads to asphyxiation and in extreme cases it leads to death.
5. Carbon monoxide poisoning occurs in closed rooms with incompletely burning substances such as stove burners or furnaces and garages having running automobile engines.
6. Person suffering from carbon monoxide poisoning has to be administered with oxygen-carbon dioxide mixture, so that high levels of CO₂ makes carbon monoxide dissociated from haemoglobin.

Question 4.
Artificial ventilation.
Answer:
(1) Artificial ventilation is the artificial respiration. It is the method of inducing breathing in a person when natural respiration has ceased or is faltering. If used properly and quickly, it can prevent death due to drowning, choking, suffocation, electric shock, etc.

(2) The process involves two main steps:
a. Establishing and maintaining an open air passage from the upper respiratory tract to the lungs.
b. Force inspiration and expiration as in mouth to mouth respiration or by mechanical means like ventilator.

(3) A ventilator is a machine that supports breathing and is used during surgery, treatment for serious lung diseases or other conditions when normal breathing fails.

Question 5.
Erythrocytes.
Answer:

1. Erythrocytes or red blood corpuscles. They are circular, biconcave, enucleated cells.
2. The RBC size : 7 pm in diameter and 2.5 pm in thickness.
3. The RBC count : 5.1 to 5.8 million RBCs/ cu mm of blood in an adult male and 4.3 to 5.2 million/cu mm in an adult female.
4. The average life span of RBC : 120 days.
5. RBCs are formed by the process of erythropoiesis. In foetus, RBC formation takes place in liver and spleen whereas in adults it occurs in red bone marrow.
6. The old and worn out RBCs are destroyed in liver and spleen.
7. Polycythemia is an increase in number of RBCs while erythrocytopenia is decrease in their (RBCs) number.

Functions of RBCs:

1. Transport of oxygen from lungs to tissues and carbon dioxide from tissues to lungs with the help of haemoglobin.
2. Maintenance of blood pH as haemoglobin acts as a buffer.
3. Maintenance of the viscosity of blood.

Question 6.
Heartbeat.
Answer:

1. The rhythmic contraction and relaxation of the heart is called heartbeat.
2. Each heartbeat includes one systole and one diastole. During systole the heart contracts and during diastole it relaxes.
3. The rate with which the heart beats is called heart rate. The normal heart rate is 72 beats per minute.
4. Tachycardia means faster heart rate of about more than 100 beats per minute.
5. Bradycardia means slower heart rate that is below 60 beats per minute.

Question 7.
Pulse.
Answer:

1. A pressure wave that travels through the arteries after each ventricular systole is called a pulse.
2. The pulse can be felt in any artery that lies near the surface of the body.
3. The radial artery at the wrist is most commonly used to feel the pulse.
4. The pulse rate per minute indicates the heart rate. Since each heartbeat generates one pulse in the arteries, the pulse rate is same as that of heart rate, i.e. 72 times per minute.

Question 8.
Peacemaker.
Answer:

1. Pacemaker is the region in tile heart which initiates the beating.
2. The natural pacemaker of the heart is sinoatrial node (SA node).
3. The pacemaker is autorhythmic, it is able to repeatedly and rhythmically generate impulses.
4. SA node is responsible for initiation of cardiac excitation. Therefore, it is called a pacemaker.

Question 9.
Blood pressure.
Answer:

1. Blood pressure is the pressure exerted by the flowing blood on the walls of arteries.
2. Blood pressure described in two terms viz. systolic blood pressure and diastolic blood pressure. Systolic blood pressure is the maximum pressure of blood when heart undergoes ventricular systole. It is responsible for flow of blood in the arteries. Normal systolic pressure is 120 mm Hg.
3. Diastolic blood pressure is the minimum pressure of blood when heart undergoes diastole. Normal diastolic pressure is 80 mm Hg.
4. Blood pressure is represented as 120/80 mm Hg for a normal human being.

Question 10.
Hypertension.
Answer:

1. In a normal healthy person the blood pressure values are 120 mm Hg (systolic)/ 80 mm Hg (diastolic).
2. When the blood pressure is persistently more than 140 mm Hg systolic pressure and 90 mm Hg diastolic arterial blood pressure then it is said to be hypertension or high blood pressure.
3. Excessively high blood pressure is very dangerous as high blood pressure of about 220/120 mm Hg may cause rupturing of blood vessels.
4. Rupture of eye blood vessels can lead to blindness.
5. If blood vessels of kidney are affected then nephritis is caused.
6. Hemorrhage occurring in the brain can lead to stroke or paralysis. Therefore, hypertension is commonly called silent killer. It may be present for years with no distinct symptoms.
7. The factors causing hypertension are arteriosclerosis (reduction of elasticity of blood vessels), atherosclerosis (deposition, of cholesterol inside the blood vessels wall), obesity, physical or emotional stress, alcoholism, smoking, cholesterol rich diet, increased secretion of renin, epinephrine or aldosterone, etc.

Question 11.
Coronary artery disease (CAD).
Answer:

1. Coronary artery disease is a condition caused due to problems like atherosclerosis.
2. In this disease, coronary arteries are narrowed due to deposition of fatty substances.
3. Due to this the blood flow to the heart is reduced.
4. In coronary heart disease, the heart muscle is damaged because of an inadequate amount of blood due to obstruction of its blood supply.
5. The symptoms of CAD depend upon the degree of obstruction.
6. Symptoms are mild chest pain or angina pectoris.
7. In severe cases it results in heart attack known as myocardial infarction.

Question 12.
Angina pectoris.
Answer:

1. Angina pectoris is the pain in the chest. It results from a reduction in blood supply to cardiac muscle due to narrowed and hardened coronary arteries.
2. Atherosclerosis and arteriosclerosis can cause this problem. Basically, the coronary arteries are affected during angina pectoris.
3. It causes heaviness and severe pain in the chest. The pain can also be felt at the neck, lower jaw, left arm and left shoulder.
4. Angina pectoris often occurs during exertion, when the heart demands more oxygen and narrowed blood vessels cannot supply. It disappears with rest.

Question 13.
Heart failure.
Answer:

1. Heart failure is caused due to progressive weakening of the heart muscle. This results in the failure of the heart to pump the blood effectively.
2. Hypertension increase the after load on the heart leading to significant enlargement of the heart.
3. This finally results in heart failure.
4. Factors responsible for heart failure are advanced age, malnutrition, chronic infections, toxins, severe anaemia or hyperthyroidism, etc.
5. Any problem leading to degeneration of heart muscle, may result in heart failure.

Question 14.
Atherosclerosis.
Answer:

1. Atherosclerosis is the deposition of fatty substances and cholesterol on the inner lining of eateries.
2. This deposition results in the formation of atherosclerotic plaque.
3. It results in the decrease of the lumen of the blood vessels causing increasing resistance for the blood to flow which in turn results in the hypertension.
4. Atherosclerosis of the coronary arteries results in decrease in the blood flow to the heart muscles.
5. Due to such condition, coronary heart disease is caused.

Question 15.
ECG.
Answer:

1. Electrocardiogram or ECG is the graphic v record of electrical variations produced by the heart during one heartbeat or cardiac cycle.
2. ECG is taken with the help of an instrument called electrocardiograph or ECG machine. Electrocardiograph records the action potentials generated by the heart muscles.
3. The electrical activity of heart is represented in the form of a graph plotted with time on X-axis against voltage displacement on Y-axis.
4. A normal ECG is a graph having series of ridges and furrows. There are waves such as P-wave, QRS complex and T-wave.
5. P-wave is a small upwards wave representing impulse generated by SA node. P-wave is caused by atrial depolarization that results in atrial contraction.
6. QRS-complex wave begins as a downward deflection, continues as a large upright triangular wave and ends’ as a downward wave.
7. QRS-complex wave represents spreading of impulse from SA node to AV node, then to bundle of His and Purkinje fibres. It causes ventricular depolarization resulting in ventricular contraction.
8. T-wave is a broad upward wave which represents ventricular repolarization resulting in ventricular relaxation.
9. Functions of ECG are mainly for diagnosis and also for prognosis. It is useful to detect abnormal functioning of heart as in coronary artery diseases, heart block, angina pectoris, tachycardia, ischemic heart disease, myocardial infarction, cardiac arrest, etc.

Question 16.
Angiography.
Answer:

1. Angiography is an X-ray imaging of the cardiac blood vessels to locate the position of blockages.
2. Depending upon the degree of blockage, remedial procedures like angioplasty or by¬pass surgery are performed.
3. In angioplasty a stent is inserted at the site of blockage to restore the blood supply while in by-pass surgery, the atherosclerotic region is by-passed with part of vein or artery taken from any other suitable part of the body, like hands or legs.

Question 17.
Silent Heart Attack or silent myocardial infarction.
Answer:

1. Silent heart attack is a type of heart attack that lacks the general symptoms of classic heart attack like extreme chest pain, hypertension, shortness of breath, sweating and dizziness.
2. Symptoms of silent heart attack are so mild that a person often confuses it for regular « discomfort and thereby ignores it.
3. Men are more affected by silent heart attack than women.

Question 18.
Heart Transplant.
Answer:

1. Heart transplant is the replacement of severely damaged heart by normal heart from brain-dead or recently dead donor,
2. Heart transplant is necessary in case of patients with end-stage heart disease and severe coronary arterial disease.

Short Answer Questions

Question 1.
What is meant by respiration? How is it useful in the production of energy?
Answer:

1. Respiration is the biochemical process in which organic compound such as glucose are oxidized to liberate chemical energy.
2. During respiration energy is released in gradual and step wise process. The released energy is in the form of bonds of ATP (Adenosine Tri Phosphate) molecules are shown below:
   C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O + 38 ATP
3. ATP is the biologically useful energy. ATP drives most of the life process.
4. When cell requires the energy, ATP is hydrolyzed and is converted into ADP with subsequent release of energy.
5. The respiratory system, blood and the body cells play an important role in the process of respiration.

Question 2.
How does exchange of gases take place at the alveolar level?
Answer:
1. Exchange of gases between the alveolar air and the blood is known as external respiration.

2. Simple squamous epithelial layer of alveolus is intimately associated with a similar layer lining the capillary wall. Both of these layers are thin walled and together they make up the respiratory membrane through which gaseous exchange occurs between the alveolar air and the blood.

3. Diffusion of gases will take place from an area of higher partial pressure to an area of lower partial pressure until the partial pressure in the two regions reaches equilibrium.

4. The partial pressure of carbon dioxide of blood entering the pulmonary capillaries is 45 mmHg while partial pressure of carbon dioxide in alveolar air is 40 mmHg. Due to this difference, carbon dioxide diffuses from the capillaries into the alveolus.

5. Similarly, partial pressure of oxygen of blood in pulmonary capillaries is 40 mmHg while in alveolar blood it is 104 mmHg. Due to this difference oxygen diffuses from alveoli to the capillaries.

Question 3.
What is the role of haemoglobin in the transport of oxygen in the blood?
Answer:

1. Haemoglobin is a respiratory pigment present in cytoplasm of RBCs. About 97% of oxygen is transported by these haemoglobin molecules from lungs to tissues.
2. Haemoglobin has a high affinity for Oa and combines with it to form oxyhaemoglobin. One molecule of Hb has four FeT, each of which can pick up a molecule of oxygen (O₂). Hb + 4O₂ → Hb (4O₂)
3. Oxyhaemoglobin is transported from lungs to the tissues where it readily dissociates to release O₂.
   Hb (4O₂) → Hb + 4O₂
4. In the alveoli where PPOa is high and PPCO₂ is low, oxygen binds with haemoglobin, but in tissues, where PPO₂ is lower and PPCO₂ is high, Oxyhaemoglobin dissociates and releases O₂ for diffusion into the tissue cells.

Question 4.
What is blood? What is the normal quantity of blood in an adult human being?
Answer:

1. Blood is the fluid connective tissue that circulates in the body.
2. Blood is derived from mesoderm.
3. It is bright red, slightly alkaline fluid having pH about 7.4. It is salty, viscous fluid heavier than water.
4. The average sized adult has about 5 litres of blood in his/her body which constitutes about 8% of the total body weight.

Question 5.
Describe the structure and the function of thrombocytes.
Answer:

1. Thrombocytes or platelets are non- nucleated, round and biconvex blood corpuscles.
2. They are smallest corpuscles measuring about 2.5 to 5 mm in diameter with a count of about 2.5 lakhs/cu mm of blood.
3. Their life span is about 5 to 10 days.
4. Thrombocytes are formed from megakaryocytes of bone marrow. They break from these cells as fragments during the process of thrombopoiesis.
5. Thrombocytosis is the increase in platelet count while thrombocytopenia is decrease in platelet count.
6. Thrombocytes possess thromboplastin which helps in clotting of blood.
7. Therefore, at the site of injury platelets aggregate and form a platelet plug. Here they release thromboplastin due to which further blood clotting reactions take place.

Question 6.
Describe the structure of the heart wall.
Answer:

1. The heart wall is composed of three layers, viz. outer epicardium, middle myocardium and inner endocardium.
2. Epicardium is composed of single layer of mesothelium having flat epithelial cells.
3. Myocardium is composed of cardiac muscle fibres. These muscle fibres perform the function of systole and diastole by showing contraction and relaxation of muscle wall of the heart.
4. Endocardium is composed of single layer of flat epithelial cells called endothelium.

Question 7.
Name the two heart sounds. How and when are they produced?
Answer:

1. In one normal heartbeat the heart sounds like lubb and dup are produced once each.
2. The rhythmic contraction (Systole) and relaxation (diastole) forms are heartbeat. The heart sounds are due to closure of valves.
3. Lubb sound is produced during ventricular systole when the cuspid valves close both the atrioventricular apertures preventing blood flow into atria.
4. Dub sound is produced during ventricular diastole when semilunar valves are closed, preventing backflow of blood from pulmonary trunk and systemic aorta into ventricles.

Question 8.
What is double circulation? What is its significance?
Answer:
(1) Double circulation : Movement of blood twice through the heart during one circulation cycle is called double circulation. Body → heart → lungs → heart → body is the course of double circulation.

(2) Significance of double circulation:
a. Double circulation is more effective type of circulation in which oxygenated and deoxygenated type of blood do not intermix.
b. The systemic circulation i.e. from body to heart and back to body while the pulmonary circulation i.e. from heart to lungs and back to heart circulate the blood uniformly.

(3) Coronary and hepatic portal circulation is also achieved due to double circulation.

Question 9.
Describe pulmonary and systemic circulation.
Answer:

1. In human beings, there is double circulation because blood passes twice through the heart during one cardiac cycle.
2. The blood follows two routes viz. pulmonary and systemic.
3. Pulmonary circulation is circulation between heart and lungs. Systemic circulation is the circulation between the heart and body organs (except lungs).
4. During pulmonary circulation, the blood passes from the right ventricle to the left atrium of the heart through lungs.
5. The right ventricle pumps deoxygenated blood into the pulmonary trunk which carries it to lungs for oxygenation. The oxygenated blood from the lungs is brought to left atrium by two pairs of pulmonary veins.
6. During systemic circulation, the blood from the left ventricle passes to the right atrium of heart through body organs.
7. The left ventricle pumps oxygenated blood into the systemic aorta which carries it to all body organs except lungs. The deoxygenated blood from the body organs is brought to right atrium by superior and inferior venae cavae.

Question 10.
How is cardiac activity regulated?
Answer:

1. Normal activities of the heart are auto regulated. The specialized muscles help in this regulation.
2. The heart is said to be myogenic due to this ability.
3. In the medulla oblongata of brain, there is cardiovascular centre,
4. From this centre, sympathetic and parasympathetic nerves innervate the sinoatrial node.
5. Sympathetic nerves secrete adrenaline and it stimulates and increases the heartbeat.
6. Parasympathetic nerves secrete acetylcholine and it decreases the heart rate.
7. Rate of heartbeat is controlled in response to inputs from various receptors like proprio-receptors.
8. These receptors monitor the position of limbs and muscles. There are chemoreceptors which monitor chemical change in blood and baroreceptors that monitor the stretching of main arteries and veins.

Chemical control on heart rate:

1. Hypoxia, acidosis, alkalosis cause decrease in cardiac activity.
2. Hormones like epinephrine and nor epinephrine enhance the cardiac activity.
3. Elevated blood level of K+ and Na+ decreases the cardiac activity.

Question 11.
What are the main features of respiratory surface?
Answer:
The respiratory surface, for the efficient gaseous exchange should have the following features:

1. It should have a large surface area.
2. It should be thin, highly vascular and permeable to allow exchange of gases.
3. It should be moist.

Question 12.
What is the co-relationship between activeness of organism and complexity of transport system?
Answer:

1. As the size of an organism increases, its surface area to volume ratio decreases. This means it has relatively less surface area available for substances to diffuse through.
2. Large multicellular organisms therefore cannot rely on diffusion alone to supply their cells with Substances such as food and oxygen and to remove waste products. Large multicellular organisms require specialized transport system.
3. In short for the organisms to become active, they must be having complex transport system to bring about their vital functions rapidly.

Question 13.
What are the granules in granulocytes?
Answer:

1. Neutrophils : Granules . of neutrophils contain cationic proteins and other proteins that are used to kill bacteria, some enzymes to breakdown bacterial proteins, lysozymes to breakdown bacterial cell wall. etc.
2. Eosinophils : Granules of eosinophil contains a unique toxic basic protein receptors that bind to IgE used to help in killing parasites.
3. Basophils : Granules of basophils contain abundant histamine, heparin and platelet activating factor.

Question 14.
What is haemoglobin count in normal human beings? What is the function of haemoglobin?
Answer:

1. The normal haemoglobin in adult male is 13-18 mg/100 ml of blood.
2. In a normal adult female, it is about 11.5-16.5 mg/100 ml of blood.
3. In anaemic individuals there is lesser amount of haemoglobin.
4. Functions of haemoglobin is to transport oxygen from lungs to tissues and carbon dioxide from tissues of lungs.
5. Haemoglobin acts as a buffer and maintains the blood pH.

Question 15.
Why has the heart-recipient to rely upon life-time supply of immunosuppressants?
Answer:
Person who has undergone heart transplant needs lifetime supply of immunosuppressants because in these persons organ rejection is a constant threat. Keeping away the immune system from attacking the transplanted organ requires constant supply of immunosuppressant drugs. These drugs help prevent immune system from attacking (“rejecting”) the donor organ. Typically, these drugs are taken for the life-time for maintaining transplanted organ.

Question 16.
Why is it difficult to hold one’s breath beyond a limit?
Answer:

1. It is difficult to hold one’s breath beyond a limit because the pressure of oxygen and carbon dioxide in blood changes as one holds his breath.
2. When the breath is held beyond a limit, the urge to breath becomes irresistible.
3.  When the breath is held forever, body becomes starved of oxygen and person may fall unconscious and the instinct to breath would take over.

Question 17.
Why and when do the leucocytes perform diapedesis?
Answer:

1. Diapedesis is the movement of leucocytes through the wall of blood capillaries into the tissue space.
2. Leucocytes perform diapedesis as an important part of their reaction to tissue injury or infection.
3. This process forms the part of the innate immune response, involving recruitment of non-specific leucocytes.
4. Monocytes also use this process during their development into macrophages.
5. Diapedesis helps leucocytes to perform their functions like phagocytosis, production of antibodies, secretion of inflammatory response chemicals, etc.

Question 18.
Why are obese persons prone to hypertension?
Answer:
(1) Being overweight or obese is a major cause of hypertension, accounting for 65% to 75% of the risk for human primary hypertension.

(2) Following factors play an important role in initiating obesity hypertension:

1. Physical compression of the kidneys by fat.
2. Activation of the renin-angiotensin – aldosterone system
3. Increased sympathetic nervous system activity.
4. Obesity means more body-surface area. In order to supply blood to these parts. Heart and blood vessels work more resulting into hypertensions.

(3) Blood pressure rises as body weight increases and therefore obese persons are prone to hypertension.

Question 19.
Why does the transplanted heart beats at higher rate than normal?
Answer:

1. The transplanted heart beats at higher rate than normal (about 100 to 110 beats per minute) because the nerves leading to the heart are cut during the operation. These nerves stimulate the pacemaker i.e. Sinoatrial node.
2. The new heart also responds more slowly to exercise and does not increase its rate as quickly as before.

Question 20.
Why do large animals cannot carry out respiration without the help of circulatory system?
Answer:

1. Large animals have various organ systems which always work in a coordinated manner.
2. These animals provide large respiratory surfaces (numerous alveoli) for the exchange of gases. But these respiratory gases must be carried to the cells of tissues which are away from the respiratory surfaces.
3. To carry these gases to tissues, there is need of transport system. These gasses are transported from respiratory surfaces to the cells of tissues through blood as a transporting medium.
4. Therefore, large animals cannot carry out respiration without the help of circulatory system.

Question 21.
What is immunity? Name its types.
Answer:

1. Immunity is the general ability of a body to recognize, neutralize or destroy and eliminate foreign substances or resist a particular infection or disease.
2. There are two basic types of immunity, viz. innate immunity and acquired immunity, Acquired immunity is further divided into four types, i.e. Natural acquired active immunity, Natural acquired passive immunity, Artificial acquired active passive immunity and Artificial acquired passive immunity.

Question 22.
Why does the platelet count decrease in dengue patient?
Answer:

1. The causative pathogen of dengue is dengue virus which induces bone marrow suppression. Since in bone marrow blood cells are formed its suppression causes the deficiency of blood cells leading to low platelet count.
2. The dengue virus also links with platelets in the blood when there is a virus-specific antibody present in the human body.
3. When vascular endothelial cell which are infected with dengue virus gets combined with platelets, they tend to destroy platelets. This is one of the major causes of low platelet count in dengue fever.
4. Even the antibodies that are produced after infection of the dengue virus also cause the destruction of platelets, thus lowering the platelet count.

Question 23.
Why does our immune system fail against pathogens like Trypanosoma cruzi and Plasmodium?
Answer:

1. Microbes have evolved a diverse range of strategies to destroy the host immune system. The protozoan parasite Trypanosoma cruzi and Plasmodium show similar such adaptations to disturb host defence mechanism.
2. This parasite attacks host tissues including both peripheral and central lymphoid tissues.
3. This causes systemic acute response in host body which the parasite tries to overcome. The parasite in fact weakens both innate and acquired immunity.
4. It interferes with the antigen presenting function of dendritic cells via an action on hosts like lectin receptors. These receptors also induce suppression of CD₄⁺ T cells responses. Therefore, our immune system fail against such pathogens.

Question 24.
What is the relation between immunity and organ transplantation?
Answer:

1. Those who undergo an organ transplant face the possibility that their immune system will reject their new organ and that they will always be at a higher risk for infections.
2. The immune system is able to recognize the difference between cells that belong to our body and those that do not by learning to identity protein markers (antigens) that are found on cell and infection surfaces.
3. In people, the antigens or markers that identity their immune system are referred to as the human leukocyte antigen (HLA).
4. Antigens that are recognized as unfriendly invaders stimulate an immune response to destroy them.
5. Therefore, when organ transplantation is done, the immune responses are temporarily stalled. This helps in acceptance of the graft in the recipient’s body.

Question 25.
How do monocytes perform amoeboid movement and phagocytosis?
Answer:
Monocytes can perform phagocytosis. They do this by using intermediary or opsonising proteins such as antibodies or complement that coat the pathogen. They also bind to the microbe directly via pattern-recognition receptors that recognize pathogens. In this way they perform amoeboid movement and indulge in phagocytosis.

Question 26.
How do monocytes modify into macrophages?
Answer:
Monocytes upon having inflammation, selectively travel to the sites of inflammation. Here they produce inflammatory cytokines and contribute to local and systemic inflammation. They are highly infiltrative. They differentiate into inflammatory macrophages, which then remove PAMPs or pathogen-associated molecular patterns and cell debris.

Chart based/Table based questions

Question 1.
Complete the following:

Organism	Habitat	Respiratory surface/organ
1. Insects	Terrestrial	—————
2. Amphibian tadpoles of frog, salamanders	—————-	—————-
3. Fish	Aquatic	————–
4. Reptiles, Birds and Mammals	—————-	—————-

Answer:

Organism	Habitat	Respiratory surface/organ
1. Insects	Terrestrial	Tracheal tubes and spiracles
2. Amphibian tadpoles of frog, salamanders	Aquatic	External gills
3. Fish	Aquatic	Internal gills
4. Reptiles, Birds and Mammals	Terrestrial	Lungs

Question 2.

Partial pressure of gases	Alveolar air	Pulmonary, capillaries
PPO2	—————	—————
PPCO2	—————-	—————-

Answer:

Partial pressure of gases	Alveolar air	Pulmonary, capillaries
PPO2	104 mm Hg	40 mm Hg
PPCO2	40 mm Hg	45 mm Hg

Question 3.

Answer:

Question 4.
Complete the following flow chart

Answer:

Question 5.
Complete the following Table

Waves on ECG	Heart Activity	Caused due to
P wave	Atrial contraction	—————-
QRS wave	—————–	Ventricular depolarization
T wave	—————–	Ventricular repolarization

Answer:

Waves on ECG	Heart Activity	Caused due to
P wave	Atrial contraction	Atrial depolarization
QRS wave	Ventricular contraction	Ventricular depolarization
T wave	Ventricular contraction	Ventricular repolarization

Question 6.

Cardiovascular disorders	Symptom
Coronary Artery Diseases (CAD)	Deposition of calcium, fat, cholesterol and ———————
—————-	Pain in chest resulting from reduction in the blood supply to the cardiac muscles.
Silent Heart Attack	Myocardial infarction without ———————-

Answer:

Cardiovascular disorders	Symptom
Coronary Artery Diseases (CAD)	Deposition of calcium, fat, cholesterol and fibrous tissues in blood vessels.
Angina pectoris	Pain in chest resulting from reduction in the blood supply to the cardiac muscles.
Silent Heart Attack	Myocardial infarction without showing symptoms of classical heart attack.

Question 7.

Instrument / Technique	Purpose of use
Sphygmomanometer	———————
—————-	X-ray imaging of the cardiac blood vessels to locate the position of blockages.
—————	To measure ECG.

Answer:

Instrument / Technique	Purpose of use
Sphygmomanometer	To measure blood pressure.
Angiography	X-ray imaging of the cardiac blood vessels to locate the position of blockages.
Electrocardiograph	To measure ECG.

Diagram based questions

Question 1.
Give the name and function of A and ‘B’ from the diagram given below

Answer:

Name	Function
(A) Epiglottis	Epiglottis prevents the entry of food into trachea.
(B) Tracheal cartilage	Tracheal cartilage prevents collapse of trachea and always keeps it open.

