A gas can push a piston, so it can do work. In Work as Area Under a Force Graph work was the area under a force–distance graph. For a gas, the same idea becomes the area under a pressure–volume graph, using the gas law to connect the states.
Why it works
A gas at pressure $P$ pushes on a piston of area $A$ with force $PA$. If the piston moves out a small distance $\Delta x$, the gas does work $PA\,\Delta x = P\,\Delta V$ on it. That energy leaves the gas, so the work done on the gas is $W = -P\,\Delta V$. If the pressure changes as the volume changes, add up thin strips: the size of the work is the area under the path on the P–V diagram.
Key idea
Work done on a gas: $W = -P\Delta V$ at constant pressure; in general $|W|$ is the area under the path on a P–V diagram. Expansion: $W < 0$ (the gas does work). Compression: $W > 0$. Constant volume: $W = 0$. Handy: $1\,\text{kPa} \cdot \text{L} = 1\,\text{J}$.
| This course (and AP Physics 2) | Science Lab and OpenStax College Physics | |
|---|---|---|
| First law | $\Delta U = Q + W$ | $\Delta U = Q - W_{\text{by}}$ |
| The work in it | $W$ = work done on the gas $= -P\Delta V$ | $W_{\text{by}}$ = work done by the gas $= +P\Delta V$ |
| Gas expands | $W < 0$ | $W_{\text{by}} > 0$ |
| Heat $Q$ | positive into the gas | positive into the gas |
The Energy and the First Law chapter of the Science Lab uses the work done by the gas, as in the right-hand column. Both are correct; just never mix them in one problem.
Worked examples
Worked example: Expanding at constant pressure
A gas at a constant $200\,\text{kPa}$ expands from $2.0\,\text{L}$ to $6.0\,\text{L}$. Find the work done on it.
- $\Delta V = 4.0\,\text{L} = 4.0 \times 10^{-3}\,\text{m}^3$.
- $W = -P\Delta V = -(2.00 \times 10^{5}\,\text{Pa})(4.0 \times 10^{-3}\,\text{m}^3) = -800\,\text{J}$.
- Negative: the gas did $800\,\text{J}$ of work on the piston. (Shortcut: $200\,\text{kPa} \times 4.0\,\text{L} = 800\,\text{J}$.)
Answer: $W = -800\,\text{J}$
Worked example: Compressing
A gas is compressed at a constant $150\,\text{kPa}$ from $5.0\,\text{L}$ to $2.0\,\text{L}$. Find the work done on it.
- $\Delta V = 2.0 - 5.0 = -3.0\,\text{L}$.
- $W = -P\Delta V = -(150\,\text{kPa})(-3.0\,\text{L}) = +450\,\text{J}$. The piston gave the gas energy.
Answer: $W = +450\,\text{J}$
Worked example: A sloped path
A gas expands along a straight line on the P–V diagram from $(100\,\text{kPa}, 1.0\,\text{L})$ to $(300\,\text{kPa}, 3.0\,\text{L})$. Find the work done on it.
- The area under a straight line is a trapezoid: $\tfrac{1}{2}(100 + 300)\,\text{kPa} \times 2.0\,\text{L} = 400\,\text{J}$.
- The gas expanded, so $W = -400\,\text{J}$.
Answer: $W = -400\,\text{J}$
Common mistake
Getting the sign of the work wrong by mixing up the work done on the gas with the work done by it. Decide first: is the gas expanding (it does work, so $W$ on it is negative) or being compressed ($W$ on it is positive)? Then check that your formula agrees.