Thermodynamics — Competition Zone

The volume (V) of a monoatomic gas varies with its temperature (T), as shown in the graph. The ratio of work done by the gas, to the heat absorbed by it, when it undergoes a change from state A to state B, is :

1. $\dfrac{2}{5}$

2. $\dfrac{2}{7}$

3. $\dfrac{1}{3}$

4. $\dfrac{2}{3}$

Answer

$\dfrac{2}{5}$

Reason —

From the graph the volume V varies linearly with the temperature T and the straight line passes through the origin, so

$\text V \propto \text T$

This means that the pressure of the gas remains constant, that is, the process from A to B is isobaric.

Heat absorbed by the gas : For an isobaric process,

$\text{dQ} = \mu\text C_p\text{dT} = \mu\left(\dfrac{5}{2}\text R\right)\text{dT}$

since the gas is monoatomic, for which $\text C_p = \dfrac{5}{2}\text R$.

Work done by the gas : For an isobaric process,

$\text{dW} = \text P\text{dV} = \mu\text R\text{dT}$

Therefore the required ratio is

$\dfrac{\text{dW}}{\text{dQ}} = \dfrac{\mu\text R\text{dT}}{\mu\left(\dfrac{5}{2}\text R\right)\text{dT}} = \dfrac{2}{5}$

A sample of an ideal gas is taken through the cyclic process abca as shown in the figure. The change in the internal energy of the gas along the path ca is −180 J. The gas absorbs 250 J of heat along the path ab and 60 J along the path bc. The work done by the gas along the path abc is :

1. 120 J
2. 130 J
3. 100 J
4. 140 J

Answer

130 J

Reason — Given,

• Change in internal energy along the path ca, dU_ca = − 180 J
• Heat absorbed along the path ab, dQ_ab = 250 J
• Heat absorbed along the path bc, dQ_bc = 60 J

In the process b → c : From the graph the volume remains constant, so

$\text{dW}_{bc} = \text P \times \text{dV} = 0$

By the first law of thermodynamics,

$\text{dQ}_{bc} = \text{dU}_{bc} + \text{dW}_{bc} \quad \Rightarrow \quad 60 = \text{dU}_{bc} + 0$

$\text{dU}_{bc} = 60\ \text J$

For the complete cyclic process abca : The internal energy returns to its initial value, so

$\text{dU}_{ab} + \text{dU}_{bc} + \text{dU}_{ca} = 0$

$\text{dU}_{ab} + 60 + (-180) = 0$

$\text{dU}_{ab} = 120\ \text J$

In the process a → b : By the first law of thermodynamics,

$\text{dQ}_{ab} = \text{dU}_{ab} + \text{dW}_{ab} \quad \Rightarrow \quad 250 = 120 + \text{dW}_{ab}$

$\text{dW}_{ab} = 130\ \text J$

Work done along the path abc :

$\text{dW}_{abc} = \text{dW}_{ab} + \text{dW}_{bc} = 130 + 0 \\[1em] = 130\ \text J$

The given figure shows two processes A and B for gas. If ΔQ_A and ΔQ_B are the amount of heat absorbed by the system in two cases and ΔU_A and ΔU_B are changes in internal energies respectively then :

1. ΔQ_A > ΔQ_B, ΔU_A > ΔU_B
2. ΔQ_A < ΔQ_B, ΔU_A < ΔU_B
3. ΔQ_A > ΔQ_B, ΔU_A = ΔU_B
4. ΔQ_A = ΔQ_B, ΔU_A = ΔU_B

Answer

ΔQ_A > ΔQ_B, ΔU_A = ΔU_B

Reason —

Both the processes A and B take the gas from the same initial state i to the same final state f.

Change in internal energy : The internal energy is a state function, so its change depends only upon the initial and the final states, which are the same for both the processes. Hence

$\Delta \text U_{\text A} = \Delta \text U_{\text B}$

Work done : The work done is the area enclosed between the curve and the volume-axis. From the graph the curve A lies above the curve B, so the area under A is greater than the area under B,

$\text W_{\text A} \gt \text W_{\text B}$

Heat absorbed : By the first law of thermodynamics,

$\Delta \text Q_{\text A} = \Delta \text U_{\text A} + \text W_{\text A} \quad \text{and} \quad \Delta \text Q_{\text B} = \Delta \text U_{\text B} + \text W_{\text B}$

Since ΔU_A = ΔU_B and W_A > W_B, we get

$\Delta \text Q_{\text A} \gt \Delta \text Q_{\text B}$

The given diagram shows four processes i.e., isochoric, isobaric, isothermal and adiabatic. The correct assignment of the processes, in the same order is given by :

1. dabc
2. adbc
3. dacb
4. adcb

Answer

dabc

Reason —

The four processes are to be identified in the order isochoric, isobaric, isothermal and adiabatic.

Isochoric process : The volume remains constant, so the curve is a straight line parallel to the pressure-axis. From the graph this is the curve d.

Isobaric process : The pressure remains constant, so the curve is a straight line parallel to the volume-axis. From the graph this is the curve a.