Question 2.
Label parts A’ and ‘B’ from the following diagram and answer the following questions

(a) What is the partial pressure of O₂ in part ‘B’?
(b) What is the partial pressure of CO₂ in part A’?
(c) How many alveoli are present in the lungs?
Answer:
Label A → Blood from pulmonary artery Label B → Alveolus
(a) The partial pressure of O₂ in alveolar air is 104 mm Hg.
(b) The partial pressure of CO₂ in pulmonary capillaries is 45 mmHg.
(c) There are about 700 million alveoli in the lungs.

Question 3.
Label parts A’ and ‘B’ from the given diagram and give their functions.

Answer: Part A → Sinoatrial [SA] node
Function : SA node acts as pacemaker of heart because it has the power of generating a new wave of contraction and making the pace of contraction.
Part B → Atrioventricular [AV] node
Function : Atrioventricular [AV] node acts as pace-setter of heart.

Question 4.
Sketch and label the dorsal (posterior) view of human heart.
Answer:

Question 5.
Sketch and label the ventral (anterior) view of human heart.
Answer:

Question 6.
Sketch and label – Electrocardiogram or ECG.
Answer:

Question 7.
Sketch and label – T.S. of Artery, Vein and Capillary.
Answer:

Question 8.
Observe the diagrams of blood cells and answer the following questions

(a) Which of the above is agranulocyte ?
(b) Describe its origin and structure.
(c) Mention its types.
(d) Explain its function.
Answer:
(a) The figure ‘D’ is agranulocyte.
(b) Structure : Agranulocytes do not show cytoplasmic granules and their nucleus is not lobed.
(c) Agranulocytes are of two types, viz. lymphocytes and monocytes.
(d) Functions of agranulocytes : Agranulocytes are responsible for immune response of body by producing antibodies and monocytes are phagocytic in function.

Question 9.
Observe the diagrammatic representation of double circulation and answer the given questions.

1. Why the circulation shown in the above diagram is called double circulation?
2. What are the two main routes of double circulation?
3. Which blood vessels carry oxygented blood to heart and deoxygenated blood to lungs?
4. Which blood vessels carry deoxygented blood to heart and oxygenated blood to body organs?

Answer:

1. During circulation, blood passes twice through the heart, therefore it is called double circulation.
2. (1) Pulmonary circulation which is from heart to lungs and back from lungs to heart.
   (2) Systemic circulation which is from heart to body and back from all body organs to the heart.
3. Oxygenated blood is carried to the heart by pulmonary veins. Dexoygenated blood is carried to the lungs by pulmonary artery.
4. Deoxygenated blood is carried to heart by superior and inferior vena cavae. Oxygenated blood is carried to the body organs by systemic or dorsal aorta.

Question 10.
Observe the cardiac cycle given below and answer the following questions

1. Which phases of cardiac cycle are shown in the above diagrammatic representation ?
2. How much time is taken for entire heart to be in diastole?
3. How much longer is ventricular systole as compared to atrial systole?

Answer:

1. There are four main phases of cardiac cycle shown in the above diagram. They are
   (1) AS : Arterial systole. (2) AD : Atrial diastole. (3) VS : Ventricular systole.
   (4) VD : Ventricular diastole which is along with joint diastole.
2. Diastole of entire heart is called joint diastole, which is for about 0.4 second.
3. Ventricular systole is almost for the double time than the atrial systole. Atrial systole is for 0.15 second whereas ventricular systole is for 0.3 second.

Long Answer Questions

Question 1.
Describe the respiratory system of human.
Answer:
Respiratory system of human : Human respiratory system consists of nostrils, nasal chambers, pharynx, larynx, trachea, bronchi, bronchioles, lungs, diaphragm and intercostal muscles.
1. Nostrils and nasal chambers:

1. Oxygen rich air is taken in the body through the nostrils or external nares. They are external opening of the nose. Carbon dioxide and water vapour are also released out of the body through the same passage i.e. the nostrils.
2. Internal nares open into the pharynx. The space between external and internal nares is knows as nasal chamber which is lined internally by mucous membrane and ciliated epithelium.
3. Nasal chamber is divided into two parts by a cartilage called mesethmoid. Each part of these halves is further divided into three regions, viz. vestibule, respiratory part and sensory part.
4. Vestibule is the anteriormost part of nasal chamber. In the vestibule fine hairs are present. They filter out the dust particles and prevent them from going inside.
5. Respiratory part is the second region which is richly supplied with the capillaries. Air is made warm and moist in this region.
6. Sensory part is lined by sensory epithelium. It is concerned with the detection of smell.

2. Pharynx:

1. Pharynx is a short and vertical tube measuring about 12 cm in length. In pharynx the respiratory and food passages cross each other.
2. The upper part of pharynx is known as naso-pharynx which conducts the air. The lower part is called laryngo-pharynx or oro¬pharynx which conducts food to the oesophagus.
3. Tonsils that are made up of lymphatic tissue are present in the pharynx. They kill the bacteria by trapping them in the mucus.

3. Larynx:

1. Larynx produces sound. In males it increases in size at puberty. This is termed as Adam’s apple. It is clearly seen in the neck region.
2. From pharynx air enters the larynx. The opening through which it enters is called glottis. Glottis is guarded by a flap called epiglottis.
3. Epiglottis prevents the entry of food particles into the trachea.
4. TWo folds of elastic tissue called vocal cords are seen along the side of glottis. When they vibrate the sound is produced.

4. Trachea:

1. The trachea or wind pipe is about 12 cm long and 2.5 cm wide.
2. It is situated in front of the oesophagus and runs downwards in the thorax through the neck.
3. The trachea is made up of fibrous muscular tissue wall which is supported by ‘C’-shaped cartilages. These cartilaginous rings are 16 to 20 in number.
4. Internally the tracheal wall bears ciliated epithelium and mucous glands.
5. When any foreign particle enters the trachea inadvertently. It is thrown out by coughing action.
6. Mucous and ciliary action remove the dust particles and push them upwards to the larynx. These particles are then gulped and taken into the oesophagus.

5. Bronchi and bronchioles:

1. At the distal end, the trachea divides into two bronchi (Singular – bronchus). Bronchi lie below the sternum or breast bone.
2. Each bronchus has a complete ring of cartilage for support. The two bronchi enter into the lungs on either side.
3. After entering into the lungs each bronchus divides into secondary and tertiary bronchi. The tertiary bronchi divide and re-divide to form minute bronchioles.
4. Bronchioles do not have cartilages in their walls. Each bronchiole ends into a balloon like alveolus.
5. Owing to the presence of alveoli the lungs become spongy and elastic.

6. Lungs:

1. Lungs are principal respiratory organs located in the thoracic cavity.
2. They are pinkish, soft, hollow, paired, elastic and distensible organs.
3. Each lung is enclosed in a pleural sac which consists of two membranes, viz. an outer parietal and inner visceral.
4. The parietal and visceral membranes enclose pleural cavity which is filled with pleural fluid. The pleural fluid lubricates and prevents friction when pleural membranes slide on each other.
5. Lungs are highly vascular as they are richly supplied with blood capillaries.
6. The left lung has two lobes while the right lung has three lobes. Each lobe has many bronchioles and alveolar sacs. The alveolar sacs are spherical and thin walled.
7. Each alveolar sac contains about 20 alveoli. The alveoli appear as a bunch of grapes. The lobule in the lung thus consists of alveolar ducts, alveolar sacs and alveoli.
8. Each alveolus has thin and elastic walls. It is about 0.1 mm in diameter. Alveoli are covered by network of capillaries from pulmonary artery and pulmonary vein. A network of pulmonary capillaries supply the alveolus.
9. The alveolar wall is 0.0001 mm thick and made up of simple, non-ciliated, squamous epithelium. It has collagen and elastin fibres.
10. Every lung has about 700 million alveoli. They increase the surface area of the lungs for exchange of gases.

Question 2.
Describe the process of respiration in man.
OR
Describe the mechanism of respiration in human beings.
Answer:
Respiration includes breathing, external respiration, internal respiration and cellular respiration.
A. Breathing : During breathing air comes in and goes out of the lungs. The rate of gaseous exchange is speeded up by breathing. Breathing involves two processes, viz. inspiration and expiration
1. Inspiration:

1. Inspiration is the process in which the air containing oxygen is taken inside the lung.
2. Inspiration is the active process which is possible due to intercostal muscles, sternum and diaphragm.
3. During inspiration, intercostal muscles contract, ribs are pulled outward as a result of which the space in the thoracic cavity is increased.
4. At the same time the lower part of the breast bone is raised and diaphragm flattens by contraction.
5. The volume of thoracic cavity is thus increased.
6. Pressure in the lungs decreases as the lungs expand and their volume is increased. Owing to this the atmospheric air enters inside the body through respiratory passage and reaches the lungs.

2. Expiration:

1. Expiration is the process in which air containing carbon dioxide and water vapour is expelled out of the lungs.
2. Expiration is a passive process.
3. During expiration, intercostal muscles relax and ribs are pulled inwards.
4. The diaphragm relaxes and becomes dome shaped. Intercostal muscles contract simultaneously and due to these events, the volume of the thoracic cavity is reduced.
5. The pressure on the lungs is thus increased as a result of which they are compressed.
6. Due to this, air rushes out of the lungs and is expelled out through the nose.
   

B. External respiration : The process of external respiration takes place in the lungs where oxygen from the lungs diffuses into the blood capillaries present in the lung tissue. Similarly carbon dioxide from the blood capillaries diffuses out and enters in the alveoli in the lungs.

C. Internal respiration : Internal respiration takes place in the cells of the body. Oxygen brought by the blood is given to the cells and the tissues during internal respiration. Similarly carbon dioxide passes into the blood cells from the cells and tissues.

D. Cellular respiration : Cellular respiration takes place in the mitochondria of the cell, where oxygen is utilised to liberate energy in the form of ATP molecules.

Question 3.
Describe the respiratory disorders.
Answer:

1. Following are some respiratory disorders : Emphysema is caused due to alveolar abnormalities. Chronic bronchitis results into coughing and shortness of breath.
2. Viral and bacterial respiratory diseases. Acute bronchitis, sinusitis, laryngitis and pneumonia are some of the inflammatory diseases caused either due to virus or due to bacteria.
3. Allergens like pollen or pet dander can cause asthma. In asthma constriction of bronchioles takes place causing periodic wheezing and difficulty in breathing.
4. Occupational hazards cause respiratory diseases like silicosis or asbestosis. In these disorders there is inflammation fibrosis leading to lung damage.

Treatments of respiratory diseases:

1. Bacterial diseases can be completely cured by specific antibiotics.
2. Viral diseases need to be taken care of by using vaporizers and decongestants.
3. Asthma needs treatment by inhalers and nebulizers.
4. For occupational disorders proper mask and other protective gear is a must.
5. Lethal diseases like pneumonia should be controlled by medication and rest.

Question 4.
How does transport of O₂ and CO₂ take place in man?
Answer:
1. Transport of O₂:

1. Only 3% of the total oxygen is transported in a dissolved state by the plasma.
2. The remaining 97% is transported in the form of oxyhaemoglobin in the RBCs.
3. Hemoglobin present is RBCs combines with oxygen to form oxyhaemoglobin.
   Hb + 4O₂ → Hb (4O₂)
4. Oxyhaemoglobin is transported from lungs to the tissues where it readily dissociates to release O₂.
5. Binding of oxygen with haemoglobin in the alveoli and release of oxygen into the tissue cells depends upon the difference in partial pressure of O₂ and CO₂.

2. Transport of CO₂: Carbon dioxide is transported by RBCs and plasma in three different forms.

1. By plasma in solution form (7%) : About 7% of CO₂ is transported in a dissolved form as carbonic acid (which can be broken down into CO₂ and H₂O).
   CO₂ + H₂O = H₂CO₃.
2. By bicarbonate ions (70%) : Nearly 70% of carbon dioxide is transported in the form of sodium bicarbonate/potassium bicarbonate in the plasma.
3. RBCs contains an enzyme, carbonic anhydrase. In the presence of this enzyme CO₂ combines with water to form carbonic acid.
4. Carbonic anhydrase also brings about dissociation of carbonic acid immediately tending to large accumulation of HCO^3- ions inside the RBCs.
   CO₂ + H₂O Carbonic anhydrese H₂CO₂ Carbonic anhydrase H₊ + HCO^3-
5. The bicarbonate ions moves out of RBCs and this would bring about imbalance of the charge inside the RBCs.
6. To maintain the ionic balance, Cl^– ions diffuse from plasma into the RBCs. This movement of chloride ions is known as chloride shift or Hamburger’s phenomenon.
7. HCO^3- ions from the plasma then joins to Na⁺/K⁺ forming NaHCO₃/KHCO₃ (to maintain PH of blood).
   HCO^3- + Na⁺ → NaHCO₃ Sodium bicarbonate
8. H⁺ is taken up by haemoglobin to form Reduced Hb (HHb).
9. At the level of the lungs due to the low partial pressure of the alveolar air, hydrogen ion and bicarbonate ions recombine to form carbonic acid and in presence of carbonic anhydrase it again yields carbon dioxide and water.
   H⁺ + HCO_3- Carbonic anhydrase H₂CO₃ Carbonic anhydrase CO₂ + H₂O.

3. By red blood cells (23%):

1. Carbon dioxide binds with the amino group of the haemoglobin and form a loosely bound compound carbaminohaemoglobin Hb + CO₂ = HbCO₂
2. Due to low partial pressure of CO₂ at alveolus carbaminohaemoglobin decomposes releasing the carbon dioxide.

Question 5.
Describe different types of leucocytes.
OR
Describe five types of leucocytes, with the help of diagrams. Add a note on their functions.
Answer:

1. Leucocytes or White Blood Corpuscles (WBCs) are colourless, nucleated, amoeboid and phagocytic cells.
2. Their size ranges between 8 to 15 pm. Total WBC count is 5000 to 9000 WBCs/cu mm of blood. The average life span of a WBC is about 3 to 4 days.
3. They are formed by leucopoiesis in red bone marrow, spleen, lymph nodes, tonsils, thymus and Payer’s patches, whereas the dead WBCs are destroyed by phagocytosis in blood, liver and lymph nodes.
4. Leucocytes are mainly divided into two types, viz., granulocytes and agranulocytes.
5. Granulocytes : Granulocytes are cells with granular cytoplasm and lobed nucleus. Based on their staining properties and shape of nucleus, they are of three types, viz. neutrophils, eosinophils and basophils.

(I) Neutrophils:

1. In neutrophils, the cytoplasmic granules take up neutral stains.
2. Their nucleus is three to five lobed.
3. It may undergo changes in structure hence they are called polymorphonuclear leucocytes or polymorphs.
4. Neutrophils are about 70% of total WBCs.
5. They are phagocytic in function and engulf microorganisms.

(II) Eosinophils or acidophils:

1. Cytoplasmic granules of eosinophils take up acidic dyes such as eosin. They have bilobed nucleus.
2. Eosinophils are about 3% of total WBCs.
3. They are non-phagocytic in nature.
4. Their number increases (i.e. eosinophilia) during allergic conditions.
5. They have antihistamine property.

(III) Basophils:

1. The cytoplasmic granules of basophils take up basic stains such as methylene blue.
2. They have twisted nucleus.
3. In size, they are smallest and constitute about 0.5% of total WBCs.
4. They too are non-phagocytic.
5. Their function is to release heparin which acts as an anticoagulant and histamine that is involved in inflammatory and allergic reaction.

Agranulocytes : There are two types of agranulocytes, viz. monocytes and lymphocytes. Agranulocytes do not show cytoplasmic granules and their nucleus is not lobed. They are of two types, viz. lymphocytes and monocytes.
(I) Lymphocytes:

1. Agranulocytes with a large round nucleus are called lymphocyte.
2. They are about 30% of total WBCs.
3. Agranulocytes are responsible for immune response of the body by producing antibodies.

(II) Monocytes:

1. Largest of all WBCs having large kidney shaped nucleus are monocytes. They are about 5% of total WBCs.
2. They are phagocytic in function.
3. They can differentiate into macrophages for engulfing microorganisms and removing cell debris. Hence they are also called scavengers.
4. At the site of infections they are seen in more enlarged form.

Question 6.
Give an account of external features of the human heart.
Answer:
(1) The heart is hollow, muscular, conical organ about the size of one’s fist with broad base and narrow apex tilted towards left measuring about 12 cm in length. 9 cm in breadth and weighing about 250 to 300 grams.

(2) The human heart has four chambers, two atria which are superior, small, thin walled receiving chambers and two ventricles which are inferior, large, thick walled, distributing chambers.

(3) Externally there is a transverse groove between the atria and the ventricles which is known as atrioventricular groove or coronary sulcus.

(4) Between the right and left ventricles there is interventricular sulcus (pi. sulci). In these sulci the coronary arteries and coronary veins are present.

(5) Oxygenated blood to the heart is supplied by coronary arteries while coronary veins collect deoxygenated blood from the heart. The coronary veins join to form coronary sinus which opens into the right atrium.

(6) Right atrium is larger in size than the left atrium. Deoxygenated blood from all over the body is brought through superior vena cava and inferior vena cava and poured into right atrium. Oxygenated blood from lungs is brought to heart by two pairs of pulmonary veins which carry it to the left atrium.

(7) Pulmonary trunk is seen arising from the right ventricle, which carries deoxygenated blood to lungs. While systemic aorta arises from the left ventricle and carries oxygenated blood to all parts of the body.

(8) The pulmonary trunk and systemic aorta are connected by ligamentum arteriosum that represents remnant of ductus arteriosus of foetus.

Question 7.
With the help of well labelled diagram describe the internal structure of human heart.
OR
Sketch and label internal view of heart.
Answer:

The heart shows four chambers with two atria and two ventricles.

I. Atria:
1. Right atrium:

1. There are two atria which are separated from each other by interatrial septum. They are thin walled receiving chambers on the upper side.
2. The right atrium receives deoxygenated blood from upper part of body through superior vena cava and from the lower part of the body by inferior vena cava. In the right atrium opens the coronary sinus.
3. Eustachian valve guards the opening of inferior vena cava while opening of coronary sinus is guarded by Thebesian valve.
4. On the right side of interatrial septum is seen an oval depression called the fossa ovalis. In the interatrial septum of the foetus there is an oval opening called foramen ovale. Fossa ovalis is remnant of this foramen ovale.
5. Right atrium opens into the right ventricle.

2. Left atrium:

1. The oxygenated blood from the lungs is brought into left atrium through four openings of pulmonary veins.
2. Left atrium opens into the left ventricle.

II. Ventricles:

1. There are two ventricles which are separated from each other by interventricular septum. They are two thick walled distributing chambers situated on the lower side of the heart.
2. Left ventricle has thickest wall as it pumps blood to all parts of the body.
3. The inner surface of the ventricle is thrown into a series of irregular muscular ridges called columnae carnae or trabeculae carnae.
4. Each atrium opens into the ventricle of its side through atrioventricular aperture. These apertures are guarded by valves made up of connective tissue. The right atrioventricular valve has three flaps hence called tricuspid valve. Left atrioventricular valve has two flaps hence called bicuspid valve or mitral valve.
5. Bicuspid and tricuspid valves are attached to papillary muscles of ventricles by chordae tendinae. The chordae tendinae prevent the valves from turning back into the atria during the contraction of ventricles.
6. From the right ventricle arises pulmonary trunk which carries deoxygenated blood to lungs for oxygenation.
7. From the left ventricle arise systemic aorta which distributes oxygenated blood to all parts of the body.
8. Pulmonary aorta and systemic aorta has three semilunar valves at the base which prevent backward flow of blood during ventricular diastole.

Question 8.
With the help of suitable diagram, describe the conducting system of human heart.
Answer:
(1) The human heart is myogenic.

(2) Conducting system of the heart consists of sinoatrial node, atrioventricular node, bundles of His and Purkinje’s fibre system.

(3) The pacemaker of the heart is sinoatrial node because here the heartbeat originates. Pacemaker has power of generation of wave of contraction. This is modified cardiac tissue, also called a nodal tissue.

(4) SA node is situated in the wall of right atrium near the opening of superior vena cava. The wave of contraction generated by SA node is conducted by cardiac muscle fibres to both the atria. This results in contraction resulting into atrial systole.

(5) The atrioventricular node (AV node) is located in the wall of right atrium near the opening of coronary sinus. AV node receives the wave of contraction generated by SA node through intermodal pathways.

(6) Bundle of His arises from AV node and divides into right and left bundle branches. These are located in the interventricular septum.

(7) The bundle branches further form Purkinje fibres which penetrate into myocardium of ventricles.

(8) The bundle of His and Purkinje fibres conduct the wave of contraction from AV node to myocardium of ventricles causing ventricular systole.

Question 9.
Describe the detail cardiac cycle.
OR
Explain the working of heart.
Answer:
(1) The working of heart or cardiac cycle is formed by atrial systole, ventricular systole and joint diastole. It takes place in 0.8 second.

(2) During atrial systole from right atrium, the deoxygenated blood is poured into right ventricle through atrioventricular aperture. Similarly, from left atrium, the oxygenated blood enters the left ventricle through atrioventricular aperture. This entire atrial systole lasts for 0.1 second.

(3) After auricular systole follows the ventricular systole. During ventricular systole, the deoxygenated blood from the right ventricle enters the pulmonary trunk, which carries blood to lungs for oxygenation. At the same time, the oxygenated blood from the left ventricle enters the aorta which is then supplied to all parts of the body. The ventricular systole lasts for 0.3 second.

(4) Joint diastole or complete cardiac diastole is the phase taking place after the systole, when the entire heart undergoes relaxation, for 0.4 second.

(5) During joint diastole, the right atrium receives deoxygenated blood from all parts of the body through superior vena cava, inferior vena cava and coronary sinus. The left atrium receives oxygenated blood from the lungs through two pairs of pulmonary veins.

Question 10.
What is blood clotting? How and when does it occur?
Answer:

1. Blood clotting is coagulation of blood in order to stop the blood flow and resuting blood loss at the time of injury.
2. When the blood vessel is intact, blood does not clot due to the presence of active anticoagulants like heparin and antithrombin. But when there is an injury causing rupture of a blood vessel, bleeding starts.
3. This bleeding is stopped by the process of blood clotting during which liquid blood is converted into semisolid jelly.

The events occurring during blood clotting are as follows:

1. Release of thromboplastin from thrombocytes and injured tissue.
2. Formation of enzyme prothrombinase in the blood due to initiation of thromboplastin.
3. Conversion of inactive prothrombin into active thrombin by prothrombinase in the presence of Ca ions.
4. Conversion of soluble fibrinogen into insoluble fibrin by thrombin.
5. Formation of a clot by enmeshing platelets, other blood cells and plasma in the fibrin fibres enmesh.

These reactions occur in 2 to 8 minutes. Therefore, clotting time is said to be 2 to 8 minutes.

Question 11.
What is repolarization and depolarization ?
Answer:
Repolarization is a stage of an action potential in which the cell experiences a reduction of voltage due to the efflux of potassium (K⁺) ions along its electrochemical gradient. This phase occurs after the cell reaches its highest voltage from depolarization.

Depolarization occurs in the four chambers of the heart : both atria first and then both ventricles. The SA node sends the depolarization wave to the atrioventricular (AV) node which-with about a 100 minutes delay to let the atria finish contracting-then causes contraction in both ventricles, seen in the QRS wave.

Question 12.
What is the correlation between depolarization and repolarization as well as contraction and relaxation of the heart?
Answer:
Depolarization and Repolarization:

1. When cardiac cells are at rest, they are polarized, meaning no electrical activity takes place.
2. The cell membrane of the cardiac muscle cell separates different concentrations of ions, such as sodium, potassium, and calcium. This is called the resting potential.
3. Electrical impulses are generated by specialized cardiac cells automatically.
4. Once an electrical cell generates an electrical impulse, this electrical impulse causes the ions to cross the cell membrane and causes the action potential, also called depolarization.
5. The movement of ions across the cell membrane through sodium, potassium and calcium channels, is the drive that causes contraction of the cardiac cells/muscle.
6. Depolarization with corresponding contraction of myocardial muscle moves as a wave through the heart. Depolarization thus corresponds with contraction of heart.
7. Repolarization is the return of the ions to their previous resting state, which corresponds with relaxation of the myocardial muscle. Repolarization thus corresponds with relaxation of heart.
8. Depolarization and repolarization are electrical activities which cause muscular activity.
9. The electrical changes in the myocardial cell during the depolarization – repolarization cycle is detected on ECG.

Question 13.
How are the signals detected and amplified by electrocardiograph?
Answer:

1. The action potential created by contractions of the heart wall spreads electrical currents from the heart throughout the body.
2. The spreading electrical currents create different potentials at points in the body, which can be sensed by electrodes placed on the skin.
3. The electrodes are made of metals and salts and they act as biological transducers.
4. Ten electrodes are attached to different points on the body while taking ECG.
5. There Eire three main leads responsible for measuring the electrical potential difference between arms and legs.
6. Electrical potential difference between electrodes is recorded.
7. As in all ECG lead measurements, the electrode connected to the right leg is considered the ground node.
8. These ECG signals are acquired using a biopotential amplifier and then displayed using instrumentation software. This is recorded on ECG machine or electrocardiograph. The recorded ECG is anailysed by an expert.

Balbharti Maharashtra State Board 12th Commerce Book Keeping & Accountancy Important Questions Chapter 7 Bills of Exchange Important Questions and Answers.

