Physics 9702 · Paper 4

pV = nRT

Squeeze a gas and the pressure rises; warm it and it pushes harder. One equation carries all of it — provided the gas is “ideal”, and you can say what that means.

What ideal means (two marks)

An is one that obeys $pV \propto T$ at all values, where $T$ is the thermodynamic temperature. The proportionality is one mark; naming the temperature scale is the other (9702/41/M/J/24 Q3(a)(i)) (9702/42/F/M/23 Q2(a)).
Real gases at ordinary pressures behave very closely to this. They stop behaving ideally when squeezed hard or cooled near condensation — then the molecules sit close enough that their own volume and their attractions stop being negligible (9702/41/O/N/23 Q3(d)).

The equation of state

pV = nRT = NkT

Symbols

$p$ = pressure of the gas (Pa)
$V$ = volume it occupies (m³)
$n$ = amount of gas (mol)
$R$ = molar gas constant, 8.31 (J K⁻¹ mol⁻¹)
$N$ = number of molecules (—)
$k$ = Boltzmann constant, 1.38 × 10⁻²³ (J K⁻¹)
$T$ = thermodynamic temperature (K)

Two spellings of one law: use $nR$ when the question talks in moles, $Nk$ when it counts molecules. They agree because $N_A = R/k$ (last lesson).
Non-negotiables: $T$ in kelvin, $V$ in m³, $p$ in Pa. Almost every dropped mark in this topic is a Celsius temperature or a cm³ volume smuggled into the equation.

Worked example

A bottle of air, fully processed

A 1.5 litre bottle ( $1.5 \times 10^{-3}\ \text{m}^3$ ) holds air at $1.0 \times 10^{5}$ Pa and 300 K. Find the amount of gas, the number of molecules, and estimate the average spacing between molecules.

$n = \dfrac{pV}{RT} = \dfrac{1.0 \times 10^{5} \times 1.5 \times 10^{-3}}{8.31 \times 300} = 0.060\ \text{mol}$ .
$N = nN_A = 0.060 \times 6.02 \times 10^{23} = 3.6 \times 10^{22}$ molecules.
Each molecule “owns” a cube of volume $V/N$ , so the spacing is about $\sqrt[3]{V/N} = \sqrt[3]{\dfrac{1.5\times 10^{-3}}{3.6\times 10^{22}}} \approx 3.5 \times 10^{-9}\ \text{m}$ .

Answer

n \approx 0.060\ \text{mol},\ N \approx 3.6\times 10^{22},\ \text{spacing} \approx 3.5\ \text{nm}

The cube-root spacing estimate is a favourite final part (9702/42/O/N/25 Q2(c)) — about ten molecular diameters, which is why the assumptions in the next lesson hold.

Your turn— tap to reveal the worked answer (9702-style check)

A gas sample is at $1.0 \times 10^{5}$ Pa, volume $2.4 \times 10^{-3}$ m³, temperature 27 °C. Later it is at $1.6 \times 10^{5}$ Pa, $1.8 \times 10^{-3}$ m³, 87 °C. Show the data are consistent with an ideal gas.

Convert first: 300 K and 360 K — examiners award a mark for visibly using kelvin (9702/42/F/M/23 Q2(b)(i)). Then test $pV/T$ both times: $\dfrac{1.0\times10^5 \times 2.4\times10^{-3}}{300} = 0.80$ and $\dfrac{1.6\times10^5 \times 1.8\times10^{-3}}{360} = 0.80\ \text{J K}^{-1}$ . Equal values → consistent with $pV \propto T$ .

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