Isothermal and adiabatic processes : Both b and c are falling curves. We know that

$\dfrac{\text{slope of adiabatic curve}}{\text{slope of isothermal curve}} = \gamma$

and γ is always greater than 1, so the adiabatic curve is steeper than the isothermal curve. From the graph the curve c is steeper than the curve b. Hence b is isothermal and c is adiabatic.

Hence the correct sequence is d, a, b, c, that is, dabc.

In which of the following processes, heat is neither absorbed nor released by a system?

1. Adiabatic
2. Isobaric
3. Isochoric
4. Isothermal

Answer

Adiabatic

Reason — An adiabatic process is one during which no exchange of heat takes place between the system and the surroundings, that is, Q = 0. For such a process to occur, the system must be perfectly insulated from the surroundings, or the process must be carried out very rapidly.

In the isobaric, isochoric and isothermal processes heat is exchanged with the surroundings.

Half mole of an ideal monoatomic gas is heated at constant pressure of 1 atm from 20°C to 90°C. Work done by the gas is close to : (R = 8.31 J/mol-K)

1. 291 J
2. 581 J
3. 146 J
4. 73 J

Answer

291 J

Reason — Given,

• Number of moles, n = 0.5 (half mole)
• Initial temperature = 20°C, final temperature = 90°C, so ΔT = 70°C = 70 K
• Constant pressure = 1 atm
• R = 8.31 J (mol-K)^-1

At constant pressure, the work done by the gas is

$\Delta \text W = \text P\Delta \text V$

From the ideal gas equation PV = nRT, at constant pressure

$\Delta(\text{PV}) = \Delta(\text{nRT}) \quad \Rightarrow \quad \text P\Delta \text V = \text n\text R\Delta \text T$

Therefore

$\Delta \text W = \text n\text R\Delta \text T = \dfrac{1}{2} \times 8.31 \times (90 - 20)$

$= \dfrac{1}{2} \times 8.31 \times 70 = \dfrac{581.7}{2}$

$= 290.85\ \text J \approx 291\ \text J$

A gas has n degrees of freedom. The ratio of specific heat of gas at constant volume to the specific heat of gas at constant pressure will be :

1. $\dfrac{n}{n + 2}$

2. $\dfrac{n + 2}{n}$

3. $\dfrac{n}{2n + 2}$

4. $\dfrac{n}{n - 2}$

Answer

$\dfrac{n}{n + 2}$

Reason — Given, the gas has n degrees of freedom.

For a gas having n degrees of freedom, the molar specific heat at constant volume is

$\text C_V = \dfrac{n\text R}{2}$

By Mayer's relation, the molar specific heat at constant pressure is

$\text C_P = \text C_V + \text R = \dfrac{n\text R}{2} + \text R$

Therefore the required ratio is

$\dfrac{\text C_V}{\text C_P} = \dfrac{\dfrac{n\text R}{2}}{\dfrac{n\text R}{2} + \text R}$

Dividing the numerator and the denominator by R,

$= \dfrac{\dfrac{n}{2}}{\dfrac{n}{2} + 1} = \dfrac{\dfrac{n}{2}}{\dfrac{n + 2}{2}} = \dfrac{n}{n + 2}$

7 mol of a certain monoatomic ideal gas undergoes a temperature increase of 40 K at constant pressure. The increase in the internal energy of the gas in this process is : (Given R = 8.3 JK^-1 mol^-1)

1. 5810 J
2. 3486 J
3. 11620 J
4. 6972 J

Answer

3486 J

Reason — Given,

• Number of moles, n = 7
• Rise in temperature, ΔT = 40 K
• The gas is monoatomic, so $\text C_V = \dfrac{3}{2}\text R$
• R = 8.3 J K^-1 mol^-1

The change in internal energy depends only upon the temperature change, whatever be the process. Hence

$\Delta \text U = \text n\text C_V\Delta \text T = \text n \times \dfrac{3\text R}{2} \times \Delta \text T$

Substituting the values,

$\Delta \text U = 7 \times \dfrac{3 \times 8.3}{2} \times 40$

$= 7 \times 12.45 \times 40 \\[1em] = 3486\ \text J$

The efficiency of an ideal heat engine working between the freezing point and boiling point of water is :

1. 6.25%
2. 20%
3. 26.8%
4. 12.5%

Answer

26.8%

Reason — Given,

• Temperature of the source (boiling point of water), T₁ = 100°C = 373 K
• Temperature of the sink (freezing point of water), T₂ = 0°C = 273 K

The efficiency of an ideal heat engine is

$\eta = \left(1 - \dfrac{\text T_2}{\text T_1}\right) \times 100$

Substituting the values,

$\eta = \left(1 - \dfrac{273}{373}\right) \times 100 = \dfrac{373 - 273}{373} \times 100$

$= \dfrac{100}{373} \times 100 \\[1em] = 26.8$

Three Carnot engines operate in series between a heat source at a temperature T₁ and a heat sink at temperature T₄. There are two other reservoirs at temperature T₂ and T₃ as shown with T₁ > T₂ > T₃ > T₄. The three engines are equally efficient if :