Maharashtra State Board 12th Commerce BK Important Questions Chapter 7 Bills of Exchange

Objective Questions

A. Select the correct option and rewrite the sentence:

Question 1.
A bill of exchange is called a _____________ by one who is entitled to receive the amount due on it.
(a) Bills Payable
(b) Draft
(c) Bills Receivable
(d) Promissory Note
Answer:
(c) Bills Receivable

Question 2.
The person who draws a bill of exchange is called _____________
(a) Payee
(b) Drawee
(c) Endorsee
(d) Drawer
Answer:
(d) Drawer

Question 3.
A bill of exchange is required to be _____________ by drawee.
(a) drafted
(b) discounted
(c) accepted
(d) endorsed
Answer:
(c) accepted

Question 4.
A person who accepts the bill is called _____________
(a) Drawer
(b) Acceptor
(c) Payee
(d) Creditor
Answer:
(b) Acceptor

Question 5.
The person to whom the amount of the bill is made payable is called _____________
OR
_____________ is a person to whom the amount on a bill is payable.
(a) Endorsee
(b) Drawer
(c) Drawee
(d) Payee
Answer:
(d) Payee

Question 6.
When the acceptor accepts the bill with certain conditions, the acceptance is called _____________ Acceptance.
(a) Qualified
(b) General
(c) Clean
(d) Special
Answer:
(a) Qualified

Question 7.
The drawee becomes an _____________ on acceptance of a bill.
(a) acceptor
(b) owner
(c) endorser
(d) drawer
Answer:
(a) acceptor

Question 8.
Borrowing money from a bank on the security of a bill of exchange is called _____________
(a) Honouring
(b) Endorsing
(c) Discounting
(d) Retiring
Answer:
(c) Discounting

Question 9.
A bill of exchange is/can be discounted with the _____________
(a) bank
(b) payee
(c) money lenders
(d) government
Answer:
(a) bank

Question 10.
Transferring a bill of exchange before maturity to a third party is called _____________ of a bill of exchange.
(a) honouring
(b) endorsement
(c) retirement
(d) discounting
Answer:
(b) endorsement

Question 11.
The person who endorses the bill of exchange is known as _____________
(a) Drawer
(b) Endorsee
(c) Endorser
(d) Drawee
Answer:
(c) Endorser

Question 12.
A person to whom a bill of exchange is endorsed is called _____________
(a) Endorsee
(b) Drawer
(c) Endorser
(d) Payee
Answer:
(a) Endorsee

Question 13.
If a bill falls due on 15th August, payment on it must be made on _____________
(a) 14th August
(b) 16th August
(c) 13th August
(d) 17th August
Answer:
(a) 14th August

Question 14.
A bill drawn on 12th June, 2020 at two months would be payable on _____________
(a) 12th August 2020
(b) 14th August 2020
(c) 15th August 2020
(d) 16th August 2020
Answer:
(b) 14th August 2020

Question 15.
If a bill is drawn on 3rd July, 2020 for 40 days, its payment must be made on _____________
(a) 14th August 2020
(b) 15th August 2020
(c) 13th August 2020
(d) 16th August 2020
Answer:
(a) 14th August 2020

Question 16.
A bill is drawn on 23rd September, 2019 at 4 months would be payable on _____________
(a) 24th January 2020
(b) 25th January 2020
(c) 26th January 2020
(d) 25th January 2019
Answer:
(b) 25th January 2020

Question 17.
A bill is drawn on 23rd October 2016 payable after 3 months, the due date of the bill will be _____________
(a) 25th January 2017
(b) 26th January 2017
(c) 24th January 2017
(d) 25th January 2016
Answer:
(a) 25th January 2017

Question 18.
_____________ means payment of the bill before due date.
(a) Discounting of Bill
(b) Retirement of Bill
(c) Renewal of Bill
(d) Endorsement of Bill
Answer:
(b) Retirement of Bill

Question 19.
A bill of one month duration is accepted on 12th July, 2020, its due date will be _____________
(a) 12th August 2020
(b) 16th August 2020
(c) 14th August 2020
(d) 15th August 2020
Answer:
(c) 14th August 2020

Question 20.
When a bill Is dishonoured, the _____________ is held responsible for the noting charges.
(a) holder
(b) drawee
(c) drawer
(d) endorser
Answer:
(b) drawee

Question 21.
Fees charged by the Notary Public for noting facts or reasons of dishonour of bill are called _____________
(a) Discount
(b) Rebate
(c) Noting Charges
(d) Commission
Answer:
(c) Noting Charges

Question 22.
Noting charges are paid when a bill is _____________
(a) honoured
(b) dishonoured
(c) renewed
(d) retired
Answer:
(b) dishonoured

Question 23.
_____________ is done in respect of dishonour of foreign bill of exchange.
(a) Discounting
(b) Endorsement
(c) Noting
(d) Protesting
Answer:
(d) Protesting

B. Give one word/phrase/term which can substitute each of the following statements:

Question 1.
A bill of exchange is drawn and accepted for a value received.
Answer:
Trade bill

Question 2.
A person who draws a bill of exchange.
Answer:
Drawer

Question 3.
A person on whom a bill of exchange is drawn.
Answer:
Drawee

Question 4.
Payment in accordance with the apparent tenor of the bill.
Answer:
Honour

Question 5.
Non-payment in accordance with the apparent tenor of the bill.
Answer:
Dishonour

Question 6.
Acceptance without making any change in the terms of a bill.
Answer:
General acceptance

Question 7.
Acceptance with some changes as regards the terms of a bill.
Answer:
Qualified acceptance

Question 8.
A bill of which payment is to be made after the fixed period.
Answer:
After date bill

Question 9.
The bill is drawn in one country and payable in other countries.
Answer:
Foreign bill

Question 10.
Encashment of the bill before the due date.
Answer:
Discounting

Question 11.
Transfer of title of the bill from the debtor to the creditor.
Answer:
Endorsement

Question 12.
Payment of the bill before the due date.
Answer:
Retirement of bill

Question 13.
A document consists of a written order signed by the maker, directing a certain person to pay a certain sum of money on-demand or on a certain future date.
Answer:
Bill of Exchange

Question 14.
A person who accepts the bill.
Answer:
Drawee or Acceptor

Question 15.
The period for which a bill is drawn.
Answer:
Term/Tenure of a bill of exchange

Question 16.
A bill of exchange that does not contain the period for its payment.
Answer:
Demand bill/Bill at sight

Question 17.
The date on which period of the bill gets expired.
Answer:
Nominal due date

Question 18.
The date on which the payment of the bill is to be made.
Answer:
Due date or Date of maturity

Question 19.
Drawee’s signature on the face of the bill to show his consent to pay the amount of the bill.
Answer:
Acceptance of a bill of exchange

Question 20.
A bill of exchange before its acceptance.
Answer:
Draft

Question 21.
A bill is drawn, accepted, and made payable within the territory of one and the same country.
Answer:
Inland bill of exchange

Question 22.
Selling a bill to the bank before its due date for an amount slightly less than its face value.
Answer:
Discounting of a bill of exchange

Question 23.
Act of signing the bill on its back by its holder to transfer its title to a third Person.
Answer:
Endorsement of a bill of exchange

Question 24.
Discount is given by holder to acceptor on the retirement of the bill of exchange.
Answer:
Rebate

Question 25.
Non-payment of the bill on the due date.
Answer:
Dishonour of a bill of exchange

Question 26.
Recording the facts of dishonour of a bill of exchange by a Notary Public.
Answer:
Noting

Question 27.
The request by the acceptor of the bill to the drawer for issuing a new bill after canceling the old bill.
Answer:
Renewal of the bill of exchange

Question 28.
The account to which the bill is sent for collection is debited.
Answer:
Bill sent for Collection Account

Question 29.
Payment of a bill on the due date.
Answer:
Honouring of a bill of exchange

Question 30.
Drafting a new bill in cancellation of the old bill at the request of drawee.
Answer:
Renewal of the bill of exchange

Question 31.
Certificate is given by Notary Public for the fact of dishonour of the bill.
Answer:
Certificate of protest

Question 32.
The account is to be debited in case of dishonour of bill in the books of the drawer.
Answer:
Drawee’s Account

Question 33.
A foreign bill accompanied by shipping documents.
Answer:
Documentary bill

C. State True or False with reasons:

Question 1.
Bills payable are a liability.
Answer:
This statement is True.
Bill is always drawn on the debtor by the creditor. The debtor i.e. drawee has to pay the money on a future date. For the drawee or acceptor of the bill, payment of the number of Bills payable is certain and therefore, for drawee, Bills payable is a liability.

Question 2.
Drawee has no right to discount the bill with the bank.
Answer:
This statement is True.
Drawee means acceptor of a bill and for him, it is bills payable. Once he accept it, sign it, and returned it to the drawer he don’t have any bill with him to discount it with the bank. He is not the owner of the bill and hence, he has no right to discount the bill with the bank.

Question 3.
A bill of exchange needs acceptance.
Answer:
This statement is True.
A bill of exchange is drawn by the creditor on the debtor. It is signed by drawer as well as by drawee. The drawee has to give his assent to the terms and conditions of the bill by putting his signature on it. A bill without acceptance is called a draft. It becomes a valid document only when the drawee accepts it.

Question 4.
A bill can’t be deposited into the bank for collection.
Answer:
This statement is False.
When a drawer or holder of the bill needs money on the due date, he has to deposit the bill into the bank for collection purposes. So that bank can collect the amount in time. This means a bill can be deposited into the bank for collection.

Question 5.
Noting charges are payable to the Notary Public in honour of a bill.
Answer:
This statement is False.
Noting charges are payable to the Notary Public at the time of registration of a dishonoured bill, not at the time of honour of a bill.

Question 6.
A bill of exchange is a negotiable instrument.
Answer:
This statement is True.
Being a negotiable instrument, a bill of exchange is a written acknowledgment of debts and also a promise to pay the debt according to the terms of the bill and can be transferred from one person to another.

Question 7.
A bill of exchange is signed by the person on whom it is drawn.
Answer:
This statement is True.
A bill of exchange is signed by the person who draws or makes it, and the person on whom it is drawn accepts it.

Question 8.
Acceptance with some change as regards the terms of a bill is called general acceptance.
Answer:
This statement is False.
General acceptance means acceptance of a draft without any change or conditions regards the terms of a bill.

Question 9.
Drawee is a person who holds the title of the bill in due course.
Answer:
This statement is False.
A drawer is a person who holds the title of the bill in due course as drawee accepts it and returns it to the drawer.

Question 10.
A payee is an official person appointed by the Central government for noting of dishonour of the bill.
Answer:
This statement is False.
A notary public is an official person appointed by the Central government for noting of dishonour of the bill and making it legal.

D. Complete the sentences:

Question 1.
A person to whom or as per his order, amount of bill is payable is a _____________
Answer:
Payee

Question 2.
The inland bill is drawn and payable in the _____________ country.
Answer:
same

Question 3.
Discounting means encashment of the bill before its _____________
Answer:
due date

Question 4.
_____________ can transfer the ownership of the bill.
Answer:
Drawer

Question 5.
Noting charges are payable to the Notary public on _____________ of a bill.
Answer:
dishonour

Question 6.
A bill of exchange is a _____________
Answer:
negotiable instrument

Question 7.
If a discounted bill is honoured, the _____________ does not record this transaction.
Answer:
drawer

Question 8.
Days of grace are not allowed in the case of a _____________
Answer:
demand bill

Question 9.
Noting charges should be borne by _____________
Answer:
drawee

E. Answer in one sentence:

Question 1.
State the types of Bills of Exchange.
Answer:
The bills of exchange may be classified as

1. Inland bills of exchange
2. Foreign bills of exchange.

Question 2.
What is the Inland bill of exchange?
Answer:
A bill of exchange that is drafted, accepted, and made payable between the parties from one and the same country is called an Inland bill of exchange.

Question 3.
What is a Foreign bill of exchange?
Answer:
A bill of exchange that is drafted and accepted in one country and made payable in another country is called a Foreign bill of exchange.

Question 4.
Which are the parties to a bill of exchange?
Answer:
There are three parties to a bill of exchange, viz.,

1. Drawer
2. Drawee
3. Payee

Question 5.
Who is the Drawer?
Answer:
The Drawer of a bill is the person who draws or makes the bill.

Question 6.
Who is the Drawee?
Answer:
The Drawee of a bill is the person on whom the bill is drawn.

Question 7.
What is a Draft?
Answer:
A bill of exchange is called a Draft before its acceptance.

Question 8.
What is an Acceptance of the Bill of Exchange?
Answer:
The act of signing the bill of exchange by the drawee with a date to show his consent to pay the amount of the bill is called an Acceptance of the Bill of Exchange.

Question 9.
What do you mean by Clean or General Acceptance?
Answer:
A Clean or General Acceptance is an acceptance where the drawee does not make any change in the terms of the bill before accepting it.

Question 10.
What is Qualified Acceptance?
Answer:
If the drawee of a bill of exchange accepts it on condition that the time or amount of the bill is changed or adds some other conditions to the bill, his acceptance is called a Qualified Acceptance.

Question 11.
What is the term of the bill of exchange?
Answer:
The period for which the bill of exchange is drawn and accepted is called the term of the bill of exchange.

Question 12.
What is the Nominal Due Date?
Answer:
The date on which the term i.e. the period of a bill of exchange gets expired is called Nominal Due Date.

Question 13.
What is the Due Date of a Bill?
Answer:
The Due Date of a Bill of Exchange is the date on which it is falling due for payment by the drawee.

Question 14.
What is Endorsing of a Bill?
Answer:
Endorsing of a Bill is the holder’s signing on its back with the intention of transferring its title or ownership to another person.

Question 15.
Who is an Endorser?
Answer:
The drawer or the holder of a bill of the exchange who transfers or endorses the same in favour of a third party is called Endorser.

Question 16.
Who is an Endorsee?
Answer:
An Endorsee is a person to whom or in whose favour a bill is endorsed or transferred.

Question 17.
What is Retirement of a Bill?
Answer:
A bill of exchange is said to be retired if its acceptor makes payment of it before its due date, usually after deducting some discount or rebate.

Question 18.
When is a bill said to be honoured?
Answer:
A bill of exchange is said to be honoured or met when the acceptor or drawee makes payment on its due date.

Question 19.
When is the bill said to be dishonoured?
Answer:
A bill of exchange is said to be dishonoured if its acceptor or drawee fails to make payment on its due date.

Question 20.
Which account is credited in the books of the drawer when the discounted bill is dishonoured?
Answer:
Cash/Bank A/c is credited in the books of the drawer when the discounted bill is dishonoured.

Question 21.
Who is a Notary Public?
Answer:
An officer appointed by the Government to certify dishonour of bills of exchange is called Notary Public.

Question 22.
Who bears the noting charges on dishonour of a bill?
Answer:
An acceptor or drawee bears the noting charges on dishonour of a bill of exchange.

Question 23.
Who pays the noting charges?
Answer:
The holder of the bill of exchange pays the noting charges.

Question 24.
What do you mean by Renewal of a Bill?
Answer:
Renewal of a Bill of Exchange means cancellation of the original bill and drafting a new bill in exchange for that by a drawer at the request of drawee.

F. Do you agree or disagree with the following statements:

Question 1.
A bill of which payment to be made after the fixed period is after date bill.
Answer:
Agree

Question 2.
Drawee can transfer the ownership of the bill.
Answer:
Disagree

Question 3.
Endorsee is a person in whose favour the bill is transferred.
Answer:
Agree

Question 4.
The drawer and payee of a bill of exchange may be one and the same person.
Answer:
Agree

Question 5.
A bill of exchange can be endorsed only once.
Answer:
Disagree

Question 6.
Days of grace are allowed in the case of demand bills.
Answer:
Disagree

Question 7.
The noting of dishonoured bills is compulsory.
Answer:
Disagree

Question 8.
The endorser is a creditor to the endorsee.
Answer:
Disagree

Question 9.
Bills payable are a liability.
Answer:
Agree

Question 10.
A bill of exchange is a negotiable instrument.
Answer:
Agree

Solved Problems

Question 1.
On 1st April 2019 Parth draws a bill for ₹ 50,000 on Zalak for 4 months period. The bill is accepted and returned to Parth. On the same date, Parth discounted the bill with his bank @ 12% p.a.
Before the due date Zalak finds herself unable to meet the bill, hence requested Parth to renew the bill for a further period of 2 months. Parth agreed and he took the bill back from the bank and received new acceptance for ₹ 52,000 including interest. This new bill is duly honoured by Zalak on the due date.
Write Journal of Parth and Zalak for the above bill transactions.
Solution:
In the books of Parth
Journal Entries

In the books of Zalak
Journal Entries

Question 2.
Prahran owes Keyur ₹ 75,000. Keyur draws a bill for ₹ 60,000 on Prihaan for 4 months period and received the cheque for the balance. The bill is duly accepted and returned by Prahran. On the same date, Keyur endorsed Prihaan’s acceptance to Monil.
On the due date, Monil informed Keyur that Prihaan dishonored his acceptance and ₹ 1,905 paid as noting charges. Keyur then drew a new bill for 3 months on Prihaan for the amount due including noting charges and interest of ₹ 2,400. On the due date, the bill was duly honoured by Prihaan.
Write Journal Entries in the books of Keyur and prepare Keyur’s account in the books of Prihaan.
Solution:
In the books of Keyur
Journal Entries

Working Notes:
1. Amount paid by cheque = Total amount due from Prihaan – the amount of Bill accepted
= 75,000 – 60,000
= ₹ 15,000

2. Amount for which new bill is drawn = Amount of bill dishonoured + Noting charges + Interest
= 60,000 + 1,905 + 2,400
= ₹ 64,305

Note: For easy understanding, the students are advised to draft the journal entries in the journal of Prihaan first. So that they will be able to understand ledger entries of Keyur’s A/c opened in the ledger of Prihaan.

Question 3.
On 1st June, 2020 Bela draws a bill for ₹ 1,00,000 on Premila for 4 months period. The bill is duly accepted and returned to Bela. One month after the date, Bela discounted the bill with bank @ 18% p.a.
On the due date, Premila dishonoured her acceptance. Bank paid noting charges ₹ 2,250. Premila requested Bela to renew the bill for a further period of 2 months. Bela agreed and took the bill back from the bank and received new acceptance for 40% amount of the bill with the full amount of noting charges and a cheque for 60% balance plus interest @ 12% p.a.
Before the due date, Premila was declared as insolvent and 30% of the amount due could be recovered from her private estate.
Write Journal of Bela and Premila for the above bill transactions.
Solution:
In the books of Bela
Journal Entries

In the books of Premila
Journal Entries

Working Notes:
1. Discount charged by the bank on discounting 1st bill = 1,00,000 × \(\frac{3}{12} \times \frac{18}{100}\) (Period of bill is 4 months, but it is discounted 1 month later) = ₹ 4,500

2. Amount paid by Premila to Bela in Part payment = 60% of Bill amount
= 60% of 1,00,000
= ₹ 60,000

3. Balance amount still due from Premila to Bela = 40% of Bill amount
= 40% of 1,00,000
= ₹ 40,000

4. Interest is to be calculated on total amount due from Premila = Balance due + Unpaid amount noting charges
= 40,000 + 2,250
= ₹ 42,250
Interest due = Balance amount × Unexpired period × Rate of interest
= 42,250 × \(\frac{2}{12} \times \frac{12}{100}\)
= ₹ 845

5. Amount paid by Premila to Bela = 60,000 + 845 = ₹ 60,845.

6. Amount for which new bill is drafted and accepted = ₹ 42,250.

7. Amount recovered by Bela from the property of Premila = 30% of total amount due
= \(\frac{30}{100}\) × 42,250
= ₹ 12,675

8. Bad debts incurred by Bela = Total Amount due – Amount recovered
= 42,250 – 12,675
= ₹ 29,575

Balbharti Maharashtra State Board 12th Chemistry Important Questions Chapter 4 Chemical Thermodynamics Important Questions and Answers.

Maharashtra State Board 12th Chemistry Important Questions Chapter 4 Chemical Thermodynamics

Question 1.
Define the term energy.
Answer:
The energy of a system is defined as its capacity to perform the work. A system with higher energy can perform more work.

Question 2.
What are different forms of energy ?
Answer:
The energy of a system has many different forms as follows :

• Kinetic energy which arises due to motion, like rotational, vibrational and translational.
• Potential energy which arises due to position and state of a matter. If depends upon the temperature of the system.
• Heat energy (or thermal energy) which is transferred from the hotter body to the colder body.
• Radiant energy which is associated with electro-magnetic or light radiation.
• Electrical energy produced in the galvanic cells.
• Chemical energy stored in chemical substances.

All these various forms of energy can be converted from one form to another without any loss.

Question 3.
Explain the concept of interconversion of different forms of energy.
Answer:
There are various forms of energy like kinetic energy, potential energy, heat or thermal energy, radiant energy, electrical energy and chemical energy.

All these forms of energy are interconvertible. For example, a body at very high level possesses higher potential energy. When it falls down, potential energy is converted into kinetic energy. Falling of water from high level is used to drive turbines converting potential energy into kinetic energy which is further converted into electrical energy.

In galvanic cells, chemical energy is converted into electrical energy.
In electrolytic cells, chemical energy is converted into electrical energy. But during interconversion, the energy can neither be created nor destroyed, and there is a conservation of energy.

Question 4.
What is thermodynamics ? What are its drawbacks?
Answer:
Thermodynamics : It is concerned with the energy changes in physical and chemical changes.

Drawbacks :

• It does not give information on the rates of physical or chemical changes.
• It does not explain mechanisms involved in physical and chemical processes.

Question 5.
Define and explain :
(1) System (2) Surroundings (3) Boundary.
Answer:
(1) System : The portion of the universe under thermo-dynamic consideration to study thermodynamic properties is called a system.
Explanation :

• As such any portion of the universe under thermodynamic consideration is a system. The thermodynamic consideration involves the study of thermodynamic parameters like pressure, volume, temperature, energy, etc.
• The system may be very small or very large.
• The system is confined by a real or an imaginary boundary.

(2) Surroundings : The remaining portion of the universe other than under thermodynamics study i.e„ the system is called the surroundings.
Explanation :

• Surroundings represent a large stock of mass and energy and can be exchanged with the system when allowed.
• For a liquid in an open vessel, the surrounding atmosphere around it represents the surroundings.

(3) Boundary : The wall or interface separating the system from its surrounding is called a boundary.
Explanation :

• This boundary may be either real or imaginary.
• Through this boundary, exchange of heat and matter between the system and surroundings can take place, e.g. when a liquid is placed in a beaker the walls of beaker represent real boundaries while open portion of the beaker is imaginary boundary.
• Everything outside the boundary represents surroundings.

Question 6.
What are the types of systems ?
Answer:
Following are the types of systems :

1. Open system
2. Closed system
3. Isolated system
4. Homogeneous system
5. Heterogeneous system.

Question 7.
Define and explain the following :
(1) Open system
(2) Closed system
(3) Isolated system.
Answer:
(1) Open system :it is defined as a system which can exchange both matter and energy with its surroundings, e.g. a beaker containing water. The water continuously absorbs energy from the surroundings and forms vapour which diffuse in the surroundings. So that this system exchanges energy and matter (or mass), with the surroundings.

Fig. 4.1 : Open system

(2) Closed system : it is defined as a system which can exchange only energy but not the matter with its surroundings, e.g. A closed vessel containing hot water so that only heat is lost to the surroundings and not the matter.

Fig. 4.2 : Closed system

(3) Isolated system : it is defined as a system which can neither exchange energy nor matter with its surroundings, e.g. hot water filled in a thermally insulated closed vessel like thermos flask.

In actual practice, perfectly isolated system is not possible.
Universe represents an isolated system.

Question 8.
‘Universe is an isolated system’. Explain.
Answer:
Universe represents an isolated system due to the following reasons :

• The total mass and energy of the universe remain constant.
• The universe has no boundary.
• The universe has no surroundings.

Question 9.
Define and explain :
(1) Homogeneous system
(2) Heterogeneous system.
Answer:
(1) Homogeneous system : A system consisting of only one uniform phase is called a homogeneous system.
Explanation :
(1) The properties of homogeneous system are uniform throughout the phase or system.
(2) The homogeneous systems are :

• Solutions of miscible liquids (water and alcohol) or soluble solids in liquids, (NaCl in water), etc.
• Mixture of gases. H₂ and N₂, NH₃ and H₂, etc.

(2) Heterogeneous system : A system consisting of two or more phases separated by interfacial boundaries is called a heterogeneous system.
Explanation : These systems are :

• Mixture of two or more immiscible liquids. E.g. Water and benzene.
• Solid in equilibrium with liquid.
  E.g. Ice ⇌ water.
• Liquid in equilibrium with vapour.
  E.g. Water ⇌ vapour.

Question 10.
Explain : (A) Extensive property (B) Intensive property of a system.
OR
What is the difference between extensive and intensive properties?
Answer:
The properties of a system are classified as (A) Extensive property and (B) Intesive property.
(A) Extensive property : It is defined as a property of a system whose magnitude depends on the amount of matter present in the system.
Explanation :

• More the quantity (or amount) of the matter of the system, more is the magnitude of extensive property, e.g., mass, volume, heat, energy, enthalpy, etc.
• The extensive properties are additive.

(B) Intensive property : It is defined as a property of a system whose magnitude is independent of the amount of matter present in the system.
Explanation :

• Intensive property is characteristic of the system, e.g., refractive index, density, viscosity, temperature, pressure, boiling point, melting point, freezing point of a pure liquid, surface tension, etc.
• The intensive properties are not additive.

Question 11.
Select extensive and intensive properties in the following :
Moles, molar heat capacity, entropy, heat capacity.
Answer:
Extensive property : Moles, entropy, heat capacity.
Intensive property : Molar heat capacity.

Question 12.
What is a state function ? Give examples.
Answer:
State function : The property which depends on the state of the system and independent of the path followed by the system to attain the final state is called a state function.
For example, pressure, volume, temperature, etc.

Question 13.
Classify the following properties as intensive or extensive :
(i) Temperature (ii) Density (iii) Enthalpy (iv) Mass (v) Energy (vi) Refractive index (vii) Pressure (viii) Viscosity (ix) Volume (x) Weight.
Answer:
(1) Intensive properties : Temperature, Density, Refractive index. Pressure, Viscosity.
(2) Extensive properties : Enthalpy, Mass, Energy, Volume, Weight.

Question 14.
What are path functions?
Answer:
Path functions : The properties which depend on the path of the process are called path functions. For example, work (W) and heat (Q).

Question 15.
Define thermodynamic equilibrium. Mention different types of thermodynamic equilibria.
Answer:
Thermodynamic equilibrium : A system is said to have attained a state of thermodynamic equilibrium if there is no change in any thermodynamic functions or state functions like energy, pressure, volume, etc. with time.

For a system to be in thermodynamic equilibrium, it has to attain following three types of equilibrium :

• Thermal equilibrium
• Chemical equilibrium
• Mechanical equilibrium

Question 16.
Distinguish between :
(1) Open system and Closed system :
Open system:

1. An open system can exchange both matter and energy with the surroundings.
2. In this, the total amount of matter does not remain constant.
3. Example : Hot water kept in an open beaker.

Closed system:

1. A closed system can exchange only energy, but not matter with the surroundings.
2. In this, the total amount of matter remains constant.
3. Example : Hot water kept in a closed glass flask.

