How Do You Calculate Work Done By or On a Gas (W = PΔV)?

HL only (B.4 Thermodynamics): For a gas changing volume at constant pressure, calculate the work done by the gas using $W=P\Delta V$, where $\Delta V=V_f-V_i$. Expansion gives positive work by the gas; compression gives negative work by the gas.

The Reasoning and Mechanism

Pressure is force per unit area: $P=F/A$. If a piston of area $A$ moves a distance $x$, the work is

$$W=Fx=(PA)x=P(Ax)=P\Delta V.$$

Since (Ax) is the change in volume, this produces $W=P\Delta V$. Use pressure in pascals and volume in cubic metres, giving $\text{Pa m}^3=\text{J}$.

For example, a gas expands at $2.0\times10^5\,\text{Pa}$ from $3.0\times10^{-3}\,\text{m}^3$ to :

$$\Delta V=(5.0-3.0)\times10^{-3}=2.0\times10^{-3}\,\text{m}^3$$ $$W=P\Delta V=(2.0\times10^5)(2.0\times10^{-3})=4.0\times10^2\,\text{J}.$$

Therefore, the gas does $400\,\text{J}$ of work. For the same volume change as compression, $W=-400\,\text{J}$, meaning $400\,\text{J}$ of work is done on the gas.

The IB convention uses the first law of thermodynamics $Q=\Delta U+W$, with $W$ defined as work done by the system. A common misconception is to make work done on the gas positive while still using this equation; that mixes sign conventions.

The equation applies directly only when pressure remains constant. If pressure changes, divide the process into narrow volume intervals and add the work contributions, or determine the area under the process curve on a pressure-volume diagram. The signed area gives the work done by the gas. Expansion moves right and gives positive area; compression moves left and gives negative area. A higher path between the same volumes represents more work, showing that work depends on the process, not only the initial and final states.

Exam Technique

For a Calculate question, show $\Delta V=V_f-V_i$, convert volumes to $\text{m}^3$, substitute into , and interpret the sign. State explicitly whether work is done by or on the gas.