Physics 9702 · syllabus 16
Cambridge A Level Physics 9702 — Thermodynamics topical past papers
Internal energy as the sum of molecular kinetic and potential energies, the work done when a gas changes volume, and the first law of thermodynamics.
- Questions
- 74
- Years
- 2009–2026
- With mark scheme
- 100%
- Questions 2022–2026
- 23
Anyone can browse these pages. Opening the questions and their mark schemes takes a free Momentum account.
What Thermodynamics covers
The internal energy of a system is the sum of the random kinetic and potential energies of its molecules, and it depends only on the state of the system, not on how it reached that state. Raising the temperature increases the average kinetic energy of the molecules and so the internal energy. During a change of state the temperature stays constant, but the internal energy still changes because the potential energy of the molecules changes.
In an ideal gas there are no forces between the molecules, so its internal energy is entirely kinetic and proportional to thermodynamic temperature. When a gas expands at constant pressure it does work W = pΔV on its surroundings; when it is compressed, the surroundings do work on it. Keeping track of which way the work goes is most of the topic.
The first law of thermodynamics, ΔU = q + W, is conservation of energy for a system: the increase in internal energy equals the energy supplied by heating plus the work done on the system. In this convention W is the work done on the gas, so W is negative when the gas expands. This is A Level content, examined on Paper 4, and it builds on temperature and ideal gases.
In the syllabus
The headings this topic has in the Cambridge International 2025–2027 syllabus, as the board prints them.
- 16 Thermodynamics
- 16.1 Internal energy
- 16.2 The first law of thermodynamics
What you should be able to do
- Explain internal energy as the sum of random molecular kinetic and potential energies
- Relate a rise in temperature to an increase in internal energy, and explain why melting changes internal energy at constant temperature
- Calculate W = pΔV and decide whether work is done on or by the gas
- Apply ΔU = q + W with consistent signs to heating, expansion and compression
- Analyse a cycle of changes using the fact that the internal energy returns to its starting value
Key ideas
- Internal energy U = sum of the random kinetic and potential energies of the molecules
- Work done at constant pressure W = pΔV, with 1 Pa × 1 m³ = 1 J
- First law: ΔU = q + W, where q is the energy supplied to the system by heating and W is the work done on the system
- For an ideal gas U is all kinetic and proportional to T; round a complete cycle ΔU = 0
How Thermodynamics has been examined
From 2022 to 2026, Thermodynamics was on Paper 4 · A Level Structured Questions in 8 of 11 exam series, worth about 7% of that paper’s marks.
| Paper | Questions | Share of marks | Series |
|---|---|---|---|
| Paper 1 · Multiple Choice | 1 | 1% | 1 of 11 |
| Paper 4 · A Level Structured Questions | 22 | 7% | 8 of 11 |
Momentum’s own count, over the questions in its bank for these years: a series is one exam session, every zone included, and a question filed under two chapters counts in both. Median marks per question: 9. Some of these questions sit in a bank chapter that also covers a neighbouring topic, so the counts can run a little high.
Filed in Momentum under: Thermal Properties of Materials.
A worked example
A gas at a constant pressure of 2.0 × 10⁵ Pa expands from 3.0 × 10⁻³ m³ to 5.0 × 10⁻³ m³ while 1000 J of energy is supplied to it by heating. Find the change in its internal energy.
- Work done by the gas: pΔV = 2.0 × 10⁵ × (5.0 − 3.0) × 10⁻³ = 400 J
- The gas expands, so the work done on it is W = −400 J
- Energy supplied by heating: q = +1000 J
- First law: ΔU = q + W = 1000 + (−400) = +600 J
Answer: The internal energy increases by 600 J, so the temperature of the gas rises
Written by Momentum for this page, with its own numbers; it is not taken from a past paper.
Mistakes that cost marks
- Using the sign convention in which W is the work done by the gas — here W is the work done on it, so for an expansion W = −pΔV and ΔU = q − pΔV
- Taking q as positive when the gas loses energy by heating — q is negative when energy leaves the system
- Including potential energy in the internal energy of an ideal gas — with no intermolecular forces it is all kinetic, which is why it depends only on temperature
- Using pΔV when the pressure changes during the process — the work is then the area under the p–V graph
Exam technique
- Set up a small table with columns for q, W and ΔU for each stage of a cycle, and fill in the signs before the numbers
- When stating the first law, say what each symbol means, including the direction of each energy transfer; the equation alone is incomplete
- On p–V graphs, find work from the area under the line, converting axis units such as kPa and cm³ or litres into Pa and m³
Practising in Momentum
- Every question in a chapter as one list, from every paper and year in the bank, newest first or shuffled.
- Each question turns over to its own part of the mark scheme, where the board published one.
- Narrow by paper, year, session, zone and marks, and star the questions worth keeping.
- Pick chapters, a number of questions and a time, and the exam builder draws a mock to sit and mark.
Past-paper questions and mark schemes are the copyright of Cambridge International. Momentum shows them to signed-in learners for revision and does not publish them on these pages.