Represent thermodynamic processes on a pressure-volume diagram, calculate the work done by a gas as the area under the curve, and analyse a simple cycle

What this dot point is asking

SEAB wants you to represent thermodynamic processes on a pressure-volume diagram, to calculate the work done by or on a gas as the area under the curve, and to analyse a simple closed cycle using the first law. This brings together the first law, the gas laws and graphical reasoning.

The answer

Work as the area under a pV curve

When a gas changes volume, the work done by the gas is the area under the process on a pressure-volume (pV) diagram. For a small change at pressure $p$:

For a varying pressure, the total work is the area under the curve. An expansion ($\Delta V > 0$) means the gas does positive work on the surroundings; a compression ($\Delta V < 0$) means work is done on the gas.

The four standard processes on a pV diagram

• Isobaric (constant pressure): a horizontal line. Work done by the gas is $p\Delta V$ (the rectangular area).
• Isochoric (constant volume): a vertical line. No volume change, so no work is done.
• Isothermal (constant temperature): a curve along $pV = \text{constant}$. For an ideal gas $\Delta U = 0$, so all heat supplied equals the work done by the gas.
• Adiabatic (no heat exchange): a steeper curve than the isotherm. With $Q = 0$, $\Delta U = -W_{\text{by}}$, so an adiabatic expansion cools the gas.

Combining with the first law

Each process is analysed with $\Delta U = Q - W_{\text{by}}$ (using work done by the gas):

• Isochoric: $W_{\text{by}} = 0$, so $\Delta U = Q$.
• Isothermal: $\Delta U = 0$, so $Q = W_{\text{by}}$.
• Adiabatic: $Q = 0$, so $\Delta U = -W_{\text{by}}$.

A closed cycle

A closed cycle returns the gas to its starting state, so over one complete cycle $\Delta U = 0$. The net work done in the cycle equals the area enclosed by the loop on the pV diagram:

• A clockwise loop means net work is done by the gas (a heat engine extracts useful work from a heat input).
• An anticlockwise loop means net work is done on the gas (a refrigerator or heat pump moves heat against the temperature gradient).

Because $\Delta U = 0$ over a cycle, the net heat input equals the net work output for a clockwise engine cycle.

Examples in context

Example 1. A petrol engine cycle. The idealised four-stroke cycle is a clockwise loop on a pV diagram: the area enclosed is the net work delivered per cycle. Engineers maximise this area (within material limits) to increase the work output, while the first law ensures the net heat supplied equals that work over a complete cycle.

Example 2. A refrigerator. A refrigerator runs an anticlockwise cycle, with net work done on the working gas by the compressor. That work, plus heat drawn from the cold interior, is dumped to the warmer room. The anticlockwise sense on the pV diagram is the graphical signature of a device that moves heat against the temperature gradient.

Try this

Q1. State what the area under a process on a pressure-volume diagram represents. [1 mark]

• Cue. The work done by the gas during that process.

Q2. A gas expands at a constant pressure of $1.5 \times 10^5\ \text{Pa}$ from $2.0 \times 10^{-3}\ \text{m}^3$ to $6.0 \times 10^{-3}\ \text{m}^3$. Find the work done by the gas. [2 marks]

• Cue. $W_{\text{by}} = p\Delta V = 1.5 \times 10^5 \times 4.0 \times 10^{-3} = 600\ \text{J}$.

Q3. Explain why the net change in internal energy of a gas is zero over one complete closed cycle. [2 marks]

• Cue. Internal energy is a state function depending only on the state; a cycle returns the gas to its starting state, so $\Delta U = 0$ and net heat input equals net work output.