Entropy Questions and Answers

Q: 32 kg of oxygen is mixed with 28 kg of nitrogen at the same temperature. The gases are at the same pressure of 103 kPa before and after mixing. If \(\bar{R}\) is the universal gas constant in kJ/kmol.K, the change in entropy of the mixture is

The entropy change is \(1.38\bar{R}\).32 kg of O2 is 1 kmol and 28 kg of N2 is 1 kmol, so each mole fraction is 0.5. Mixing ideal gases at the same temperature and pressure gives\[ \begin{aligned} \Delta S &= -\bar{R}\sum n_{i}\ln x_{i} = -\bar{R}\,[\ln 0.5 + \ln 0.5] \\ &= 2\bar{R}\ln 2 = 1.386\bar{R} \end{aligned} \]The option \(0.69\bar{R}\) counts only one gas's contribution, \(\ln 2\). \(\bar{R}\) and \(0.34\bar{R}\) do not follow from the formula. The pressure value, 103 kPa, does not matter, since it is the same before and after.

Q: The entropy of a reversible process always:

The answer given is that it remains constant.A reversible process generates no entropy. The total entropy of the system and its surroundings therefore stays constant, \(\Delta S_{universe} = 0\), and in a reversible adiabatic process the entropy of the system itself is constant, which is why such a process is called isentropic.An increase in total entropy is the mark of an irreversible process, and a decrease is impossible for an isolated system. 'Zero' confuses the change in entropy with the entropy itself, which is not zero.

Q: The entropy of an isolated system or universe always:

The entropy of an isolated system or the universe increases.This is the principle of increase of entropy, a statement of the second law: for an isolated system \(\Delta S \geq 0\), with equality only for an ideal reversible process. Every real process is irreversible, so the entropy of the universe always increases.A decrease would violate the second law. Remaining constant holds only in the reversible limit, which real processes never reach, and the entropy is not zero.

Q: The net entropy generation (considering the system and the thermal reservoir together) during the process is closest to :[given : 1 kJ energy transfer takes place at 400° C, while the increase in the entropy of system is 10 J/K]

The net entropy generation is closest to 8.51 J/K.The reservoir at 400°C, that is 673 K, gives up 1 kJ, so its entropy falls. The system's entropy rises by 10 J/K.\[ \begin{aligned} \Delta S_{res} &= -\frac{1000}{673} = -1.49\ \text{J/K} \\ S_{gen} &= 10 - 1.49 = 8.51\ \text{J/K} \end{aligned} \]The temperature must be in kelvin: using 400 directly would give \(10 - 2.5 = 7.50\) J/K, which is the trap among the options. 10 J/K counts only the system and ignores the reservoir's loss of entropy.

Q: When a perfect gas enclosed in cylinder - piston device executes a reversible adiabatic expansion process :

The key's answer is its entropy change will be zero.A reversible adiabatic process is isentropic. Entropy changes by \(dS = \frac{\delta Q_{rev}}{T}\); with no heat transfer and no irreversibility (no friction, no unrestrained expansion), the gas's entropy stays exactly the same.Entropy would increase only in an irreversible adiabatic process, and it can never decrease in an adiabatic process, since that would violate the second law. The statement about the surroundings is also true in itself, because no heat reaches them, but it describes the surroundings rather than the gas; the defining property of the gas's own process is zero entropy change, which is what the key accepts.

Q: Two identical blocks of a metal with heat capacity C are initially at temperatures \( T_{1} \) and \( T_{2} \), \( (T_{1} \lt T_{2}) \). They are brought in thermal contact. When the system reaches equilibrium the change in entropy would be :

The change in entropy is \( C \ln\left(\frac{(T_{1} + T_{2})^{2}}{4T_{1}T_{2}}\right) \).Identical blocks reach the mean temperature \( T_{f} = \frac{T_{1} + T_{2}}{2} \). Each block's entropy changes by \( C\ln(T_{f}/T_{i}) \):\[ \begin{aligned} \Delta S &= C \ln\frac{T_{f}}{T_{1}} + C \ln\frac{T_{f}}{T_{2}} \\ &= C \ln\frac{T_{f}^{2}}{T_{1}T_{2}} \\ &= C \ln\frac{(T_{1} + T_{2})^{2}}{4T_{1}T_{2}} \end{aligned} \]Since the arithmetic mean exceeds the geometric mean, the argument is greater than 1 and \( \Delta S \gt 0 \), as an irreversible process requires. The option with the inverted fraction is negative; the ones with 1 or 2 in place of 4 forget that \( T_{f} \) is a half-sum, squared.

