Understanding p-V Diagrams Revision Notes | A Level AQA Physics

1. Why p–V Diagrams Are the Language of Engines

Every heat engine — from a car engine to a steam turbine — works by taking a gas through a cycle of pressure and volume changes. To analyse any engine, physicists need a way to visualise those changes at a glance. That tool is the p–V diagram.

Think of a p–V diagram as a map of a gas's journey. The horizontal axis shows the volume of the gas (in m³), and the vertical axis shows its pressure (in Pa). Every time the gas changes state — expanding, being compressed, heating up, or cooling down — it traces a path across this map.

The real power of this diagram is that the area under any curve directly tells you the work done. You do not need to know the internal mechanism of the engine; the graph encodes the energy for you. That is why mastering p–V diagrams is central to the AQA A Level engineering physics unit.

2. Reading a p–V Diagram: States and Processes

A single point on a p–V diagram represents one specific state of the gas — a particular combination of pressure and volume (and, by the ideal gas law, temperature). A curve or line connecting two points represents a process: the path the gas takes as it moves from one state to another.

For an ideal gas, the product $pV$ is constant during an isothermal process. This means if the volume doubles, the pressure halves — the curve bends smoothly rather than forming a straight line. Isothermals at higher temperatures sit further from the origin, because a hotter gas at the same volume exerts greater pressure.

A useful way to distinguish the two curves: if you see two curves crossing at the same point on a p–V diagram, the steeper one is always the adiabatic. The isothermal curve is shallower because heat can flow in to partially compensate for the pressure drop during expansion.

The table above is your reference for identifying any process on a p–V diagram. In an exam, you will often be given a diagram and asked to name the process — use the shape and direction of the line as your first clue, then check which quantity is held constant.

3. Representing Processes: Drawing and Interpreting Curves

When you draw a process on a p–V diagram, the direction of the arrow matters. An arrow pointing to the right (increasing volume) represents expansion; an arrow pointing to the left (decreasing volume) represents compression. This direction tells you immediately whether the gas is doing work on its surroundings or having work done on it.

For a complete engine cycle, the gas returns to its starting state — so the path forms a closed loop on the p–V diagram. The direction of travel around the loop (clockwise or anticlockwise) tells you whether the cycle is a heat engine (net work output, clockwise) or a refrigerator/heat pump (net work input, anticlockwise).

When sketching these curves in an exam, you do not need to plot precise values. What matters is the relative steepness: the adiabatic must always be drawn steeper than the isothermal at the same point. Getting this wrong is one of the most common errors in this topic.

4. Work Done: The Area Under the Curve

Here is the central idea of this topic: the work done by a gas during any process equals the area under the p–V curve for that process. This comes directly from the definition of work done against a pressure: $W = p \Delta V$ for a constant-pressure process, which generalises to the area under the curve for any process.

For processes where pressure is not constant — such as isothermal or adiabatic processes — you cannot use $W = p \Delta V$ directly. Instead, you estimate the area under the curve by treating it as a geometric shape or by counting grid squares on the diagram. This is the estimation method AQA expects you to apply.

The sign convention is straightforward: if the gas expands (volume increases, arrow points right), the area is positive and the gas does work on its surroundings. If the gas is compressed (volume decreases, arrow points left), the area is negative and work is done on the gas.

Worked example: Calculating Work Done in an Isobaric Expansion

A gas expands at a constant pressure of $2.0 \times 10^5$ Pa. Its volume increases from $1.5 \times 10^{-3}$ m³ to $4.5 \times 10^{-3}$ m³. Calculate the work done by the gas.

1. Constant pressure $\Rightarrow$ isobaric process. Use $W = p \Delta V$. — On a p–V diagram, this is a horizontal line. The area underneath is a rectangle: width = $\Delta V$, height = $p$. [M1]
2. $\Delta V = V_2 - V_1 = 4.5 \times 10^{-3} - 1.5 \times 10^{-3} = 3.0 \times 10^{-3}$ m³ — Always subtract initial volume from final volume. A positive result confirms expansion. [M1]
3. $W = p \Delta V = 2.0 \times 10^5 \times 3.0 \times 10^{-3}$ — Both values are now in SI units (Pa and m³), so the result will be in joules directly.
4. $W = 2.0 \times 10^5 \times 3.0 \times 10^{-3} = 2.0 \times 3.0 \times 10^{5-3} = 6.0 \times 10^2 = 600$ J — Multiply the coefficients (2.0 × 3.0 = 6.0) and add the exponents (5 + (−3) = 2). [A1]
5. $W = +600$ J — Positive because the gas expanded. The gas transferred 600 J of energy to its surroundings by doing work.

