AP Physics 2 Thermodynamics — Worked Answer Explanations

Unit 1 · 18% of the AP exam · 12 questions explained

Below is a complete answer key for our AP Physics 2 Thermodynamics practice questions. For each question you'll find the correct choice, a full written explanation of how to get there, and — for every wrong answer — a short note on exactly why it's tempting and where it goes wrong. Reading these straight through is one of the fastest ways to find the gaps in a unit before exam day.

Prefer to test yourself first? Take the timed Thermodynamics practice test and come back here to review, or head back to the Thermodynamics unit overview.

In-content ad

Question 1 · Easy
Which of the following best describes the zeroth law of thermodynamics?
- A
  Energy cannot be created or destroyed in an isolated system.
  Why not A: This is the first law of thermodynamics, not the zeroth.
- B
  If two systems are each in thermal equilibrium with a third system, they are in thermal equilibrium with each other.Correct
- C
  Heat flows spontaneously from cold objects to hot objects.
  Why not C: Heat flows from hot to cold spontaneously; this reversal violates the second law.
- D
  The entropy of a perfect crystal at absolute zero is zero.
  Why not D: This is the third law of thermodynamics.
Explanation
The zeroth law establishes thermal equilibrium as a transitive relation: if $A \leftrightarrow C$ (equilibrium) and $B \leftrightarrow C$ , then $A \leftrightarrow B$ . This underpins the concept of temperature — two objects in mutual equilibrium share the same temperature. The first law addresses energy conservation, the second addresses entropy and spontaneous direction of heat flow, and the third addresses absolute zero entropy.

Key takeaway
The zeroth law defines thermal equilibrium transitivity, which is the basis for temperature measurement.
Question 2 · Easy
A gas undergoes an isothermal process. Which of the following is true about the change in internal energy $Δ U$ of an ideal gas during this process?
- A
  $Δ U > 0$ because work is done on the gas.
  Why not A: Whether work is done on or by the gas depends on compression vs expansion; the key is that isothermal means constant temperature.
- B
  $Δ U < 0$ because the gas must release heat to maintain constant temperature.
  Why not B: Whether heat is released depends on direction (expansion vs compression). The internal energy depends only on temperature for an ideal gas.
- C
  $Δ U = 0$ because internal energy of an ideal gas depends only on temperature.Correct
- D
  $Δ U = Q$ because no work is done during an isothermal process.
  Why not D: Work is done during an isothermal process (volume changes); $Δ U = 0 = Q - W$ , so $Q = W$ , not $Δ U = Q$ .
Explanation
For an ideal gas, internal energy depends only on temperature: $U = \frac{f}{2} n R T$ where $f$ is degrees of freedom. An isothermal process keeps $T$ constant, so $Δ T = 0$ and thus $Δ U = 0$ . From the first law: $Δ U = Q - W = 0$ , which means $Q = W$ — any heat added equals work done by the gas (or vice versa for compression).

Key takeaway
For an ideal gas, $\Delta U$ depends only on $\Delta T$; isothermal means $\Delta T = 0$, so $\Delta U = 0$.
Question 3 · Easy
An ideal gas is held at constant volume while heat $Q$ is added to it. Which expression correctly gives the change in internal energy $Δ U$ ?
- A
  $Δ U = Q - W$ , where $W = P Δ V > 0$
  Why not A: At constant volume $Δ V = 0$ , so $W = P Δ V = 0$ ; no work is done by the gas.
- B
  $Δ U = Q$ Correct
- C
  $Δ U = - Q$
  Why not C: The sign is wrong; adding heat increases internal energy, not decreases it.
- D
  $Δ U = 0$ , because temperature does not change at constant volume.
  Why not D: Temperature does change when heat is added at constant volume; internal energy increases.
Explanation
The first law states $Δ U = Q - W$ . Work done by an ideal gas is $W = P Δ V$ . At constant volume, $Δ V = 0$ , so $W = 0$ . Therefore $Δ U = Q - 0 = Q$ . All the added heat goes into raising the internal energy (and thus the temperature) of the gas. This is exactly why the molar heat capacity at constant volume $C_{V}$ is defined by $Q = n C_{V} Δ T$ .

