The first law and work of expansion

The first law of thermodynamics

The last lesson showed two ways to change $U$: heat and work. Put them together and you have the first law of thermodynamics:

The energy of an isolated system is constant.

This is the law of conservation of energy. Energy can move between the system and the surroundings, and change form. It cannot be made or destroyed.

For an isolated system, $q = 0$ and $w = 0$, so $\Delta U = 0$. Notice that $q$ and $w$ each depend on the path. Their sum, $\Delta U$, does not.

Work of expansion

Most work in chemistry is done by gases that expand or are compressed. Picture a gas in a cylinder under a light piston that moves without friction. The outside pressure on the piston is $p_\text{ex}$.

When the gas pushes the piston out, it does work against $p_\text{ex}$. Force is pressure times area, and the piston moves a distance $l$. The work done by the gas is $p_\text{ex} \times A \times l = p_\text{ex}\Delta V$. With our sign rule, work done by the gas is negative:

• When the gas expands, $\Delta V > 0$, so $w < 0$.
• When the gas is compressed, $\Delta V < 0$, so $w > 0$.

Units of work

If pressure is in bar and volume in litres, the work comes out in L bar. Convert it to joules:

With atmospheres, $1\ \text{L atm} = 101.3\ \text{J}$.

Take a gas that expands from 1.5 L to 4.5 L against a constant 2.0 bar.

• $\Delta V = 4.5 - 1.5 = 3.0$ L.
• $w = -2.0 \times 3.0 = -6.0$ L bar.
• $w = -6.0 \times 100 = -600$ J.

Work as an area

On a graph of pressure against volume, $p_\text{ex}\Delta V$ is the area of a rectangle. Its height is $p_\text{ex}$ and its width is $\Delta V$. So the work done by a gas is an area under the path on a p-V graph.

Lower the outside pressure from 4.5 bar to 1 bar. The gas expands further, but against a smaller push. Which setting gives the most work in one step? Compare it with the reversible value each time.

Free expansion

Let a gas expand into a vacuum. Nothing pushes back, so $p_\text{ex} = 0$:

This is free expansion. For an ideal gas the temperature does not change either, as there are no forces between its molecules. So $\Delta U = 0$ and $q = 0$ as well.

Reversible expansion

Now let the gas expand very slowly. Keep $p_\text{ex}$ just a tiny bit less than the pressure of the gas at every moment. The change is so gentle that a tiny increase in $p_\text{ex}$ would push it back. Such a change is reversible.

Along a reversible path the gas pushes against the largest pressure it can. So it does the most work possible between the same two states. On the p-V graph, the work is the whole area under the curve.

For an ideal gas at constant temperature, the area under the curve gives:

Here $R = 8.314$ J K$^{-1}$ mol$^{-1}$, $T$ is in kelvin, and log means the logarithm to base 10.

Isothermal expansion of an ideal gas

The internal energy of an ideal gas depends only on its temperature. At constant temperature, $\Delta U = 0$. The first law then gives:

All the work the gas does is paid for by heat it absorbs from the surroundings.

Take 1.00 mol of an ideal gas at 300 K. It doubles its volume reversibly.

• $w = -2.303 \times 1.00 \times 8.314 \times 300 \times \log 2 = -1729$ J.
• $\Delta U = 0$, so $q = +1729$ J.

Practise with new numbers

Try it in Python

The program finds the work of expansion in three ways: against a constant pressure, into a vacuum, and reversibly at constant temperature.

Common mistakes

• Dropping the minus sign in $w = -p_\text{ex}\Delta V$. An expanding gas loses energy as work.
• Using the pressure of the gas instead of $p_\text{ex}$ for an expansion against a constant outside pressure.
• Leaving work in L bar. Multiply by 100 to get joules.
• Using the natural log with 2.303. The factor 2.303 goes with the base-10 log.
• Thinking $q = 0$ when $T$ is constant. For an isothermal ideal gas, $\Delta U = 0$, so $q = -w$.

Remember these

• First law: $\Delta U = q + w$. The energy of an isolated system is constant.
• Against a constant outside pressure: $w = -p_\text{ex}\Delta V$, and $1$ L bar $= 100$ J.
• Free expansion: $w = 0$.
• Reversible, isothermal, ideal gas: $w = -2.303\,nRT\log\dfrac{V_f}{V_i}$, the most work possible.
• Isothermal ideal gas: $\Delta U = 0$ and $q = -w$.

Check yourself

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

1. A gas is compressed from 6.0 L to 2.0 L by a constant outside pressure of 3.0 bar. What is $w$ in joules?

2. An ideal gas expands into an evacuated flask at constant temperature. What are $w$, $q$ and $\Delta U$?

3. Two samples of the same gas expand between the same two states, one reversibly and one in a single step. Which does more work?

4. An ideal gas absorbs 500 J of heat while it expands at constant temperature. What is $w$?