Q: One block of certain metal is at temperature \(T_{1}\) and a second identical block is at a temperature \(T_{2}\). These blocks are brought in contact and the system is thermally isolated from the surroundings. Assume that the heat capacity at constant volume C of each block is independent of temperature T. The increase in the entropy of the universe, when the system comes to an equilibrium is :

The entropy increase is C ln\(\left[(T_{1} + T_{2})^{2}/4T_{1}T_{2}\right]\).Identical blocks with equal heat capacity reach the final temperature \(T_{f} = (T_{1} + T_{2})/2\). Each block's entropy change is \(C\ln(T_{f}/T_{i})\), so:\[ \begin{aligned} \Delta S &= C\ln\frac{T_{f}}{T_{1}} + C\ln\frac{T_{f}}{T_{2}} = C\ln\frac{T_{f}^{2}}{T_{1}T_{2}} \\ &= C\ln\frac{(T_{1} + T_{2})^{2}}{4T_{1}T_{2}} \end{aligned} \]Because the arithmetic mean exceeds the geometric mean, the argument is greater than 1 and \(\Delta S \gt 0\), as required for an irreversible process. Option D is the inverse ratio, which would give a decrease.

Q: One block at temperature \(T_{1}\) and another identical block at temperature \(T_{2}\) are placed in thermal contact and come to an equilibrium, the system being isolated from surroundings. If the heat capacities at constant volume C depend on temperature as C = aT, the entropy change of the system would be :

The entropy change is \(a\Big[2\Big(\frac{T_{1}^{2}+T_{2}^{2}}{2}\Big)^{1/2}-T_{1}-T_{2}\Big]\).Energy is conserved, so the heat lost by one block equals the heat gained by the other. With \(C = aT\):\[ \begin{aligned} \int_{T_{1}}^{T_{f}} aT\,dT + \int_{T_{2}}^{T_{f}} aT\,dT &= 0 \\ T_{f} &= \left(\frac{T_{1}^{2} + T_{2}^{2}}{2}\right)^{1/2} \end{aligned} \]Then \(dS = C\,dT/T = a\,dT\), so each block changes entropy by \(a(T_{f} - T_{i})\):\[ \Delta S = a(2T_{f} - T_{1} - T_{2}) \]The ln forms belong to a constant heat capacity, where \(T_{f}\) is the arithmetic mean; they do not apply when C grows with T.

Q: In adiabatic expansion of a system in which temperature changes from T1 to T2 the entropy :

The entropy will remain unchanged.In a reversible adiabatic process, no heat is exchanged, \(dQ = 0\), so \(dS = \frac{dQ_{rev}}{T} = 0\). The process is isentropic: the temperature changes from T1 to T2 because work is done, but the entropy stays the same.Entropy rises only in an irreversible adiabatic process, such as free expansion, which is not the standard meaning of 'adiabatic expansion' here. It never decreases in an adiabatic process, and the ratio of temperatures does not decide the sign.

Q: A solid with heat capacity \(C_{A}\) at temperature \(T_{A}\) is placed in contact with the heat bath having P heat capacity \(C_{B}\) at temperature \(T_{B}\) such that \(C_{B} \gg C_{A}\). The change in entropy of the combined system after a solid reached a thermal equilibrium with a heat bath is \(\Delta S\) = .....

The entropy change is \(C_{A}\left[\ln\frac{T_{B}}{T_{A}} + \frac{T_{A}}{T_{B}} - 1\right]\).Because \(C_{B} \gg C_{A}\), the bath stays at \(T_{B}\) and the solid ends at \(T_{B}\). The solid absorbs heat \(Q = C_{A}(T_{B} - T_{A})\), which the bath gives up at constant temperature:\[ \begin{aligned} \Delta S_{A} &= \int_{T_{A}}^{T_{B}}\frac{C_{A}dT}{T} \\ &= C_{A}\ln\frac{T_{B}}{T_{A}} \\ \Delta S_{B} &= -\frac{C_{A}(T_{B} - T_{A})}{T_{B}} \\ &= C_{A}\left(\frac{T_{A}}{T_{B}} - 1\right) \end{aligned} \]Adding them gives the keyed result, which is never negative (since \(\ln x \ge 1 - 1/x\)), as the second law requires.The options with \(C_{B}\ln\) terms treat the bath as if its temperature changed. The option with \(T_{B}/T_{A}\) inverts the bath's ratio and does not follow from \(Q/T_{B}\).