Key takeaway
At constant volume, no $PdV$ work is done, so $\Delta U = Q$ by the first law.
Question 4 · Easy
The average kinetic energy of molecules in an ideal monatomic gas at temperature $T$ (in kelvin) is given by:
- A
  $\overset{ˉ}{K} = k T$
  Why not A: This is missing the $\frac{3}{2}$ factor from the equipartition theorem for three translational degrees of freedom.
- B
  $\overset{ˉ}{K} = \frac{1}{2} k T$
  Why not B: This represents the energy per single degree of freedom, not the total for a monatomic gas.
- C
  $\overset{ˉ}{K} = \frac{3}{2} k T$ Correct
- D
  $\overset{ˉ}{K} = \frac{3}{2} n R T$
  Why not D: The $n R$ form gives the total internal energy of $n$ moles, not the average per molecule. Per-molecule uses Boltzmann constant $k$ .
Explanation
The equipartition theorem assigns $\frac{1}{2} k T$ of energy per degree of freedom. A monatomic ideal gas has 3 translational degrees of freedom (x, y, z), so the average kinetic energy per molecule is:
$\overset{ˉ}{K} = \frac{3}{2} k T$
where $k = 1.38 \times 1 0^{- 23} J/K$ is Boltzmann's constant. For $n$ moles of gas the total internal energy is $U = \frac{3}{2} n R T$ .

Key takeaway
Each translational degree of freedom contributes $\frac{1}{2}kT$; a monatomic ideal gas has 3, giving $\bar{K} = \frac{3}{2}kT$.
Question 5 · Medium
An ideal gas expands adiabatically and does $W = 500 J$ of work on its surroundings. What is the change in internal energy of the gas?
- A
  $Δ U = + 500 J$
  Why not A: This confuses the sign; the gas loses internal energy when it does positive work adiabatically.
- B
  $Δ U = 0$
  Why not B: $Δ U = 0$ would require an isothermal process; adiabatic expansion changes temperature.
- C
  $Δ U = - 500 J$ Correct
- D
  Cannot be determined without knowing the pressure.
  Why not D: The first law gives $Δ U$ directly from $Q$ and $W$ ; pressure is not needed here.
Explanation
An adiabatic process has $Q = 0$ (no heat exchange). Applying the first law:
$Δ U = Q - W = 0 - 500 J = - 500 J$
The gas does positive work on its surroundings, so its internal energy decreases. This corresponds to a drop in temperature — adiabatic expansion cools a gas. This is the principle behind refrigeration and why air cools when it expands rapidly.

Key takeaway
Adiabatic means $Q=0$; all work comes from internal energy, so $\Delta U = -W$ for expansion.
Question 6 · Medium
A heat engine operates between a hot reservoir at $T_{H} = 600 K$ and a cold reservoir at $T_{C} = 300 K$ . What is the maximum (Carnot) efficiency of this engine?
- A
  $25%$
  Why not A: This would result from using Celsius temperatures incorrectly: $(600 - 300) /1200$ .
- B
  $50%$ Correct
- C
  $75%$
  Why not C: This inverts the ratio: $T_{C} / T_{H}$ instead of $1 - T_{C} / T_{H}$ .
- D
  $100%$
  Why not D: 100% efficiency is impossible by the second law; some heat must be rejected to the cold reservoir.
Explanation
The Carnot efficiency gives the theoretical maximum for any heat engine operating between two reservoirs:
$e_{Carnot} = 1 - \frac{T _{C}}{T _{H}}$
Temperatures must be in kelvin. Substituting:
$e = 1 - \frac{300}{600} = 1 - 0.5 = 0.50 = 50%$
No real engine can exceed this efficiency. The second law guarantees that $Q_{C} > 0$ ; perfect conversion of heat to work is impossible.

