AP Physics 2 - 9.6 Entropy and the Second Law of Thermodynamics- Exam Style questions- MCQs
Entropy and the Second Law of Thermodynamics AP Physics 2 MCQ
Unit 9: Thermodynamics
Weightage : 15–18%
Question

The figure shows a sample of an ideal gas enclosed within a cylinder that has been fitted with a movable piston. The piston and the sides of the cylinder are thermally insulated. The bottom of the cylinder is in contact with a thermal reservoir. The gas is compressed isothermally while thermal equilibrium is maintained with the reservoir. The piston, gas, and reservoir form a closed, isolated system. True statements about entropy for this reversible situation include which of the following? Select two answers.
(B) The entropy of the reservoir increases because thermal energy is transferred to it from the gas.
(C) The system has a net increase in entropy as a result of the process.
(D) The change in entropy of the system depends on how quickly the isothermal compression occurs.
▶️ Answer/Explanation
Correct Answers: \( \boxed{\mathrm{A \;and\; B}} \)
During an isothermal compression of an ideal gas, the temperature remains constant. Since work is done on the gas, an equal amount of thermal energy is transferred from the gas to the thermal reservoir to keep the temperature unchanged.
For a reversible isothermal process,
\( \Delta S=\dfrac{Q_{\mathrm{rev}}}{T}. \)
Option A: The gas loses thermal energy to the reservoir, so \(Q_{\mathrm{gas}}<0\). Therefore, the entropy of the gas decreases.
Option B: The reservoir gains exactly the same amount of thermal energy that the gas loses. Thus, the entropy of the reservoir increases.
Option C: Because the process is reversible, the entropy decrease of the gas is exactly balanced by the entropy increase of the reservoir. Therefore,
\( \Delta S_{\mathrm{total}}=0, \)
so there is no net increase in entropy.
Option D: For a reversible isothermal process, the entropy change depends only on the initial and final equilibrium states, not on how quickly the compression occurs.
Question

Two identical samples of helium gas are in identical sealed flasks at room temperature. One flask is just sitting in the room, and the other is inside a large, insulated vacuum container. Both flasks are opened and the samples are released, so the helium in one flask spreads throughout the room and the helium in the other flask spreads in the sealed container, as shown in the figures. Which of the following is true of the change in entropy that occurs in each case when the flasks are opened?
(B) The entropy increases as the helium mixes with air but not as it spreads out into the sealed container.
(C) There is no change in entropy for either case because the temperature of the helium does not change.
(D) The entropy changes in both cases, but there is not sufficient information to determine whether it increases or decreases.
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{A}} \)
When the flasks are opened, the helium gas spontaneously expands into a much larger available volume. In the room, the helium also mixes with the surrounding air, while in the vacuum container it freely expands to occupy the entire container.
In both situations, the number of accessible microscopic arrangements (microstates) increases. According to Boltzmann’s relation,
\( S=k_{\mathrm{B}}\ln\Omega, \)
where \(S\) is entropy and \(\Omega\) is the number of possible microscopic states. Since \(\Omega\) increases when the gas spreads out, the entropy also increases.
Although the temperature remains essentially constant, entropy depends on the number of accessible microstates, not solely on temperature. Therefore, both processes result in an increase in entropy.
Hence, the correct answer is (A).
Question
Which of the following processes is not involved in an ideal Carnot cycle?
(B) Isobaric expansion
(C) Adiabatic expansion
(D) Adiabatic compression
▶️ Answer/Explanation
Correct Answer: \( \boxed{\mathrm{B}} \)
An ideal Carnot cycle consists of four reversible thermodynamic processes:
• Isothermal expansion
• Adiabatic expansion
• Isothermal compression
• Adiabatic compression
During the isothermal processes, heat is exchanged with the hot and cold reservoirs while the temperature remains constant. During the adiabatic processes, no heat is transferred:
\( Q=0. \)
An isobaric (constant-pressure) process is not part of the Carnot cycle.
Therefore, the correct answer is (B).
