Isothermal, isobaric and isochoric processes

Three simple processes

We can change a gas while holding one thing steady.

Isothermal

The gas sits in a large bath at one temperature and changes slowly. On a pressure-volume graph it follows a curve $pV =$ constant, an isotherm. Every joule of work it does is matched by a joule of heat from the bath; squeezed, it gives its work back to the bath as heat.

You should find that squeezing to a third of the volume needs work $nRT\ln 3$, all of which leaves as heat: the internal energy does not change.

Isobaric and isochoric

At a steady pressure, the line on the graph is flat and the work is a rectangle. At a steady volume the line is upright and no work is done: all the heat changes the internal energy.

Watch out. 'No change of temperature' does not mean 'no heat'. In an isothermal expansion heat flows in steadily.

A worked example

$0.60$ mol of gas at $290\ \text{K}$ is squeezed slowly from $15$ to $5.0$ litres at a steady temperature. Find $W$, $Q$ and $\Delta U$.

1. Work. $W = 0.60\ \text{mol} \times 8.31\ \text{J mol}^{-1}\text{K}^{-1} \times 290\ \text{K} \times \ln\dfrac{5.0\ \text{L}}{15\ \text{L}} = -1589\ \text{J}$.
2. Internal energy. Steady temperature: $\Delta U = 0$.
3. Heat. $Q = W = -1589\ \text{J}$: the gas gives that much heat to the bath.

Practise with new numbers

Try it in Python

The program works out the squeeze of the example.

Remember these

• Isothermal: $\Delta U = 0$, $Q = W = nRT\ln\dfrac{V_2}{V_1}$.
• Isobaric: $W = p\,\Delta V$, $Q = nC_p\,\Delta T$.
• Isochoric: $W = 0$, $Q = \Delta U = nC_v\,\Delta T$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. $2.0$ mol of gas at $350\ \text{K}$ expands isothermally to $e$ times its volume ($\ln e = 1$). How much heat does it take in?

2. A gas in a rigid tank takes in $300\ \text{J}$. What is $\Delta U$?

3. On a pressure-volume graph, what shape is an isochoric process?

4. Why must an isothermal process be slow?