(2) Closed system and Isolated system :
Closed system:

1. A closed system can exchange only energy, but not matter with the surroundings.
2. In this, the total amount of energy does not remain constant.
3. Example : Hot water kept in a sealed glass flask.

Isolated system:

1. An isolated system can exchange neither matter nor energy with the surroundings.
2. In this, the total amount of energy remains constant.
3. Example : Hot water kept in a thermos flask.

(3) Open system and Isolated system :
Open system:

1. An open system can exchange matter with the surroundings.
2. It can exchange energy with the surroundings.
3. In this, the total amount of energy does not remain constant.
4. Example : Hot water kept in an open beaker.

Isolated system

1. An isolated system cannot exchange matter with the surroundings.
2. It cannot exchange energy with the surroundings.
3. In this, the total amount of energy remains constant.
4. Example : Hot water kept in a thermos flask.

Question 17.
What is a thermodynamic process ? What are different types of processes ?
Answer:
(i) Thermodynamic process : It is defined as a transition by which a state of a system changes from initial equilibrium state to final equilibrium state.

The process is carried out by changing the state functions or thermodynamic variables like pressure, volume and temperature. During the process one or more properties of the system change.

(ii) Types of processes :

• Isothermal process
• Isobaric process
• Isochoric process
• Adiabatic process
• Reversible process
• Irreversible (spontaneous) process.

Question 18.
Define and explain different types of processes.
Answer:
There are following types of processes :
(1) Isothermal process : It is defined as a process in which the temperature of the system remains constant throughout the change of a state of the system.
In this, ΔT = 0.

Features :

• In this process, the temperature at initial state, final state and throughout the process remains constant.
• In this process, system exchanges heat energy with its surroundings to maintain constant temperature. E.g., in case of exothermic process liberated heat is given to the surroundings and in case of endothermic process heat is absorbed from the surroundings so that temperature of the system remains constant and ΔT = 0.
• Isothermal process is carried out with a closed system.
• Internal energy (U) of the system remains constant, hence, Δ U = 0.
• In this process, pressure and volume of a gaseous system change.

(2) Isobaric process : It is defined as a process which is carried out at constant pressure. Hence, Δ P = 0.
Features :

• In this process, the volume (of gaseous system) changes against a constant pressure.
• Since the external atmospheric pressure remains always constant, all the processes carried out in open vessels, or in the laboratory are isobaric processes.
• In this volume and temperature change.
• Internal energy of a system changes, hence, ΔU ≠ 0.

(3) Isochoric process : It is defined as a process which is carried out at constant volume of the system.
Features :

• In this process, temperature and pressure of the system change but volume remains constant.
• Since ΔV = 0, no mechanical work is performed.
• In this internal energy (U) of the system changes. The example of this process in cooking takes place in a pressure cooker.

(4) Adiabatic process : It is defined as a process in which there is no exchange of heat energy between the system and its surroundings. Hence, Q = 0.
Features :

• An adiabatic process is carried out in an isolated system.
• In this process, temperature and internal energy of a system change, ΔT ≠ 0, Δ U ≠ 0.
• During expansion, temperature and energy decrease and during compression, temperature and energy increase.
• If the process is exothermic, the temperature rises and if the process is endothermic the temperature decreases in the adiabatic process.

(5) Reversible process : A process carried out in such a manner that at every stage, the driving force is only infinitesimally greater than the opposing force and it can be reversed by an infinitesimal increase in force and the system exists in equilibrium with its surroundings throughout, is called a reversible process.
Features :

• This is a hypothetical process.
• Driving force is infinitesimally greater than the opposing force throughout the change.
• The process can be reversed at any point by making infinitesimal changes in the conditions.
• The process takes place infinitesimally slowly involving infinite number of steps.
• At the end of every step of the process, the system attains mechanical equilibrium, hence, throughout the process, the system exists in temperature-pressure equilibrium with its surroundings.
• In this process, maximum work is obtained.
• Temperature remains constant throughout the isothermal reversible process.

(6) Irreversible process : it is defined as the unidirectional process which proceeds in a definite direction and cannot be reversed at any stage and in which driving force and opposing force differ in a large magnitude. It is also called a spontaneous process.
Features :

• It takes place without the aid of external agency.
• All irreversible processes are spontaneous.
• All natural processes are irreversible processes.
• Equilibrium is attained at the end of process.
• They are real processes and are not hypothetical.

Examples :

• Flow of heat from a matter at higher temperature to a matter at lower temperature.
• Flow of a gas from higher to lower pressure.
• Flow of water from higher level to lower level.
• Flow of a solvent into a solution through a semipermeable membrane due to osmosis.
• Flow of electricity from higher potential terminal to lower potential terminal.

Question 19.
Distinguish between :
(1) Isothermal process and Adiabatic process.
(2) Reversible and irreversible processes.
Answer:
Isothermal process:

1. In an isothermal process, the temperature of the system remains constant. ΔT = 0
2. In this process, the system exchanges heat with the surroundings. Q ≠ 0 (Closed system)
3. The total internal energy of the system remains constant.
4. In this process, the system is not thermally isolated.
5. In this process, Q = -W as ΔU = 0.
6. ΔH = 0.

Adiabatic process:

1. In an adiabatic process, the temperature of the system changes. ΔT ≠ 0
2. In this process, the system does not exchange heat with the surroundings. Q = 0 (Isolated system)
3. The total internal energy of the system changes. ΔU ≠ 0
4. In this process, the system is thermally isolated.
5. In this process, W = ΔU.
6. ΔH ≠ 0.

(2) Reversible and irreversible processes.
Reversible process:

1. The process whose direction can be reversed at any stage by an infinitesimal increase in the opposing force is called a reversible process.
2. Such a process is not spontaneous and takes place infinitesimally slowly and takes infinite time for completion.
3. In this process, the thermodynamic equilibrium is always maintained between the system and the surroundings at every step.
4. The opposing force is infinitesimally less than the driving force.
5. It is an ideal or hypothetical process.
6. Maximum work can be derived from such a process.

Irreversible process:

1. The process whose direction cannot be reversed by an infinitesimal increase in the opposing force is called an irreversible process.
2. Such a process is spontaneous and takes finite time for completion.
3. The thermodynamic equilibrium is attained only at the end of the process.
4. The opposing force is significantly less than the driving force.
5. It is a practical or real and spontaneous process.
6. Work derived from such a process is always less than the maximum work.

Question 20.
Show that pressure times volume (PV) is equal to work.
Answer:
The work is defined as the energy by which a body is displaced through a distance d by applying a force f.
∴ W = f × d
If area is A = d² and volume V = d³ then,
PV = \(\frac{f}{A}\) × d³ = \(\frac{f}{d^{2}}\) × d³ = f × d = W
∴ The term PV represents the pressure-volume work.

Question 21.
Explain the process of (A) expansion and (B) compression with suitable examples.
Answer:
(A ) Expansion : Consider an ideal cylinder fitted with a piston and filled with H₂O_2(l).
2H₂O_2(l) → 2H₂O_(l) + O_2(l)

Fig. 4.5 : Decomposition of H₂O₂
The oxygen gas produced pushes the piston upwards lifting the mass. Thus, the system performs the work on the surroundings and loses energy by expansion. In this work is done by the system.

(B) Compression : Consider an ideal cylinder fitted with a piston and containing gaseous NH_3(g) and HCl_(g).

Fig. 4.6 : Reaction between NH_3(g) and HCl_(g)
NH_3(g) + HCl_(g) → NH₄Cl_(s)
As the reaction proceeds, due to consumption of gases, the volume decreases and there is work due to compression. In this work is done on the system by surroundings and the system gains energy.

Question 22.
What are the sign conventions for Q and W in (A) expansion, (B) compression?
Answer:
(A) For expansion, work is done by the system hence,
Q = -ve and W = -ve
(B) For compression, work is done on the system hence,
Q = -ve and W = +ve

Fig. 4.7 : Sign conventions

Question 23.
Explain sign convention of work during expansion and compression.
OR
Explain +W and -W.
Answer:
(A) Expansion of a gas :
(1) When a gas expands against a constant pressure, P_ex changing the volume from initial volume V₁ to final volume V₂,
Change in volume, Δ V = V₂ – V₁
The mechanical work = W = -P_ex × Δ V
= -P_ex (V₂ – V₁)

(2) During expansion V₂ > V₁. The work is said to be performed by the system on the surroundings. This results in the decrease in the (work) energy of the system. Hence the work is negative, i.e. W is -ve.

(B) Compression of a gas : During compression, V₂ < V₁. The work is said to be performed on the system by the surroundings. This results in the increase in the (work) energy of the system. Hence the work is positive, i.e. W is + ve.

Question 24.
What are different units of energy and work ?
Answer:

Question 25.
What are the characteristics of maximum work?
Answer:
(1) The process is carried out at constant temperature.
(2) During the complete process, driving force is infinitesimally greater than opposing force.
(3) Throughout the process, the system exists in equilibrium with its surroundings.
(4) The work obtained is maximum. This is given by,
W_max = -2.303 nRT log₁₀ \(\frac{V_{2}}{V_{1}}\)
OR
W_max = -2.303 nRT log₁₀ \(\frac{P_{1}}{P_{2}}\)
where n, P, V and T represent number of moles, pressure, volume and temperature respectively.
(5) ΔU = 0, ΔH = 0.
(6) The heat absorbed in reversible manner
Q_rev, is completely converted into work.
Q_rev = -W_max.
Hence work obtained is maximum.

Solved Examples 4.3

Question 26.
Solve the following :

(1) 2.5 moles of an ideal gas are expanded isothermally from 12 dm³ to 25 dm³ against a pressure of 3.0 bar. Calculate the work obtained.
Solution :
Given : n = 2.5 mol; V₁ = 12 dm³;
V₂ = 25 dm³
P_ext = 3.0 bar; W = ?
W = -P_ext × (V₂ – V₁)
= – 3 × (25 – 12)
= -39 dm³ bar
∵ V₁ dm³ = 100 J
∴ W = -39 × 100 = -3900 J = -3.9kJ
Ans. W= -3.9 kJ

(2) When 2.2 moles of an ideal gas are expanded from 3.5 dm³ to 12 dm³ against a constant pressure, the work obtained is 3910 J. Estimate the external pressure.
Solution :
Given : n = 2.2 mol; V₁ = 3.5 dm³; V₂ = 12 dm³
W= -3910 J = \(\frac{-3910}{100}\)dm³ bar
W = -P_ex (V₂ – V₁)

= 4.6 bar
Ans. P_ex = 4.6 bar

(3) Three moles of an ideal gas are expanded isothermally from a volume of 300 cm³ to 2.5 dm³ at 300 K against a pressure of 1.9 bar. Calculate the work done in L atm and joules.
Solution :
Given : Number of moles of a gas = n = 3 mol
Initial volume = V₁ = 300 cm³
= 0.3 dm³
Final volume = V₂ = 2.5 dm³
External pressure = P_ex =1.9 bar
Temperature = T = 300 K
∵ W = -P_ex (V₂ – V₁)
= -1.9 (2.5 – 0.3)
= -4.18 dm³ bar
Now, 1 dm³ bar = 100 J
∴ W = -4.18 × 100
= -4180 J
Ans. Work of expansion = W = -4180 J

(4) Calculate the constant external pressure needed to compress an ideal gas from 25 dm³ to 15 dm³. The amount of work done in the compression process is 3500 joules.
Solution :
As the compression of the gas takes place against a constant pressure, the work done is given by
W = -P_ex(V₂ – V₁)
W = Work done by the gas against the external pressure = 3500 J
∴ W = \(\frac{3500}{100}\) = 35 dm³ bar
P = Constant external pressure = ?
V₂ = Final volume = 15 dm³
V₁ = Initial volume = 25 dm³
∴ 35= -P × (15 – 25)
∴ 35 = 10 × P
∴ P = \(\frac{35}{10}\) = 3.5 bar
Ans. External pressure = 3.5 bar

(5) Three moles of an ideal gas are compressed isothermally and reversibly to a volume of 2 dm³. The work done is 2.983 kJ at 22°C. Calculate the initial volume of the gas.
Solution :
Given : Number of moles of a gas = n = 3 mol
Final volume = V₂ = 2 dm³
Initial volume = V₁ = ?
For compression,
Wmax = +2.983 kJ = 2983 J
Temperature = T = (273 + 22) K = 295 K
Wmax= -2.303 nRT log₁₀\(\frac{V_{2}}{V_{1}}\)
∴ \(\frac{2983}{2.303 \times 3 \times 8.314 \times 295}\)
= -[log₁₀2 – log₁₀V₁]
0.1760 = -log₁₀2 + log₁₀ V₁
= -0.3010 + log₁₀V₁
∴ log₁₀ V₁ = 0.1760 + 0.3010 = 0.4770
∴ V₁ = Antilog 0.4770
= 3.0 dm³
Ans. Initial volume of the gas = 3.0 dm³

(6) A chemical reaction takes place in a container of cross sectional area 100 cm². As a result of the reaction, a piston is pushed out through 10 cm against an external pressure of 1.0 bar. Calculate the work done by the system.
Solution :
Given : Cross sectional area = A = 100 cm²
Displacement of a piston = 1 = 10 cm
External pressure = P = 1.0 bar
Work = W = ?
Volume change = A × l
∴ ΔV = 100 × 10
= 1000 cm³
= 1 dm³
Work = W = -P × ΔV
= -1 × 1
= -1 dm³ bar
= – 1 × 100 J
= -100 J
Ans. Work = W = -100 J

(7) 5 moles of helium expand isothermally and reversibly from a pressure 4 atm to 0.4 at 300 K. Calculate the work done, change in internal energy and heat absorbed during the expansion. (R = 8.314 J k^-1 mol^-1)
Solution :
Given : n = 5 mol
P₁ = 4 atm
P₂ = 0.4 atm
T = 300 K
W_max = ?, ΔU = ?, Q = ?
W_max = -2.303 nRT log₁₀ (\(\frac{P_{1}}{P_{2}}\))
= -2.303 × 5 × 8.314 × 300 log₁₀ \(\frac{4}{0.4}\)
= -2.303 × 5 × 8.314 × 300 × 1
= -28720 J
= -28.72 kJ
For an isothermal process, ΔU = 0
By first law, ΔU = Q + Wmax
∴ Q = -W_max
= – (-28.72) = 28.72 kJ
Ans. W_max = – 28.72 kJ; ΔU = 0;
Q = 28.72 kJ

(8) 2.8 × 10^-2 kg of nitrogen is expanded isothermally and reversibly at 300 K from 15.15 × 10⁵ Pa when the work done is found to be -17.33 kJ. Find the final pressure.
Solution :
Given : Mass of nitrogen = m = 2.8 × 10^-2 kg
Temperature = T = 300 K
Work obtained in expansion = Wmax = -17.33 kJ
= – 17330 J
Initial pressure = P₁ = 15.15 × 10⁵ Pa
= 1.515 × 10⁶ Pa
Molar mass of nitrogen (N₂) = M_N2
= 28 × 10^-3 kg mol^-1
Final pressure = P₂ = ?
Number of moles of N₂ = n = \(\frac{m}{M_{\mathrm{N}_{2}}}\)
= \(\frac{2.8 \times 10^{-2}}{28 \times 10^{-3}}=1 \mathrm{~mol}\)
W_max = -2.303 × nRT log₁₀ \(\frac{P_{1}}{P_{2}}\)
17330 = 2.303 × 1 × 8.314 × 300 × \(\log _{10} \frac{1.515 \times 10^{6}}{P_{2}}\)
∴ \(\frac{17330}{2.303 \times 1 \times 8.314 \times 300}\)
= [log₁₀ 1.515 × 10⁶ – log₁₀P₂]
3.017 = 6.1804 – log₁₀P₂
∴ log₁₀P₂ = 6.1804 – 3.017 = 3.1634
∴ P₂ = Antilog 3.1634
= 1456.8 Pa
Ans. Final pressure = 1456.8 Pa

(9) Carbon monoxide expands isothermally and reversibly at 300 K doing 4.754 kJ of work. If the initial volume changes from 10 dm³ to 20 dm³, calculate the number of moles of carbon monoxide. (R = 8.314 JK^-1 mol^-1)
Solution :
W_max = -2.303 nRT log₁₀ \(\frac{V_{2}}{V_{1}}\)
W_max = Maximum work done = -4.754 kJ
= -4754 J, n = Number of moles = ?,
= 8.314 JK^-1 mol^-1
T = 300 K, V₁ = Initial volume of carbon monoxide = 10 dm³
V₂ = Final volume of carbon monoxide = 20 dm³
∴ -4754 = 22.303 × n × 8.314 × 300 log₁₀ \(\frac {20}{10}\)
∴ -4754 = – 2.303 × n × 8.314 × 300 × log₁₀2
∴ – 4754 = – 2.303 × n × 8.314 × 300 × 0.3010
∴ n = \(\frac{-4754}{-2.303 \times 8.314 \times 300 \times 0.3010}\)
= 2.75 mol
Ans. Number of mol = 2.75 mol

(10) Given that the work done in isothermal and reversible expansion is 6.4 kJ when 2 moles of an ideal gas expanded to double its volume. Calculate the temperature at which expansion takes place.
Solution :
Given : Work = W_max = -6.4 kJ (For expansion)
= – 6400 J
Number of moles = 2
If ₁ = x L
V₂ = 2 L
Temperature = T = ?
For isothermal reversible expansion,

(11) 300 mmol of perfect gas occupies 13 dm³ at 320 K. Calculate the work done in joules when the gas expands :
(a) Isothermally against a constant external pressure of 0.20 bar,
(b) Isothermal and reversible process,
(c) Into vacuum until the volume of gas is increased by 3 dm³. (R = 8.314 J mol^-1K^-1)
Solution :
Given : Number of moles of a gas = n
= 300 mmol = 0.3 mol
Initial volume = V₁ = 13 dm³
Increase in volume = ΔV = 3 dm³
Pressure = P_ex = 0.2 atm
Temperature = 320 K

(a) Expansion against constant pressure is an irreversible process.
∴ W = -P_ex × ΔV
= -0.2 × 3
= -0.6 dm³ bar
= -0.6 × 100 J
= -60 J

(b) For isothermal reversible process,
W_max = 2.303 nRT log₁₀ \(\frac{V_{2}}{V_{1}}\)
Now, V₂ = V₁ + ΔV = 13 + 3 = 16 dm³
W_max = – 2.303 × 0.3 × 8.314 × 320 log₁₀ \(\frac {16}{13}\)
= -165.4 J

(c) In vacuum, P_ex = 0
∴ W = -P_ex × ΔV
= -0 × 3
= 0
Ans. (a) W= -60.78 J
(b) W_max = -165.4 J
(c) W = 0.

Question 27.
Define and explain the term internal energy.
Answer:
Internal energy : It is defined as the total energy constituting potential energy and kinetic energy of the molecules present in the system.
Explanation :

• The internal energy of a system is a state function and thermodynamic function. It is denoted by U.
• Its value depends on the state of a system.
• The change in internal energy (Δ U) depends only on the initial state and the final state of the system.
  Δ U = U₂ – U₁
• It is an extensive property of the system.
• It has same unit as heat and work.
• Total internal energy U of the system is,
  Total energy = Potential energy + Kinetic energy

Question 28.
Explain the formulation of first law of thermodynamics.
OR
Deduce mathematical equation for the first law of thermodynamics. Justify its expression.
Answer:
(1) The first law of thermodynamics is based on the principle of conservation of energy.
(2) If Q is the heat absorbed by the system and if W is the work done by surroundings on the system then the internal energy of the system will increase by Δ U.
(3) From the conservation of energy we can write,
Increase in internal energy of the system = Quantity of heat absorbed by the system + Work done on the system
∴ ΔU = Q + W
(4) For an infinitesimal change,
dU = dQ + dW

Question 29.
Deduce the mathematical expression of first law of thermodynamics for the following processes :
(1) Isothermal process
(2) Isobaric process
(3) Isochoric process
(4) Adiabatic process.
Answer:
(1) Isothermal process :This is a process which is carried out at constant temperature. Since internal energy, U of the system depends on temperature there is no change in the internal energy U of the system. Hence ΔU = 0.
By first law of thermodynamics,
ΔU = Q +W
∴ 0 = Q + W
∴ Q = -W or W = -Q.

• Hence in expansion, the heat absorbed by the system is entirely converted into work on the surroundings.
• In compression, the work done on the system is converted into heat which is transferred to the surroundings by the system, keeping temperature constant.

(2) Isobaric process : In this, throughout the process pressure remains constant. Hence the system performs the work of expansion due to volume change ΔV.
W= -P_ext × ΔV
Let Q_P be the heat absorbed by the system at constant pressure.
By first law of thermodynamics,
ΔU = Q_P + W.
∴ ΔU = Q_P – P_exΔV
or Q_P = ΔU + P_exΔV
In this process, the heat absorbed Q_P is used to increase the internal energy (ΔU) of the system.

(3) Isochoric process : In this process the volume of the system remains constant. Hence ΔV = 0. Therefore, the system does not perform mechanical work.
∴ W = -PΔV = -P × (0) = 0
Let Q_V be the heat absorbed at constant volume.
By first law of thermodynamics,
ΔU = Q + W
∴ ΔU = Q_V.

(4) Adiabatic process : In this process, the system does not exchange heat, Q with its surroundings.
∴ Q = 0.
Since by first law of thermodynamics,
ΔU = Q + W
∴ ΔU = W_ad.
Hence,
(i) the increase in internal energy ΔU is due to the work done on the system by surroundings. This results in increase in energy and temperature of the system.
(ii) if the work is done by the system on surroundings, like expansion, then there is a decrease in internal energy (-ΔU) and temperature of the system decreases.

Question 30.
What are the IUPAC sign conventions of Q, U and W?
Answer:
In thermodynamics, the sign conventions are adopted according to IUPAC convention, based on acquisition of energy.
(i) Heat absorbed = +Q
Heat evolved = -Q
(ii) Internal energy change :
Increase in energy = + Δ U
Decrease in energy = – Δ U
(iii) Work done by the system = – W
Work done on the system = + W

Question 31.
Define and explain the term enthalpy.
OR
What is meant by enthalpy of a system ?
Answer:
Enthalpy (H) : It is defined as the total energy of a system consisting of internal energy (U) and pressure – volume (P × V) type of energy, i.e. enthalpy represents the sum of internal energy U and product PV energy. It is denoted by H and is represented as

Explanation :

• Enthalpy represents total heat content of the system.
• Enthalpy is a thermodynamic state function.
• Enthalpy is an extensive property.
• The absorption of heat by a system increases its enthalpy. Hence enthalpy is called heat content of the system.

Question 32.
Derive the expression, ΔH = ΔU + PΔV.
Answer:
Enthalpy (H) of a system is defined as
H = U + PV
where U is internal energy
P is pressure and V is volume.
Consider a process in which a state of a system changes from an initial state A to a final state B. Let H₁, U₁, P₁, V₁ and H₂, U₂, P₂, V₂ be the state functions of the system in initial and final states.

Then,
H₁ = U₁ + P₂V₂ and H₂ = U₂ + P₂V₂
The enthalpy change ΔH is given by,
ΔH = H₂ – H₁
= (U₂ + P₂V₂) – (U₁ + P₁V₁)
= (U₂ – U1) + (P₂V₂ – P₁V₁)
= ΔU + ΔPV
where ΔU = U₂ – U₁
At constant pressure, P₁ = P₂ = P
∴ P₂V₂ – P₁V₁ = PV₂ – PV₁
= P(V₂ – V₁)
= P × ΔV
Hence, ΔH = ΔU + PΔV
This is a relation for enthalpy change.

Question 33.
Show that the heat absorbed at constant pressure is equal to the change in enthalpy of the system.
OR
Why is enthalpy called heat content of the system?
Answer:
By the first law of thermodynamics,
ΔU = Q + W
where ΔU is the change in internal energy
Q is heat supplied to the system
W is the work obtained.
∴ Q = ΔU – W
If Q_P is the heat absorbed at constant pressure by the system, so that the volume changes by Δ V against constant pressure P then,
W = -PΔV
∴ Q_P = ΔU – (-PΔV)
∴ Q_P = ΔU + PΔV ……… (1)
If ΔH is the enthalpy change for the system, then
ΔH = ΔU + PΔV ……….. (2)
By comparing above equations, (1) and (2), we can write, Q_P = ΔH
Hence heat absorbed at constant pressure is equal to the enthalpy change for the system.
Since by increase in enthalpy heat content of the system increases, enthalpy is also called as the heat content of the system.

Question 34.
What are the conditions under which ΔH = ΔU?
Answer:

1. For any thermodynamic process or a chemical reaction at constant volume, Δ V = 0.
2. Since ΔH = ΔU + PΔV, at constant volume ΔH = ΔU.
3. In the reactions, involving only solids and liquids, Δ V is negligibly small, hence ΔH = ΔH.
4. In a chemical reaction, in which number of moles of gaseous reactants and gaseous products are equal, then change in number of moles, Δ n = n₂ – n₁ = 0. Since ΔH = ΔU + ΔnRT, as Δn = 0, ΔH = ΔU.

