The first law of thermodynamics
Cambridge A-Level Physics (9702) · Unit 16: Thermodynamics · 8 flashcards
The first law of thermodynamics is topic 16.2 in the Cambridge A-Level Physics (9702) syllabus , positioned in Unit 16 — Thermodynamics , alongside Internal energy. In one line: The work done (W) during a constant pressure (p) volume change (ΔV) is given by: W = pΔV. Remember to use consistent units (Pascals for pressure and m³ for volume to get Joules for work).
Marked as A2 Level: examined at A Level in Paper 4 (A Level Structured Questions) and Paper 5 (Planning, Analysis and Evaluation). It is not tested on the AS-only papers (Papers 1, 2 and 3).
The deck below contains 8 flashcards — 2 definitions, 3 key concepts and 3 calculations — covering the precise wording mark schemes reward. Use the 2 definition cards to lock down command-word answers (define, state), then move on to the concept and calculation cards to handle explain, describe, calculate and compare questions.
Some cards in this deck are on other syllabus topics: 16.1 Internal energy. Each card is labelled with its topic.
The formula for work done during a constant pressure volume change
The work done (W) during a constant pressure (p) volume change (ΔV) is given by: W = pΔV. Remember to use consistent units (Pascals for pressure and m³ for volume to get Joules for work).
What the Cambridge 9702 syllabus says
2025-2027 syllabus, in our words · A2 LevelWhat the syllabus expects you to be able to do in this topic, in plain English. For the exact wording, see the official syllabus (PDF).
- Recall the work done at constant pressure as a gas changes volume, W = pΔV, and use it; understand how work done on the gas differs from work the gas itself does.
- Recall the first law of thermodynamics, ΔU = q + W, and use it, where ΔU is the rise in internal energy, q is energy given to the system through heating and W is work done on it.
Also in this deck: 16.1 Internal energy
- Understand that a system's internal energy depends on its state, and that it can be written as the sum of the randomly distributed kinetic and potential energies of its molecules.
- Relate an increase in an object's internal energy to a rise in its temperature.
Cambridge syllabus keywords to use in your answers
These are the official Cambridge 9702 terms tagged to this section. Mark schemes credit responses that use the exact term — weave them into your answers verbatim rather than paraphrasing.
Tips to avoid common mistakes in these topics
- › Use the convention ΔU = q + w, where q is energy supplied TO the system and w is work done ON the system.
- › For ideal gases, state clearly that internal energy is the sum of only the kinetic energies of the atoms/molecules.
- › Internal energy must be defined as the sum of a random distribution of kinetic and potential energies associated with the molecules.
Topic 16.2 · The first law of thermodynamics
What is the formula for work done during a constant pressure volume change?
The work done (W) during a constant pressure (p) volume change (ΔV) is given by: W = pΔV. Remember to use consistent units (Pascals for pressure and m³ for volume to get Joules for work).
Topic 16.2 · The first law of thermodynamics
What is the sign convention for work done 'by' the gas versus work done 'on' the gas?
In the Cambridge first law, ΔU = q + W, W is the work done ON the gas. When the gas is compressed, work is done on it, so W is positive. When the gas expands, it does work on the surroundings, so W is negative (W = –pΔV).
Topic 16.2 · The first law of thermodynamics
State the first law of thermodynamics in terms of internal energy, heating, and work.
The first law of thermodynamics states: ΔU = q + W, where ΔU is the increase in internal energy of the system, q is the energy transferred to the system by heating, and W is the work done on the system.
Topic 16.2 · The first law of thermodynamics
If a gas expands and does 500J of work while absorbing 300J of heat, what is the change in internal energy?
Since the gas does work, W = -500J (work is done 'by' the gas). q = +300J (heat absorbed 'by' the system). Therefore, ΔU = 300J + (-500J) = -200J. The internal energy decreases.
Topic 16.1 · Internal energy
Explain the physical meaning of internal energy (U) in the context of thermodynamics.
Internal energy (U) represents the total energy of the particles (atoms, molecules) within a system. It includes kinetic energy (due to motion) and potential energy (due to intermolecular forces). ΔU represents the change in this total energy.
Topic 16.2 · The first law of thermodynamics
A gas is compressed at a constant pressure of 2.0 x 10⁵ Pa from a volume of 0.50 m³ to 0.25 m³. Calculate the work done on the gas.
ΔV = 0.25 m³ – 0.50 m³ = –0.25 m³. Work done on the gas W = –pΔV = –(2.0 × 10⁵ Pa)(–0.25 m³) = +5.0 × 10⁴ J. It is positive because the gas is compressed, so work is done on it.
Topic 16.2 · The first law of thermodynamics
What does a positive value of 'q' signify in the first law of thermodynamics?
A positive value of 'q' indicates that energy is being transferred to the system by heating. This leads to an increase in the internal energy of the system, assuming work remains constant.
Topic 16.2 · The first law of thermodynamics
A system's internal energy increases by 400 J, and 100 J of work is done on it. How much heat was transferred, and was it added to or removed from the system?
Using ΔU = q + W, we have 400 J = q + 100 J. Solving for q, we get q = 300 J. Since q is positive, 300 J of heat was added to the system.
Review the material
Read full revision notes on The first law of thermodynamics — definitions, equations, common mistakes, and exam tips.
Read NotesMore topics in Unit 16 — Thermodynamics
The first law of thermodynamics sits alongside these A-Level Physics decks in the same syllabus unit. Each uses the same spaced-repetition system, so progress in one informs the next.
Key terms covered in this The first law of thermodynamics deck
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