1. $\text T_2 = (\text T_1^3 \text T_4)^{1/4},\ \text T_3 = (\text T_1 \text T_4^3)^{1/4}$

2. $\text T_2 = (\text T_1^2 \text T_4)^{1/3},\ \text T_3 = (\text T_1 \text T_4^2)^{1/3}$

3. $\text T_2 = (\text T_1 \text T_4)^{1/2},\ \text T_3 = (\text T_1^2 \text T_4)^{1/3}$

4. $\text T_2 = (\text T_1 \text T_4^2)^{1/3},\ \text T_3 = (\text T_1^2 \text T_4)^{1/3}$

Answer

$\text T_2 = (\text T_1^2 \text T_4)^{1/3},\ \text T_3 = (\text T_1 \text T_4^2)^{1/3}$

Reason —

Given, the three Carnot engines operate in series between T₁ and T₄, with T₁ > T₂ > T₃ > T₄, and the three engines are equally efficient.

The efficiency of a Carnot engine is

$\eta = 1 - \dfrac{\text T_{\text{sink}}}{\text T_{\text{source}}}$

For the three engines to be equally efficient,

$1 - \dfrac{\text T_2}{\text T_1} = 1 - \dfrac{\text T_3}{\text T_2} = 1 - \dfrac{\text T_4}{\text T_3}$

$\dfrac{\text T_2}{\text T_1} = \dfrac{\text T_3}{\text T_2} = \dfrac{\text T_4}{\text T_3}$

From the first two ratios :

$\text T_2^2 = \text T_1\text T_3 \qquad \text{...(i)}$

From the last two ratios :

$\text T_3^2 = \text T_2\text T_4 \qquad \text{...(ii)}$

Substituting the value of T₃ from equation (ii) into equation (i),

$\text T_2^2 = \text T_1\sqrt{\text T_2\text T_4}$

Squaring both the sides,

$\text T_2^4 = \text T_1^2\text T_2\text T_4 \quad \Rightarrow \quad \text T_2^3 = \text T_1^2\text T_4$

$\text T_2 = (\text T_1^2\text T_4)^{1/3}$

Substituting this value in equation (ii),

$\text T_3^2 = (\text T_1^2\text T_4)^{1/3}\text T_4 = \text T_1^{2/3}\text T_4^{4/3}$

$\text T_3 = \text T_1^{1/3}\text T_4^{2/3} = (\text T_1\text T_4^2)^{1/3}$

Two Carnot engines A and B are operated in series. The first engine A, receives heat at T₁ (= 600 K) and rejects to a reservoir at temperature T₂. The second engine B receives heat rejected by the first engine and in turn rejects to a heat reservoir at T₃ (= 400 K). If work outputs of two engines are equal then temperature T₂ will be :

1. 600 K
2. 500 K
3. 400 K
4. 300 K

Answer

500 K

Reason — Given,

• Engine A receives heat at T₁ = 600 K and rejects to a reservoir at T₂
• Engine B receives the heat rejected by A and rejects to a reservoir at T₃ = 400 K
• The work outputs of the two engines are equal

Let Q₁ be the heat received by engine A at T₁, Q₂ the heat rejected by A and received by engine B, and Q₃ the heat rejected by engine B.

Work output of engine A :

$\text W_{\text A} = \text Q_1 - \text Q_2$

Work output of engine B :

$\text W_{\text B} = \text Q_2 - \text Q_3$

Since the work outputs are equal,

$\text Q_1 - \text Q_2 = \text Q_2 - \text Q_3 \quad \Rightarrow \quad \text Q_1 + \text Q_3 = 2\text Q_2$

Dividing throughout by Q₂,

$\dfrac{\text Q_1}{\text Q_2} + \dfrac{\text Q_3}{\text Q_2} = 2$

For a Carnot engine the heat exchanged is proportional to the absolute temperature, so $\dfrac{\text Q_1}{\text Q_2} = \dfrac{\text T_1}{\text T_2}$ and $\dfrac{\text Q_3}{\text Q_2} = \dfrac{\text T_3}{\text T_2}$. Therefore

$\dfrac{\text T_1}{\text T_2} + \dfrac{\text T_3}{\text T_2} = 2 \quad \Rightarrow \quad \text T_2 = \dfrac{\text T_1 + \text T_3}{2}$

Substituting the values,

$\text T_2 = \dfrac{600 + 400}{2} = \dfrac{1000}{2} \\[1em] = 500\ \text K$

C_p and C_v are specific heats at constant pressure and constant volume respectively. It is observed that :

C_p − C_v = a for hydrogen gas

C_p − C_v = b for nitrogen gas

The correct relation between a and b is :

1. a = 28b
2. $a = \dfrac{1}{14} b$
3. a = b
4. a = 14b

Answer

a = 14b

Reason — Here C_p and C_v are the specific heats, that is, the heat capacities per unit mass. They are related to the molar specific heats C_P and C_V by

