Cyclic Processes and Work Done Revision Notes | A Level AQA Physics

1. Why Cyclic Processes Matter

Every heat engine — from a car engine to a steam turbine — works by repeatedly taking a gas through the same sequence of pressure and volume changes. That repetition is a cyclic process, and understanding it is the key to understanding how engines extract useful work from heat.

The elegant result you will use throughout this topic is this: the net work done per cycle equals the area enclosed by the loop on a pressure-volume (p-V) diagram. Once you can read a p-V diagram confidently, you can extract quantitative information about any engine cycle without needing a separate formula for every stage.

2. The p-V Diagram: Reading the Language of Thermodynamics

Picture a graph with volume $V$ on the horizontal axis and pressure $p$ on the vertical axis. Any single point on this graph represents one specific state of the gas — a particular combination of pressure and volume. A curve connecting two points represents a process: the gas changing from one state to another.

The area under a curve on a p-V diagram has a precise physical meaning: it equals the work done by the gas during that process. For an expansion (volume increasing, moving right), the gas does positive work on its surroundings. For a compression (volume decreasing, moving left), the surroundings do positive work on the gas — so the work done by the gas is negative.

The AQA specification requires you to use $W = p\Delta V$ for the constant-pressure case only. You are not expected to derive or apply integral expressions for other process shapes (such as isothermal or adiabatic curves). Keep that boundary clear in your revision.

3. Cyclic Processes and the Enclosed Area

A cyclic process is represented on a p-V diagram as a closed loop: the gas starts at a point, follows a path through several states, and returns exactly to its starting point. Because the gas returns to its original state, its internal energy is unchanged over one complete cycle — the net change in internal energy is zero.

Here is the central result: the net work done by the gas per cycle equals the area enclosed by the loop. To see why, consider that the gas does work (positive, area under the upper path) during expansion, and work is done on the gas (negative, area under the lower path) during compression. The difference between these two areas is the area of the enclosed loop.

The direction of traversal around the loop determines the sign of the net work. A clockwise loop means the upper (expansion) path is at higher pressure than the lower (compression) path, so the gas does more work expanding than is done on it during compression. The net result is positive work done by the gas — this is how an engine produces a useful work output each cycle.

An anticlockwise loop reverses this: more work is done on the gas during compression than the gas does during expansion. The net work done by the gas is negative, meaning net work is done on the gas per cycle. This is the situation in a refrigerator or heat pump.

4. Applying $W = p\Delta V$: The Constant-Pressure Case

On a p-V diagram, a constant-pressure process appears as a horizontal line. This is the one process shape for which AQA requires you to calculate work done using a formula directly: $W = p\Delta V$. For all other curve shapes, you read the work done from the area under the curve (often given as a grid square count or stated in the question).

Notice that the unit check works out cleanly: Pa $\times$ m³ = N m$^{-2}$ $\times$ m³ = N m = J. Always carry units through your substitution — it is a reliable way to catch errors before they cost marks.

Worked example: Calculating Work Done in a Constant-Pressure Expansion

A gas expands at a constant pressure of $1.5 \times 10^5$ Pa. Its volume increases from $2.0 \times 10^{-3}$ m³ to $5.0 \times 10^{-3}$ m³. Calculate the work done by the gas during this expansion. [2 marks]

1. $W = p\Delta V$ — [M1] Selecting the correct expression for constant-pressure work done.
2. $\Delta V = V_{\text{final}} - V_{\text{initial}} = 5.0 \times 10^{-3} - 2.0 \times 10^{-3} = 3.0 \times 10^{-3}$ m³ — Always subtract initial from final. A positive result confirms expansion.
3. $W = 1.5 \times 10^5 \times 3.0 \times 10^{-3}$ — [M1] Correct substitution with consistent SI units.
4. $W = 1.5 \times 3.0 \times 10^{5-3} = 4.5 \times 10^{2} = 450$ J — [A1] Multiply coefficients: $1.5 \times 3.0 = 4.5$. Combine powers: $10^5 \times 10^{-3} = 10^{5+(-3)} = 10^2$. Final answer: 450 J.
5. $W = +450$ J — Positive value confirms the gas does work on its surroundings during expansion. Units: Pa $\times$ m³ = J. ✓

Answer: $W = 450$ J (work done by the gas on its surroundings)

5. Net Work Done Per Cycle: Putting It Together

Now bring both ideas together. In a realistic engine cycle, the gas may expand along one path and be compressed along a different path. Each path contributes a work term — positive for expansion, negative for compression. The net work done by the gas over the complete cycle is the algebraic sum of all these contributions.