Solved Examples 4.6 – 4.8

Question 35.
Solve the following :

(1) Calculate the change in internal energy when a gas is expanded by supplying 1500 J of heat energy. Work done in expansion is 850 J.
Solution :
Given : Q = 1500 J
W = -850 J (For expansion work is negative)
ΔU = ?
By first law of thermodynamics,
ΔU = Q + W
= 1500 + (- 850)
= 650 J
Ans. Change in internal energy = Δ U = 650 J

(2) A system absorbs 520 J of heat and performs work of 210 J. Calculate the change in internal energy.
Solution :
Given : Since the heat is absorbed by the system, the work is of expansion.
Q = 520 J
W= -210 J
ΔV = ?
ΔU = Q + W
= 520 + (- 210)
= 310 J
Ans. Internal energy change = Δ U = 310 J

(3) A gas expands from 6 litres to 20 dm³ at constant pressure 2.5 atmosphere. If the system is supplied with 5000 J of heat, calculate W and ΔU.
Solution :
Given : V₁ = 6 dm³
V₂ = 20 dm³
P = 2.5 atm
Q = 5000 J
W = ?; ΔU = ?
For expansion,
W = -P_ex(V₂ – V₁)
= -2.5 (20 – 6)
= – 35
= – 35 × 100 J
= – 3500 J
ΔU = Q + W
= 5000 + (-3500)
= -1500 J
Ans. W= -3500 J; ΔU = – 1500 J

(4) An ideal gas expands against a constant pressure of 2.026 × 10⁵ Pa from 5 dm³ to 15 dm³. If the change in the internal energy is 418 J, calculate the change in enthalpy.
Solution :
As the expansion takes place at a constant pressure, the change in enthalpy is given by
ΔH = ΔU + P(V₂ – V₁)
ΔH = Change in enthalpy = ?
ΔU = Change in internal energy = 418 J
P = Constant pressure = 2.026 × 10⁵ Pa
V₂ = 15 dm³ = 15 × 10^-3 m³
V₁ = 5 dm³ = 5 × 10^-3 m³
∴ ΔH = 418 + 2.026 × 10⁵ × (15 × 10^-3 – 5 × 10^-3)
= 418 + 2026
∴ ΔH = 2.444 × 10³ J = 2.444 kJ
Ans. Change in enthalpy = 2.444 × 10³ J
= 2.444 kJ

(5) In a reaction, 2.5 kJ of heat is released from the system and 5.5 kJ of work is done on the system. Find ΔU.
Solution :
Given : Q = -2.5 kJ (since heat is released)
W= + 5.5 kJ (since the work will be of compression)
ΔU = ?
ΔU = Q + W
= -2.5 + 5.5
= +3 kJ
Internal energy of the system will increase by 3 kJ.
Ans. Δ U = 3 kJ

(6) A chemical reaction is carried out by supplying 8 kJ of heat. The system performs the work of 2.7 kJ. Calculate ΔH and ΔU.
Solution :
Given : Q = + 8 kJ (since heat is absorbed by the system)
W = -2.7 kJ (It will be a work of expansion)
ΔH = ?, ΔU = ?
ΔU = Q + W = 8 + (-2.7) = 5.3 kJ
Internal energy of the system will increase by 5.3 kJ.
Due to expansion, Δ V > 0,
∴ PΔV = +2.7 kJ
ΔH = ΔU + PΔV = 5.3 + 2.7 = 8 kJ
Enthalpy of the system will increase by 8 kJ
Ans. ΔU = 5.3 kJ, ΔH = 8 kJ

(7) A sample of gas absorbs 4000 kJ of heat, (a) if volume remains constant. What is ΔU? (b) Suppose that in addition to absorption of heat by the sample, the surroundings does 2000 kJ of work on the sample. What is Δ U ? (c) Suppose that as the original sample absorbs heat, it expands against atmospheric pressure and does 600 kJ of work on its surroundings. What is ΔU ?
Solution :
Given : Q = + 4000 kJ (since heat is absorbed)
(a) Since volume remains constant, Δ V = 0.
W = -P_ex (V₂ – V₁)
= -P_exΔV = -P_ex(0) = 0
∴ ΔU = Q + W = 4000 + 0 = 4000 kJ

(b) Q = + 4000 kJ
W = + 2000 kJ (Work done on the system)
ΔU = Q + W = 4000 + 2000 = 6000 kJ

(c) W = -600 kJ (Work of expansion)
ΔU = Q + W
ΔU = 4000 + (-600) = 3400 kJ
Ans. (a) Δ U = 4000 kJ
(b) Δ U = 6000 kJ
(c) Δ U = 3400 kJ

(8) Calculate the internal energy change at 298 K for the following reaction :
\(\frac{1}{2} \mathbf{N}_{2(\mathrm{~g})}+\frac{3}{2} \mathbf{H}_{2(\mathrm{~g})} \rightarrow \mathbf{N H}_{3(\mathrm{~g})}\)
The enthalpy change at constant pressure is -46.0 kJ mol^-1. (R = 8.314 JK^-1 mol^-1)
Solution :
Given : \(\frac{1}{2} \mathbf{N}_{2(\mathrm{~g})}+\frac{3}{2} \mathbf{H}_{2(\mathrm{~g})} \rightarrow \mathbf{N H}_{3(\mathrm{~g})}\)
ΔH= -46.0 kJ mol^-1
ΔH = Heat of formation of NH₃ at constant pressure
= -46.0 kJ mol^-1 = -4600 J mol^-1
Δ U = Change in internal energy = ?
Δn = (Number of moles of ammonia) – (Number of moles of hydrogen + Number of moles of nitrogen)
= [1 – (\(\frac {1}{2}\) + \(\frac {3}{2}\))]= -1 mol
R= 8.314 JK^-1 mol^-1
T = Temperature in kelvin = 298 K
ΔH = Δ U + ΔnRT
∴ -46000 = ΔU + (-1 × 8.314 × 298)
∴ -46000 = ΔU – 2477.0
∴ ΔU = -46000 + 2477.0
= -43523 J
= -43.523 kJ
Ans. Change in internal energy = -43.523 kJ

(9) 5 moles of helium expand isothermally and reversibly from a pressure 40 × 10^-5 Nm^-2 to 4 × 10^-5 Nm^-2 at 300 K. Calculate the work done, change in internal energy and heat absorbed during the expansion. (R = 8.314 JK^-1 mol^-1)
Solution :
As the expansion takes place isothermally and reversibly, the work done is given by
Wmax = -2.203 nRT log\(\frac{P_{1}}{P_{2}}\)
Wmax = Maximum work done
n = Number of moles of helium = 5 moles
R = Gas constant = 8.314 JK^-1 mol^-1
T = 300 K
P₁ = Initial pressure = 40 × 10^-5 Nm^-2
P₂ = Final pressure = 4 × 10^-5 Nm^-2
∴ W_max = -2.303 × 5 × 8.314 × 300 × log \(\frac {40}{4}\)
= – 2.303 × 5 × 8.314 × 300 × log 10
= -2.303 × 5 × 8.314 × 300 × 1
= – 28720J
As the expansion takes place isothermally, i.e., at the same temperature, there is no change in the internal energy of the system.
Q = ΔU + W
∴ Q = – W= + 28720 J as ΔU = 0
Ans. Work done = -28720 J, Heat absorbed = 28720 J, Change in internal energy = 0

(10) Calculate the work done in each of the following reactions. State whether work is done on or by the system.
(a) The oxidation of one mole of SO₂ at 50°C.
2SO_2(g) + O_2(g) → 2SO_3(g)
(b) Decomposition of 2 moles of NH₄NO₃ at 100°C
NH₄NO_3(s) → N₂O_(g) + 2H₂O_(g)
Solution :
(a) Given reaction :
2SO_2(g) + O_2(g) → 2SO_3(g)
For 1 mole of SO₂,
SO_2(g) + \(\frac {1}{2}\)O₂(g) → SO_2(g)
∴ Δn = (n₂)_{gaseous products} – (n₁)_{gaseous reactants}
= 1 – (1 + \(\frac {1}{2}\))
= -0.5 mol
Since there is decrease in number of moles of gases, there will be compression, hence, the work will be done on the system by the surroundings.
Work is given by,
∴ W = – ΔnRT
= – (- 0.5) × 8.314 × (273 + 50)
= + 1342.7 J

(b) Given reaction :
NH₄NO_3(s) → N₂O_(g) + 2H₂O_(g)
For 2 moles of NH₄NO₃,
2NH₄NO_3(s) → 2N₂O_(g) + 4H₂O_(g)
∴ Change in number of moles,
Δn = (n₂)_{gaseous products} – (n₁)_{gaseous reactants}
= (2 + 4) – 0
= 6 mol
Since there is an increase in number of moles of gases, work of expansion is done by the system on the surroundings.
∴ W = -ΔnRT
= – 6 × 8.314 × (273+ 100)
= – 18606 J
= – 18.606 kJ
Ans. (a) W = 1342.7 J (b) W= -18.606 kJ

(11) The amount of heat evolved for the combustion of ethane is -900kJ mol^-1 at 300K and 1 atm.
C₂H_6(g) + \(\frac {7}{2}\)O_2(g) → 2CO_2(g) + 3H₂O_(l)
Calculate W, ΔH and ΔU for the combustion of 12 × 10^-3 kg ethane.
Solution :
Given : ΔH = -900 kJ mol^-1
Temperature = T = 300 K
Pressure = P = 1 atm
Mass of ethane = m = 12 × 10^-3 kg
Molar mass of ethane (C₂H₆) = 30 × 10^-3 kg mol^-1
ΔH = ? Δ U = ? for given ethane.
Number of moles of C₂H₆ = n = \(\frac{m \mathrm{~kg}}{M \mathrm{~kg} \mathrm{~mol}^{-1}}\)
= \(\frac{12 \times 10^{-3}}{30 \times 10^{-3}}\)
= 0.4 mol
For the given reaction,
C₂H_6(g) + \(\frac {7}{2}\)O_2(g) → 2CO_2(g) + 3H₂O_(l)
Δn = (n₂)_{gaseous products} – (n₁)_{gaseous reactants}
= 2 – (1 + \(\frac {7}{2}\)) = -2.5 mol
For 1 mol of C₂H₆, Δn = -2.5 mol
∴ For 0.4 mol of C₂H₆, Δn = -2.5 × 0.4
= -1 mol
Since there is a decrease in number of moles, the work is of compression on the system.
W = -ΔnRT
= – (-1) × 8.314 × 300
= + 2494 J
= + 2.494 kJ
For 1 mol of C₂H₆ ΔH = -900 kJ
∴ For 0.4 mol of C₂H₆, ΔH= – 900 × 0.4
= – 360 kJ
ΔH = ΔU + ΔnRT
ΔU = ΔH – ΔnRT
= -360 – (-1) × 8.314 × 300× 10^-3
= – 360 + 2.494
= – 357.506 kJ
Ans. W = + 2.494 kJ, ΔH = -360 kJ;
ΔU= – 357.506 kJ

(12) The latent heat of evaporation of water is 80 kJ mol^-1. If 100 g water are evaporated at 100 °C and 1 atm, calculate W, ΔH, ΔU and Q.
Solution :
Given : Latent heat of evaporation = ΔH
= 80 kJ mol^-1 of water
Temperature = T = 273 + 100 = 373 K
Pressure = P = 1 atm
Mass of water = m = 100 g
Molar mass of water = 18 g mol^-1
W = ?, ΔH = 1, U = ?, Q = ?
Number of moles of water = \(\frac{m}{M}=\frac{100}{18}\) = 5.556 mol
H₂O_(l) → H₂O_(g)
5.556 mol 5.556 mol
Change in number of moles = Δn = 5.556 – 0
= 5.556 mol
For evaporation of 1 mol H₂O, ΔH = 80 kJ
For 5.556 mol H₂O, ΔH= 80 × 5.556 = 444.5 kJ
In this reaction, the work will be of expansion.
W= -ΔnRT
= -5.556 × 8.314 × 373
= – 17230 J
= -17.23 kJ
Now,
ΔH = ΔU + ΔnRT
ΔU = ΔH – ΔnRT
= 444.5 – 5.556 × 8.314 × 373 × 10^-3
= 444.5 – 17.23
= 427.27 kJ
In this, Q = Q_P = ΔH = 444.5 kJ
Ans. W= -17.23 kJ; ΔH = 444.5 kJ
ΔU= 427.21 kJ, Q = 444.5 kJ

(13) Oxidation of propane is represented as
C₃H_8(g) + 5O_2(g) → 3CO_2(g) + 4H₂O_(g), ΔH⁰ = -2043 kJ
How much pressure volume work is done and what is the value of AU at constant pressure of 1 bar when the volume change is + 22.4 dm³.
Solution :
Given :
ΔH⁰ = – 2043 kJ
Change in volume = ΔV = +22.4 L
ΔU = ?
C₃H_8(g) + 5O_2(g) → 3CO_2(g) + 4H₂O_(g)
Δn = (n₂)_{gaseous products} – (n₁)_{gaseous reactants}
= (3 + 4) – (1 + 5)
= 1 mol
Since there is an increase in number of moles, the work will be of expansion.
W = -P × ΔV dm³ bar
= – 1 × 22.4
= – 1 × 22.4 × 100 J
= – 2240 J
= -2.240
ΔH = ΔU + PΔV
ΔU = ΔH – PΔV
= – 2043 – (2.24)
= – 2040.7 kJ
Ans. W = -2.27 kJ, ΔU = -2040.7 kJ

(14) How much heat is evolved when 12 g of CO reacts with NO₂ ? The reaction is :
4CO_(g) + 2NO_2(g) → 4CO_2(g) + N_2(g),
Δ_rH⁰ = -1200 kJ
Solution :
Given : 4CO_(g) + 2NO_2(g) → 4CO_2(g) + N_2(g)
Molar mass of CO = 28 g mol^-1
Δ_rH⁰ = – 1200 kJ;
Molar mass of CO = 28 g mol^-1
m_co = 12 g, ΔH = ?
From the reaction,
∵ For 4 × 28 g CO, ΔH⁰ = – 1200 kJ
∵ For 12g CO ΔH⁰ = \(\frac{(-1200) \times 12}{4 \times 28}\)
= -128.6 kJ
Ans. Heat evolved = 128.6 kJ

Question 36.
What is phase transformation?
Answer:
Phase transformation (or transition) :

• The transition of one phase (physical state) of a matter to another phase at constant temperature and pressure without change in chemical composition is called phase transformation.
• During phase transformation, both the phases exist at equilibrium.
  Solids ⇌ Liquid; Liquid ⇌ Vapour.

Question 37.
Mention different types of phase transitions.
Answer:
The following are the types of phase changes :
(1) Fusion : This involves the change of a matter from solid state to liquid state. In this heat is absorbed, hence it is endothermic (ΔH > 0).
H₂O_(s) → H₂O_(l)
(2) Vaporisation or evaporation : This involves the change of a matter from liquid state to gaseous state. In this heat is absorbed, hence it is endothermic (ΔH > 0).
H₂O_(l) → H₂O_(g)
(3) Sublimation : This involves the change of matter from solid state directly into gaseous state. In this heat is absorbed, hence it is endothermic (ΔH > 0).
Camphor_(s) → Camphor_(g)

Question 38.
Define and explain enthalpy of freezing.
Answer:
Enthalpy of freezing (Δ_freezH) : The enthalpy change that accompanies the solidification of one mole of a liquid into solid at constant temperature and pressure is called enthalpy of freezing.
For example,
\(\mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})} \stackrel{1 \mathrm{~atm}, 273 \mathrm{~K}}{\longrightarrow} \mathrm{H}_{2} \mathrm{O}_{(\mathrm{s})}\)
Δ_freezH = -6.01 kJ mol^-1
This equation describes that when one mole of water freezes (solidifies) at 0 °C (273 K) and 1 atmosphere, 6.01 kJ of heat will be released to the surroundings.

Question 39.
Define and explain the following :
(A) Enthalpy of vaporisation.
(B) Enthalpy of sublimation.
Answer:
(A) Enthalpy of vaporisation (Δ_vapH) : The enthalpy change that accompanies the vaporisation of one mole of a liquid at constant temperature and pressure is called heat of vaporisation or evaporation. For example,
\(\mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})} \stackrel{1 \mathrm{~atm}, 373 \mathrm{~K}}{\longrightarrow} \mathrm{H}_{2} \mathrm{O}_{(\mathrm{g})}\)
Δ_vapH = +40.7 kJ mol^-1
This equation describes that when one mole of water is evaporated at 100 °C (373 K) and 1 atmosphere, 40.7 kJ of heat will be absorbed.

(B) Enthalpy of sublimation (Δ_subH) : The enthalpy change or the amount of heat absorbed that accompanies the sublimation of one mole of a solid directly into its vapour at constant temperature and pressure is called enthalpy of sublimation.
For example,
\(\mathrm{CO}_{2(\mathrm{~s})} \stackrel{1 \mathrm{~atm}, 195 \mathrm{~K}}{\longrightarrow} \mathrm{CO}_{2(\mathrm{~g})}\)
Δ_subH = 25.2 kJ mol^-1
This equation describes that when 1 mole of dry solid carbon dioxide, CO_2(s) sublimes forming gaseous CO_2(g), 25.2 kJ of heat will be absorbed.

Question 40.
Explain process of sublimation and enthalpy of sublimation ?
OR
How is enthalpy of sublimation related to enthalpy of fusion and enthalpy of vaporisation ?
Answer:
(1) The sublimation involves the conversion of a solid into vapour at constant temperature and pressure. For example,
H₂O_(s) → H₂O_(g), Δ_subH = 51.08 kJ mol^-1 at 0°C.
(2) This conversion takes place in two steps, first melting of solid into liquid and second vaporisation of the liquid.

Hence we can write,
Δ_subH = Δ_fusH + Δ_vapH

Question 41.
Arrange the following in order of increasing enthalpy :
H₂O_(s), H₂O_(g), H₂O_(l)
Answer:
The increasing order of enthalpy of the given substance will be,
H_H2O(g), < H_H2O(l), < H_H2O(s)
This is because the conversion of H₂O_(s) to H₂O_(l) and further to H₂O_(g) involves absorption of heat.

Question 42.
Define and explain :
(A) Enthalpy of atomisation
(B) Enthalpy of ionisation.
Answer:
(A) Enthalpy of atomisation (Δ_atoH) : The enthalpy change or amount of heat absorbed accompanying the dissociation of the molecules in one mole of a gaseous substance into free gaseous atoms at constant temperature and pressure is called enthalpy of atomisation.
For example,
Cl_2(g) → 2Cl_(g), Δ_atoH = 242 kJ mol^-1
CH_4(g) → C_(g) + 4H_(g), Δ_atoH = 1660 kJ mol^-1.

(B) Enthalpy of ionisation (Δ_ionH) : The enthalpy change or amount of heat absorbed accompanying the removal of one electron from each atom or ion in one mole of gaseous atoms or ions is called enthalpy of ionisation.
For example,
Na_(g) → Na⁺_(g) + e^– Δ_ionH = 494 kJ mol^-1
This equation describes that when one mole of gaseous sodium atoms, Na_(g) are ionised forming gaseous ions, Na⁺_(g), the energy required is 494 kJ.

Question 43.
Define and explain electron gain enthalpy.
Answer:
Electron gain enthalpy (Δ_egH) : It is defined as the enthalpy change, when mole of gaseous atoms of an element accept electrons to form gaseous ion.
E.g. Cl_(g) + e^– → Cl^– _(g) Δ_egH = – 349 kJ mol^-1.
It is the reverse of ionisation process.

Question 44.
Define enthalpy of solution.
Answer:
Enthalpy of solution : It is defined as the enthalpy change in a process when one mole of a substance is dissolved in specified amount of a solvent.
NaCl_(s) + aq ⇌ NaCl_(aq) Δ_solnH = 4 kJ mol^-1

Question 45.
Define enthalpy of solution at infinite dilution.
Answer:
Enthalpy of solution (Δ_solnH) : It is defined as the enthalpy change when one mole of a substance is dissolved in a large excess of a solvent, so that further dilution will not change the enthalpy at constant temperature and pressure.
For example,
HCl_(g) + aq → HCl_(aq) Δ_solnH
= -75.14 kJ mol^-1

Question 46.
Explain the enthalpy of solution of an ionic compound.
Answer:
An ionic compound dissolves in a polar solvent like water in two steps as follows :
Step-I : The ions are separated from the molecule involving crystal lattice enthalpy Δ_LH.
\(\mathrm{MX}_{(\mathrm{s})} \rightarrow \mathrm{M}_{(\mathrm{g})}^{+}+\mathrm{X}_{(\mathrm{g})}^{-} \quad \Delta_{\mathrm{L}} H\)
Δ_LH is always positive.

Step-II : The gaseous ions are hydrated with water molecules involving hydration energy, Δ_hydH.
\(\mathbf{M}_{(\mathrm{g})}^{+}\) + xH₂O_(l) → [M(H₂O)_x]⁺
\(\mathrm{X}_{(\mathrm{g})}^{-}\) + yH₂O_(l) → [X(H₂O)_y]^–
Δ_hydH is always negative.
The enthalpy change ΔsolnH of solution is given by,
Δ_solnH = Δ_LH + Δ_hydH
For example, consider enthalpy of solution of NaCl(s).
Δ_LH_NaCl = 790 kJ mol^-1
Δ_hydH_NaCl = -786 kJ mol^-1
Hence enthalpy change for solution of NaCl(s) is,
Δ_solnH = Δ_LH_NaCl + Δ_hydH_NaCl
= 790 + (-786)
= + 4 kJ mol^-1
Therefore dissolution of NaCl in water is an endothermic process.

Solved Examples 4.9

Question 47.
Solve the following :

(1) Heat of fusion of ice at 0 °C and 1 atmosphere is 6.1 kJ mol^-1 and heat of evaporation of water at 100 °C is 40.7 kJ mol^-1. Calculate the enthalpy change for the conversion of 1 mole of ice at 0 °C into vapour at 100 °C. Heat capacity of water is 4.184 JK^-1 mol^-1.
Solution :
Given : Heat of fusion of ice = Δ_fusH = 6.01 kJ mol^-1
Heat of evaporation of water = Δ_vapH = 40.7 kJ mol^-1
Temperature of ice = 273 K
Temperature of vapour = (273 + 100) K = 373 K
Heat capacity of water = 4.184 JK^-1 g^-1
Heat capacity of 1 mole of water
= C_H2O = 4.184 × 18
= 75.312 JK^-1 mol^-1
The conversion of ice at 0 °C to water at 100°C takes place in three steps as follows:

ΔH₁ = n_ice × Δ_fusH = 1_mol × 6.01_{kJ mol^-1} = 6.01 kJ
ΔH₂ is the enthalpy change for raising the temperature from 273 K to 373 K.
ΔH₂ = n_water × C_H2O × (T₂ – T₁)
= 1_mol × 75.312_{JK^-1 mol^-1} × (373 – 273)K
= 7531 J
= 7.531 kJ
ΔH₃ = n_water × Δ_vapH
= 1 × 40.7
= 40.7 kJ
Hence total enthalpy change will be ΔH = ΔH₁ + ΔH₂ + ΔH₃
= 6.01 + 7.531 + 40.7
= 54.241 kJ
Ans. ΔH = 54.241 kJ

(2) Heat of sublimation of ice at 0°C and 1 atmosphere is 6.01 kJ mol^-1 and heat of evaporation of water at 0 °C and 1 atmosphere is 45.07 kJ mol^-1. Calculate the heat of sublimation of one mole of ice at 0 °C and 1 atmosphere. Write the equation for the same.
Solution :
The sublimation of ice can be represented by following equation,
\(\mathrm{H}_{2} \mathrm{O}_{(\mathrm{s})} \stackrel{1 \mathrm{~atm}, 0^{\circ} \mathrm{C}}{\longrightarrow} \mathrm{H}_{2} \mathrm{O}_{(\mathrm{g})}\)
This is a process of two steps.

ΔH₁ = n_H2O × Δ_fusH
= 1 × 6.01 =6.01 kJ
ΔH₂ = n_H2O × Δ_vapH
= 1 × 45.07 = 45.07
Heat of sublimation = ΔH = ΔH₁ + ΔH₂
= 6.01 + 45.07
= 51.08 kJ
Ans. Heat of sublimation of ice = 51.08 kJ

(3) Heat of evaporation of ethyl alcohol at 78.5 °C and 1 atmosphere is 38.6 kJ mol^-1. If 100 g ethyl alcohol vapour is condensed, what will be ΔH ?
Solution :
Given : Δ_vapH_C2H5OH = 38.6 kJ mol-1
Mass of C₂H₅OH = m = 100 g
Molar mass of C₂H₅OH = M = 46 g mol-1
Δ_condH_C2H5OH = ?
Number of moles of C₂H₅OH = \(\frac{m}{M}=\frac{100}{46}\)
= 2.174 mol
Heat of condensation = Δ_condH = -38.6 kJ mol^-1
∴ Δ_condH = n × Δ_condH
= 2.174_mol × (-38.6)_{kJ mol^-1} kJ = – 83.9 kJ
Ans. Heat of condensation = Δ_condH = -83.9 kJ

(4) The hydration enthalpies of Li⁺_(g), and Br^–_(g) are -500 kJ mol^-1 and -350 kJ mol^-1 respectively and the lattice energy of LiBr_(s) is 807 kJmol^-1. Write the thermochemical equations for enthalpy of solution of LiBr(s) and calculate its value.
Solution :
Given : Enthalpy of hydration of Li⁺_(g)
= Δ_hydH1
= -500 kJmol^-1
Enthalpy of hydration of Br^–_(g) = Δ_hydH₂
= -350 kJ mol^-1
Lattice energy of LiBr_(s) = Δ_LH₃ = 807 kJ mol^-1
Enthalpy of solution of LiBr_(s) = Δ_solnΔH = ?
The thermochemical equation for the dissolution of LiBr_(s) forming a solution is,
LiBr_(s) + aq → Li⁺_(aq) + Br^–_(aq) (I) Δ_solH = ?
This takes place in two steps as follows :
(i) LiBr_(s) → Li⁺_(g) + Br^–_(g) Δ_LH₃
(ii) (a) Li⁺_(g) + aq → Li⁺_(aq) Δ_hydH₁
(b) Br^–_(g) + aq → Br^–_(aq) Δ_hydH₂
Hence by adding equations (i) and (ii) (a) and (b) we get equation I.
∴ Δ_solH = Δ_LH₃ + Δ_hydH₁ + Δ_hydH₂
= 807 + (-500) + (-350)
= -43 kJ mol^-1
Ans. Heat of solution of LiBr_(s) = Δ_solH = -43 kJ mol^-1

(5) Heat of solution of NaCl is 3.9 kJ mol^-1. If the lattice energy of NaCl is 787 kJ mol^-1, calculate the hydration energies of ions of the salt.
Solution :
Given : Heat of solution of NaCl
= Δ_solnH⁰
= ΔH₁ = 3.9 kJmol^-1
Lattice energy of NaCl = ΔLH
= ΔH₂ = 787 kJ mol^-1
Hydration energy of Na⁺_(g) and Cl^–_(g)
= Δ_hydH(Na⁺ + Cl^–)
= ΔH₃ = ?
Thermochemical equation for dissolution of NaCl(s) is;
NaCl_(s) + aq → Na⁺_(aq) + Cl^–_(aq)…ΔH₁
NaCl_(s) → Na⁺_(g) + Cl^–_(g)… ΔH₂
Na⁺_(g) + Cl^–_(g) + aq → Na⁺_(aq) + Cl^–_(aq) ΔH₃
∴ ΔH₁ = ΔH₂ + AH₃
3.9 = 787 + AH3
∴ ΔH₃ = -787 + 3.9= -783.1 kJmol^-1
Ans. Hydration energy of Na⁺_(g) and Cl^–_(g)
= -783.1 kJmol^-1

