Physics 9702 · Paper 4

Work = pΔV

The $W$ in the first law needs a number. When the pressure stays steady, one line of geometry provides it.

Where pΔV comes from

The gas pushes the piston with force F = pA through distance s. Work done = pAs, and As is the volume gained, so W = pΔV (for steady pressure).

The gas pushes the piston (area $A$ ) with force $F = pA$ .
The piston moves out a distance $s$ , so the gas does work $F s = pAs$ .
But $As$ is exactly the volume gained: $W = p\,\Delta V$ — valid while $p$ stays constant (true for anything expanding against the atmosphere).

Sign rule, worth two marks on its own: when the system expands, the volume increases and the system does work pushing the surroundings back — so the work done on it is negative, $W = -p\,\Delta V$ (9702/41/O/N/23 Q2(b)(ii)) (9702/42/M/J/25 Q3(b)(iv)). Compression: positive.
A 2025 paper demanded the sign in the answer itself: gas expanding from 0.18 to 0.32 m³ at $1.6 \times 10^{5}$ Pa gives $W = 1.6\times10^{5} \times (0.18 - 0.32) = -2.2 \times 10^{4}\ \text{J}$ — magnitude alone loses the answer mark (9702/42/F/M/25 Q3(b)(i)).

At constant pressure, the work exchanged is the area under the p–V line: W = pΔV. Moving right (expansion) the gas does work on the surroundings, so the W in ΔU = q + W is negative.

On a p–V graph, $p\,\Delta V$ is the area under the line. That picture pays off in the cycles lesson.

Gas, liquid, solid: same formula

Worked example

A liquid vaporising (2024 paper)

$7.2 \times 10^{-5}$ m³ of liquid (72 mL) boils away completely at atmospheric pressure $1.0 \times 10^{5}$ Pa, becoming 0.017 m³ of vapour. Show the work done on the substance is about −1.7 kJ (9702/42/O/N/24 Q3(b)(i)).

The liquid's starting volume is negligible next to the vapour's: $\Delta V \approx 0.017\ \text{m}^3$ .
$|W| = p\,\Delta V = 1.0\times10^{5} \times 0.017 = 1.7\ \text{kJ}$ , negative because the substance expands.

Answer

W \approx -1.7\ \text{kJ}

The same machinery works on a solid — the surprise of a 2025 question. An aluminium block (9.752 kg) heated through 500 °C expands by only $0.110 \times 10^{-3}$ m³, so $|W| = 1.01\times10^{5} \times 0.110\times10^{-3} = 11\ \text{J}$ — against $\Delta U = 4.38$ MJ. The work is negligible, which is itself the final mark: doubling the pressure changes nothing measurable (9702/42/M/J/25 Q3(b)(iii)) (9702/42/M/J/25 Q3(c)).
Scale of the effect: for gases, expansion work is a serious share of the energy; for liquids turning to gas it is a few per cent (1.7 out of 19.3 kJ); for solids it is a rounding error. That ordering follows directly from how much the volume changes.

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