Practice a College Board-style free response question on Thermodynamics. Write your response, then reveal the model answer to see exactly what earns each point.
Free Response Question · Unit 1 · PV Diagrams & the First Law
A fixed amount of ideal gas undergoes the cyclic process A → B → C → A shown below on a PV diagram. The cycle consists of three steps:
Step
Description
Pressure
Volume
A → B
Isobaric expansion
Constant at 2.0 × 10⁵ Pa
Increases from 1.0 × 10⁻³ m³ to 3.0 × 10⁻³ m³
B → C
Isochoric cooling
Decreases from 2.0 × 10⁵ Pa to 1.0 × 10⁵ Pa
Constant at 3.0 × 10⁻³ m³
C → A
Compression back to A along a straight line on the PV diagram
Increases from 1.0 × 10⁵ Pa to 2.0 × 10⁵ Pa
Decreases from 3.0 × 10⁻³ m³ to 1.0 × 10⁻³ m³
A
Calculate the work done BY the gas during step A → B, and state whether this work is positive or negative for the gas.
✓ Model answer (earns the point)
Since A → B is isobaric, work done by the gas equals W = PΔV. Here, P = 2.0 × 10⁵ Pa and ΔV = (3.0 × 10⁻³ − 1.0 × 10⁻³) m³ = 2.0 × 10⁻³ m³.
W = (2.0 × 10⁵ Pa)(2.0 × 10⁻³ m³) = 400 J. Because the gas expands (volume increases), this work is positive — the gas does positive work on its surroundings.
Why it scores: Identifies the correct formula for an isobaric process (W = PΔV), correctly substitutes values with correct units, arrives at the numerical answer, AND states the correct sign with physical reasoning (expansion → positive work by the gas).
B
Determine the work done by the gas during step B → C. Justify your answer using the properties of an isochoric process.
✓ Model answer (earns the point)
The work done by the gas during B → C is zero. Step B → C is an isochoric (constant volume) process — the volume remains fixed at 3.0 × 10⁻³ m³ throughout. Since work done by a gas equals the area under the PV curve (∫PdV), and dV = 0 at every point during this step, no work is done by or on the gas, regardless of how much the pressure changes.
Why it scores: States the correct numerical answer (W = 0) AND justifies it by connecting the constant-volume condition directly to the work integral/area-under-curve reasoning, rather than just asserting the rule without explanation.
C
The net work done by the gas over the full cycle A → B → C → A equals the area enclosed by the triangular loop on the PV diagram, which is 100 J. Using the first law of thermodynamics, determine the net heat exchanged by the gas over the complete cycle, and state whether heat is absorbed or released overall.
✓ Model answer (earns the point)
Over one complete cycle, the gas returns to its initial state, so its internal energy is unchanged: ΔU_cycle = 0. Applying the first law, ΔU = Q − W, so 0 = Q − 100 J, which gives Q = 100 J. Since Q is positive, the gas absorbs a net 100 J of heat over the full cycle — exactly enough to supply the 100 J of net work it performs on its surroundings, since none of that energy comes from a change in internal energy.
Why it scores: Recognizes that ΔU = 0 for any full cycle of an ideal gas (because U depends only on T, and T returns to its initial value), correctly applies the first law to solve for Q, gets the correct numerical value, AND states the correct direction (heat absorbed, not released).
How to score points on AP Physics 2 FRQs
Show your equation before you substitute numbers. Graders award points for correctly identifying the relevant relationship (PΔV, ΔU = Q − W, etc.), not just for a final number.
State the sign and explain why. "Positive because the gas expands" earns more than a bare number — physics FRQs reward reasoning, not just arithmetic.
Connect to the underlying physical law. For cyclic processes, explicitly stating that ΔU = 0 because the gas returns to its initial temperature is the key justification graders look for.
Watch your units throughout. Pa·m³ = J — keep units consistent and convert before calculating, especially with temperature (always kelvin).
Answer exactly what's asked. "Calculate and justify" needs both a number and the reasoning; "determine" alone usually still wants you to show your work.