Direction for [ Question No: 11 To 11 ] :

Note: Usual notations have been used for thermodynamic variables.

Useful Data: Unless otherwise specified, the following data may be assumed. Universal gas constant, \(\bar{R}\) = 8.314 kJ/kmol.K; Acceleration due to gravity, g = 9.81 m/s²; Molecular mass of air, M_air = 29 kg/kmol; Specific heat of air at constant pressure, c_p = 1.005 kJ/kg.K; Ratio of specific heats of air, γ = 1.4. Assume air to be a perfect gas unless specified otherwise.

11.

32 kg of oxygen is mixed with 28 kg of nitrogen at the same temperature. The gases are at the same pressure of 103 kPa before and after mixing. If \(\bar{R}\) is the universal gas constant in kJ/kmol.K, the change in entropy of the mixture is

A.\(1.38\bar{R}\)
B.\(0.69\bar{R}\)
C.\(\bar{R}\)
D.\(0.34\bar{R}\)

Exam Year(s)

GATE XE - 2012,

Answer: A

The entropy change is \(1.38\bar{R}\).

32 kg of O₂ is 1 kmol and 28 kg of N₂ is 1 kmol, so each mole fraction is 0.5. Mixing ideal gases at the same temperature and pressure gives

\[ \begin{aligned} \Delta S &= -\bar{R}\sum n_{i}\ln x_{i} = -\bar{R}\,[\ln 0.5 + \ln 0.5] \\ &= 2\bar{R}\ln 2 = 1.386\bar{R} \end{aligned} \]

The option \(0.69\bar{R}\) counts only one gas's contribution, \(\ln 2\). \(\bar{R}\) and \(0.34\bar{R}\) do not follow from the formula. The pressure value, 103 kPa, does not matter, since it is the same before and after.

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12.

The entropy of a reversible process always:

A.Remain constant
B.Increases
C.Decreases
D.Zero

Exam Year(s)

RSMSSB JE - 2025, RSMSSB - 2025,

Answer: A

The answer given is that it remains constant.

A reversible process generates no entropy. The total entropy of the system and its surroundings therefore stays constant, \(\Delta S_{universe} = 0\), and in a reversible adiabatic process the entropy of the system itself is constant, which is why such a process is called isentropic.

An increase in total entropy is the mark of an irreversible process, and a decrease is impossible for an isolated system. 'Zero' confuses the change in entropy with the entropy itself, which is not zero.

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13.

The entropy of an isolated system or universe always:

A.Increases
B.Decrease
C.Remain constant
D.Zero

Exam Year(s)

RSMSSB JE - 2025, RSMSSB - 2025,

Answer: A

The entropy of an isolated system or the universe increases.

This is the principle of increase of entropy, a statement of the second law: for an isolated system \(\Delta S \geq 0\), with equality only for an ideal reversible process. Every real process is irreversible, so the entropy of the universe always increases.

A decrease would violate the second law. Remaining constant holds only in the reversible limit, which real processes never reach, and the entropy is not zero.

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14.

The net entropy generation (considering the system and the thermal reservoir together) during the process is closest to :

[given : 1 kJ energy transfer takes place at 400° C, while the increase in the entropy of system is 10 J/K]

A.10 J/K
B.7.50 J/K
C.7.70 J/K
D.8.51 J/K

Exam Year(s)

RSMSSB JE - 2025, RSMSSB - 2025,

Answer: D

The net entropy generation is closest to 8.51 J/K.

The reservoir at 400°C, that is 673 K, gives up 1 kJ, so its entropy falls. The system's entropy rises by 10 J/K.

\[ \begin{aligned} \Delta S_{res} &= -\frac{1000}{673} = -1.49\ \text{J/K} \\ S_{gen} &= 10 - 1.49 = 8.51\ \text{J/K} \end{aligned} \]

The temperature must be in kelvin: using 400 directly would give \(10 - 2.5 = 7.50\) J/K, which is the trap among the options. 10 J/K counts only the system and ignores the reservoir's loss of entropy.

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15.

