The questions are on the first law thermodynamics test. Take it under real conditions before reading the mark scheme.
Quick answer: The first law of thermodynamics states ΔU = Q + W: the internal energy of a gas changes by the heat added plus the work done on it. This mark scheme gives the model answers and the marks for the 8 questions of the AP Physics 2 test on first law thermodynamics, from first law with numbers to engine or refrigerator?, with the equivalent answers to accept.
The model answers show the expected steps. Award marks for the correct sign convention and for reasoning in terms of energy, not only for numerical answers. Apply follow-through marking when a later part reuses an earlier wrong value correctly.
Level: AP Physics 2 (algebra-based) · Curriculum: AP P2 Unit 9
Question 1 First law with numbers (3 marks)
- Q = +500 J and W = −200 J.
- ΔU = 500 − 200 = +300 J.
- ΔT = 300/(1.5 × 0.10 × 8.31) = 241 K.
Marking: 1 mark per part; sign of W needed in part a.
Question 2 A two-step path (3 marks)
- pΔV = 1.5 × 10⁵ × 2.0 × 10⁻³ = 300 J.
- pAVA = 1.5 × 10⁵ × 1.0 × 10⁻³ = 150 J and pCVC = 0.5 × 10⁵ × 3.0 × 10⁻³ = 150 J. Equal pV means equal T.
- ΔU = 0, and B to C does no work, so Q = −W = +300 J.
Marking: 1 mark per part.
Question 3 Process types (2 marks)
- W = 0, so Q = ΔU.
- ΔU = 0 because T is constant, so the energy comes from heat flowing in from the surroundings (Q = −W).
Marking: 1 mark each part.
Question 4 An engine claim (3 marks)
- W = 1200 − 900 = 300 J.
- e = 300/1200 = 0.25.
- emax = 1 − 300/600 = 0.50. Since 0.25 < 0.50, the engine is possible.
Marking: 1 mark per part.
Question 5 Maximum efficiency (2 marks)
- emax = 1 − 300/450 = 0.333.
- Wmax = 300 J, so QC = 900 − 300 = 600 J.
Marking: 1 mark each part.
Question 6 Entropy and heat flow (3 marks)
- Hot body: −900/450 = −2.0 J/K. Cold body: +900/300 = +3.0 J/K.
- The total change is +1.0 J/K, which is positive, so total entropy increases as the second law requires.
Marking: Part a: 1 mark per body. Part b: 1 mark.
Question 7 Compressing a gas quickly (2 marks)
- The cylinder is insulated and the process is fast, so Q = 0.
- ΔU = Q + W with Q = 0 and W > 0, so ΔU is positive; for an ideal gas, a rise in U means a rise in T.
Marking: 1 mark each part.
Question 8 Engine or refrigerator? (2 marks)
- The gas does net work on the surroundings, so the net work done on the gas is negative.
- Net work is done on the gas and heat is moved from cold to hot, so it represents a refrigerator (or heat pump).
Marking: 1 mark each part.