$\text C_P = \text Mc_p \quad \text{and} \quad \text C_V = \text Mc_v$

where M is the molar mass. By Mayer's relation,

$\text M(c_p - c_v) = \text C_P - \text C_V = \text R$

$c_p - c_v = \dfrac{\text R}{\text M}$

For hydrogen gas, M = 2, so

$a = c_p - c_v = \dfrac{\text R}{2}$

For nitrogen gas, M = 28, so

$b = c_p - c_v = \dfrac{\text R}{28}$

Taking the ratio,

$\dfrac{a}{b} = \dfrac{\text R/2}{\text R/28} = \dfrac{28}{2} = 14$

$a = 14b$

Two moles of an ideal monoatomic gas occupies a volume V at 27°C. The gas expands adiabatically to a volume 2V. Calculate (i) the final temperature of the gas and (ii) change in its internal energy.

1. (i) 189 K (ii) −27 kJ
2. (i) 195 K (ii) −2.7 kJ
3. (i) 189 K (ii) 27 kJ
4. (i) 195 k (ii) 2.7 kJ

Answer

(i) 189 K (ii) −27 kJ

Reason — Given,

• Number of moles of the monoatomic gas, μ = 2
• Initial temperature, T₁ = 27°C = 300 K
• Initial volume = V, final volume = 2V
• For a monoatomic gas, $\gamma = \dfrac{5}{3}$

(i) Final temperature of the gas : For an adiabatic change,

$\text T_1\text V_1^{\gamma - 1} = \text T_2\text V_2^{\gamma - 1} \quad \Rightarrow \quad \text T_2 = \text T_1\left(\dfrac{\text V_1}{\text V_2}\right)^{\gamma - 1}$

Here $\gamma - 1 = \dfrac{5}{3} - 1 = \dfrac{2}{3}$, so

$\text T_2 = 300 \times \left(\dfrac{\text V}{2\text V}\right)^{2/3} = \dfrac{300}{2^{2/3}}$

$= \dfrac{300}{1.587} = 189\ \text K$

(ii) Change in internal energy : For a monoatomic gas the degrees of freedom f = 3, and

$\Delta \text U = \dfrac{1}{2}\mu f\text R\Delta \text T$

Substituting the values,

$\Delta \text U = \dfrac{1}{2} \times 2 \times 3 \times 8.31 \times (189 - 300)$

$= 3 \times 8.31 \times (-111) \\[1em] = -2767\ \text J = -2.7\ \text{kJ}$

Note: The change in internal energy works out to − 2.7 kJ, and the textbook's own hint for this question also obtains − 2.7 kJ. The value "− 27 kJ" printed in option (a) therefore appears to be a misprint for "− 2.7 kJ". Option (a) is the intended answer, since it alone carries the correct final temperature of 189 K.

When heat Q is supplied to a diatomic gas of rigid molecules, at constant volume, its temperature increases by ΔT. The heat required to produce the same change in temperature, at a constant pressure is :

1. $\dfrac{2}{3}\text Q$

2. $\dfrac{5}{3}\text Q$

3. $\dfrac{3}{2}\text Q$

4. $\dfrac{7}{5}\text Q$

Answer

$\dfrac{7}{5}\text Q$

Reason — Given, heat Q is supplied to a diatomic gas of rigid molecules at constant volume, producing a temperature rise ΔT.

At constant volume : The heat supplied is

$\text Q = \text n\text C_V\Delta \text T$

At constant pressure : For the same temperature change, the heat required is

$\text Q' = \text n\text C_P\Delta \text T$

Dividing the second by the first,

$\dfrac{\text Q'}{\text Q} = \dfrac{\text n\text C_P\Delta \text T}{\text n\text C_V\Delta \text T} = \dfrac{\text C_P}{\text C_V} = \gamma$

For a diatomic gas of rigid molecules, $\gamma = \dfrac{7}{5}$. Therefore

$\text Q' = \dfrac{7}{5}\text Q$

n moles of an ideal gas with constant volume heat capacity C_v undergo an isobaric expansion by certain volume. The ratio of the work done in the process, to the heat supplied is :

1. $\dfrac{4n\text R}{\text C_v + n\text R}$

2. $\dfrac{4n\text R}{\text C_v - n\text R}$

3. $\dfrac{n\text R}{\text C_v - n\text R}$

4. $\dfrac{n\text R}{\text C_v + n\text R}$

Answer

$\dfrac{n\text R}{\text C_v + n\text R}$

Reason — Given, n moles of an ideal gas with constant volume heat capacity C_v undergo an isobaric expansion.