(6) Enthalpies of solution are given as follows :
CuSO_4(s) + 10H₂O → CUSO₄(10H₂O)
ΔH₁ = -54.5 kJ mol^-1
CuSO₄(s) + 100 H₂O → CUSO₄(100H₂O)
ΔH₂ = -68.4 kJ mol^-1
A solution contains 1 mol of CuSO₄ in 180 g water at 25 °C. If it is diluted by adding 1620 g water, calculate the enthalpy of dilution.
Solution :
Given : Enthalpy of solution of CuSO₄ in 10 mol H₂O
= Δ_solnH = ΔH₁ = -54.5 kJ mol^-1
Enthalpy of solution of CuSO₄ in 100 mol
H₂O = ΔH₂
= -68.4 kJmol^-1
Mass of water = 1620 g
For dilution, Δ_dilH = ?
Now 180 g H₂O = \(\frac {180}{18}\) = 10 mol H₂O
And, 1620 g H₂O = \(\frac {1620}{18}\) = 90 mol H₂O
Hence for heat of dilution,
CUSO₄(10H₂O) + 90H₂O_(l) → CUSO₄(100H₂O) Δ_dilH = ?
∴ Δ_dilH = ΔH₂ -ΔH₁
= -68.4 – (54.5)
= -13.9 kJmol^-1
Ans. Heat of dilution = Δ_dilH = -13.9 kJ mol^-1

(7) Heat of solution and heat of hydration of AgF are -20.5kJmol^-1 and -930kJmol^-1 respectively. Calculate lattice energy of AgF.
Solution :
Given : Heat of solution of AgF = Δ_solnH
= -ΔH₁ = -20.5 kJmol^-1
Heat of hydration of AgF = Δ_hydH = ΔH₂
= -930 kJ mol^-1
Lattice energy of AgF = Δ_LH = ΔH₃ = ?
For heat of solution, AgF_(s) + aq → AgF_(aq) ΔH₁
For heat of hydration,
Ag⁺_(g) + F^–_(g) + aq → Ag⁺_(aq) + F^–_(aq) ΔH₂
For Lattice energy, Ag⁺_(g) + F^–_(g) → AgF_(s)
Δ_LH = ?
From above equations,
∴ ΔH₃ = ΔH₂ – ΔH₁
= -930 – (-20.5)
= -909.5 kJmol^-1
Ans. Lattice energy of AgF_(s) = -909.5 kJ mol^-1

(8) Bond enthalpy of H₂ is 436kJmol^-1 while hydration energy of hydrogen ion is -1075 kJ mol^-1. Calculate the enthalpy of formation of H⁺_(aq). (Ionisation energy of hydrogen is 1312 kJ mol^-1
Solution :
Given : Bond enthalpy of H_2(g) = ΔH⁰H_2(g)
= 436 kJ mol^-1
Hydration energy of H⁺_(g) = ΔH₂ = -1075 kJmol^-1
Ionisation energy of H_(g) = ΔH₃ = 1312 kJ mol^-1
Enthalpy of formation of H⁺_(aq) = Δ_fH = ?
Thermochemical equation for the formation of H⁺_(aq)
\(\frac {1}{2}\)H_2(g) + aq → H⁺_(aq)Δ_fH
This takes place in three steps as follows :

Hence heat of formation of H⁺_(aq) is
Δ_fH = ΔH₁ + ΔH₂ + ΔH₃
= \(\frac {1}{2}\) × 436 + (-1075) + 1312
= 218 – 1075 + 1312.
= 455 kJ mol^-1
Ans. Enthalpy of formation of H⁺_(aq) = 455 kJmol^-1

(9) Calculate lattice energy of crystalline sodium chloride from the following data :

Solution :
Given : Bond enthalpy of Cl₂ = ΔH⁰ = 244 kJmol^-1
Thermochemical equation for the formation of 1 mole of NaCl_(s),
\(\mathrm{Na}_{(\mathrm{s})}+\frac{1}{2} \mathrm{Cl}_{2} \longrightarrow \mathrm{NaCl}_{(s)} \quad \Delta_{\mathrm{f}} H_{\mathrm{NaCl}}^{0}=-411 \mathrm{~kJ}\)
Lattice energy, Δ_LH = ?
Since enthalpy is a state function, this reaction can be written in various steps as follows :

By Hess’s law,
Δ_LH⁰ = Δ_subH⁰+ Δ_ionH⁰ + \(\frac {1}{2}\)ΔH⁰cl₂ + ΔegH⁰ + Δ_LH⁰
-411 = 109 + 496 + \(\frac {1}{2}\) × 244 + (-348) + Δ_LH
= 109 + 496 + 122 – 348 + Δ_LH
∴ Δ_LH= -790 kJ mol^-1.
Ans. Lattice energy of NaCl_(s) = -790 kJ mol^–

(10) Calculate the internal energy at 298 K for the formation of one mole of ammonia, if the enthalpy change at constant pressure is -42.0 kJ mol^-1.
(Given : R = 8.314 J K^-1 mol^-1)
Solution :
Given : ΔH = -42.0 kJ mol^-1, T = 298 K, ΔU = ?
\(\frac{1}{2} \mathrm{~N}_{2(\mathrm{~g})}+\frac{3}{2} \mathrm{H}_{2(\mathrm{~g})} \longrightarrow \mathrm{NH}_{3(\mathrm{~g})}\)
Δn = 1 – (\(\frac {1}{2}\) + \(\frac {3}{2}\)) = -1 mol
ΔH = ΔU + ΔnRT
∴ ΔU = ΔH – ΔnRT
= -42 – (- 1) × 8.314 × 298 × 10^-3
= -42 + 2.477
= -39.523 kJ
Ans. ΔU = -39.523 kJ

Question 48.
What is thermochemistry ? Explain.
Answer:
Thermochemistry : Thermodynamic study of heat changes or enthalpy changes during the chemical reactions is called thermochemistry.
Consider a reaction, Reactants → Products
The heat changes ΔH for the reaction may be represented as,
ΔH_reaction = Σ H_products – Σ H_reactants
where H represents enthalpy.
The energy released or absorbed during a chemical change appears in the form of heat energy.

Question 49.
Define and explain the term, enthalpy or heat of reaction.
Answer:
Enthalpy or heat of reaction : The enthalpy of a chemical reaction is the difference between the sum of the enthalpies of products and that of the reactants with every substance in a definite physical state and in the amounts (moles) represented by the coefficients in the balanced equation.
Explanation : Consider the following general reaction,
aA + bB → cC + dD
The heat of the reaction ΔH is the difference in sum of enthalpies of products and sum of enthalpies of reactants.
∴ ΔH = Σ H_products – Σ H_reactants
= [cH_C + dH_D] – [aH_A + bH_B]
= Σ_PH – Σ_RZ
where H represents enthalpy of the substance.
For endothermic reaction, ΔH is positive, (ΔH > 0).
For exothermic reaction, ΔH is negative, (ΔH < 0).

Question 50.
Explain the sign convention used for ΔH.
Answer:
The change in enthalpy or heat of reaction ΔH is given by,

(i) If the sum of enthalpies of products, ΣH and reactants, Σ_RH are equal then ΔH for the reaction is zero, (ΔH = 0).
i. e. Σ_PH = Σ_RH
∴ ΔH = Σ_PH – Σ_RH = 0
(ii) If the sum of enthalpies of products Σ_PH is greater than the sum of enthalpies of reactants Σ_RH, then ΔH is positive, (ΔH > 0). Since such reactions take place with the absorption of heat from surroundings, they are called endothermic reactions.
∴ ΣH_products > ΣH_reactants
∴ ΔH > 0
(iii) If the sum of enthalpies of products Σ_PH is less than the sum of enthalpies of reactants, Σ_RH then ΔH is negative, (ΔH < 0). Since in such reactions heat is given out to the surroundings, they are called exothermic reactions.
∴ Σ_PH < Σ_RH
∴ ΔH < 0

Question 51.
Define : (i) Exothermic process (ii) Endothermic process.
Answer:
(i) Exothermic process : A process taking place with the evolution of heat is called exothermic process.
For this process, Q is -ve, ΔH is -ve.
(ii) Endothermic process : A process taking place with the absorption of heat (from the surroundings) is called endothermic process.
For this process, Q is +ve, ΔH is +ve.

Question 52.
Distinguish between Endothermic reaction and Exothermic reaction.
Answer:
Endothermic reaction:

• In endothermic reaction heat is absorbed from suroundings.
• Sum of enthalpies of products is greater than sum of enthalpies of reactants i.e. Σ_PH > Σ_RH
• Heat of reaction, ΔH is positive.
• Products are less stable than reactants.
• C_(s) + O_2(g) → CO_2(g)
  ΔH = -394 kJ
• This reaction requires supply of thermal energy.

Exothermic reaction:

• In exothermic reaction heat is given out to surroundings.
• Sum of enthalpies of products is less than sum of enthalpies of reactants.
  i.e. Σ_PH < Σ_RH
• Heat of reaction, ΔH is negative.
• Products are more stable than reactants.
• N_2(g) + O_2(g) → 2NO
  ΔH = + 180 kJ
• This reaction does not require supply of thermal energy.

Question 53.
Explain the standard state of an element.
Answer:
Standard state of an element : It is defined as the most stable state of an element at 298 K and 1 atmosphere (or 1 bar).
In this state, the enthalpy of the element is assumed to be zero.
∴ H⁰_element or in general H_element = 0
∴ H⁰_graphite = H_H2(g) = 0; H_Na(s) = 0; H_Hg(l) = 0

Question 54.
What is a thermochemical equation? Explain with an example.
Answer:
Thermochemical equation : It is defined as a balanced chemical equation along with the corresponding heat of reaction (ΔH) and physical states and number of moles of all reactants and all products appropriately mentioned.
E.g. C₆H₁₂O_6(s) + 6O_2(g) = 6CO_2(g) + 6H₂O_(l)
ΔH = -2808 kJ mol^-1

Question 55.
What are the guidelines followed for writing thermochemical equations?
Answer:
According to IUPAC conventions, while writing thermochemical equations, following rules must be followed :
(1) Reaction is represented by balanced chemical equation for the number of moles of the reactants and the products. E.g.
CH_4(g) + 2O_2(g) = CO_2(g) + 2H₂O_(l)
ΔrH°= -890 kJ mol^-1
(2) The physical states of all the substances in the reaction must be mentioned. E.g. (s) for solid, (l) for liquid and (g) for gas.
(3) Heat or enthalpy changes are measured at 298 K and 1 atmosphere (or 1 bar).
(4) ΔH⁰ is written at right hand side of thermochemical equation.
(5) Proper sign must be indicated for ΔH⁰. For endothermic reaction ΔH⁰ is positive, (+ΔH⁰) and for exothermic reaction ΔH is negative, (-ΔH⁰).
(6) The enthalpy of the elements in their standard states is taken as zero. (H⁰_Element = 0; H⁰_C(s) = 0, H⁰_H2(g) = 0)
(7) When all the substances taking part in the reaction are in their standard states, the enthalpy change is written as ΔH⁰.
(8) The enthalpy of any compound is equal to its heat of formation.
(9) In case of elements, the allotropic form must be mentioned. E.g. C_(graphite), S_(rhombic), Sn_(white)
(10) For the reverse reaction, ΔH⁰ value has equal magnitude but opposite sign.

Question 56.
Define the following terms giving examples :
(1) Standard enthalpy of reaction.
(2) Standard enthalpy of formation or standard heat of formation
(3) Standard enthalpy of combustion or standard heat of combustion.
Answer:
(1) Standard enthalpy of reaction : it is defined as the difference between the sum of enthalpies of products and that of the reactants with every substance in its standard state at constant temperature (298 K) and pressure (1 atm).
Reactants → Products
ΔH⁰_reaction = ΣH⁰_products – ΣH⁰_reactants

(2) Standard enthalpy of formation or standard heat of formation (Δ_fH⁰) : It is defined as the enthalpy change ΔH⁰ when one mole of a pure compound is formed in its standard state from its constituent elements in their standard states at constant temperature (298 K) and pressure (1 atmosphere or 1 bar). It is denoted by Δ_fH⁰. E.g.
C_(s) + O_2(g) = CO_2(g) Δ_fH⁰= -394 kJ mol^-1
(Δ_fH⁰ may be positive or negative.)

(3) Standard enthalpy of combustion or standard heat of combustion : it is defined as the enthalpy change when one mole of a substance in the standard state undergoes complete combustion in a sufficient amount of oxygen at constant temperature (298 K) and pressure (1 atmosphere or 1 bar). It is denoted by Δ_cH⁰.
E.g. CH₃OH_(l) + \(\frac {3}{2}\)O_2(g) = CO_2(g) + 2H₂O
ΔCH⁰ = -726 kJ mol^-1
(Δ_cH⁰ is always negative.)
[Note : Calorific value : It is the enthalpy change or amount of heat liberated when one gram of a substance undergoes combustion.

Question 57.
Show that the standard heat of formation of a compound is equal to its enthalpy.
Answer:
Consider the formation of one mole of gaseous CO₂ in the standard state at 298 K and 1 atmosphere. The thermochemical equation for formation can be represented as,
C_(s) + O_2(g) = CO_2(g) Δ_fH⁰ = -394 kJ mol^-1
Now heat of this reaction, ΔH⁰ is,
ΔH⁰_reaction = Σ_PH⁰ – Σ_RH⁰
∴ Δ_fH⁰_co2(g) = H⁰_co2(g) – [H⁰c_(s) + H⁰O_2(g)]
Since the enthalpies of elements in their standard states are zero,
i.e.
H⁰c_(s) = o, H⁰O_2(g) = 0
∴ Δ_fH⁰co_2(g) = H⁰co_2(g) – [0 + 0]
∴ Δ_fH⁰co2 = Hco_2(g)
Therefore standard heat of formation of a compound is equal to its enthalpy.

Question 58.
Standard enthalpy of formation of various following compounds are given. Write thermochemical equation for each :

Compound	ΔfH0 KJ mol-1
Cao(s)	-635.1
Al2Cl6(s)	-1300
C2H6(g)	-84.7
CH3COOH(l)	-484.7
C2H5OH(l)	-277.7
NaNO3(s)	-950.8

Answer:
(Hint: Write one mole of the compound on right hand side and corresponding constituent elements along with their standard physical states on left hand side.)

Question 59.
Standard enthalpy of combustion of different substances are given. Write thermochemical equation for each.

Substance	ΔCH0 KJ mol-1
C(graphite)	-393.5
C6H6(l)	-3268
C2H5OH(l)	-1409
CH3CHO	-1166

Answer:
In the combustion reaction, C forms CO_2(g),
H forms H₂O_(l), etc.
(1) C_(graphite) + O_2(g) → CO_2(g)
Δ_CH⁰ = -393.5 kJ mol^-1

Question 60.
Write the thermochemical equations for enthalpy of solution of :
(1) Glucose (C₆H₁₂O₆)
(2) NaCl_(s)
(3) CaBr_2(s)
Answer:

Question 61.
How is standard enthalpy of formation useful to calculate standard enthalpy of reaction ?
Answer:
(1) The standard enthalpies of formation, Δ_fH⁰ of the compounds can be used to determine the standard enthalpy of reaction (Δ_rH⁰).
(2) Δ_rH⁰ of a reaction can be obtained by subtracting the sum of Δ_fH⁰ values of all the reactants from the sum of Δ_fH⁰ values of all the products with each Δ_fH⁰ value multiplied by the appropriate coefficient of that substance in the balanced thermochemical equation.
(3) Consider following reaction :
aA + bB → cC + dD
The standard enthalpy of the reaction is given by,

where a, b, c and d are the coefficients (moles) of the substances A, B, C and D respectively.

Question 62.
Write the balanced chemical equation that have ΔH⁰ value equal to Δ_fH⁰ for each of the following substances :
(1) C₂H_2(g)
(2) KCIO_3(s)
(3) C₁₂H₂₂O_11(s) (4) CH₃-CH₂-OH₍₁₎
Answer:
Δ_fH⁰ represents the standard enthalpy of formation of each given substance. Hence it is necessary to write thermochemical equation for the formation of each substance. ΔH⁰ of this formation reaction is equal to standard heat of formation, Δ_fH⁰.

Question 63.
Consider the chemical reaction,
OF_2(g) + H₂O_(g) → O_2(g) + 2HF_(g) ΔH⁰ = -323 kJ
What is ΔH⁰ of the reaction if (a) the equation is multiplied by 3, (b) direction of reaction is reversed?
Answer:
(a) If the given thermochemical equation is multiplied by 3 then,
Δ_rH⁰ = 3ΔH⁰ = 3 × (-323) = -969 kJ
(b) If the direction of equation is reversed, then the reaction will be,
O_2(g) + 2HF_(g) → OF_2(g) + H₂O_(g)
∴ Δ_rH⁰ = – ΔH⁰ = – (- 323) = + 323 kJ

Question 64.
Define bond enthalpy (or bond energy).
Answer:
Bond enthalpy (or Bond energy) : The enthalpy change or amount of heat energy required to break one mole of particular covalent bonds of gaseous molecules forming free gaseous atoms or radicals at constant temperature (298 K) and pressure (1 atmosphere) is called bond enthalpy or bond energy. For example, bond enthalpy of H₂ is 436.4 kJ mol^-1.

Question 65.
Explain bond enthalpy of diatomic molecules.
Answer:
In case of diatomic molecules, since there is only one bond, the bond enthalpy is equal to heat of atomisation. For example, heat of atomisation of
HCl_(g) is 431.9 kJ mol^-1.

(Bond enthalpy is generally denoted by D).

Question 66.
Explain bond enthalpy in polyatomic molecules.
Answer:
Consider bond enthalpy in H₂O. The thermochemical equation for dissociation of H₂O_(g) is,
H₂O_(g) → 2H_(g) + O_(g), Δ_rH⁰ = 927 kJ mol^-1
In this, two O – H bonds are broken. It can be represented in stepwise as follows :

In above, even if two identical O – H bonds are borken, the energies required to break each bond are different.
The average bond enthalpy of O – H bond is,
Δ_rH⁰ = \(\frac {927}{2}\) = 463.5 kJ mol^-1

Solved Examples 4.10

Question 67.
Solve the following :

(1) Standard enthalpy of formation of ethane, C₂H_6(g) is -84.7 kJ mol^-1. Calculate the enthalpy change for the formation of 0.1 kg ethane.
Solution :
Given : Enthalpy of formation of C₂H_6(g)
= Δ_fH⁰ = ΔH₁ = -84.7 kJ mol^-1
Mass of C₂H_6(g) = 0.1 kg = 100 g
Molar mass ofC₂H₆ = 30 g mol^-1
ΔH⁰ for the formation of 0.1 kg C₂H₆ = 100 g
C₂H₆ = ΔH₂ = ?

Ans. Heat of formation = -282.3 kJ

(2) When 10 g C₂H₅OH_(l) are formed, 51 kJ heat is liberated. Calculate standard enthalpy of formation of C₂H₅OH_(l).
Solution :
Given : Mass of C₂H₅OH_(l) = m = 10 g
Heat liberated = ΔH1 = -51 kJ
Molar mass of C₂H₅OH = 46 gmol^-1
Standard enthalpy of formation of C₂H₅OH_(l)
= Δ_fH = ?
Standard enthalpy of formation is the enthalpy change for the formation of 1 mole C₂H₅OH_(l) i.e., 46 g C₂H₅OH_(l).
Now,
∵ For the formation of 10 g C₂H₅OH_(l)
ΔH₁ = -51 kJ
∴ For the formation of 46 g C₂H₅OH,
Δ_fH⁰ = \(\frac{-51 \times 46}{10}\) = – 234.6 kJ mol^-1
Ans. Standard enthalpy of formation of C₂H₅OH = Δ_fH⁰ = – 234.6 kJ mol^-1

(3) Standard enthalpy of combustion of CH₃OH is -726kJ mol^-1. Calculate enthalpy change for the combustion of 0.5 kg CH₃OH.
Solution :
Given : Standard enthalpy of combustion of
CH₃OH = Δ_CH⁰ = ΔH₁ = -726 kJ mol^-1
Mass of CH₃OH = m = 0.5 kg = 500 g
Molar mass of CH₃OH = 32 g mol^-1
Enthalpy of combustion = Δ_CH = ΔH₂ = ?
Now,
Enthalpy of combustion is ΔH for the combustion of 1 mole CH₃OH = 32 g CH₃OH.
∵ For 1 mole CH₃OH = 32g CH₃OH
ΔH₁ = – 726 kJ
∴ For 500 g CH₃OH, ΔH₂ = \(\frac{-726 \times 500}{32}\)
= -11344 kJ
Ans. Enthalpy change for combustion of 0.5 kg CH₃OH = – 11344 kJ

(4) The heat evolved in a reaction of 7.5 g of Fe₂O₃ with enough CO is 1.164 kJ.
Calculate ΔH⁰ for the reaction,
Fe₂O_3(s) + 3CO_(g) → 2Fe_(s) + 3CO_2(g)
Solution :
Fe₂O_3(s) + 3CO_(g) → 2Fe_(s) + 3CO_2(g)
ΔH = -1.164 kJ
Atomic mass of Fe = 56 g mol^-1
Atomic mass of O = 16 g mol^-1
Mass of Fe₂O₃ = 7.5 g
ΔH = -1.164 kJ
ΔH⁰ for reaction = ?
Molar mass of Fe₂O₃ = 2 × 56 + 3 × 16
= 160 g mol^-1
∵ For 7.5 g Fe₂O₃ ΔH= – 1.164 kJ
∴ For 160 g Fe₂O₃
ΔH⁰ = \(\frac{-1.164 \times 160}{7.5}\) = -24.86 kJ mol^-1
Ans. ΔH⁰ for the reaction = -24.83 kJ mol^-1

(5) Calculate the standard enthalpy of the reaction,
2C (graphite) + 3H_2(g) → C₂H_6(g), ΔH⁰ = ? from the following ΔH⁰ values :

Solution :

Ans. Standard enthalpy of formation of C₂H₆ = -84.4 kJ mol^-1

(6) The enthalpy of combustion of ethane is -1300 kJ. How much heat will be evolved by combustion of 1.3 × 10^-3 kg of ethane?
Solution :
Given : Δ_CH_C2H6(g) = -1300 kJ mol^-1
ΔH = ?
Amount of C₂H_6(g) = 1.3 × 10^-3 kg
Molar mass of C₂H₆ = 30 × 10^-3 kg mol^-1
Number of moles of C₂H₆
= nC₂H₆ = \(\frac{1.3 \times 10^{-3}}{30 \times 10^{-3}}\) = 4,333 × 10^-2 mol
For, combustion of 1 mol C₂H₆ ΔH = -1300 kJ
∴ For combustion of 4.333 × 10^-2 mol C₂H₆,
ΔH = 4.333 × 10^-2 × ( -1300) = – 56.33 kJ
Ans. Heat evolved is -56.33 kJ

(7) Calculate heat of formation of pentane from the following data :
(i) C_(s) + O_2(g) = CO_2(g) ΔH⁰ = -393.51 kJ
(ii) H_2(g) + \(\frac {1}{2}\)O_2(g) = H₂O_(l) ΔH⁰ = -285.80 kJ
(iii) C₅H₁₂ + 😯_2(g) = 5CO_2(g) + 6H₂O₁ ΔH⁰ = -3537 kJ
Solution :
Given :
(i) CO_(s) + O_2(g) = CO_2(g) ….. (1)
\(\Delta H_{1}^{0}\) = -393.51 kJ mol^-1
(ii) H_2(g) + \(\frac {1}{2}\)O_2(g) = H₂O_(l) … (2)
\(\Delta H_{2}^{0}\) = – 285.80 kJ mol^-1
(iii) C₅H_12(g) + 😯_2(g) = 5CO_2(g) + 6H₂O_(l) ….. (3)
\(\Delta H_{3}^{0}\) = -3537 kJ mol^-1
Required thermochemical equation :
5C_(s) + 6H_2(g) → C₅H_12(g) – ΔH = ?
Add 5 × equation (1) and 6 × equation (2) and subtract equation (3), then we get the required equation.
∴ ΔH⁰ = 5 \(\Delta H_{1}^{0}\) + 6 \(\Delta H_{2}^{0}\) – \(\Delta H_{3}^{0}\)
= 5( -393.52) + 6( -285.8) – (-3537)
= -1967.6 – 1714.8 + 3537
= -145.4 kJ mol^-1
Ans. Δ_fH⁰C₅H₁₂ = -145.4 kJ mol^-1

(8) How much heat is evolved when 12 g of CO react with NO₂ according to the following reaction,
4CO_(g) + 2NO_2(g) → 4CO_2(g) + N_2(g), ΔH⁰ = -1198 kJ ?
Solution :
Given : Mass of CO(g) = m = 12 g
Molar mass of CO = 28 g mol^-1
4CO_(g) + 2NO_2(g) → 4CO_2(g) + N_2(g)
Δ_rH⁰ = -1198 kJ
Mass of 4 moles of CO = 4 × 28 g CO = 112 g CO

Ans. Heat evolved during combustion of 12 g CO = 128.4 kJ, or Δ_CH = -128.4 kJ

(9) The heats of formation of C₁₂H₂₂O_11(S), CO_2(g) and H₂O_(l) are -2271.82, – 393.50 and 285.76 kJ respectively. Calculate the amount of cane sugar (C₁₂H₂₂O_11(S)) which will supply 11296.8 kJ of energy.
Solution :
Given : Δ_fH_C12H22O11(S) = -2271.82 J mol^-1
Δ_fH_CO2(g) = – 393.5 kJ mol^-1
Δ_fH_H2O(l) = – 285.76 kJ mol^-1
Energy required = 11296.8 kJ
Thermochemical equation for combustion of C₁₂H₂₂O_11(S) is represented as,

= [ 12(-393.5) + 11 (-285.76] – [-2271.82 + 12(0)]
= [ -4722 – 3143.36] + 2271.82
= -5593.54 kJ mol^-1
Molar mass of C₁₂H₂₂O_11(S) = 342
To obtain 5593.5 kJ energy, C₁₂H₂₂O_11(S) required is 342 gram.
Hence for 11296.8 energy, the amount of C₁₂H₂₂O_11(S) required as = \(\frac{11296.8 \times 342}{5593.54}\)
= 690.7 g
Ans. Amount of sugar required = 690.7 g

(10) 6.24 g of ethanol are vaporized by supplying 5.89 kJ of heat energy. What is enthalpy of vaporization of ethanol?
Solution :
Given : Mass of ethanol (C₂H₅OH) = m = 6.24 g
Heat energy supplied = ΔH = 5.89 kJ
Heat of vaporisation of ethanol = Δ_vapH = ?
Molar mass of ethanol, C₂H₅OH = 46 g mol^-1
∵ For 6.24 g C₂H₅OH ΔH = 5.89kJ
∴ For 1 mole C₂H₅OH = 46 g C₂H₅OH
ΔH = \(\frac{5.89 \times 46}{6.24}\)
= 43.42 kJ mol^-1
∴ Enthalpy of vaporisation of C₂H₅OH_(l)
= 43.42 kJ
Ans. Enthalpy of vaporisation of C₂H₅OH_(l)
= 43.42 kJ

(11) Given the following equations calculate the standard enthalpy of the reaction :

Solution :

By subtracting eq. (ii) from eq. (iii), we get eq. (i)
∴ eq. (i) = eq. (iii) – eq. (ii)
\(\Delta H_{1}^{0}=\Delta H_{3}^{0}-\Delta H_{2}^{0}\)
= -1670 – (-847.6)
= – 822.4 kJ
∴ Δ_rH⁰ = ΔH⁰₁ = -822.4 kJ
Ans. Standrad enthalpy of the reaction = Δ_rH⁰ = -822.4 kJ

(12) Calculate the standard enthalpy of combustion of CH₂COOH_(l) from the following data : Δ_fH⁰(CO₂) = -393.3 kJ mol^-1
Δ_fH⁰(H₂O) = -285.8 kJ mol^-1
Δ_fH⁰(CH₃COOH) = -483.2 kJ mol^-1
Solution :


∴ ΔH₁ = 2ΔH₂ + 2ΔH₃ – ΔH₄
= 2(-393.3) + 2(-285.8) – (-483.2)
= -786.6 – 571.6 + 483.2
= -875 kJ mol^-1
Ans. Standard enthalpy of combustion of CH₃COOH = -875 kJ mol^-1.

