When a gas or other compressible system changes volume against an external pressure, energy can be transferred as pressure-volume work.

Consider a piston of area $A$ moving a distance $\Delta x$ against a constant external pressure $p_{\mathrm{ext}}$. The resisting force magnitude is

$$F=p_{\mathrm{ext}}A,$$

and the volume change is

$$\Delta V=A\Delta x.$$

Therefore the work done by the system on the surroundings is

$$\boxed{W_{pV}=p_{\mathrm{ext}}\Delta V}$$

when the external pressure is constant.

Under the thermodynamic sign convention

$$\Delta U=q-W,$$

an expansion has $\Delta V>0$ and therefore $W_{pV}>0$: the system does work on its surroundings. A compression has $\Delta V<0$, so $W_{pV}<0$: the surroundings do net pressure-volume work on the system.

Example

A gas expands from $2.0,\mathrm L$ to $5.0,\mathrm L$ against a constant external pressure of $100,\mathrm{kPa}$.

The volume change is

$$\Delta V=3.0,\mathrm L.$$

Since

$$1,\mathrm{kPa,L}=1,\mathrm J,$$

we obtain

$$W_{pV}=(100,\mathrm{kPa})(3.0,\mathrm L)=300,\mathrm J.$$

The gas has transferred $300,\mathrm J$ of energy to its surroundings as expansion work.

The external pressure matters because it is the mechanical resistance actually pushed against. When pressure varies during a process, the simple product $p_{\mathrm{ext}}\Delta V$ must be replaced by an accumulation of work over the changing volume. The constant-pressure result is the simplest case of that more general boundary-work calculation.