The p–V diagram (A-level only) — AQA A-Level Physics
Teaching note checked against the official specification.
A p-V diagram plots the pressure of a gas against its volume, with the area under the curve representing the work done by or on the gas. For AQA A-Level Physics, you must interpret these diagrams to calculate work done, identify thermodynamic processes (isothermal, adiabatic, isobaric, isochoric), and apply the first law of thermodynamics.
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The p–V diagram (A-level only) explained
The p–V diagram (A-level only) covers 2 specification points, including Representation of processes on p–V diagram.
Read the full explanation
and Estimation of work done in terms of area below…. Open a point below for its explanation and revision guidance.
Representation of processes on p–V diagram.
A p–V diagram plots pressure on the vertical axis against volume on the horizontal axis, so each point fixes a unique equilibrium state of a fixed mass of gas. A process is drawn as a continuous path joining an initial point to a final point. An isothermal change follows a curve of the form pV = constant, a straight horizontal line shows constant pressure (isobaric), a vertical line shows constant volume (isochoric), and an adiabatic change is steeper than the isotherm through the same point. Arrows show direction, so expansion moves right and compression moves left. For example, a gas expanding at constant pressure from V₁ to V₂ is a horizontal line at p, while the same expansion isothermally curves downwards.
Estimation of work done in terms of area below the graph.
For a reversible process, the work done by a gas equals the area under the p–V path, because work is the integral of pressure with respect to volume. On a diagram, estimate this area by splitting the region between the path and the volume axis into strips or simple shapes such as rectangles and triangles, then adding the parts. Count full squares and combine part-squares for a grid, or use the trapezium rule on tabulated values. For expansion the gas does work on the surroundings and the area is positive; for compression work is done on the gas and the area is negative relative to expansion. For example, a constant-pressure expansion from V₁ to V₂ has work p(V₂ − V₁), the area of a rectangle.
Your focus
- Draw and label p–V axes correctly for a fixed mass of gas.
- Represent isothermal, isobaric, isochoric and adiabatic processes with appropriate path shapes.
- Indicate the direction of a process using arrows and explain the physical condition shown by each path.
Show all 6 objectives
- Explain why the area under a p–V graph represents work done.
- Estimate work done from a p–V diagram using squares, strips or the trapezium rule.
- Apply the correct sign and units when reporting work done by or on a gas.
Which command word is this?
You must use the correct formula or area under the graph to obtain a numerical answer with appropriate units and significant figures. Show all working.
The p–V diagram (A-level only) exam tips
Topic Overview
The p-V diagram is a graphical tool used in thermodynamics to show how the pressure (p) and volume (V) of a gas change during a process. It is essential for understanding work done by or on a gas, as the area under the curve represents the work done. This topic is A-level only and builds on GCSE ideas of gas laws and the first law of thermodynamics.
In AQA A-Level Physics, you must be able to interpret and sketch p-V diagrams for isothermal, adiabatic, isobaric, and isochoric processes, and calculate work done from the area under the curve. It links to the ideal gas equation, internal energy, and the first law, forming a core part of the thermodynamics section.
Key Concepts
- →Work done by a gas during expansion or compression is equal to the area under the p-V curve; for a cycle, net work is the enclosed area.
- →Isothermal processes follow pV = constant and are represented by a smooth hyperbola; adiabatic processes follow pV^gamma = constant and are steeper.
- →Isobaric processes are horizontal lines (constant pressure), and isochoric processes are vertical lines (constant volume, no work done).
- →The first law of thermodynamics, ΔU = Q - W, relates internal energy change (ΔU), heat added (Q), and work done by the gas (W).
- →For an ideal gas, internal energy depends only on temperature, so isothermal processes have ΔU = 0, meaning Q = W.
Marking Points
- Axes are labelled pressure p and volume V with correct units, and the diagram is drawn for a fixed mass of gas.
- Each point on the diagram represents one equilibrium state, and a process is shown as a path with an arrow indicating direction.
- Isothermal processes are drawn as curves of constant pV, isobaric as horizontal lines, and isochoric as vertical lines.
- Adiabatic curves are steeper than isotherms through the same point because temperature changes during the process.
- Expansion is shown moving towards larger V and compression towards smaller V, with the path shape consistent with the stated conditions.
- Work done is identified as the area between the p–V path and the volume axis, with the correct limits V₁ and V₂.
- The area is estimated by a valid method such as counting grid squares, dividing into strips, or using the trapezium rule.
- Units are handled consistently, for example pressure in Pa and volume in m³ giving work in J.
- The sign or direction is interpreted correctly: expansion gives work done by the gas, compression gives work done on the gas.
- A stated numerical estimate is reasonable for the scale of the axes and the method used.
Examiner Tips
- 💡Sketch axes with labels and units before plotting any path, so the examiner can see the correct orientation immediately.
- 💡Use a dashed guide curve for the isotherm through a point when comparing it with an adiabatic path.
- 💡State the condition for each process, such as constant temperature or constant pressure, next to the relevant line.
- 💡Show the subdivision clearly, such as drawing strips or writing the counted squares, so the estimation method is visible.
- 💡Give the value with a sensible number of significant figures and include the unit J.
- 💡State whether the work is done by or on the gas when the question asks for direction as well as magnitude.
- 💡Always label axes with quantity and unit (p/Pa, V/m^3) and indicate the direction of the process with arrows on the curve.
- 💡When calculating work done from a graph, count squares or use area formulas; for non-linear curves, estimate area under the curve carefully.
- 💡In explanations, use precise terms: 'work done by the gas' for expansion, 'work done on the gas' for compression, and state that area under the curve represents work done.
