Learn9702Paper 4 · A Level Structured QuestionsIdeal gasesThe ideal gas equation

Ideal gases · Subtopic 1 of 2

The Ideal Gas Equation

One mole contains $N_A=6.02\times10^{23}$ particles. An ideal gas obeys $pV=nRT$ , where $n$ is the number of moles, $R=8.31$ J mol⁻¹ K⁻¹ and $T$ is in kelvin. Equivalently $pV=NkT$ , with $N$ molecules and the Boltzmann constant $k=\dfrac R{N_A}=1.38\times10^{-23}$ J K⁻¹. Boyle's law, Charles's law and the pressure law are special cases.

Cambridge 9702 Paper 4Edexcel YPH11 WPH15AQA 7408 Paper 2≈ 35 minRevision 6 · updated 2026-10-27

Section 1 of 14

Learning objectives

Use the mole, $N_A$ and molar mass to find numbers of particles.
Use $pV=nRT$ and $pV=NkT$ , with $T$ in kelvin.
Use $\dfrac{p_1V_1}{T_1}=\dfrac{p_2V_2}{T_2}$ for a fixed mass of gas.
State the conditions under which real gases behave ideally.

Pause and think

A car tyre is pumped to $2.2$ bar on a cold $5$ °C morning. After a long drive the air inside reaches $45$ °C. Roughly how much has the pressure risen?

Section 2 of 14

Core idea

Start hereTyres on a hot day

Drive on a summer motorway and your tyre pressure rises. The air inside hasn't changed in amount or volume, only temperature. One equation links pressure, volume, temperature and amount for any gas that behaves ideally.

The big idea

$pV=nRT=NkT$ ( $T$ in kelvin). $R=8.31$ J mol $^{-1}$ K $^{-1}$ ; $k=\frac R{N_A}=1.38\times10^{-23}$ J K $^{-1}$ .

1. The equation

pV=nRT=NkT

$n$: moles.
$N$: molecules ( $N=nN_A$ ).
$T$: kelvin.

Isotherms: pV = nRT at three temperatures (1 mol)

T (K) = 350

Each curve is

p\propto\dfrac1V

. Hotter gas has a higher curve: more pressure at the same volume.

Worked exampleMolar volume

1 mol at 273 K and 100 kPa.

Goal: $V$ .
Strategy: Rearrange.

$V=\frac{nRT}p=\frac{8.31\times273}{1.0\times10^5}$
Substitute.
$=0.0227\text{ m}^3=22.7\text{ L}$
Evaluate.

Answer

22.7

litres.

✓ CheckStandard textbook value ✓.

Worked exampleA warming tyre

250 kPa at 15 °C → 40 °C, fixed volume.

Goal: $p_2$ .
Strategy: $\frac pT$ constant (kelvin).

$288\text{ K}\to313\text{ K}$
Convert.
$p_2=250\times\frac{313}{288}=272\text{ kPa}$
Ratio.

Answer

272

kPa.

✓ CheckUsing °C would give 667 kPa: absurd ✓.

Worked exampleCounting molecules

$2.0\times10^{-3}$ m³ at 300 K and 150 kPa.

Goal: $N$ .
Strategy: $N=\frac{pV}{kT}$ .

$pV=150\times10^3\times2.0\times10^{-3}=300\text{ J}$
Numerator.
$N=\frac{300}{1.38\times10^{-23}\times300}=7.2\times10^{22}$
Divide.

Answer

7.2\times10^{22}

✓ Check

\approx0.12

mol ✓.

Common misconceptionCelsius Always kelvin in gas laws: they're proportionalities from absolute zero.

Summary

$pV=nRT=NkT$ .
Kelvin always.

Mole: $6.02\times10^{23}$ particles.
Ideal gas: Obeys $pV=nRT$ exactly.

Isotherms for 0.10 mol of an ideal gas

At constant temperature

pV=nRT

is constant, so doubling the volume halves the pressure. A hotter gas has a larger

pV

, so its curve lies further from the axes.

HistoryBoyle's air pump. In

1662

Robert Boyle, with his assistant Robert Hooke, trapped air in a J-shaped tube with mercury and showed that halving the volume doubled the pressure. It was one of the first physical laws expressed as a mathematical relationship from careful measurement.

Fun factA breath of Caesar. Each breath contains about

10^{22}

molecules. Atmospheric mixing over two thousand years means that, on average, every breath you take probably contains around one molecule from Julius Caesar's last breath. Avogadro's number is that big.

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