Physics 9702 · Paper 1

Work Done by a Gas

A gas pushes on everything around it. When it expands, that push moves a boundary, so the gas does work.
This is how steam engines and car engines turn heat into motion.

From W = Fs to W = pΔV

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).

W = p\,\Delta V

Symbols

$W$ = work done by the gas (J)
$p$ = pressure of the gas (kept constant) (Pa)
$\Delta V$ = increase in volume (m³)

The derivation is two substitutions. Pressure gives the force on the piston: $F = pA$ (lesson 4.09).
The piston moves a distance s, so $W = Fs = pAs$ . But $As$ is exactly the volume swept out: $\Delta V$ . So $W = p\Delta V$ .
This assumes the pressure stays constant during the change, which is true for slow changes against the atmosphere.
If the gas is compressed instead, the surroundings do work on the gas, with the same size $p\Delta V$ .

Worked example

Smallest case: a steady expansion

A gas at a constant pressure of 1.0 × 10⁵ Pa expands by 2.0 × 10⁻³ m³. Find the work done by the gas (9702-style).

$W = p\Delta V = 1.0 \times 10^{5} \times 2.0 \times 10^{-3} = 200 \text{ J}$ .

Answer

W = 200 \text{ J}

Common mistake

Pressure in Pa must be paired with volume in m³. Convert litres and cm³ first:

1 \text{ litre} = 10^{-3} \text{ m}^3

1 \text{ cm}^3 = 10^{-6} \text{ m}^3

(lesson 1.4).

A unit check, and where this goes next

$\text{Pa m}^3 = \text{N m}^{-2}\,\text{m}^3 = \text{N m} = \text{J}$ : the units confirm that $p\Delta V$ is an energy. This formula returns in A2 thermal physics, where changing pressure is handled with the area under a pressure–volume graph.

Honest note on exam frequency: no 2024 or 2025 paper asked about $p\Delta V$ directly. It is still on the syllabus, so learn the derivation and the unit care, but spend your practice time on the energy chains in the next lessons.

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