Key takeaway
Carnot efficiency $e = 1 - T_C/T_H$ uses kelvin temperatures and sets the upper bound for real engines.
Question 7 · Medium
A $P V$ diagram shows a gas undergoing a rectangular cycle (two isobaric and two isochoric processes). In which segment does the gas do the most positive work?
- A
  The isochoric (constant volume) expansion at high pressure.
  Why not A: An isochoric process has $Δ V = 0$ , so $W = P Δ V = 0$ — no work is done.
- B
  The isobaric compression at low pressure.
  Why not B: Compression is work done on the gas, so $W$ is negative for the gas.
- C
  The isobaric expansion at high pressure.Correct
- D
  The isochoric compression at low volume.
  Why not D: Isochoric means constant volume; $W = P Δ V = 0$ regardless of pressure.
Explanation
Work done by a gas is $W = P Δ V$ . In a rectangular $P V$ cycle:
- Isochoric segments: $Δ V = 0 \Rightarrow W = 0$
- Isobaric expansion at high pressure $P_{H}$ : $W = P_{H} Δ V > 0$ (largest)
- Isobaric compression at low pressure $P_{L}$ : $W = P_{L} (- Δ V) < 0$
The isobaric expansion at the higher pressure gives the largest positive work because $W = P_{H} Δ V$ and $P_{H} > P_{L}$ . The net work of the full cycle equals the area enclosed by the rectangle on the $P V$ diagram.
Key takeaway
On a $PV$ diagram, work is area under the curve; isobaric expansion at high $P$ contributes the largest positive work.
Question 8 · Medium
Which of the following processes is impossible according to the second law of thermodynamics?
- A
  A heat engine converts 40% of absorbed heat into work and rejects the rest.
  Why not A: This is allowed — it simply means the engine is less than 100% efficient, consistent with the second law.
- B
  Heat flows from a hot object to a cold object when they are in thermal contact.
  Why not B: This is the natural, spontaneous direction of heat flow — perfectly consistent with the second law.
- C
  A refrigerator removes heat from a cold interior and expels heat to a warm room, using electrical work.
  Why not C: A refrigerator does external work to move heat from cold to hot — this is allowed and is how real refrigerators operate.
- D
  A cyclic heat engine converts all absorbed heat into work with no heat rejected to a cold reservoir.Correct
Explanation
The Kelvin-Planck statement of the second law: it is impossible for any cyclic heat engine to convert heat entirely into work with no other effect (i.e., no heat rejected to a cold reservoir). Such an engine would have 100% efficiency, which violates the second law. All real engines must reject some heat $Q_{C} > 0$ to a cold reservoir. Options A, B, and C all describe physically allowed processes.

Key takeaway
The second law forbids a 100%-efficient cyclic heat engine; heat rejection to a cold reservoir is mandatory.
Question 9 · Medium
One mole of an ideal monatomic gas undergoes an isobaric expansion from volume $V_{1}$ to $2 V_{1}$ at pressure $P$ . Which expression correctly gives the heat $Q$ added to the gas? (Use $R$ for the ideal gas constant.)
- A
  $Q = P V_{1}$
  Why not A: This equals the work done $W = P Δ V = P V_{1}$ but ignores the change in internal energy.
- B
  $Q = \frac{3}{2} P V_{1}$
  Why not B: This equals $Δ U$ alone ( $\frac{3}{2} n R Δ T = \frac{3}{2} P Δ V$ ), forgetting to add the work done.
- C
  $Q = \frac{5}{2} P V_{1}$ Correct
- D
  $Q = 2 P V_{1}$
  Why not D: This would correspond to $C_{V} = 2 R$ per mole, not the correct $\frac{3}{2} R$ for a monatomic gas.
Explanation
For an isobaric process, $Q = Δ U + W$ .

Work done by the gas: $W = P Δ V = P (2 V_{1} - V_{1}) = P V_{1}$

Change in internal energy for 1 mol monatomic ideal gas: Using $P V = n R T$ , $Δ T = P Δ V / (n R) = P V_{1} / R$ . Then:
$Δ U = \frac{3}{2} n R Δ T = \frac{3}{2} (1) (R) (\frac{P V _{1}}{R}) = \frac{3}{2} P V_{1}$

Total heat: $Q = \frac{3}{2} P V_{1} + P V_{1} = \frac{5}{2} P V_{1}$

Alternatively, $Q = n C_{P} Δ T$ where $C_{P} = \frac{5}{2} R$ for monatomic ideal gas.

Key takeaway
For isobaric expansion, $Q = \Delta U + W = \frac{3}{2}P\Delta V + P\Delta V = \frac{5}{2}P\Delta V$ for a monatomic ideal gas.
Question 10 · Hard
A heat engine absorbs $Q_{H} = 800 J$ from a hot reservoir and does $W = 200 J$ of work per cycle. What is the change in entropy of the cold reservoir per cycle if it is at $T_{C} = 400 K$ ?
- A
  $Δ S_{C} = 0.5 J/K$
  Why not A: This uses $W / T_{C}$ instead of $Q_{C} / T_{C}$ ; the rejected heat, not the work, determines cold-reservoir entropy change.
- B
  $Δ S_{C} = 1.5 J/K$ Correct
- C
  $Δ S_{C} = 2.0 J/K$
  Why not C: This uses $Q_{H} / T_{C} = 800/400$ ; the heat rejected to the cold reservoir is $Q_{C} = Q_{H} - W$ , not $Q_{H}$ .
- D
  $Δ S_{C} = - 1.5 J/K$
  Why not D: The cold reservoir receives heat, so its entropy increases (positive $Δ S$ ).
Explanation
First, find the heat rejected to the cold reservoir using energy conservation:
$Q_{C} = Q_{H} - W = 800 - 200 = 600 J$

The entropy change of the cold reservoir (which absorbs $Q_{C}$ at constant temperature $T_{C}$ ):
$Δ S_{C} = \frac{Q _{C}}{T _{C}} = \frac{600}{400} = 1.5 J/K$

The cold reservoir gains entropy because it absorbs heat. Note that $Δ S_{hot} = - Q_{H} / T_{H}$ (hot reservoir loses entropy), and by the second law, the total entropy change $Δ S_{total} \geq 0$ .