On a p-V diagram, this sum is exactly the area enclosed by the loop. For a rectangular cycle (two horizontal and two vertical segments), you can calculate the enclosed area directly from the geometry. For more complex shapes, the question will either provide the area or ask you to estimate it by counting grid squares.

Worked example: Net Work Done for a Rectangular Cycle

A gas undergoes a rectangular cycle on a p-V diagram with the following four stages: • A→B: expansion at constant pressure $p_1 = 4.0 \times 10^5$ Pa, volume from $1.0 \times 10^{-3}$ m³ to $3.0 \times 10^{-3}$ m³ • B→C: pressure drops at constant volume to $p_2 = 1.0 \times 10^5$ Pa • C→D: compression at constant pressure $p_2 = 1.0 \times 10^5$ Pa, volume from $3.0 \times 10^{-3}$ m³ back to $1.0 \times 10^{-3}$ m³ • D→A: pressure rises at constant volume back to $p_1$ Calculate the net work done by the gas per cycle. [4 marks]

1. Work is done only when volume changes. Stages B→C and D→A are constant-volume: $\Delta V = 0$, so $W = 0$ for both. — [B1] Recognising that constant-volume processes involve no work done. This is a common source of marks.
2. $W_{AB} = p_1 \Delta V = 4.0 \times 10^5 \times (3.0 \times 10^{-3} - 1.0 \times 10^{-3})$ $= 4.0 \times 10^5 \times 2.0 \times 10^{-3}$ $= 8.0 \times 10^{5-3} = 8.0 \times 10^2 = 800$ J — [M1] Correct substitution for the expansion stage. Coefficients: $4.0 \times 2.0 = 8.0$. Powers: $10^5 \times 10^{-3} = 10^2$.
3. $W_{CD} = p_2 \Delta V = 1.0 \times 10^5 \times (1.0 \times 10^{-3} - 3.0 \times 10^{-3})$ $= 1.0 \times 10^5 \times (-2.0 \times 10^{-3})$ $= -2.0 \times 10^{5-3} = -2.0 \times 10^2 = -200$ J — [M1] $\Delta V$ is negative because the gas is compressed. Coefficients: $1.0 \times 2.0 = 2.0$. The negative sign must be retained.
4. $W_{\text{net}} = W_{AB} + W_{CD} = 800 + (-200) = 600$ J — [A1] The net work done by the gas per cycle is +600 J. This equals the area of the rectangle: $(p_1 - p_2) \times \Delta V = (4.0 - 1.0) \times 10^5 \times 2.0 \times 10^{-3} = 3.0 \times 10^5 \times 2.0 \times 10^{-3} = 600$ J. ✓
5. Area of rectangle $= \Delta p \times \Delta V = (4.0 \times 10^5 - 1.0 \times 10^5) \times (3.0 \times 10^{-3} - 1.0 \times 10^{-3})$ $= 3.0 \times 10^5 \times 2.0 \times 10^{-3} = 600$ J ✓ — The area method gives the same result, confirming the calculation. For any cycle shape, the net work equals the enclosed area — this is the general principle.

Answer: Net work done by the gas per cycle $= 600$ J

6. Sign Convention and Cycle Direction

The direction in which the cycle is traversed on the p-V diagram determines whether the net work is done by the gas or on the gas. This is not a straightforward detail — it is the difference between an engine and a refrigerator.

To determine the direction, trace the cycle from the expansion stage. If the expansion (rightward movement) occurs along the upper path and compression (leftward movement) along the lower path, the cycle is clockwise and the gas does net positive work. If the paths are reversed, the cycle is anticlockwise.

7. Exam Technique and Common Mistakes

AQA questions on cyclic processes tend to test three things: reading areas from p-V diagrams, applying $W = p\Delta V$ correctly, and interpreting the sign and direction of work. The marks are often straightforward to earn if you follow a consistent method — but they are easy to lose through sign errors or unit confusion.

The two specification requirements for this topic are tightly linked: the work done per cycle equals the enclosed area, and the only formula you need is $W = p\Delta V$ for constant-pressure stages. Master these two ideas together — reading areas from diagrams and applying the constant-pressure formula — and you have everything required to answer any AQA question on cyclic processes and work done.