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Thermodynamics

Practice Questions:

Question 51: easy

Assertion (A): In a free adiabatic expansion of an ideal gas, the final state is the same as the initial state.

Reason (R): As temperature of a gas increases work done by it is positive.

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)

2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)

3. (A) is true but (R) is false

4. Both (A) and (R) are false

View Answer

For free adiabatic expansion, \( Q = 0 \) and \( W = 0 \). According to the first law of thermodynamics, \( \Delta U = Q - W = 0 \). For an ideal gas, \( \Delta U = nC_v\Delta T \), so \( \Delta T = 0 \), meaning the final temperature is the same as the initial temperature. However, the final state (P,V,T) is not entirely the same as the initial state, only T is.

Thus, Assertion (A) is false. If temperature increases, work done by the gas can be positive, but it is not a direct consequence, and expansion (positive work) typically cools the gas. So, Reason (R) is also false. Therefore, both (A) and (R) are false.

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Question 52: easy

Assertion (A): There is no change in internal energy for ideal gas at constant temperature.

Reason (R): Internal energy of an ideal gas is a function of temperature only.

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)

2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)

3. (A) is true but (R) is false

4. Both (A) and (R) are false

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For an ideal gas, internal energy \(U\) depends solely on temperature \(T\). Therefore, if \(T\) is constant, \(Delta U = 0\). Reason (R) correctly explains Assertion (A).

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Question 53: easy

Assertion (A): At \(0K\), pressure of an ideal gas becomes zero.

Reason (R): At \(0K\), according to ideal gas equation \(PV = 0\), volume cannot be zero hence pressure should be zero to satisfy this equation.

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)

2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)

3. (A) is true but (R) is false

4. Both (A) and (R) are false

View Answer

From the ideal gas equation \(PV = nRT\), if \(T = 0K\), then \(PV = 0\). Since volume \(V\) cannot be zero, pressure \(P\) must be zero. Reason (R) directly explains Assertion (A).

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Question 54: easy

Assertion (A): Molar heat capacity of an ideal monoatomic gas at constant volume is a constant at all temperatures.

Reason (R): As the temperature of an monoatomic ideal gas is increased, number of degrees of freedom of gas molecules remains constant.

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)

2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)

3. (A) is true but (R) is false

4. Both (A) and (R) are false

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For an ideal monoatomic gas, \(C_v = \frac{3}{2}R\) as it only has 3 translational degrees of freedom. This number \(f=3\) remains constant with temperature. Thus, \(C_v\) is constant. Reason (R) correctly explains Assertion (A).

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Question 55: easy

Assertion (A): Experimental results indicate that the molar specific heat of hydrogen gas at constant volume below \( 50 \text{ K} \) is equal to \( 5/2 R \), where \( R \) is the universal gas constant.

Reason (R): A diatomic hydrogen molecule possesses three translational and two rotational degrees of freedom at all temperatures.

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)

2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)

3. (A) is true but (R) is false

4. Both (A) and (R) are false

View Answer

Assertion (A) is false. Below \( 50 \text{ K} \), hydrogen's rotational modes freeze out, so \( C_V \) approaches \( 3/2 R \), not \( 5/2 R \).

Reason (R) is false because degrees of freedom depend on temperature; vibrational modes activate at high T, and rotational modes freeze out at low T.

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Question 56: easy

Assertion (A): Molar heat capacity at constant pressure can be less than molar heat capacity at constant volume.

Reason (R): \( C_p – C_V = R \) is valid only for ideal monoatomic gas.

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)

2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)

3. (A) is true but (R) is false

4. Both (A) and (R) are false

View Answer

Assertion (A) is false; \( C_p \) is always greater than \( C_V \) because work is done at constant pressure.

Reason (R) is false; Mayer's relation, \( C_p - C_V = R \), is valid for all ideal gases, regardless of atomicity.

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Question 57: easy

Assertion (A): An ideal gas is enclosed within a container fitted with a piston when volume of this enclosed gas is increased at constant temperature. The pressure exerted by the gas on the piston decreases.

Reason (R): In the above situation the rate of molecules striking the piston decreases. Therefore pressure exerted by gas on piston decreases.

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)

2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)

3. (A) is true but (R) is false

4. Both (A) and (R) are false

View Answer

Assertion (A) is true by Boyle's Law (\( PV = \text{constant} \) at constant \( T \)). Reason (R) explains (A) microscopically: increasing volume at constant temperature reduces the density of molecules and thus the frequency of collisions with the piston, leading to decreased pressure.

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Question 58: easy

Assertion (A): A real gas behaves as an ideal gas at high temperature and low pressure.

Reason (R): At low pressure and high temperature intermolecular forces vanish away and volume of gas molecules is negligible.

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)

2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)

3. (A) is true but (R) is false

4. Both (A) and (R) are false

View Answer

Assertion (A) is true. Real gases approximate ideal gas behavior under conditions of high temperature (high kinetic energy overcomes intermolecular forces) and low pressure (molecules are far apart, making their own volume negligible).

Reason (R) accurately states these conditions as the underlying cause for ideal gas behavior. Thus, R is the correct explanation for A.

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Question 59: easy

Assertion (A): On a V-T graph, the slope of an isobar increases with pressure.

Reason (R): At constant temperature, for an ideal gas its volume is directly proportional to its pressure.

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)

2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)

3. (A) is true but (R) is false

4. Both (A) and (R) are false

View Answer

For an isobar, \( V = (\frac{nR}{P})T \). The slope on a V-T graph is \( \frac{nR}{P} \). As P increases, slope decreases, so (A) is false. Boyle's law states that at constant T, \( V \propto \frac{1}{P} \), i.e., V is inversely proportional to P, so (R) is false.

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Question 60: easy

Assertion (A): For an ideal gas, at constant temperature, the product of the pressure and volume is constant.

Reason (R): The mean square velocity of gas molecules is inversely proportional to mass of molecule.

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)

2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)

3. (A) is true but (R) is false

4. Both (A) and (R) are false

View Answer

Boyle's Law states that for an ideal gas at constant T, \( PV = \text{constant} \). So (A) is true. The mean square velocity \( = \frac{3kT}{m} \), so it is inversely proportional to molecular mass m.

So (R) is true. However, (R) does not explain Boyle's law (A).

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