Key takeaway
Entropy change of a reservoir: $\Delta S = Q/T$ where $Q$ is heat absorbed; find $Q_C = Q_H - W$ first.
Question 11 · Hard
An ideal gas undergoes the following four-step cycle: (1) isobaric expansion at $P_{H}$ , (2) isochoric pressure drop to $P_{L}$ , (3) isobaric compression at $P_{L}$ , (4) isochoric pressure rise back to $P_{H}$ . If the volume changes from $V_{1}$ to $V_{2} > V_{1}$ , what is the net work done by the gas per cycle?
- A
  $W_{net} = P_{H} V_{2} - P_{L} V_{1}$
  Why not A: This omits the work of compression and mixes volumes incorrectly.
- B
  $W_{net} = (P_{H} - P_{L}) (V_{1} + V_{2})$
  Why not B: The net work is the area of the rectangle, which is a product, not involving $V_{1} + V_{2}$ this way.
- C
  $W_{net} = (P_{H} - P_{L}) (V_{2} - V_{1})$ Correct
- D
  $W_{net} = P_{H} (V_{2} - V_{1}) - P_{L} (V_{2} - V_{1}) /2$
  Why not D: The factor of $1/2$ would apply to a triangular cycle, not a rectangular one.
Explanation
Net work equals the area enclosed by the cycle on a $P V$ diagram.
- Step 1 (isobaric expansion): $W_{1} = P_{H} (V_{2} - V_{1}) > 0$
- Step 2 (isochoric drop): $W_{2} = 0$
- Step 3 (isobaric compression): $W_{3} = P_{L} (V_{1} - V_{2}) = - P_{L} (V_{2} - V_{1}) < 0$
- Step 4 (isochoric rise): $W_{4} = 0$
$W_{net} = P_{H} (V_{2} - V_{1}) - P_{L} (V_{2} - V_{1}) = (P_{H} - P_{L}) (V_{2} - V_{1})$

This is the area of the rectangle with width $(V_{2} - V_{1})$ and height $(P_{H} - P_{L})$ .
Key takeaway
Net work of a rectangular $PV$ cycle equals the enclosed area: $(\Delta P)(\Delta V) = (P_H - P_L)(V_2 - V_1)$.
Question 12 · Hard
Two identical samples of an ideal gas start at the same state $(P_{0}, V_{0}, T_{0})$ . Sample A undergoes an adiabatic expansion to volume $2 V_{0}$ . Sample B undergoes an isothermal expansion to volume $2 V_{0}$ . Which sample ends at a higher pressure, and why?
- A
  Sample A, because adiabatic processes conserve energy better.
  Why not A: Adiabatic processes conserve energy (no heat transfer), but this causes temperature to drop, leading to lower pressure, not higher.
- B
  Sample B, because isothermal expansion keeps temperature constant while adiabatic expansion causes cooling.Correct
- C
  Both samples end at the same pressure, because they expand to the same volume.
  Why not C: Same final volume does not imply same final pressure; temperature differs between the two processes.
- D
  Sample A, because adiabatic expansion curves are steeper, reaching higher pressures.
  Why not D: Adiabatic curves are steeper (more negative slope) on a $P V$ diagram, which means the pressure drops more steeply for the same volume increase — ending at lower pressure.
Explanation
Using the ideal gas law $P = n R T / V$ at the final state $(2 V_{0})$ :

Sample B (isothermal): $T = T_{0}$ throughout, so:
$P_{B} = \frac{n R T _{0}}{2 V _{0}} = \frac{P _{0}}{2}$

Sample A (adiabatic): Temperature drops during expansion. For an adiabatic process, $P V^{γ} = const$ , so $P_{A} (2 V_{0})^{γ} = P_{0} V_{0}^{γ}$ :
$P_{A} = P_{0} (\frac{1}{2})^{γ}$
Since $γ > 1$ (e.g., $γ = 5/3$ for monatomic), $P_{A} < P_{0} /2 = P_{B}$ .

Sample B has higher final pressure because its temperature remains $T_{0}$ , while Sample A's temperature drops during adiabatic expansion. On a $P V$ diagram, the adiabat falls below the isotherm for the same volume increase.

Key takeaway
Adiabatic expansion cools the gas more than isothermal; same final volume means the colder gas (adiabatic) has lower final pressure.