Physics 9702 · syllabus 15
Cambridge A Level Physics 9702 — Ideal gases topical past papers
Gases at the molecular scale: the mole, the equation of state pV = nRT = NkT, and kinetic theory linking pressure and temperature to molecular motion.
- Questions
- 95
- Years
- 2009–2026
- With mark scheme
- 100%
- Questions 2022–2026
- 34
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What Ideal gases covers
Amount of substance, measured in moles, joins the SI base quantities here: one mole of any substance contains the Avogadro number of particles, 6.02 × 10²³. A gas in which pV is proportional to thermodynamic temperature is called an ideal gas, and its equation of state is pV = nRT for n moles or pV = NkT for N molecules, where the Boltzmann constant is k = R/Nₐ.
Kinetic theory explains the pressure. The molecules are modelled as a very large number of identical particles in random motion, of negligible volume, colliding elastically with each other and the walls, with no forces between them except during collisions, which last a negligible time. Each collision with a wall changes a molecule's momentum, and the force from many collisions gives the pressure. A one-dimensional model, extended to three dimensions, gives pV = ⅓Nm⟨c²⟩.
Comparing that result with pV = NkT shows that the mean translational kinetic energy of a molecule, ½m⟨c²⟩, equals (3/2)kT: temperature is a measure of the average kinetic energy of the molecules. This is A Level content, examined on Paper 4. It builds on momentum from AS and leads into internal energy in thermodynamics.
In the syllabus
The headings this topic has in the Cambridge International 2025–2027 syllabus, as the board prints them.
- 15 Ideal gases
- 15.1 The mole
- 15.2 Equation of state
- 15.3 Kinetic theory of gases
What you should be able to do
- Use the mole, molar mass and the Avogadro constant to convert between mass, amount and number of molecules
- Use pV = nRT and pV = NkT, with temperature in kelvin
- State the assumptions of the kinetic theory of gases
- Derive pV = ⅓Nm⟨c²⟩ from molecular collisions with the walls of a container
- Calculate a root-mean-square speed and use the result that the mean kinetic energy of a molecule is (3/2)kT
Key ideas
- n = N/Nₐ, with Nₐ = 6.02 × 10²³ mol⁻¹
- Equation of state pV = nRT = NkT, with R = 8.31 J K⁻¹ mol⁻¹ and k = R/Nₐ = 1.38 × 10⁻²³ J K⁻¹
- Kinetic theory: pV = ⅓Nm⟨c²⟩, where ⟨c²⟩ is the mean-square speed
- r.m.s. speed = √⟨c²⟩; mean translational kinetic energy of a molecule ½m⟨c²⟩ = (3/2)kT
How Ideal gases has been examined
From 2022 to 2026, Ideal gases was on Paper 4 · A Level Structured Questions in 11 of 11 exam series, worth about 11% of that paper’s marks.
| Paper | Questions | Share of marks | Series |
|---|---|---|---|
| Paper 4 · A Level Structured Questions | 34 | 11% | 11 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: 10.
Filed in Momentum under: Ideal Gases.
A worked example
A cylinder of volume 0.020 m³ holds helium at a pressure of 1.5 × 10⁵ Pa and a temperature of 300 K. The molar mass of helium is 4.0 × 10⁻³ kg mol⁻¹. Find the amount of gas and the root-mean-square speed of its atoms.
- n = pV/RT = 1.5 × 10⁵ × 0.020 / (8.31 × 300) = 1.20 mol
- Mass of one atom: m = 4.0 × 10⁻³ / 6.02 × 10²³ = 6.64 × 10⁻²⁷ kg
- ½m⟨c²⟩ = (3/2)kT, so ⟨c²⟩ = 3kT/m = 3 × 1.38 × 10⁻²³ × 300 / 6.64 × 10⁻²⁷ = 1.87 × 10⁶ m² s⁻²
- r.m.s. speed = √(1.87 × 10⁶) = 1.37 × 10³ m s⁻¹
Answer: 1.2 mol; r.m.s. speed 1.4 × 10³ m s⁻¹
Written by Momentum for this page, with its own numbers; it is not taken from a past paper.
Mistakes that cost marks
- Substituting a Celsius temperature into pV = nRT — at 27 °C this makes a calculated pressure about eleven times too small; use 300 K
- Using the molar mass in grams per mole as the mass of one molecule — the mass of a molecule is the molar mass in kg mol⁻¹ divided by Nₐ
- Finding the r.m.s. speed by averaging the speeds and then squaring — square each speed, average the squares, then take the square root
- Mixing n and N — pV = nRT uses moles and R; pV = NkT uses the number of molecules and k
- Leaving the volume in cm³ or litres — 1 litre is 1.0 × 10⁻³ m³ and 1 cm³ is 1.0 × 10⁻⁶ m³
Exam technique
- In the derivation of pV = ⅓Nm⟨c²⟩, state each step: change of momentum 2mc per collision, time 2L/c between collisions with the same wall, force as rate of change of momentum, then sum over N molecules and use ⟨c²⟩ = 3⟨cₓ²⟩
- When asked for assumptions, be precise: 'the volume of the molecules is negligible compared with the volume of the container' earns more than 'molecules are small'
- Check whether a question gives the mass of the whole gas or of one molecule before substituting into ½m⟨c²⟩ = (3/2)kT
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.