(13) The bond enthalpies of H_2(g), Br_2(g) and HBr_(g) are 436 kJ mol^-1, 193 kJ mol^-1 and 366 kJ mol^-1 respectively. Calculate the enthalpy change for the following reaction,
H_2(g) + Br_2(g) → 2HBr_(g).
Solution :
Given : Bond enthalpy of H_2(g) = ΔH⁰H_2(g)
= 436 kJ mol-1
Bond enthalpy of Br_2(g) = ΔH⁰Br_2(g) = 193 kJ mol^-1
Bond enthalpy of HBr_(g) = ΔH⁰HBr_(g) = 366 kJ mol^-1
Given reaction,
H_2(g) + Br_2(g) → 2HBr_(g)
OR
H-H_(g) + Br-Br_(g) → 2H-Br_(g)
The enthalpy change of the reaction is,

= [436 + 193] – 2[366]
= 629 – 732
= -103 kJ
Ans. Enthalpy change for the reaction = Δ_rH⁰
= -103 kJ

(14) Calculate Δ_rH⁰ of the reaction
CH_4(g) + O_2(g) → CH₂O_(g) + H₂O_(g)
From the following data:

Solution:
Given:

Standard enthalpy change for the reaction = Δ_rH⁰ = ?

= [ 2ΔH⁰_C-H + ΔH⁰_o=o ] – [ΔH⁰_C=o + 2ΔH⁰_o-H]
= [2 × 414 + 499] – [745 + 2 × 464]
= [828 + 499] – [745 + 928]
= -346 kJ
Ans. Standard enthalpy change for the reaction = ΔrH⁰ = -346 kJ

(15) Calculate C-Cl bond enthalpy from the following data :
CH₃Cl_(g) + Cl_2(g) → CH₂Cl_2(g) + HCl_(g) ΔH⁰ = – 104 kJ

Bond	C–H	Cl–Cl	H–Cl
ΔH0/KJ mol-1	414	243	431

(330 kJ mol^-1)
Solution :
Given :

Bond	C–H	Cl–Cl	H–Cl
ΔH0/KJ mol-1	414	243	431

For the given reaction, Δ_rH⁰ = -104 kJ
Bond enthalpy of C-Cl = ΔH⁰C–Cl] = ?

In this reaction, 1 C–H, 1 Cl–Cl bonds of the reactants are broken while 1C–Cl and 1H–Cl bonds of the products are formed.
Sum of bond enthalpies of bonds formed of the products

Ans. Bond enthalpy of C–Cl = ΔH⁰C–Cl
= 330 kJ mol^-1

(16) The enthalpy change for the atomisation of 10¹⁰ molecules of ammonia is 1.94 × 10^-11 kJ. Calculate the bond enthalpy of N-H bond.
Solution :
Given : Enthalpy change for atomisation of 10¹⁰ molecules = 1.94 × 10^-11 kJ
Number of NH₃ molecules dissociate = 10¹⁰
Bond enthalpy of N-H = ΔH = ?
1 mole of NH₃ contains 6.022 × 10²³ NH₃ molecules.
∵ For atomisation of 1010 molecules of NH₃
ΔH = 1.94 × 10^-11 kJ
∴ For atomisation of 6.022 × 10²³ molecules of NH₃,
ΔH = \(\frac{1.94 \times 10^{-11} \times 6.022 \times 10^{23}}{10^{10}}\)
= 1168 kJ mol^-1
In NH₃ three N-H bonds are broken on atomisation.
NH_3(g) → N_(g) + 3H_(g) ΔH = 1168 kJ mol^-1
∴ Average bond enthalpy of N-H bond is,
ΔH = \(\frac{1168}{3}\) = 389.3 kJ mol^-1
Ans. Bond enthalpy of N-H bond
= 389.3 kJ mol^-1

(17) Calculate the enthalpy of atomisation (or dissociation) of CH₂Br_2(g) at 25°C from the following data :

Bond enthalpies	C-H	C-Br
ΔH0 kJ mol-1	414	352

Solution :
Given : Bond enthalpies : ΔH⁰_C-H
= 414 kJ mol-1;
ΔH⁰_C-Br = 352 kJ mol^-1
Enthalpy of atomisation of CH₂Br_2(g) = ?
Thermochemical equation for atomisation (or dissociation) of CH₂Br₂ is,

ΔatomH° = sum of bond enthalpies of all bonds broken
= 2ΔH⁰_C-H + 2ΔH⁰_C-Br
= 2 × 414 + 2 × 352
= 828 + 704
= 1532 kJ mol^-1
Ans. Enthalpy of atomisation of CH₂Br_2(g)
= 1532 kJ mol^-1

(18) Enthalpy of sublimation of graphite is 716 kJ mol^-1.

Bond enthalpy	H-H	C-H
ΔH0 kJ mol-1	436.4	414

Calculate standard enthalpy of formation of CH₄.
Solution :
Given : Δ_subH⁰_graphite = 716 kJ mol^-1

Bond enthalpy	H-H	C-H
ΔH0 kJ mol-1	436.4	414

Thermochemical equation for the formation of CH₄,

= [716 + 2 × 436.4] – [4 × 414]
= [716+ 872.8] – [1656]
= 1588.8 – 1656
= -67.2 kJ mol^-1
Ans. Standard enthalpy of formation of CH4 = Δ_fH⁰_CH4(g) = -67.2 kJ mol^-1

(19) Calculate enthalpy of formation of propane from the following data :
Heat of sublimation of graphite is 716 kJ mo^-1.

Bond enthalpy	H-H	C-H	C-C
ΔH0 kJ mol-1	436.4	414	350

Solution :
Given: Enthalpy of sublimation of graphite = Δ_subH⁰_C
= 716 kJmol^-1

Bond enthalpy	H-H	C-H	C-C
ΔH0 kJ mol-1	436.4	414	350

Enthalpy of formation of propane = Δ_fH⁰ = ?
Thermochemical equation of the formation of propane, CH₃-CH₂-CH₃,

= [3 × 716 + 4 × 436.4] – [2 × 350 + 8 × 414]
= [2148 + 1745.6] – [700 + 3312]
= -118.4 kJmol^-1
Ans. Enthalpy of formation of propane (C₃H₈)
= -118.4 kJmol^-1

(20) The standard enthalpy of formation of propene, CH₃-CH = CH₂ is -13.2 kJ mol^-1. Enthalpy of sublimation (atomisation) of graphite is 716 kJmol^-1.

Bond enthalpy	H-H	C-H	C-C
ΔH0 kJ mol-1	436.4	414	350

Calculate bond enthalpy of C = C
Solution :

Bond enthalpy of C = C = ΔH⁰_C=C = ?
For the formation of propene, (CH₃ – CH = CH₂),

13.2 = [3 × 716 + 3 × 436.4] – [6 × 414 + 350 + \(\Delta H_{\mathrm{C}=\mathrm{C}}^{0}\) ]
= [2148 + 1309.2] – [2484 + 350 + \(\Delta H_{\mathrm{C}=\mathrm{C}}^{0}\) ]
= [3457.2] – [2834 + \(\Delta H_{\mathrm{C}=\mathrm{C}}^{0}\) ]
= 3457.2 – 2834 – \(\Delta H_{\mathrm{C}=\mathrm{C}}^{0}\)
\(\Delta H_{\mathrm{C}=\mathrm{C}}^{0}\) = 3457.2 – 2834 – 13.2
= 610 kJmol^-1
Ans. Bond enthalpy of C = C = \(\Delta H_{\mathrm{C}=\mathrm{C}}^{0}\)
= 610 kJ mol^-1

(21) Calculate the enthalpy of the reaction,
CH₃COOH_(g) + CH₃CH₂OH_(g) → CH₃COOCH₂CH_3(g) + H₂O_(g)
Bond enthalpies of O-H, C-O, in kJmol^-1 are 464, 351 respectively.
Solution :

In this reaction 1O-H and 1C-0 bond of the reactants are broken while 1C-0 and 1O-H bonds of the products are formed. Enthalpy of reaction,

Ans. Hence Enthalpy change for the reaction = Δ_rH⁰ = 0.

Question 68.
(1) What is a spontaneous process?
(2) What are its characteristics?
Answer:
(1) Spontaneous process : It is defined as a process that takes place on its own or without the intervention of the external agency or influence. For example, expansion of a gas or flow of a gas from higher pressure to low pressure or a flow of heat from higher temperature to lower temperature.

(2) Characteristics :

• It occurs on its own and doesn’t require external agency.
• It proceeds in one direction and can’t be completely reversed by external stimulant.
• These processes may be fast or slow.
• These processes proceed until an equilibrium is reached.

Question 69.
Give the examples of spontaneous processes.
Answer:
The examples of the spontaneous processes are as follows :

1. All natural processes are spontaneous.
2. A flow of gas from higher pressure to lower pressure.
3. Flow of water on its own from higher level to lower level.
4. Flow of heat from hotter body to colder body.
5. Acid-base neutralisation is a spontaneous reaction.

Question 70.
What is relation between spontaneity and energy of a system ?
Answer:
(1) The spontaneous process takes place in a direction in which energy of the system decreases. For example, neutralisation reaction between NaOH and HCl solution is exothermic with release of energy.

(2) The spontaneous process also takes place with the increase in energy by absorbing heat. For example,
(a) Melting of ice at 0 °C by absorption of heat
(b) Dissolution of NaCl,
NaCl_(s) + aq → NaCl_(aq) → Na⁺_(aq) + Cl^–_(aq)
ΔH⁰ = + 3.9 kJ mol^-1

Question 71.
Which of the following are spontaneous ?
(a) Dissolving sugar in hot coffee.
(b) Separation of Ar and Kr from their mixture.
(c) Spreading of fragrance when a bottle of perfume is opened.
(d) Flow of heat from cold object to hot object.
(e) Heat transfer from ice to room temperature at 25 °C.
Answer:
The spontaneous processes are :
(a) Dissolving sugar in hot coffee.
(b) Separation of Ar and Kr from their mixture.
(c) Spreading of fragrance when a bottle of perfume is opened.

The non-spontaneous processes are :
(d) Flow of heat from cold object to hot object.
(e) Heat transfer from ice to room temperature at 25 °C.

Question 72.
Explain : (a) Order in a system.
(b) Disorder in a system.
Answer:
(a) (i) When the atoms, molecules or ions constituting the system are arranged in a perfect order then the system is said to be in order. For example, in the solid state, the constituent atoms, molecules or ions are tightly placed at lattice points in the crystal lattice.
(ii) When solid melts forming a liquid or when a liquid vaporises, the constituents are separated and are in random motion imparting maximum disorder.
(iii) As energy of the system decreases order increases.

(b) Increase in entropy is a measure of disorder in the system. Consider following process :

Fig. 4.11 : Order decreases and disorder increases, Entropy increases

Question 73.
What is the change in order and entropy in the following :
(i) Dissolution of solid I₂ in water.
(ii) Dissociation of H_2(g) into atoms ?
Answer:
(i) For dissolution of solid I₂,

In the solid I2, there is ordered arrangement which collapses in solution increasing disorder and entropy, hence ΔS is positive.

(ii) In the dissociation of H_2(g)
H_2(g) → 2H_(g) (ΔS > 0)
In the molecular state, two H atoms in every molecule are together but in atomic state the disorder is increased with the increase in entropy and hence ΔS > 0.

Question 74.
How does addition of heat to a system at different temperatures changes disorder or ΔS ?
Answer:

• The amount of heat added to a system at higher temperature causes less disorder than when the heat is added at lower temperature.
• Since disorder depends on the temperature at which heat is added, ΔS relates reciprocally to temperature.
• This can also be explained from equation,
  ΔS = \(\frac{Q_{\text {rev }}}{T}\)

Question 75.
Explain the change in entropy for the following processes :
(i) 2H₂O_2(l) → 2H₂O_(l) + O_2(g)
(ii) 2H_2(g) + O_2(g) → 2H₂O_(l)
(iii) When ice melts at 0 °C and water vaporises at 100 °C.
Answer:
(i) In the following reaction,
2H₂O_2(l) → 2H₂O_(l) + O_2(g) ΔS = + 126 JK
Due to formation of O₂ gas from liquid, entropy increases.
(ii) In the reaction, entropy decreases due to formation of liquid H₂O from gaseous H₂ and O₂.
(iii) \(\text { Ice } \stackrel{0^{\circ} \mathrm{C}}{\longrightarrow} \text { water } \stackrel{100^{\circ} \mathrm{C}}{\longrightarrow} \text { vapour }\)
In these two steps, entropy increases due to increase in disorder from solid ice to liquid water and further to gaseous state.

Question 76.
How does entropy change in the following processes ? Explain.
(a) freezing of a liquid
(b) sublimation of a solid
(c) dissolving sugar in water
(d) condensation of vapour.
Answer:
(a) Freezing of a liquid results in decrease in randomness and disorder, hence entropy decreases, ΔS < 0.
(b) Sublimation of a solid converts it into vapour where the molecules or atoms are free to move randomly. Hence disorder increases accompanying increase in entropy, ΔS > 0.
(c) Dissolving sugar in water separates the molecules of sugar in the solution increasing disorder and entropy, ΔS > 0.
(d) Condensation of vapour decreases the disorder and randomness, hence entropy, ΔS < 0.

Question 77.
Predict the sign of ΔS in the following processes. Give reasons for your answer :
OR
Explain with reason sign conventions of ΔS in the following reactions.

Answer:
(a) N₂O_4(g) → 2NO_2(g)
Since 1 mole N₂O₄ on dissociation gives two moles of NO₂, the number of molecules increase, disorder increases hence entropy increases, ΔS > 0.

(b) Fe₂O_3(s) + 3H_2(g) → 2Fe_(s) + 3H₂O_(g)
In the reaction number of moles of gaseous reactants and products are same, hence ΔS = 0.

(c) N_2(g) + 3H_2(g) → 2NH_3(g)
In the reaction, 4 moles of gaseous reactants form 2 moles of gaseous products (Δn < 0). Therefore disorder decreases and hence entropy decreases, ΔS < 0.

(d) MgCO_3(s) → MgO_(s) + CO_2(g)
In this 1 mole of orderly solid MgCO₃ gives 1 mole of solid MgO and 1 mole of gaseous CO₂ (Δn > 0) with more disorder. Hence entropy increases, ΔS > 0.

(e) CO_2(g) → CO_2(s)
In this system from higher disorder in gaseous state changes to less disorder in the solid state, hence entropy decreases, ΔS < 0.

(f) Cl_2(g) → 2Cl_(g)
Since the dissociation of Cl₂ gas gives double Cl atoms, the number of atoms increases (Δn >0) increasing the disorder of the system. Hence ΔS > 0.

Question 78.
Identify which of the following pairs has larger entropy ? Why ?
(a) He_(g) in a volume of 1 L or He_(g) in a volume of 5 L both at 25 °C.
(b) O_2(g) at 1 atm or O_2(g) at 10 atm both at the same temperature.
(c) C₂H₅OH_(l) or C₂H₅OH_(g)
(d) 5 mol of Ne or 2 mol of Ne.
Answer:
(a) Atoms of He in 5 L at 25 °C occupy more volume than in 1 L. Hence, the randomness and disorder is more in 5 L. Expansion of a gas always increases its entropy. Therefore He_(g) in 5L will have larger entropy.

(b) O_2(g) at 1 atm will occupy more volume than O_2(g) at 10 atm at the same temperature. Hence at 1 atm O_2(g) will have higher disorder and hence higher entropy.

(c) The molecules of gaseous C₂H₅OH_(g) will have more disorder and randomness due to free motion of molecules than C₂H₅OH_(l). Hence entropy of C₂H₅OH_(g) will be larger.

(d) 5 mol Ne will contain more Ne atoms than 2 mol Ne. Hence disorder in 5 mol will be more. Therefore 5 mol Ne will have larger entropy.

Question 79.
Mention entropy change (ΔS) for :
(i) spontaneous process
(ii) nonspontaneous process
(iii) at equilibrium.
Answer:
(a) ΔS_total > 0, the process is spontaneous
(b) ΔS_total < 0, the process is non-spontaneous
(c) ΔS_total = 0, the process is at equilibrium.

Question 80.
Define Gibbs free energy and change in free energy. What are the units of Gibbs free energy ?
OR
Derive the relation between ΔG and ΔS Total.
Answer:
(i) Gibbs free energy, G is defined as,
G = H – TS
where H is the enthalpy, S is the entropy of the system at absolute temperature T.
Since H, T and S are state functions, G is a state function and a thermodynamic function.

(ii) At constant temperature and pressure, change in free energy ΔG for the system is represented as, ΔG = ΔH – TΔS

This is called Gibbs free energy equation for ΔG. In this ΔS is total entropy change, i.e., ΔS_Total.

(iii) The SI units of ΔG are J or kJ (or Jmol^-1 or kJmol^-1).
The c.g.s. units of ΔG are cal or kcal (or cal mol^-1 or kcal mol^-1.)

Question 81.
Explain Gibbs free energy and spontaneity of the process.
Answer:
The total entropy change for a system and its surroundings accompanying a process is given by,
ΔS_Total = ΔS_system + ΔS_surr
By second law, for a spontaneous process,
ΔS_Total > 0. If + ΔH is the enthalpy change (or enthalpy increase) for the process, or a reaction at constant temperature (T) and pressure, then enthalpy change (or enthalpy decrease) for the surroundings will be -ΔH.

By Gibbs equation,
ΔG = ΔH – TΔS
By comparing above two equations,
∴ ΔG = -TΔS_Total
As ΔS_Total increases, ΔG decreases.
For a spontaneous process, ΔS_Total > 0
which is according to second law of thermodynamics.
∴ ΔG < 0.
Hence in a spontaneous process, Gibbs free energy decreases (ΔG < 0) while entropy increases (ΔS > 0).
Therefore for a non-spontaneous process Gibbs free energy increases (Δ G > 0).
It can be concluded that for a process at equilibrium, ΔG=0.
Hence,

• For the spontaneous process, Δ G < 0
• For the non-spontaneous process, Δ G > 0
• For the process at equilibrium, Δ G = 0.

Question 82.
How does second law of thermodynamics explain the conditions of spontaneity ?
Answer:
The second law explains the conditions of spontaneity as below :
(i) ΔS_total  > 0 and ΔG < 0, the process is spontaneous.
(ii) ΔS_total  < 0 and ΔG > 0, the process is nonspontaneous.
(iii) ΔS_total = 0 and ΔG = 0, the process is at equilibrium.

Question 83.
Discuss the factors, ΔH, ΔS and ΔG for spontaneous and non-spontaneous processes.
OR
What can be said about the spontaneity of reactions when (1) ΔH and ΔS are both positive (2) ΔH and ΔS are both negative (3) ΔH is positive and ΔS is negative (4) ΔH is negative and ΔS is positive.
Answer:
For a spontaneous or a non-spontaneous process, ΔH and ΔS may be positive or negative (ΔH < 0 or ΔH > 0; ΔS < 0 or ΔS > 0). But ΔG must decrease, i.e., ΔG < 0. If ΔG > 0, the process or a reaction will definitely be non-spontaneous. This can be explained by Gibbs equation, ΔG = ΔH – TΔS.
(1) If ΔH and ΔS are both negative, then ΔG will be negative only when TΔS < ΔH or when temperature T is low. Such reactions must be carried out at low temperatures.
(2) If ΔH and ΔS are both positive then ΔG will be negative if, TΔS > ΔH; such reactions must be carried out at high temperature.
(3) If ΔH is negative (ΔH < 0) and ΔS is positive (ΔS > 0) then for all temperatures ΔG will be negative and the reaction will be spontaneous. But as temperature increase, ΔG will be more negative, hence the reaction will be more spontaneous at higher temperature.
(4) If ΔH is positive, (ΔH > 0) and ΔS is negative (ΔS < 0), ΔG will be always positive (ΔG > 0) and hence the reaction will be non-spontaneous at all temperatures.

This can be summarised in the following table :

Question 84.
Obtain a temperature condition for equilibrium.
Answer:
For a system at equilibrium, free energy change ΔG is,
ΔG = ΔH – TΔS
where ΔH is enthalpy change, ΔS is entropy change at temperature, T. Since ΔG = 0 at equilibrium,
O = ΔH – TΔS
∴ TΔS = ΔH
OR T = \(\frac{\Delta H}{\Delta S}\)
Hence at temperature T, changeover between forward spontaneous step and backward non-spontaneous step occurs and the system attains an equilibrium.
Here ΔH and ΔS are assumed to be independent of temperature.

Question 85.
Predict the signs of ΔH, ΔS and ΔG of the system when a solid melts at 1 atmosphere and at (a) -55 °C (b) -95 °C (c) -77 °C, if the normal melting point of the solid is -77 °C.
Answer:
Since the normal melting point of the solid is -77°C, to melt the solid at any temperature other than at -77 °C, the pressure is required to be changed. During the phase change, the system will be at equilibrium, hence Δ G = 0.
(a) In case a solid at -55 °C, the temperature should be decreased (ΔH < 0, ΔS < 0) to -77 °C and then it will melt, so ΔH > 0, ΔS > 0, ΔG = 0.
(b) In case of a solid at -95 °C, it represents supercooled system and the temperature should be raised to -77 °C (ΔH > 0, ΔS > 0) and then it will melt so ΔH > 0, ΔS > 0, ΔG = 0.
(c) At -77 °C, solid will melt, solid and liquid will be at equilibrium. Melting involves absorption of heat,

ΔH >0, ΔS > 0, ΔG = 0.

Solved Examples 4.11

Question 86.
Solve the following :

(1) In an isothermal reversible process, 6 kJ heat is absorbed at 27 °C. Calculate the entropy change.
Solution :
Given : Temperature = T = 273 + 21 = 300 K
Heat absorbed = Q_rev = 6 kJ = 6000 J
Entropy change = ΔS = ?
ΔS = \(\frac{Q_{\text {rev }}}{T}=\frac{6000}{300}\) = 20 JK^-1
Ans. Entropy change = ΔS = 20 JK^-1

(2) The latent heat of evaporation of water is 2.26 kJ g^-1 at 1 atm and 100 °C. Calculate the entropy change for evaporation of 1 mole of water at 100 °C.
Solution :
Given : Latent heat of evaporation = Δ_vapH⁰
= 2.26 kJ g^-1
Temperature = T = 273 + 100 = 373 K
Molar mass of water = 18 g mol^-1
ΔS = ?
For 1g H₂O_(l) Δ_vapH⁰ = 2.26 KJ
∴ For 1 mol H₂O_(l) = 18 g H₂O_(l)
Δ_vapH⁰ = 2.26 × 18
= 40.68 kJ
= 40680 J
Entropy change, ΔS is given by,
ΔS = \(\frac{\Delta_{\mathrm{vap}} H^{0}}{T}=\frac{40680}{373}\) = 109.06 JK^-1 mol^-1
Ans. Entropy change = ΔS = 109.06 JK^-1 mol^-1

(3) Calculate the standard (absolute) entropy change for the formation of CO_2(g).

Substance	C(graphite)	O2(g)	CO2(g)
Standard molar enthalpy JK-1 mol-1	5.74	205	213.7

Solution:
Given:

Substance	C(graphite)	O2(g)	CO2(g)
Standard molar enthalpy S0 JK-1 mol-1	5.74	205	213.7

For the formation of CO_2(g),

(4) The standard entropies of H_2(g), O_2(g) and H₂O_(g) in JK^-1 mol^-1 are 130, 205 and 189 respectively. The heat of formation of H₂O_(g) is -242 kJ mol^-1. Calculate ΔS for formation of H₂O_(g), for the surroundings and the universe at 298 K. Mention whether the reaction is spontaneous or non-spontaneous.
Solution :
Given :

Substance	H2(g)	O2(g)	H2O(g)
Standard entropy S0 JK-1 mol-1	130	205	189

Δ_fH⁰ = -242 kJ mol^-1
ΔS_universe = ?, ΔS_surr = ?
Thermochemical equation for the formation of H₂O_(g)
H_2(g) + \(\frac {1}{2}\)O_2(g) → H₂O_(g)
ΔS⁰ = [S⁰_H2O] – [H⁰_H2 + \(\frac {1}{2}\) H⁰_O2]
= 189 – [130 + \(\frac {1}{2}\)(205)]
= 189 – [232.5]
= -43.5 JK^-1 mol^-1
Hence, ΔSsystem = -43.5 JK-1 mol^-1
Since for the formation of H₂O_(g)
Δ_fH⁰ = -242 kJmol^-1 = -242 × 10³ Jmol^-1, the reaction is exothermic. Hence the surroundings gains heat energy 242 × 10³J. Therefore entropy of the system decreases while entropy of surroundings increases.