When a perfect gas enclosed in cylinder - piston device executes a reversible adiabatic expansion process :

A.The entropy change of the surrounding will be zero
B.Its entropy will increase
C.Its entropy will decrease
D.Its entropy change will be zero

Exam Year(s)

RSMSSB JE - 2025, RSMSSB - 2025,

Answer: D

The key's answer is its entropy change will be zero.

A reversible adiabatic process is isentropic. Entropy changes by \(dS = \frac{\delta Q_{rev}}{T}\); with no heat transfer and no irreversibility (no friction, no unrestrained expansion), the gas's entropy stays exactly the same.

Entropy would increase only in an irreversible adiabatic process, and it can never decrease in an adiabatic process, since that would violate the second law. The statement about the surroundings is also true in itself, because no heat reaches them, but it describes the surroundings rather than the gas; the defining property of the gas's own process is zero entropy change, which is what the key accepts.

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16.

Two identical blocks of a metal with heat capacity C are initially at temperatures \( T_{1} \) and \( T_{2} \), \( (T_{1} \lt T_{2}) \). They are brought in thermal contact. When the system reaches equilibrium the change in entropy would be :

A.\( C \ln \left(\frac{(T_{1} + T_{2})^{2}}{T_{1}T_{2}}\right) \)
B.\( C \ln \left(\frac{(T_{1} + T_{2})^{2}}{4T_{1}T_{2}}\right) \)
C.\( C \ln \left(\frac{T_{1}T_{2}}{(T_{1} + T_{2})^{2}}\right) \)
D.\( C \ln \left(\frac{(T_{1} + T_{2})^{2}}{2T_{1}T_{2}}\right) \)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: B

The change in entropy is \( C \ln\left(\frac{(T_{1} + T_{2})^{2}}{4T_{1}T_{2}}\right) \).

Identical blocks reach the mean temperature \( T_{f} = \frac{T_{1} + T_{2}}{2} \). Each block's entropy changes by \( C\ln(T_{f}/T_{i}) \):

\[ \begin{aligned} \Delta S &= C \ln\frac{T_{f}}{T_{1}} + C \ln\frac{T_{f}}{T_{2}} \\ &= C \ln\frac{T_{f}^{2}}{T_{1}T_{2}} \\ &= C \ln\frac{(T_{1} + T_{2})^{2}}{4T_{1}T_{2}} \end{aligned} \]

Since the arithmetic mean exceeds the geometric mean, the argument is greater than 1 and \( \Delta S \gt 0 \), as an irreversible process requires. The option with the inverted fraction is negative; the ones with 1 or 2 in place of 4 forget that \( T_{f} \) is a half-sum, squared.

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17.

One block of certain metal is at temperature \(T_{1}\) and a second identical block is at a temperature \(T_{2}\). These blocks are brought in contact and the system is thermally isolated from the surroundings. Assume that the heat capacity at constant volume C of each block is independent of temperature T. The increase in the entropy of the universe, when the system comes to an equilibrium is :

A.C ln\(\left[\frac{T_{1}T_{2}}{|T_{2} - T_{1}|}\right]\)
B.C ln\(\left[\frac{(T_{1} + T_{2})}{2T_{1}T_{2}}\right]\)
C.C ln\(\left[\frac{(T_{1} + T_{2})^{2}}{4T_{1}T_{2}}\right]\)
D.C ln\(\left[\frac{2T_{1}T_{2}}{T_{1} + T_{2}}\right]\)

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: C

The entropy increase is C ln\(\left[(T_{1} + T_{2})^{2}/4T_{1}T_{2}\right]\).

Identical blocks with equal heat capacity reach the final temperature \(T_{f} = (T_{1} + T_{2})/2\). Each block's entropy change is \(C\ln(T_{f}/T_{i})\), so:

\[ \begin{aligned} \Delta S &= C\ln\frac{T_{f}}{T_{1}} + C\ln\frac{T_{f}}{T_{2}} = C\ln\frac{T_{f}^{2}}{T_{1}T_{2}} \\ &= C\ln\frac{(T_{1} + T_{2})^{2}}{4T_{1}T_{2}} \end{aligned} \]

Because the arithmetic mean exceeds the geometric mean, the argument is greater than 1 and \(\Delta S \gt 0\), as required for an irreversible process. Option D is the inverse ratio, which would give a decrease.

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18.

