Cyclic processes, the second law and reversibility

Back to the start

An engine must work again and again, so its gas goes round a cycle and comes back to where it began. Its internal energy is back too, so over one cycle

On a pressure-volume graph the cycle is a closed loop. The net work is the area inside the loop: positive when the loop goes clockwise, as in an engine, and negative when it goes anticlockwise, as in a refrigerator.

You should find that the net work shrinks as the two pressures come closer, and is zero for a loop with no area.

The second law

The first law would allow an engine that takes heat from the sea and turns all of it into work. Nobody has ever built one. The second law says why not, in two equal forms.

• Kelvin-Planck. No engine can work in a cycle and turn all the heat it takes in into work. Some heat must always go to a colder body.
• Clausius. No machine can move heat from a colder body to a hotter one with no work done on it.

Reversible and irreversible

A process is reversible if it can be run backwards so that the gas and everything around it return to their start. It must be very slow, so the gas is always in balance, with no friction and no heat flowing across a gap in temperature.

Every real process is irreversible: friction, a gas rushing into a vacuum, two liquids mixing, heat flowing from hot to cold. A reversible process is an ideal we can come close to, never reach.

Watch out. A gas expanding at a steady temperature does turn all its heat into work. That does not break the second law, because it is not a cycle: the gas ends with a larger volume.

A worked example

A gas goes clockwise round a rectangle: it expands from $1.0\ \text{L}$ to $4.0\ \text{L}$ at $3.5 \times 10^{5}\ \text{Pa}$ and is squeezed back at $1.2 \times 10^{5}\ \text{Pa}$. Find the net work.

1. Work out. $3.5 \times 10^{5}\ \text{Pa} \times 3.0 \times 10^{-3}\ \text{m}^3 = 1050\ \text{J}$.
2. Work in. $1.2 \times 10^{5}\ \text{Pa} \times 3.0 \times 10^{-3}\ \text{m}^3 = 360\ \text{J}$.
3. Net work. $1050\ \text{J} - 360\ \text{J} = 690\ \text{J}$, so the gas also takes in $690\ \text{J}$ of net heat.

Practise with new numbers

Try it in Python

The program works out the rectangle of the example.

Remember these

• Over a cycle: $\Delta U = 0$ and $Q_\text{net} = W_\text{net}$, the area of the loop.
• The second law: no engine turns all its heat into work; heat does not flow from cold to hot by itself.
• Reversible means slow, with no friction and no heat flow across a gap in temperature.

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 goes anticlockwise round a loop whose area is $140\ \text{J}$. What net work does it do, and what is its net heat?

2. An inventor says that an engine takes heat from the sea and turns all of it into work, cycle after cycle. Which law does it break?

3. Which is nearly reversible: a gas squeezed very slowly in a large bath, or tea being stirred?

4. In one cycle a gas takes in $820\ \text{J}$ and gives out $530\ \text{J}$. How much net work does it do?