Work done in the process : For an isobaric process,

$\text W = \text P\Delta \text V = \text n\text R\Delta \text T \qquad \text{...(i)}$

Heat supplied in the process : The heat supplied at constant pressure is

$\text Q = \text n\text C_p\Delta \text T$

where C_p is the molar specific heat at constant pressure. Since C_v here is the heat capacity of the whole sample of n moles, the molar specific heat at constant volume is $\dfrac{\text C_v}{n}$, and by Mayer's relation

$\text C_p = \dfrac{\text C_v}{n} + \text R$

Therefore

$\text Q = \text n\left(\dfrac{\text C_v}{n} + \text R\right)\Delta \text T = (\text C_v + \text{nR})\Delta \text T \qquad \text{...(ii)}$

Dividing equation (i) by equation (ii),

$\dfrac{\text W}{\text Q} = \dfrac{\text{nR}\Delta \text T}{(\text C_v + \text{nR})\Delta \text T} = \dfrac{n\text R}{\text C_v + n\text R}$

In a process, temperature and volume of one mole of an ideal monoatomic gas are varied according to the relation VT = K, where K is a constant. In this process, the temperature of the gas is increased by ΔT. The amount of heat absorbed by gas is : (R = gas constant)

1. $\dfrac{1}{2}\text{KR}\Delta \text T$

2. $\dfrac{2\text K}{3}\Delta \text T$

3. $\dfrac{1}{2}\text R \Delta \text T$

4. $\dfrac{3}{2}\text R \Delta \text T$

Answer

$\dfrac{1}{2}\text R \Delta \text T$

Reason — Given,

• One mole of an ideal monoatomic gas
• The process obeys VT = K, where K is a constant
• The temperature is increased by ΔT

Finding the nature of the process : From the ideal gas equation for one mole,

$\text{PV} = \text{RT} \quad \Rightarrow \quad \text T = \dfrac{\text{PV}}{\text R}$

Substituting this in VT = K,

$\text V \times \dfrac{\text{PV}}{\text R} = \text K \quad \Rightarrow \quad \text{PV}^2 = \text{RK} = \text{constant} \qquad \text{...(i)}$

Work done : Equation (i) represents a polytropic process PV^x = constant with x = 2. The work done in such a process is

$\text W = \dfrac{\text P_2\text V_2 - \text P_1\text V_1}{1 - x} = \dfrac{\text R\text T_2 - \text R\text T_1}{1 - x} = \dfrac{\text R\Delta \text T}{1 - x}$

Substituting x = 2,

$\text W = \dfrac{\text R\Delta \text T}{1 - 2} = -\text R\Delta \text T$

Change in internal energy : For a monoatomic gas,

$\Delta \text U = \dfrac{3}{2}\text R\Delta \text T$

Heat absorbed : By the first law of thermodynamics,

$\Delta \text Q = \Delta \text U + \text W = \dfrac{3}{2}\text R\Delta \text T - \text R\Delta \text T$

$= \dfrac{1}{2}\text R\Delta \text T$

A rigid diatomic ideal gas undergoes an adiabatic process at room temperature. The relation between temperature and volume for this process is TV^x = constant, then x is :

1. $\dfrac{2}{5}$

2. $\dfrac{2}{3}$

3. $\dfrac{5}{3}$

4. $\dfrac{3}{5}$

Answer

$\dfrac{2}{5}$

Reason — Given, a rigid diatomic ideal gas undergoes an adiabatic process, and the relation is TV^x = constant.

For a rigid diatomic gas the degrees of freedom are 5, so

$\gamma = \dfrac{7}{5}$

Poisson's equation for an adiabatic process, expressed in terms of the temperature and the volume, is

$\text{TV}^{\gamma - 1} = \text{constant}$

Comparing this with the given relation TV^x = constant,

$x = \gamma - 1 = \dfrac{7}{5} - 1 \\[1em] = \dfrac{2}{5}$

A monoatomic gas at pressure P and volume V is suddenly compressed to one-eighth of its original volume. The final pressure at constant entropy will be :

1. P
2. 8P
3. 32P
4. 64P

Answer

32P

Reason — Given,

• Initial pressure = P and initial volume = V
• Final volume $= \dfrac{\text V}{8}$
• The gas is monoatomic, so $\gamma = \dfrac{5}{3}$

The gas is compressed suddenly and the entropy remains constant, so the change is adiabatic and obeys Poisson's law,

$\text{PV}^{\gamma} = \text P'\left(\dfrac{\text V}{8}\right)^{\gamma}$

Rearranging,

$\text P' = \text P\left(\dfrac{\text V}{\text V/8}\right)^{\gamma} = \text P \times (8)^{5/3}$

Now

$(8)^{5/3} = \left(\sqrt[3]{8}\right)^5 = (2)^5 = 32$

Therefore

$\text P' = 32\text P$

An ideal gas undergoes four different processes from the same initial state as shown in the figure. Those processes are adiabatic, isothermal, isobaric and isochoric. The curve which represents the adiabatic process among 1, 2, 3 and 4 is :

1. 3
2. 4
3. 1
4. 2

Answer

2

Reason —

The four processes starting from the same initial state are adiabatic, isothermal, isobaric and isochoric.

• The isobaric process (PV⁰ = constant) is a straight line parallel to the volume-axis, which is the curve 4.
• The isochoric process is a straight line parallel to the pressure-axis, which is the curve 1.
• Of the two remaining falling curves, the adiabatic curve is steeper than the isothermal curve, since

$\dfrac{\text{slope of adiabatic curve}}{\text{slope of isothermal curve}} = \gamma \gt 1$

The isothermal process obeys PV = constant and the adiabatic process obeys PV^γ = constant. From the graph the curve 2 is steeper than the curve 3.

