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Thermodynamics

Practice Questions:

Question 31: moderate

Consider the following statements

A. Work and heat are path functions in thermodynamics.

B. The internal energy of a gaseous system is state function.

C. For gaseous system, CP is greater than CV .

D. Work done by gas at constant volume is zero.

Based on above information pick the correct option.

1. Only statement (A) is correct

2. Only statements (A), (B) and (C) are correct

3. Only statements (B), (C) and (A) are correct

4. All statements (A), (B), (C) and (D) are correct

View Answer

All the given statements are correct:

A. Work and heat depend on the path taken during a process, so they are path functions.
B. Internal energy depends only on the state of the system, making it a state function.
C. For gases, $C_P > C_V$

because extra heat is required at constant pressure to do expansion work.
D. At constant volume, the gas does no work since volume does not change ( $W = P\Delta V = 0$

).

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Question 32: easy

A container of volume $200\text{ cm}^3$ contains 0.2 mole of hydrogen gas and 0.3 mole of argon gas. The pressure of the system at temperature 200 K ($R = 8.3\text{ J K}^{-1}\text{ mol}^{-1}$ ) will be

1. $4.15 \times 10^5\text{ Pa}$

2. $4.15 \times 10^6\text{ Pa}$

3. $6.15 \times 10^5\text{ Pa}$

4. $6.15 \times 10^4\text{ Pa}$

View Answer

Using the ideal gas equation $P = \frac{nRT}{V}$, with total moles $n = 0.2 + 0.3 = 0.5\text{ mol}$ and volume $V = 200 \times 10^{-6}\text{ m}^3$. Calculating gives $P = \frac{0.5 \times 8.3 \times 200}{2 \times 10^{-4}} = 4.15 \times 10^6\text{ Pa}$.

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Question 33: easy

Consider two containers A and B containing identical gases at the same pressure, volume and temperature. The gas in container A is compressed to half of its original volume isothermally while the gas in container B is compressed to half of its original value adiabatically. The ratio of final pressure of gas in B to that of gas in A is:

1. $2^{\gamma - 1}$

2. $\left(\frac{1}{2}\right)^{\gamma - 1}$

3. $\left(\frac{1}{1 - \gamma}\right)^2$

4. $\left(\frac{1}{\gamma - 1}\right)^2$

View Answer

For isothermal process in A: $P_A = 2P_0$. For adiabatic process in B: $P_B = P_0 (2)^\gamma$. The ratio of final pressures is $\frac{P_B}{P_A} = \frac{P_0 2^\gamma}{2 P_0} = 2^{\gamma - 1}$.

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Question 34: moderate

Consider an ideal gas confined in an isolated closed chamber. As the gas undergoes an adiabatic expansion, the average time of collision between molecules increases as $V^q$, where V is the volume of the gas. The value of q is $\left(\gamma = \frac{C_p}{C_v}\right)$

1. $\frac{3\gamma + 5}{6}$

2. $\frac{3\gamma - 5}{6}$

3. $\frac{\gamma + 1}{2}$

4. $\frac{\gamma - 1}{2}$

View Answer

Mean free path $\lambda \propto V$ and $v_{\text{rms}} \propto \sqrt{T}$. For an adiabatic process, $T \propto V^{-(\gamma - 1)}$, so $v_{\text{rms}} \propto V^{-(\gamma - 1)/2}$. Average time $\tau = \frac{\lambda}{v_{\text{rms}}} \propto V^{1 + (\gamma - 1)/2} = V^{(\gamma+1)/2}$, so $q = \frac{\gamma+1}{2}$.

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Question 35: easy

If $10\text{ J}$ of heat energy is supplied to a gas sample and $5\text{ J}$ of its internal energy decreases during the process, then work done by the gas will be

1. 10 J

2. 5 J

3. 15 J

4. 20 J

View Answer

Using the First Law of Thermodynamics, $Delta Q = Delta U + W$. Here, $Delta Q = 10\text{ J}$ and $Delta U = -5\text{ J}$ (decrease). Substituting these values, $10 = -5 + W$, which gives $W = 15\text{ J}$.

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Question 36: easy

Two moles of helium gas is mixed with three moles of hydrogen gas (taken to be rigid). The molar specific heat of mixture at constant volume will be

1. $2.1R$

2. $1.2R$

3. $5.7R$

4. $7.5R$

View Answer

The molar specific heat of a mixture at constant volume is $C_{v,\text{mix}} = \frac{n_1 C_{v1} + n_2 C_{v2}}{n_1 + n_2}$. For monatomic helium, $C_{v1} = 1.5R$, and for rigid diatomic hydrogen, $C_{v2} = 2.5R$. Substituting gives $C_{v,\text{mix}} = \frac{2(1.5R) + 3(2.5R)}{5} = 2.1R$.

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Question 37: easy

The work done by 3 moles of gas at 47°C to triple its volume at constant pressure is ($R = 2$ cal mol$^{-1}$ °C$^{-1}$):

1. 3402 cal

2. 3428 cal

3. 3832 cal

4. 3840 cal

View Answer

At constant pressure, $W = nR\Delta T$. Since volume triples, temperature also triples (from $320$ K to $960$ K), so $\Delta T = 640$ K. Work $W = 3 \times 2 \times 640 = 3840$ cal.

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Question 38: easy

Consider statements (A) and (B) given below:

A. Thermodynamics deals with the process of conversion of heat into work only.

B. Heat given to a system and work done by the system are state variables in thermodynamics.

Choose the correct option.

1. Statement (A) is correct but statement (B) is incorrect

2. Statement (A) is incorrect but statement (B) is correct

3. Both statements are correct

4. Both statements are incorrect

View Answer

Statement A is incorrect because thermodynamics also deals with other energy conversions. Statement B is incorrect because heat and work are path functions, not state variables.

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Question 39: easy

Two moles of an ideal monoatomic gas undergoes an adiabatic process from temperature $300\text{ K}$ to $600\text{ K}$. Work done by this ideal gas in the process is

1. 600R

2. –200R

3. –450R

4. –900R

View Answer

Work done in an adiabatic process is $W = \frac{nR(T_1 - T_2)}{\gamma - 1}$. For monoatomic gas, $\gamma = 5/3$. Substituting the parameters: $W = \frac{2R(300 - 600)}{5/3 - 1} = \frac{-600R}{2/3} = -900R$.

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Question 40: easy

110 J of heat is added to a gaseous system whose internal energy is increased by 40 J then amount of external work done is

1. 150 J

2. 70 J

3. 110 J

4. 40 J

View Answer

According to the first law of thermodynamics, $\Delta Q = \Delta U + W$. Substituting the values, $110\text{ J} = 40\text{ J} + W \Rightarrow W = 70\text{ J}$.

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