Answer: $W = 600$ J (work done by the gas on its surroundings)

For a curved process, the approach is the same in principle — you are still finding the area under the graph — but you must use an estimation method. AQA expects you to either approximate the curve as a trapezium, or count grid squares and multiply by the value each square represents.

Worked example: Estimating Work Done Under a Curved p–V Process

A gas expands along a curved path from state A ($p = 4.0 \times 10^5$ Pa, $V = 1.0 \times 10^{-3}$ m³) to state B ($p = 1.0 \times 10^5$ Pa, $V = 4.0 \times 10^{-3}$ m³). Estimate the work done by the gas using the trapezium approximation.

1. Approximate the area under the curve as a trapezium with parallel sides equal to the pressures at A and B, and width equal to $\Delta V$. — The trapezium rule gives a reasonable estimate when the curve is smooth. It slightly overestimates for a concave-down curve and underestimates for a concave-up curve. [M1]
2. $W \approx \frac{1}{2}(p_A + p_B) \times \Delta V$ — This is the standard trapezium area: half the sum of the parallel sides multiplied by the perpendicular width.
3. $\Delta V = V_B - V_A = 4.0 \times 10^{-3} - 1.0 \times 10^{-3} = 3.0 \times 10^{-3}$ m³ — Positive $\Delta V$ confirms the gas is expanding, so work done by the gas will be positive.
4. $p_A + p_B = 4.0 \times 10^5 + 1.0 \times 10^5 = 5.0 \times 10^5$ Pa — Both pressures are in Pa, so no unit conversion is needed here.
5. $W \approx \frac{1}{2} \times 5.0 \times 10^5 \times 3.0 \times 10^{-3}$ $= 0.5 \times 5.0 \times 3.0 \times 10^{5-3}$ $= 0.5 \times 15.0 \times 10^{2}$ $= 7.5 \times 10^{2} = 750$ J — Work through the powers of ten carefully: 10^5 × 10^(−3) = 10^2. Then 0.5 × 15.0 = 7.5. [A1]
6. $W \approx 750$ J — The word 'estimate' in the question means you should acknowledge this is an approximation. The actual value (for a true isothermal curve here) would be slightly different, but 750 J is a valid exam answer using the trapezium method.

Answer: $W \approx 750$ J (estimated work done by the gas during expansion)

5. Net Work in a Complete Thermodynamic Cycle

When a gas completes a full cycle on a p–V diagram — returning to its starting state — the net work done equals the area enclosed by the loop. This is because the work done during expansion (area under the upper curve, going right) is partially cancelled by the work done on the gas during compression (area under the lower curve, going left).

For a clockwise cycle, the expansion path sits above the compression path. The area under the expansion curve is larger, so the net work done by the gas is positive — this is a heat engine doing useful work. For an anticlockwise cycle, the compression path sits above the expansion path, meaning net work is done on the gas — this describes a refrigerator or heat pump.

In the diagram above, the upper curve represents the expansion stroke (gas does work on surroundings) and the lower curve represents the compression stroke (surroundings do work on gas). The net work output per cycle equals the area of the enclosed region between the two curves.

6. Comparing Processes: Which Does More Work?

A common exam question asks you to compare the work done in two different processes connecting the same two states. The key insight is that the work done depends on the path taken, not just the start and end points. Two processes can connect the same states A and B but enclose very different areas under their curves.

Consider an isobaric expansion and an isothermal expansion, both starting at the same state A and ending at the same volume $V_B$. The isobaric process is a horizontal line — it maintains the higher starting pressure throughout. The isothermal curve drops in pressure as volume increases. The isobaric process therefore has a larger area underneath it and represents more work done by the gas.

7. Exam Technique and Mark-Winning Strategies

AQA questions on p–V diagrams tend to follow predictable patterns. You will either be asked to sketch a process, identify a process from a given diagram, calculate or estimate work done, or compare two processes. Each of these has a specific approach that earns marks efficiently.

When sketching a process, always label your axes with quantity and unit ($p$ / Pa and $V$ / m³), mark the start and end states clearly, and draw the arrow in the correct direction. For an adiabatic curve, explicitly make it steeper than any isothermal you draw on the same diagram — examiners look for this distinction.

When estimating work done from a graph, state your method clearly. If you use the trapezium rule, write the formula and show the substitution. If you count grid squares, state the value of each square in joules. Examiners award method marks for showing your approach, even if your final numerical answer is slightly off due to the estimation.

The worked cycle above captures everything this topic demands: identifying each process, applying the correct work formula (or recognising zero work for isochoric stages), summing contributions with correct signs, and interpreting the result physically. Practise this structure with different cycle shapes and you will be well prepared for any p–V diagram question AQA sets.