Hence the curve 2 represents the adiabatic process.

0.08 kg air is heated at constant volume through 5°C. The specific heat of air at constant volume is 0.17 kcal/kg° C and J = 4.18 J/cal. The change in its internal energy is approximately :

1. 318 J
2. 298 J
3. 284 J
4. 142 J

Answer

284 J

Reason — Given,

• Mass of air, m = 0.08 kg
• Rise in temperature, ΔT = 5°C
• Specific heat of air at constant volume, c_v = 0.17 kcal (kg-°C)^-1
• J = 4.18 J cal^-1

Since the air is heated at constant volume, no work is done,

$\text W = \text P\text{dV} = \text P \times 0 = 0$

By the first law of thermodynamics,

$\Delta \text U = \text Q - \text W = \text Q$

The heat taken by the air is

$\text Q = \text mc_v\Delta \text T = 0.08 \times 0.17 \times 5$

$= 0.068\ \text{kcal} = 68\ \text{cal}$

Converting into joule,

$\Delta \text U = 68 \times 4.18 \\[1em] = 284.2\ \text J \approx 284\ \text J$

In an adiabatic process, which of the following statements is true?

1. The molar heat capacity is infinite
2. Work done by the gas equals the increase in internal energy
3. The molar heat capacity is zero
4. The internal energy of the gas decreases as the temperature increases

Answer

The molar heat capacity is zero

Reason — In an adiabatic process no heat is exchanged with the surroundings, so

$\text{dQ} = 0$

The molar heat capacity is defined as

$\text C = \dfrac{\text{dQ}}{\mu\text{dT}}$

Since dQ = 0 while dT is not zero, we get

$\text C = 0$

Hence the molar heat capacity in an adiabatic process is zero.

Option 2 is incorrect, because dQ = 0 gives dU = − dW, so the work done by the gas equals the decrease in internal energy. Option 4 is incorrect, because $\text{dU} = \dfrac{f}{2}\mu\text R\text{dT}$, so the internal energy increases when the temperature increases.

Two gases A and B are filled at the same pressure in separate cylinders with movable pistons of radius r_A and r_B, respectively. On supplying an equal amount of heat to both the systems reversibly under constant pressure, the pistons of gas A and B are displaced by 16 cm and 9 cm, respectively. If the change in their internal energy is the same, then the ratio $\dfrac{\text r_A}{\text r_B}$ is equal to:

1. 4/3

2. 3/4

3. $\dfrac{2}{\sqrt{3}}$

4. $\dfrac{\sqrt{3}}{2}$

Answer

3/4

Reason — Given,

• The two gases are at the same pressure, P_A = P_B = P
• Equal amounts of heat are supplied, Q_A = Q_B
• The changes in internal energy are the same, ΔU_A = ΔU_B
• Displacement of the piston of gas A, x_A = 16 cm
• Displacement of the piston of gas B, x_B = 9 cm

By the first law of thermodynamics,

$\text Q = \Delta \text U + \text W$

Since the heat supplied and the change in internal energy are the same for both the gases, the work done must also be the same,

$\text W_{\text A} = \text W_{\text B}$

The process takes place at constant pressure, so

$\text P\Delta \text V_{\text A} = \text P\Delta \text V_{\text B} \quad \Rightarrow \quad \Delta \text V_{\text A} = \Delta \text V_{\text B}$

The change in volume is the area of cross-section of the piston multiplied by its displacement,

$\left(\pi \text r_{\text A}^2\right)\text x_{\text A} = \left(\pi \text r_{\text B}^2\right)\text x_{\text B}$

$\dfrac{\text r_{\text A}^2}{\text r_{\text B}^2} = \dfrac{\text x_{\text B}}{\text x_{\text A}} = \dfrac{9}{16}$

Taking the square root,

$\dfrac{\text r_{\text A}}{\text r_{\text B}} = \sqrt{\dfrac{9}{16}} = \dfrac{3}{4}$

Note: The textbook answer key gives option (a), i.e., 4/3, but the correct option is 3/4. Since the pistons sweep out equal volumes, the piston moving a greater distance must have a smaller radius. The textbook’s hint has interchanged the displacements.

The efficiency of a Carnot engine operating with a hot reservoir kept at a temperature of 1000 K is 0.4. It extracts 150 J of heat per cycle from the hot reservoir. The work extracted from this engine is being fully used to run a heat pump which has a coefficient of performance 10. The hot reservoir of the heat pump is at a temperature of 300 K. Which of the following statements is/are correct ?

1. Work extracted from the Carnot engine in one cycle is 60 J.
2. Temperature of the cold reservoir of the Carnot engine is 600 K.
3. Temperature of the cold reservoir of the heat pump is 270 K.
4. Heat supplied to the hot reservoir of the heat pump in one cycle is 540 J.

