Common Error: Misassigning signs yields \(Q = -700\text{ J}\) (Choice D) or \(+700\text{ J}\) (Choice C).
Question 6 Correct: (B)
Rationale: In the reference frame of the container, molecules collide elastically with the moving piston. Because the piston has inward velocity, particles rebound with increased speed, increasing average translational kinetic energy and macroscopic temperature.
Common Error: Attributing ideal gas temperature changes to intermolecular potential energy.
AP Physics 2 — Topic 9.4 Scoring Guidelines Page 1 of 3
Section II: Short Answer Scoring Rubrics & Solutions 9 Points Total
Question 7: Path Independence & State Functions [4 Points] Conceptual Justification
(a) Comparing \(\Delta U\) [2 Points]:
Answer: \(\Delta U_A = \Delta U_B\).
(b) Comparing \(W_{\text{on}}\) [2 Points]:
Answer: \(W_{\text{on}, B} < W_{\text{on}, A}\) (more negative).
Exemplary Student Response: "Internal energy is a state function that depends only on the current state \((P, V, T)\) of the gas. Since both Path A and Path B begin at the same state \(i\) and end at the same state \(f\), the net change is identical: \(\Delta U_A = \Delta U_B\). For work, the area under the curve on a \(PV\) diagram represents the work done by the gas. Path B remains at the high initial pressure before depressurizing, enclosing more area than the curved isotherm of Path A. Because both processes are expansions, work done on the gas is negative; therefore, \(W_{\text{on}, B}\) is more negative than \(W_{\text{on}, A}\) (\(W_{\text{on}, B} < W_{\text{on}, A}\))."
Question 8: Analytical Calculation & PV Graphing [5 Points] Quantitative & Graphical
Work Calculation [2 Points]:
\(W_{\text{on}, 1} = -P\Delta V = -(1.0 \times 10^5\text{ Pa})(3.0 - 1.0)\times 10^{-3}\text{ m}^3\)
\(W_{\text{on}, 1} = -(1.0 \times 10^5)(2.0 \times 10^{-3}) = -200\text{ J}\)
PV Diagram Sketch [3 Points]:
Target Student Graph 1.0 0.5 1.0 2.0 3.0 State 1 State 2 State 3 Step 1: Isobaric | Step 2: Isochoric
AP Physics 2 — Topic 9.4 Scoring Guidelines Page 2 of 3
AP Exam Format — Qualitative / Quantitative Translation & Thermal Cycle Analysis
14 Points Total
Part (a): Temperatures at A and B [3 Points] Ideal Gas Law
State A: \(T_A = \frac{P_A V_A}{nR} = \frac{(1.0\times 10^5\text{ Pa})(2.0\times 10^{-3}\text{ m}^3)}{(0.050\text{ mol})(8.314\text{ J/(mol}\cdot\text{K)})} = \frac{200}{0.4157} \approx 481\text{ K}\)
State B: \(T_B = \frac{P_B V_B}{nR} = 3 \times T_A \approx 1443\text{ K}\) (accept 1440–1445 K).
Part (b): Work Done on the Gas [3 Points] Area Integration
Process \(B \to C\) [2 Points]: Area of trapezoid under linear curve: \(\text{Area} = \frac{P_B + P_C}{2}\Delta V = \frac{3.0\times 10^5 + 1.0\times 10^5}{2}(4.0\times 10^{-3}) = 800\text{ J}\). Gas expands \(\implies W_{\text{on}, B \to C} = -800\text{ J}\).
Process \(C \to A\) [1 Point]: Isobaric compression: \(W_{\text{on}, C \to A} = -P\Delta V = -(1.0\times 10^5)(2.0 - 6.0)\times 10^{-3} = +400\text{ J}\).
Part (c): Thermodynamic Cycle Table Solution [4 Points] 1 pt per fully correct row
| Process | \(\Delta U\) (J) | \(Q\) (J) | \(W_{\text{on}}\) (J) | Scoring Note / Derivation |
|---|---|---|---|---|
| \(A \to B\) | +600 | +600 | 0 | Isochoric: \(W=0\), \(\Delta U = \frac{3}{2}\Delta(PV) = \frac{3}{2}(400) = +600\text{ J}\) |
| \(B \to C\) | 0 | +800 | -800 | \(P_B V_B = P_C V_C = 600\text{ J} \implies \Delta T = 0 \implies \Delta U = 0\); \(Q = -W\) |
| \(C \to A\) | -600 | -1000 | +400 | \(\Delta U = \frac{3}{2}(200 - 600) = -600\text{ J}\); \(Q = \Delta U - W = -600 - 400 = -1000\text{ J}\) |
| Full Cycle | 0 | +400 | -400 | State function: \(\sum\Delta U = 0\); \(W_{\text{net, by}} = \text{area of enclosed triangle} = +400\text{ J}\) |
Part (d): AP Paragraph-Length Argument Rubric [4 Points] Scientific Argumentation
1 pt (First Law Application): Clearly identifies that although work is done on the gas during compression (\(W_{\text{on}} = +400\text{ J}\)), heat is simultaneously released to the surroundings (\(Q = -1000\text{ J}\)).
1 pt (Internal Energy & Temperature): Reasons that because \(|Q_{\text{released}}| > |W_{\text{done on}}|\), the net internal energy changes negatively (\(\Delta U = -600\text{ J}\)), requiring the temperature to decrease (\(T_C \to T_A\)).
1 pt (Microscopic Collision Mechanism): Explains that while gas particles gain speed upon striking the inward-moving piston, they collide with cooler cylinder walls / cooling reservoir, transferring more kinetic energy away than gained from the piston.
1 pt (AP Coherence & Structure): Connects macroscopic energy transfers to particle mechanics in an internally consistent, well-reasoned paragraph without contradiction.
Exemplary AP Paragraph Response: "The student's claim is incorrect because it accounts only for the mechanical work done on the gas while ignoring thermal energy transfer. According to the First Law of Thermodynamics, \(\Delta U = Q + W_{\text{on}}\). During the isobaric compression from \(C\) to \(A\), \(+400\text{ J}\) of work is performed on the gas. However, to maintain a constant pressure as the volume decreases by a factor of three, the gas must be cooled by removing \(1000\text{ J}\) of heat (\(Q = -1000\text{ J}\)). Because the magnitude of heat removed far exceeds the mechanical work done, the net change in internal energy is negative (\(\Delta U = -600\text{ J}\)), meaning the temperature drops from \(1443\text{ K}\) to \(481\text{ K}\). Microscopically, although gas molecules momentarily gain kinetic energy upon rebounding from the advancing piston, they continually collide with the cooler container walls and lose substantially more kinetic energy to the external thermal reservoir than they acquire from the moving piston. Consequently, the average translational kinetic energy of the molecules decreases, causing the temperature to decrease."
AP Physics 2 — Topic 9.4 Scoring Guidelines Page 3 of 3