Hence, ΔS_sys < 0 but ΔS_universe > 0, hence the reaction is spontaneous.
Ans. ΔS_H2O(g) = -43.5 JK^-1 mol^-1
ΔS_surr = 813 JK^-1 mol^-1
ΔS_universe = 769.5 JK^-1.

(5) Calculate ΔS_Total and hence show whether the following reaction is spontaneous at 25 °C.
Hg_(s) + O_2(g) → Hg_(l) + SO_2(g) ΔH⁰ = – 238.6 kJ
ΔS⁰ = +36.7 JK^-1
Solution :
Given : Hg_(s) + O_2(g) → Hg_(l) + SO_2(g)
Δ_rH⁰ = – 238.6 kJ
ΔS⁰ = +36.7 JK^-1
T = 273 + 25 = 298 K
ΔS_Total = ?
ΔS_Total = ΔS_sys + Δ S_surr
Now, ΔS_sys = 36.7 JK^-1
Since the reaction is exothermic, system loses heat to surroundings. Hence the entropy of the surroundings increases.
ΔH_surr = + 238.6 kJ = 238600 J

∵ ΔS_Total > 0, the reaction is spontaneous.
Ans. ΔS_Total = 837.4 JK^-1
The reaction is spontaneous.

(6) What is the value of AS_surr for the following reaction at 298 K ?
6CO_2(g) + 6H₂O_(l) → C₆H₁₂O_6(s) + 6O_2(g),
ΔG⁰ = 2879 kJ mol^-1, ΔS⁰ = -210 JK^-1 mol^-1.
Solution :
Given :
6CO_2(g) + 6H₂O_(l) → C₆H₁₂O_6(s) + 6O_2(g),
ΔG⁰ = 2879 kJmol^-1;
ΔS⁰ = -210 JK^-1mol^-1 = -0.210 kJ K^-1 mol^-1
T = 298 K ΔH⁰ = ?
ΔG⁰ = ΔH⁰ – TΔS⁰
∴ ΔH⁰ = ΔG⁰ + TΔS⁰
= 2879 + 298(-0.210)
= 2879 – 62.58
= 2816.42 kJ mol^-1
Since ΔH⁰ > 0, the reaction is endothermic, and system absorbs heat from surroundings. Hence surroundings loses heat,

Ans. ΔS⁰surr = -9.45 kJ K^-1

(7) Calculate ΔS_surr when on mole of methanol (CH₃OH) is formed from its elements under standard conditions if Δ_fH⁰(CH₃OH) = -238.9 J mol^-1.
Solution :
Given : Number of moles of ethanol,
(C₂H₅OH) = n = 1 mol
Δ_fH⁰(CH₃OH) = -238.9 kJ mol^-1
= -238.9 × 10³J mol^-1
Temperature = T = 298 K
ΔS = ?
ΔS_surr = ?
Since Δ_fH⁰ is negative, the reaction for the formation of one mole of C₂H₅OH is exothermic.
As heat is released to the surroundings,
ΔH⁰_surr = + 238.9 kJ mol^-1
∴ ΔS_surr = \(\frac{\Delta H_{\text {surr }}^{0}}{T}=\frac{+238.9 \times 10^{3}}{298}\)
= +801.7 JK^-1
Thus entropy of the surroundings increases.
Ans. ΔS_surr = +801.7 JK^-1

(8) What is the value of ΔSsurr for the following reaction at 298 K –
6CO_2(g) + 6H₂O_(l) → C₆H₁₂O_6(s) + 6O_2(g)
Given that: ΔG° = 2879 KJ mol^-1
ΔS⁰ = -210 J K^-1 mol^-1
Solution :
Given : ΔG⁰ = 2879 KJ mol^-1 = 2879 × 10³ J mol^-1
ΔS⁰ = -210 JK^-1 mol^-1
ΔS_surr = ?
ΔG⁰ = ΔH⁰ – TΔS⁰
∴ ΔH⁰ = ΔG⁰ + TΔS⁰
= 2879 × 10³ + 298 × (- 210)
= 2879 × 10³ – 62580
= 2816420 J
Since, for a system, ΔH⁰ is +2816420 J, the surrounding loses heat to system,
∴ ΔH⁰_surr = – 2816420 J
∴ ΔS⁰_surr = \(\frac{\Delta H_{\text {surr }}^{0}}{T}\)
= \(\frac{-2816420}{298}\)
= -9451 JK^-1
= -9.451 kJ K^-1
Ans. ΔS_surr = -9.451 kJ K^-1

(9) Determine whether the reactions with the following ΔH and ΔS values are spontaneous or non-spontaneous. State whether they are exothermic or endothermic.
(a) ΔH = -110 kJ and ΔS = +40 JK^-1 at 400 K
(b) ΔH = +50 kJ and ΔS = -130 JK^-1 at 250 K.
Solution :
(a) Given : ΔH= -110 kJ ΔS = 40 JK^-1 = 0.04 kJK^-1
Temperature = T = 400 K ΔG = ?
Since ΔH is negative, the reaction is exothermic
ΔG = ΔH – TΔS
= -110 – 400 × 0.04
= -110 – 16
= -126 kJ
Since ΔG is negative, the reaction is spontaneous.

(b) Given : ΔH=50 kJ,
ΔS= -130 JK^-1 = -0.13 kJ K^-1
Temperature = T = 250 K
ΔG = ?
Since ΔH is positive, the reaction is endothermic.
ΔG = ΔH – TΔS
= 50 – 250 × (-0.13)
= 50 + 32.5
= 82.5 kJ
Since ΔG > 0, the reaction is non-spontaneous.
Ans. (a) ΔG = -126 kJ; The reaction is exothermic and spontaneous.
(b) ΔG = 82.5 kJ; The reaction is endothermic and non-spontaneous.

(10) For a certain reaction, ΔH⁰ = -224 kJ and ΔS⁰ = -153 JK^-1. At what temperature will it change from spontaneous to non-spontaneous ?
Solution :
Given : ΔH⁰ = – 224 kJ = – 224000 J
ΔS⁰ = – 153 JK^-1
Temperature (T) at which, reaction changes from spontaneous to non-spontaneous = ?
Find the temperature at equilibrium, where ΔG⁰ = 0
ΔG⁰ = ΔH⁰ – TΔS⁰
0 = ΔH⁰ – TΔS⁰
∴ TΔS⁰ = ΔH⁰
∴ T = \(\frac{\Delta H^{0}}{\Delta S^{0}}\)
= \(\frac{224000}{153}\)
= 1464 K.
Hence reaction will be spontaneous below 1464 K. It will be at equilibrium at 1464 K and non-spontaneous above 1464 K.
Ans. Change over temperature from spontaneous to non-spontaneous = 1464 K.

(11) Determine whether the reactions with the following ΔH and ΔS values are spontaneous or non-spontaneous. State whether the reactions are exothermic or endothermic.
(a) ΔH = -110 kJ, ΔS = +40 JK^-1 at 400 K
(b) ΔH = + 40 kJ, ΔS = – 120 JK^-1 at 250 K
Solution :
(a) Given : ΔH = -110 kJ, ΔS = +40 JK^-1 at T = 400K
ΔG = ΔH – TΔS

= -110 – 16
= -126 kJ
Since ΔG is negative, the reaction is spontaneous.
Since ΔH is negative, the reaction is exothermic.

(b) Given : ΔH = + 40 kJ, ΔS = -120 JK^-1 at T = 250 K
ΔG = ΔH – TΔS

= 40 + 30
= 70 kJ
Since ΔG is negative, the reaction is spontaneous.
Since ΔH is negative, the reaction is exothermic.

(12) Determine whether the following reaction will be spontaneous or non-spontaneous under standard conditions.
Zn_(s) + Cu²⁺ → Zn²⁺ +Cu_(s) ΔH⁰ = -219 kJ, ΔS⁰ = -21 JK^-1
Solution :
Given : ΔH⁰ = -219 kJ;
ΔS⁰ = -21 JK^-1= 0.021 kJ K^-1
ΔG⁰ = ?
For standard conditions : Pressure = 1 atm
Temperature = T = 298 K
ΔG⁰ = ΔH⁰ – TΔS⁰
= -219 – 298 × (-0.021)
= -219 + 6.258
= -212.742 kJ
Since ΔG < 0, the reaction is spontaneous.
Ans. The reaction is spontaneous.

(13) The equilibrium constant for a gaseous reversible reaction at 200 °C is 1.64 × 10³ atm². Calculate ΔG° for the reaction.
Solution :
Given : Equilibrium constant = KP = 1.64 × 10³ atm²
Temperature = T = 273 + 200 = 473 K
ΔG⁰ = ?
ΔG⁰ = -2.303 RTlog₁₀ K_p
= – 2.303 × 8.314 × 473 × log₁₀ 1.64 × 10³
= – 2.303 × 8.314 × 473 × (3.2148)
= -29115 J
= -29.115 kJ
Ans. ΔG⁰ = -29.115 kJ

(14) Calculate ΔG for the reaction at 25°C
CO_(g) + 2H_2(g) ⇌ CH₃OH_(g)
ΔG⁰ = -24.8 kJ mol^-1.
if P_co = 4 atm, P_H2 = 2 atm, P_CH3OH = 2 atm.
Solution :
Given : Partial pressures : pco = 4 atm,
P_H2 = 2 atm,
P_CH3OH = 2 atm
Temperature = T = 273 + 25 = 298 K
ΔG⁰ = -24.8 kJ mol^-1
CO_(g) + 2H_2(g) ⇌ CH₃OH_(g)
The reaction quotient, Q is,

= – 24.8 + 2.303 × 8.314 × 298 × log₁₀ 0.125
= – 24.8 + 2.303 × 8.314 × 298 × (\(\overline{1} \cdot 09691\))
= – 24.8 + 2.303 × 8.314 × 298 × (- 0.90709)
= – 24.8 – 2.303 × 8.314 × 298 × 0.90709 × 10^-3
= -24.8 – 5.176
= -29.976 kJ mol^-1.
Ans. ΔG = – 29.976 kJ mol^-1

(15) Calculate K_P for the reaction,
C₂H_4(g) + H_2(g) ⇌ C₂H_6(g),
ΔG⁰ = -100 kJ mol^-1, at 25°C.
Solution :
Given : ΔG⁰ = – 100 kJ mol^-1 = – 100 × 10³ J mol^-1
= -1 × 10⁵ Jmol^-1
Temperature = T = 273 + 25 = 298 K
Equilibrium constant = K_P = ?

(16) K_P for the reaction,
MgCO_3(s) → MgO_(s) + CO_2(g) is 9 × 10^-10.
Calculate ΔG⁰ for the reaction at 25 °C.
Solution :
Given : K_P = 9 Δ 10^-10
Temperature = T = 273 + 25 = 298 K
ΔG⁰ = ?
ΔG⁰ = -2.303 RTlog₁₀K_P
= -2.303 × 8.314 × 298 × log109 × 10^-10
= -2.303 × 8.314 × 298 × \([\overline{10} \cdot 9542]\)
= – 2.303 × 8.314 × 298 × [ – 9.0458]
= 51683 Jmol^-1
= 51.683 kJmol^-1
Ans. ΔG⁰ = 51.653 kJ mol^-1

(17) Calculate ΔH, ΔS and ΔG for melting of 10 g ice at 0 °C and 1 atm. (Δ_fusH⁰ = 6.02 kJ mol^-1 for ice)
Solution :
Given : Δ_fusH⁰ = 6.02 kJ mol^-1 = 6.02 × 10³ Jmol^-1
Temperature = T = 273 + 0 = 273 K
Mass of ice = 10 g
Molar mass of H₂O = 18 g mol^-1
ΔH= ?, ΔS = ?, ΔG = ?
For melting of ice,
H₂O_(s) ⇌ H₂O_(l)
For 1 mol ice = 18 g ice Δ_fusionH = 6.05 kJ
∴ For 10 g ice
ΔH = \(\frac{6.02 \times 10}{18}\)
= 3.344 kJ
ΔH = 3.344 kJ = 3.344 × 10³ J
∴ ΔS = \(\frac{\Delta H}{T}=\frac{3.344 \times 10^{3}}{273}\) = 12.25 JK^-1
ΔG = ΔH – TΔS
= 3.344 – 273 × 12.25 ×10^-3 kJ
= 3.344 – 3.344
= 0
Since ΔG = 0, the system is at equilibrium.
Ans. ΔH = 3.344 kJ; ΔS = 12.25 JK^-1; ΔG = 0

(18) Calculate Kp, ΔG⁰ for the reaction,
C_(s) + H₂O_(g) ⇌ CO_(g) + H_2(g)
at 990 K if the equilibrium concentrations are as follows :
[H₂O] = 1.10 mol dm-^-3,
[CO] = [H₂] = 0.2 mol dm^-3,
R = 0.08206 L atm K^-1 mol^-1.
Solution :
Given : [H₂O] = 1.1 mol dm^-3,
[CO] = 0.2 mol dm^-3,
[H₂] = 0.2 mol dm^-3, T = 990 K,
R = 0.08206 L atm K^-1 mol^-1
K_P = ? ΔG⁰ = ?
C_(s) + H₂O_(g) ⇌ CO_(g) + H_2(g)

K_P = K_C × (RT)^Δn
= 0.03636 × (0.08206 × 990)
= 2.954 atm
ΔG⁰ = -2.303 RTlog₁₀K_P
= -2.303 × 8.314 × 990 × log₁₀ 2.954
= -2.303 × 8.314 × 990 × 0.4704
= -8917 J
= -8.917 kJ
Ans. K_P = 2.954 atm; ΔG⁰ = -8.917 kJ

Multiple Choice Questions

Question 87.
Select and write the most appropriate answer from the given alternatives for each subquestion :

1. For an isochoric process, the change in
(a) pressure is zero
(b) volume is negative
(c) volume is zero
(d) temperature is zero
Answer:
(c) volume is zero

2. Which of the following is an extensive property ?
(a) Surface tension
(b) Refractive index
(c) Energy
(d) Temperature
Answer:
(c) Energy

3. Which of the following is an intensive property ?
(a) Enthalpy
(b) Weight
(c) Refractive index
(d) Volume
Answer:
(c) Refractive index

4. Which of the following pairs is an intensive property ?
(a) Density, viscosity
(b) Surface tension, mass
(c) Viscosity, internal energy
(d) Heat capacity, volume
Answer:
(a) Density, viscosity

5. The property which is not intensive is
(a) freezing point
(b) viscosity
(c) temperature
(d) free energy
Answer:
(d) free energy

6. Which of the following is not an extensive property ?
(a) molarity
(b) molar heat capacity
(c) mass
(d) volume
Answer:
(b) molar heat capacity

7. Which of the following is NOT a state function ?
(a) Work
(b) Enthalpy
(c) Temperature
(d) Pressure
Answer:
(a) Work

8. In an adiabatic process
(a) ΔT ≠ 0
(b) ΔU ≠ 0
(c) Q = 0
(d) All of these
Answer:
(d) All of these

9. For an isothermal and reversible process
(a) P₁V₁ = P₂V₂
(b) P₁V₁ ≠ P₂V₂
(c) ΔV ≠ 0
(d) ΔH ≠ 0
Answer:
(a) P₁V₁ = P₂V₂

10. For the process to occur under adiabatic conditions, the correct condition is :
(a) ΔT = 0
(b) Δp = 0
(c) Q = 0
(d) W = 0
Answer:
(c) Q = 0

11. What is true for an adiabatic process ?
(a) ΔT = 0
(b) ΔU
(c) ΔH = ΔU
(d) Q = 0
Answer:
(d) Q = 0

12. ΔU = 0 is true for
(a) Adiabatic process
(b) Isothermal process
(c) Isobaric process
(d) Isochoric process
Answer:
(b) Isothermal process

13. When a gas expands in vacuum, the work done by the gas is
(a) maximum
(b) zero
(c) less than zero
(d) greater than zero
Answer:
(b) zero

14. When a sample of an ideal gas is allowed to expand at constant temperature against an atomospheric pressure,
(a) surroundings does work on the system
(b) ΔU = 0
(c) no heat exchange takes place between the system and surroundings
(d) internal energy of the system increases
Answer:
(b) ΔU = 0

15. In what reaction of the following work is done by the system on the surroundings ?
(a) Hg_(l) → Hg_(g)
(b) 3O_2(g) → 2O_3(g)
(c) H_2(g) + Cl_2(g) → 2HCl_(g)
(d) N_2(g) + 3H_2(g) → 2NH_3(g)
Answer:
(a) Hg_(l) → Hg_(g)

16. A gas does 0.320 kJ of work on its surroundings and absorbs 120 J of heat from the surroundings. Hence, ΔU is
(a) 440 kJ
(b) 200 J
(c) 120.32 J
(d) -200J
Answer:
(d) -200J

17. For an isothermal and reversible expansion of 0.5 mol of an ideal gas, Wmax is -3.918 kJ. The value of ΔU is
(a) 3.918 kJ
(b) zero
(c) 1.959 kJ
(d) 3918 J
Answer:
(b) zero

18. The mathematical expression of the first law of thermodynamics for an adiabatic process is
(a) W = Q
(b) W = -ΔU
(c) W = +ΔU
(d ) W = -Q
Answer:
(c) W = +ΔU

19. A gaseous system absorbs 600 kJ of heat and performs the work of expansion equal to 130 kJ. The internal energy change is
(a) 730 kJ
(b) -470 kJ
(c) -730 kJ
(d) 470 kJ
Answer:
(d) 470 kJ

20. When a gas is compressed, the work obtained is 360 J while the heat transferred is 190 J. Hence the change in internal energy is
(a) -170 J
(b) 170 J
(c) 550 J
(d) -550 J
Answer:
(b) 170 J

21. For the reaction N_2(g) + 3H_2(g) = 2NH_3(g); Which of the following is valid ?
(a) ΔH = ΔU
(b) ΔH < ΔU
(c) ΔH > ΔU
(d) ΔH = 2ΔH
Answer:
(b) ΔH < ΔU

22. For which reaction ΔH = ΔU ?

Answer:
(b) \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6(\mathrm{~s})}+6 \mathrm{O}_{2(\mathrm{~g})} \rightarrow 6 \mathrm{CO}_{2(\mathrm{~g})}+6 \mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})}\)

23. For the following reaction at 298 K
H_2(g) + \(\frac {1}{8}\)O_2(g) = H₂O_(l)
Which of the following alternative is correct ?
(a) ΔH = ΔU
(b) ΔH > ΔU
(c) ΔH < ΔU
(d) ΔH = 1.5 ΔU
Answer:
(c) ΔH < ΔU

24. The heat of combustion of carbon is 394 kJ mol^-1. The heat evolved in combustion of 6.023 × 10²¹ atoms of carbon is
(a) 3940 kJ
(b) 3940.0 kJ
(c) 3.94 kJ
(d) 0.394 kJ
Answer:
(c) 3.94 kJ

25. Which of the reactions defines the heat of formation ?

Answer:
(d) \(\frac{1}{2} \mathrm{H}_{2(\mathrm{~g})}+\frac{1}{2} \mathrm{Cl}_{2(\mathrm{~g})} \rightarrow \mathrm{HCl}_{(\mathrm{g})}\)

26. ΔU^o of combustion of methane is -X kJ mol^-1. The value of ΔH⁰ is
(a) = ΔU^o
(b) > ΔU^o
(c) < ΔU^o
(d) =0
Answer:
(c) < ΔU^o

27. The enthalpy of combustion of 5(rhomic)is -297.4 kJ mol^-1. The amount of sulphur required to produce 29.74 kJ of heat is
(a) 32 × 10^-2 kg
(b) 3.2 × 10^-3 kg
(c) 3.2 × 10^-2 kg
(d) 6.4 × 10^-3 kg
Answer:
(b) 3.2 × 10^-3 kg

28. The heat of formation of SO_2(g) and SO_3(g) are -269 kJ mol^-1 and -395 kJ mol^-1 respectively the value of ΔH for the reaction
SO_2(g) + \(\frac {1}{2}\)O_2(g) → SO_3(g) is
(a) -664 kJ mol^-1
(b) -126 kJ mol^-1
(c) 63 kJ mol^-1
(d) 126 kJ mol^-1
Answer:
(b) -126 kJ mol^-1

29. The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, -890.3 kJ mol^-1, -393.5 kJ mol^-1 and -285.8 kJ mol^-1 respectively. Enthalpy of formation of CH_4(g) will be
(a) – 74.8 kJ mol^-1
(b) – 52.27 kJ mol^-1
(c) + 74.8 kJ mol^-1
(d) + 52.26 kJ mol^-1
Answer:
(a) – 74.8 kJ mol^-1

30. The enthalpies of formation of N₂O_(g) and NO_(g) are 82 kJ mol^-1 and 90 kJ mol^-1 respectively. Then enthalpy of a reaction 2N₂O_(g) + O_2(g) → 4NO_(g) is …………
(a) 8 kJ
(b) -16 kJ
(c) 88 kJ
(d) 196 kJ
Answer:
(d) 196 kJ

31. The heat of combustion of naphthalene (C₁₀H₈) to CO₂ gas and water vapour at 298 K and at constant pressure is -5.1567 × 10⁶ J. The heat of combustion at constant volume at 298 K is (R = 8.314 JK^-1 mol^-1)
(a) -5.1567 × 10⁶ J
(b) -5.6161 × 10⁶ J
(c) -5.1616 × 10⁶ J
(d) -5.7161 × 10⁶ J
Answer:
(c) -5.1616 × 10⁶ J

32. Given the reaction,
2NH_3(g) → N_2(g) + 3H_2(g) ΔH = 92.6 kJ
The enthalpy of formation of NH₃ is
(a) -92.6 kJ
(b) 92.6 kJ mol^-1
(c) -46.3 kJmol^-1
(d) -185.2 kJmol^-1
Answer:
(c) -46.3 kJmol^-1

33. Calculate the heat of reaction at 298 K for the reaction C₂H_4(g) + H_2(g) = C₂H_6(g)
Given that the heats of combustion of ethylene, hydrogen and ethane are 337.0, 68.4 and 373.0 kcal respectively.
(a) 23.4 kcal
(b) 62.2 kcal
(c) 32.4 kcal
(d) 34.2 kcal
Answer:
(c) 32.4 kcal

34. Entropy change for a process is given by,
(a) Q_rev × T
(b) Q_rev/T
(c) \(\frac{T}{Q_{\text {rev }}}\)
(d) ΔH_rev × T
Answer:
(b) Q_rev/T

35. For a spontaneous process, total entropy change for a system and its surroundings is
(a) ΔS_total < 0
(b) ΔS_total = 0
(c) ΔS_total > 0
(d) ΔS_total ≤ 0
Answer:
(c) ΔS_total > 0

36. For a system at equilibrium,
(a) ΔS_total = 0
(b) ΔS_total > 0
(c) ΔS_total < 0
(d) ΔS_total ≥ 0
Answer:
(a) ΔS_total = 0

37. If the enthalpy of vaporisation of water at 100°C is 186.5 J·mol^-1, the entropy of vaporization will be-
(a) 4.0 J·K^-1 mol^-1
(b) 3.0 J·K^-1 mol^-1
(c) 1.5 J·K^-1 mol^-1
(d) 0.5 J·K^-1 mol^-1
Answer:
(d) 0.5 J·K^-1 mol^-1

38. Heat of fusion of ice is 6.02kJmol-1 at 0 °C. If 100 g water is frozen at 0 °C, entropy change will be
(a) -0.1225 JK^-1
(b) 310.6 JK^-1
(c) -122.6 JK^-1
(d) 92.8 JK^-1
Answer:
(c) -122.6 JK^-1

39. If for a reaction ΔH is negative and ΔS is positive then the reaction is
(a) spontaneous at all temperatures
(b) non-spontaneous at all temperatures
(c) spontaneous only at high temperatures
(d) spontaneous only at low temperature
Answer:
(a) spontaneous at all temperatures

40. The relationship between ΔG^o of a reaction and its equilibrium constant is

Answer:
(c) \(\frac{R T \ln K}{\Delta G^{0}}=-1\)

41. Which of the following has highest entropy?
(a) Al_(s)
(b) CaCO_3(s)
(c) H₂O_(l)
(d) CO_2(g)
Answer:
(d) CO_2(g)

42. The entropy change for the formation of 3.5 mol NO_(g) from the following data will be,

Answer:
(b) 42.875 JK^-1

43. Gibbs free energy change at equilibrium is
(a) ΔG = 0
(b) ΔG > 0
(c) ΔG < 0
(d) ΔG ≤ 0
Answer:
(a) ΔG = 0

44. For spontaneous process,
(a) ΔG = 0
(b) ΔG > 0
(c) ΔG < 0
(d) ΔG ≤ 0
Answer:
(c) ΔG < 0

45. A substance which shows highest entropy is
(a) SrCO_3(S)
(b) Cu_(S)
(c) NaC_(aq)
(d) Cl_2(g)
Answer:
(d) Cl_2(g)

46. For which of the following reactions ΔS is negative ?

Answer:
(a) \(\mathrm{Mg}_{(s)}+\mathrm{Cl}_{2(\mathrm{~g})} \rightarrow \mathrm{MgCl}_{2(s)}\)

47. For a reaction, at 300K enthalpy is 138 kJ and entropy change is 115 JK^-1. Hence the free energy change of the reaction is
(a) 130.5 kJ
(b) 103.5 kJ
(c) 82.8 kJ
(d) – 60.5 kJ
Answer:
(b) 103.5 kJ

48. Bond enthalpies of H_2^-1, I_2(g) and HI are 436, 151 and 298 kJ mol^-1 respectively. Hence enthalpy of formation of HI_(g) is
(a) -9 kJmol^-1
(b) -4.5kJmol^-1
(c) 4.5 kJ mol^-1
(d) 9 kJ mol^-1
Answer:
(b) -4.5kJmol^-1

49. The average bond energy of C-H bond is 410 kJmol^-1. The enthalpy change of atomisation of 3.2 g CH_4(g) is
(a) 1312 kJ
(b) 29.8 kJ
(c) 328 kJ
(d) 120 kJ
Answer:
(c) 328 kJ

50. For a chemical reaction ΔS = -0.035 kJ/K and ΔH = -20 kJ. At what temperature does the reaction turn non-spontaneous ?
(a) 5.14 K
(b) 57.14 K
(c) 571.4 K
(d) 5714.0 K
Answer:
(c) 571.4 K

51. For a certain reaction, ΔH = -50 kJ and ΔS = -80 JK^-1, at what temperature does the reaction turn from spontaneous to non-spontaneous.
(a) 6.25 K
(b) 62.5 K
(c) 625 K
(d) 6250 K
Answer:
(c) 625 K