Last updated: 2026-07-01

Work Done by an Ideal Gas

When a gas expands it pushes on whatever surrounds it — a piston, the atmosphere, the walls of a balloon — and that push acting over a distance is mechanical work. The amount of work depends not just on the starting and ending volumes but on the path the gas takes between them. This calculator finds the work done by an ideal gas for the two most common quasi-static paths: an isothermal process at constant temperature, and an isobaric process at constant pressure. The article walks through the default isothermal mode; the isobaric mode uses the constant-pressure formula described below.

Where the formula comes from

For any slow volume change, the work done by the gas is the area under its curve on a pressure–volume (PV) diagram:

The shape of that curve — and therefore the area — depends on what is held fixed.

Isothermal process. At constant temperature an ideal gas obeys $PV = nRT$, so the pressure falls as the volume grows: $P = nRT/V$. Substituting and integrating gives

The natural logarithm appears because it is the integral of $1/V$.

Isobaric process. At constant pressure the curve is a horizontal line, and the area under it is a simple rectangle:

Formula

The gas constant is $R = 8.314463\ \text{J/(mol·K)}$. Temperatures enter the ratio in kelvin, and with volumes in cubic metres and pressure in pascals the work comes out in joules.

Worked example

One mole of an ideal gas expands isothermally at $T = 300\ \text{K}$ from $V_1 = 1\ \text{L}$ to $V_2 = 2\ \text{L}$:

The gas does about $1729\ \text{J}$ of work on its surroundings. For comparison, holding the same gas at a constant $101.325\ \text{kPa}$ while it expands by the same $1\ \text{L}$ (from 1 L to 2 L) would give an isobaric work of $P\,\Delta V = 101{,}325 \times 0.001 = 101.3\ \text{J}$ — much less, because the isobaric path sits at a lower average pressure over that expansion.

Sign convention

The result is the work done by the gas. Expansion ($V_2 > V_1$) gives a positive value because the gas transfers energy outward; compression gives a negative value. Some textbooks instead report the work done on the gas, which carries the opposite sign — the first law is then written either $\Delta U = Q - W$ (work by gas) or $\Delta U = Q + W$ (work on gas), so it is worth checking which convention a given source uses.

Limitations

Both formulas assume an ideal gas undergoing a reversible, quasi-static process — slow enough that the gas stays near equilibrium and a single pressure and temperature describe it throughout. Real gases deviate from $PV = nRT$ at high pressure or near condensation, and rapid or turbulent changes are not quasi-static. Other paths, such as adiabatic (no heat exchange) or isochoric (constant volume, where the gas does no work at all), follow different expressions and are outside the scope of these two modes.