One block at temperature \(T_{1}\) and another identical block at temperature \(T_{2}\) are placed in thermal contact and come to an equilibrium, the system being isolated from surroundings. If the heat capacities at constant volume C depend on temperature as C = aT, the entropy change of the system would be :

A.\(a\Big[2\Big(\frac{T_{1}^{2}+T_{2}^{2}}{2}\Big)^{1/2}-T_{1}-T_{2}\Big]\)
B.\(a\left[\frac{T_{1} + T_{2}}{2}\right]\)
C.\(a\,\text{ln}\left[\frac{(T_{1} + T_{2})^{2}}{4T_{1}T_{2}}\right]\)
D.\(a\,\text{ln}\left[\frac{2T_{1}T_{2}}{T_{1} + T_{2}}\right]\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: A

The entropy change is \(a\Big[2\Big(\frac{T_{1}^{2}+T_{2}^{2}}{2}\Big)^{1/2}-T_{1}-T_{2}\Big]\).

Energy is conserved, so the heat lost by one block equals the heat gained by the other. With \(C = aT\):

\[ \begin{aligned} \int_{T_{1}}^{T_{f}} aT\,dT + \int_{T_{2}}^{T_{f}} aT\,dT &= 0 \\ T_{f} &= \left(\frac{T_{1}^{2} + T_{2}^{2}}{2}\right)^{1/2} \end{aligned} \]

Then \(dS = C\,dT/T = a\,dT\), so each block changes entropy by \(a(T_{f} - T_{i})\):

\[ \Delta S = a(2T_{f} - T_{1} - T_{2}) \]

The ln forms belong to a constant heat capacity, where \(T_{f}\) is the arithmetic mean; they do not apply when C grows with T.

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19.

In adiabatic expansion of a system in which temperature changes from T₁ to T₂ the entropy :

A.will increase
B.will decrease
C.will remain unchanged
D.may increase or decrease depending on the ratio of T\(_{1}\)/T\(_{2}\)

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: C

The entropy will remain unchanged.

In a reversible adiabatic process, no heat is exchanged, \(dQ = 0\), so \(dS = \frac{dQ_{rev}}{T} = 0\). The process is isentropic: the temperature changes from T₁ to T₂ because work is done, but the entropy stays the same.

Entropy rises only in an irreversible adiabatic process, such as free expansion, which is not the standard meaning of 'adiabatic expansion' here. It never decreases in an adiabatic process, and the ratio of temperatures does not decide the sign.

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20.

A solid with heat capacity \(C_{A}\) at temperature \(T_{A}\) is placed in contact with the heat bath having P heat capacity \(C_{B}\) at temperature \(T_{B}\) such that \(C_{B} \gg C_{A}\). The change in entropy of the combined system after a solid reached a thermal equilibrium with a heat bath is \(\Delta S\) = .....

A.\(C_{A}\ln\frac{T_{B}}{T_{A}} + C_{B}\ln\frac{T_{B}}{T_{A}}\)
B.\(C_{A}\left[\ln\frac{T_{B}}{T_{A}} + \frac{T_{A}}{T_{B}} - 1\right]\)
C.\(C_{A}\left[\ln\frac{T_{B}}{T_{A}} + \frac{T_{B}}{T_{A}} - 1\right]\)
D.\(C_{A}\ln\frac{T_{B}}{T_{A}} + C_{B}\ln\left(\frac{T_{A}}{T_{B}} - 1\right)\)

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: B

The entropy change is \(C_{A}\left[\ln\frac{T_{B}}{T_{A}} + \frac{T_{A}}{T_{B}} - 1\right]\).

Because \(C_{B} \gg C_{A}\), the bath stays at \(T_{B}\) and the solid ends at \(T_{B}\). The solid absorbs heat \(Q = C_{A}(T_{B} - T_{A})\), which the bath gives up at constant temperature:

\[ \begin{aligned} \Delta S_{A} &= \int_{T_{A}}^{T_{B}}\frac{C_{A}dT}{T} \\ &= C_{A}\ln\frac{T_{B}}{T_{A}} \\ \Delta S_{B} &= -\frac{C_{A}(T_{B} - T_{A})}{T_{B}} \\ &= C_{A}\left(\frac{T_{A}}{T_{B}} - 1\right) \end{aligned} \]

Adding them gives the keyed result, which is never negative (since \(\ln x \ge 1 - 1/x\)), as the second law requires.

The options with \(C_{B}\ln\) terms treat the bath as if its temperature changed. The option with \(T_{B}/T_{A}\) inverts the bath's ratio and does not follow from \(Q/T_{B}\).

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Entropy Questions and Answers

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