Answer

1. Work extracted from the Carnot engine in one cycle is 60 J.

2. Temperature of the cold reservoir of the Carnot engine is 600 K.

3. Temperature of the cold reservoir of the heat pump is 270 K.

Reason — Given,

• Efficiency of the Carnot engine, η = 0.4
• Temperature of the hot reservoir of the engine, T₁ = 1000 K
• Heat extracted per cycle, Q₁ = 150 J
• Coefficient of performance of the heat pump, (COP)_HP = 10
• Temperature of the hot reservoir of the heat pump, T₃ = 300 K

Checking option 1 : The work extracted from the Carnot engine is

$\text W = \eta \times \text Q_1 = 0.4 \times 150 \\[1em] = 60\ \text J$

Hence option 1 is correct.

Checking option 2 : The efficiency of the Carnot engine is

$\eta = 1 - \dfrac{\text T_2}{\text T_1} \quad \Rightarrow \quad 0.4 = 1 - \dfrac{\text T_2}{1000}$

$\dfrac{\text T_2}{1000} = 0.6 \quad \Rightarrow \quad \text T_2 = 600\ \text K$

Hence option 2 is correct.

Checking option 3 : For a heat pump the coefficient of performance is

$\text{COP} = \dfrac{\text T_3}{\text T_3 - \text T_4}$

Substituting the values,

$10 = \dfrac{300}{300 - \text T_4} \quad \Rightarrow \quad 300 - \text T_4 = 30$

$\text T_4 = 270\ \text K$

Hence option 3 is correct.

Checking option 4 : The work extracted from the engine is fully used to run the heat pump, so the work input to the pump is 60 J. The heat delivered to the hot reservoir of the heat pump is

$\text Q_4 = \text{COP} \times \text W = 10 \times 60 = 600\ \text J$

Since this is not 540 J, option 4 is incorrect.

One mole of a monoatomic ideal gas undergoes an adiabatic expansion in which its volume becomes eight times its inital value. If the initial temperature of the gas is 100 K and the universal gas constant R = 8.0 J mol^-1 K^-1, the decrease in its internal energy, in J, is ...... . Calculate up to second decimal place.

Answer

Given,

• Number of moles of the monoatomic gas, μ = 1
• Initial temperature, T₁ = 100 K
• Final volume, V₂ = 8V₁
• R = 8.0 J mol^-1 K^-1
• For a monoatomic gas, $\gamma = \dfrac{5}{3}$

Final temperature : For an adiabatic process,

$\text T_1\text V_1^{\gamma - 1} = \text T_2\text V_2^{\gamma - 1}$

Substituting V₂ = 8V₁ and $\gamma - 1 = \dfrac{2}{3}$,

$100 \times \text V_1^{2/3} = \text T_2 \times (8\text V_1)^{2/3}$

$\text T_2 = \dfrac{100}{(8)^{2/3}}$

Now

$(8)^{2/3} = \left(\sqrt[3]{8}\right)^2 = (2)^2 = 4$

Therefore

$\text T_2 = \dfrac{100}{4} = 25\ \text K$

Change in internal energy : For an ideal gas,

$\Delta \text U = \mu\text C_v\Delta \text T = \dfrac{\text R}{\gamma - 1}(\text T_2 - \text T_1)$

Substituting the values,

$\Delta \text U = \dfrac{8.0}{\dfrac{5}{3} - 1}(25 - 100) = \dfrac{8.0}{\dfrac{2}{3}} \times (-75)$

$= \dfrac{3}{2} \times 8.0 \times (-75) = 12 \times (-75)$

$= -900\ \text J$

Hence, the decrease in the internal energy of the gas is 900.00 J.

An ideal monatomic gas of n moles is taken through a cycle WXYZW consisting of consecutive adiabatic and isobaric quasi-static processes, as shown in the schematic V-T diagram. The volume of the gas at W, X and Y points are 64 cm³, 125 cm³ and 250 cm³, respectively. If the absolute temperature of the gas T_W at the point W is such that nRT_W = 1J (R is the universal gas constant), then the amount of heat absorbed (in J) by the gas along the path XY is ......... .

Answer

Given,

• Volume of the gas at W, V_W = 64 cm³
• Volume of the gas at X, V_X = 125 cm³
• Volume of the gas at Y, V_Y = 250 cm³
• nRT_W = 1 J
• The gas is monatomic, so $\gamma = \dfrac{5}{3}$ and $\text C_p = \dfrac{5}{2}\text R$

From the V-T diagram, WX is an adiabatic process and XY is an isobaric process.

Temperature at X : For the adiabatic process WX,

$\text T_{\text W}\text V_{\text W}^{\gamma - 1} = \text T_{\text X}\text V_{\text X}^{\gamma - 1}$

$\text T_{\text X} = \text T_{\text W}\left(\dfrac{\text V_{\text W}}{\text V_{\text X}}\right)^{\gamma - 1} = \text T_{\text W}\left(\dfrac{64}{125}\right)^{2/3}$

Now

$\left(\dfrac{64}{125}\right)^{2/3} = \left(\dfrac{4}{5}\right)^2 = \dfrac{16}{25}$

Therefore

$\text T_{\text X} = \dfrac{16}{25}\text T_{\text W}$

Temperature at Y : The process XY is isobaric, so by Charles' law

$\dfrac{\text V_{\text X}}{\text T_{\text X}} = \dfrac{\text V_{\text Y}}{\text T_{\text Y}} \quad \Rightarrow \quad \text T_{\text Y} = \text T_{\text X} \times \dfrac{\text V_{\text Y}}{\text V_{\text X}}$

$= \text T_{\text X} \times \dfrac{250}{125} = 2\text T_{\text X}$

Heat absorbed along the path XY : For an isobaric process,

$\text Q = \text n\text C_p\Delta \text T = \text n \times \dfrac{5}{2}\text R(\text T_{\text Y} - \text T_{\text X})$

$= \dfrac{5}{2}\text{nR}(2\text T_{\text X} - \text T_{\text X}) = \dfrac{5}{2}\text{nR}\text T_{\text X}$

Substituting $\text T_{\text X} = \dfrac{16}{25}\text T_{\text W}$,

$\text Q = \dfrac{5}{2} \times \dfrac{16}{25} \times \text{nR}\text T_{\text W} = \dfrac{80}{50}\text{nR}\text T_{\text W}$

Since nRT_W = 1 J,

$\text Q = 1.6 \times 1 \\[1em] = 1.6\ \text J$

Hence, the heat absorbed by the gas along the path XY is 1.60 J.

Directions :

• Each set has One Multiple Choice Question.
• Each set has Two lists : List-I and List-II.
• List-I has Four entries (P), (Q), (R) and (S) and List-II has Five entries (1), (2), (3), (4) and (5).
• Four options are given in each Multiple Choice Question based on List-I and List-II and only one of these four options satisfies the condition asked in the Multiple Choice Question.
• Answer to each question will be evaluated according to the following marking scheme :
• Full Marks : +3 only if the option corresponding to the correct combination is chosen; Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered); Negative Marks: −1 In all other cases.

One mole of a monatomic ideal gas undergoes the cyclic process J → K → L → M → J, as shown in the P-T diagram. Match the quantities mentioned in List-I with their values in List-II and choose the correct option. [R is the gas constant.]

1. P → 1; Q → 3; R → 5; S → 4
2. P → 4; Q → 3; R → 5; S → 2
3. P → 4; Q → 1; R → 2; S → 2
4. P → 2; Q → 5; R → 3; S → 4.

Answer

P → 4; Q → 3; R → 5; S → 2

Reason —

From the P-T diagram the four states of one mole of the monatomic ideal gas are

$\text J(\text P_0, \text V_0, \text T_0), \quad \text K(\text P_0, 3\text V_0, 3\text T_0), \quad \text L\left(2\text P_0, \dfrac{3\text V_0}{2}, 3\text T_0\right), \quad \text M\left(2\text P_0, \dfrac{\text V_0}{2}, \text T_0\right)$

where P₀V₀ = RT₀. The process JK is isobaric, KL is isothermal, LM is isobaric and MJ is isothermal.

(P) Work done in the complete cyclic process :

JK is isobaric, so

$\text W_{\text{JK}} = \text P_0(3\text V_0 - \text V_0) = 2\text P_0\text V_0 = 2\text{RT}_0$

KL is isothermal at 3T₀, and since PV is constant the volume falls to half as the pressure doubles,

$\text W_{\text{KL}} = \text R(3\text T_0)\ln\left(\dfrac{1}{2}\right) = -3\text{RT}_0\ln 2$

LM is isobaric, so

$\text W_{\text{LM}} = 2\text P_0\left(\dfrac{\text V_0}{2} - \dfrac{3\text V_0}{2}\right) = -2\text P_0\text V_0 = -2\text{RT}_0$

MJ is isothermal at T₀, and the volume doubles as the pressure halves,

$\text W_{\text{MJ}} = \text{RT}_0\ln 2$

Adding all the four,

$\text W = 2\text{RT}_0 - 3\text{RT}_0\ln 2 - 2\text{RT}_0 + \text{RT}_0\ln 2 \\[1em] = -2\text{RT}_0\ln 2$

Hence P → 4.

(Q) Change in the internal energy in the process JK : For one mole of a monatomic gas,

$\Delta \text U = \dfrac{3}{2}\text R\Delta \text T = \dfrac{3}{2}\text R(3\text T_0 - \text T_0)$

$= \dfrac{3}{2}\text R(2\text T_0) = 3\text{RT}_0$

Hence Q → 3.

(R) Heat given to the gas in the process KL : The process is isothermal, so ΔU = 0 and by the first law

$\text Q = \text W_{\text{KL}} = -3\text{RT}_0\ln 2$

Hence R → 5.

(S) Change in the internal energy in the process MJ : The process is isothermal, so the temperature does not change and

$\Delta \text U = 0$

Hence S → 2.