Common Mistakes
- Drawing a process as a straight diagonal line when the statement says isothermal; the correction is to draw a curve of constant pV.
- Labelling the vertical axis as volume and the horizontal axis as pressure; the correction is pressure on the vertical axis and volume on the horizontal axis.
- Omitting arrows so the direction of the process is unclear; the correction is to add an arrow along the path from initial to final state.
- Using the area under a straight line joining the endpoints rather than under the actual curved path; the correction is to estimate the area beneath the drawn path.
- Forgetting to convert volume from cm³ to m³ before multiplying by pressure; the correction is to convert to m³ so the product is in joules.
- Treating the area as always positive regardless of direction; the correction is to relate the sign to expansion or compression.
- Students often think that work done is always positive, but work done on the gas during compression is negative when considering work done by the gas. Always specify direction.
- Many confuse isothermal and adiabatic curves, thinking both are the same shape. Isothermal curves are less steep than adiabatic curves because temperature is constant, while adiabatic involves temperature change.
- Some believe that if volume is constant, work done is non-zero. In fact, for isochoric processes, no work is done because there is no displacement.
Revision Plan
- 1Week 1: Revise the ideal gas equation and gas laws, then learn the shapes of isothermal, adiabatic, isobaric, and isochoric processes on p-V diagrams.
- 2Week 1: Practice calculating work done from simple p-V diagrams (rectangular and triangular areas) and for isothermal processes using W = nRT ln(V2/V1).
- 3Week 2: Apply the first law of thermodynamics to each process type, linking ΔU, Q, and W.
- 4Week 2: Attempt exam-style questions, including cyclic processes, and mark your answers using the mark scheme to identify gaps.
- 5Week 2: Create a summary sheet with key formulas, graph shapes, and common pitfalls, then test yourself with active recall.
Exam Question Types
- 📋Calculation of work done from a p-V diagram: often involves finding area under a curve or enclosed area for a cycle. Advice: identify the shape, use appropriate area formula, and state units.
- 📋Explanation of thermodynamic processes: e.g., 'Explain why the work done is zero for an isochoric process.' Advice: refer to volume constant, no displacement, so W = pΔV = 0.
- 📋Sketching and interpreting p-V diagrams: e.g., 'Sketch the p-V diagram for an isothermal expansion and an adiabatic expansion starting from the same point.' Advice: label axes, show correct relative steepness, and indicate direction.
- 📋Applying the first law to a cycle: e.g., 'For the cycle shown, calculate the net work done and state whether it is a heat engine or refrigerator.' Advice: net work = enclosed area, and if work done by gas is positive, it's a heat engine.
Command Word Expectations (AQA)
You must use the correct formula or area under the graph to obtain a numerical answer with appropriate units and significant figures. Show all working.
You must provide a reasoned argument linking cause and effect, using correct physics terminology. For example, explain why work done is zero in an isochoric process by stating volume is constant, so no displacement occurs.
You must draw a graph with correct axes labels and units, and show the correct shape and relative positions of curves. No precise data points are needed, but key features must be clear.
How Students Lose Marks (Examiner Pitfalls)
Step-by-Step Worked Solutions
Question: A fixed mass of ideal gas undergoes an isothermal expansion from a volume of 2.0 x 10^-3 m^3 to 5.0 x 10^-3 m^3 at a constant temperature of 300 K. The initial pressure is 1.5 x 10^5 Pa. Calculate the work done by the gas during this expansion. (Gas constant R = 8.31 J mol^-1 K^-1, and assume the gas is ideal.)
- 1.Step 1: Identify given facts: initial volume V1 = 2.0 x 10^-3 m^3, final volume V2 = 5.0 x 10^-3 m^3, initial pressure p1 = 1.5 x 10^5 Pa, temperature T = 300 K (constant).
- 2.Step 2: For an isothermal process, work done by the gas is W = nRT ln(V2/V1). First, find number of moles n using p1V1 = nRT: n = p1V1 / (RT) = (1.5 x 10^5 x 2.0 x 10^-3) / (8.31 x 300) = 300 / 2493 = 0.1203 mol.
- 3.Step 3: Calculate work done: W = nRT ln(V2/V1) = 0.1203 x 8.31 x 300 x ln(5.0/2.0) = 300 x ln(2.5) = 300 x 0.9163 = 274.9 J. Round to 2 significant figures: 2.7 x 10^2 J.
- 4.Step 4: State final conclusion with units: The work done by the gas is 2.7 x 10^2 J.
Question: A gas undergoes a cyclic process ABCA on a p-V diagram. From A to B, the gas expands isobarically at 2.0 x 10^5 Pa from 1.0 x 10^-3 m^3 to 3.0 x 10^-3 m^3. From B to C, the gas is cooled isochorically to a pressure of 1.0 x 10^5 Pa. From C to A, the gas is compressed isobarically back to its original volume. Calculate the net work done by the gas in one complete cycle.
- 1.Step 1: Identify the processes: A to B is isobaric expansion, B to C is isochoric cooling, C to A is isobaric compression.
- 2.Step 2: Calculate work done for each process. For A to B: W_AB = p ΔV = 2.0 x 10^5 x (3.0 x 10^-3 - 1.0 x 10^-3) = 2.0 x 10^5 x 2.0 x 10^-3 = 400 J (work done by gas).
- 3.Step 3: For B to C: W_BC = 0 J because volume is constant (isochoric). For C to A: W_CA = p ΔV = 1.0 x 10^5 x (1.0 x 10^-3 - 3.0 x 10^-3) = 1.0 x 10^5 x (-2.0 x 10^-3) = -200 J (work done on gas, so work done by gas is -200 J).
- 4.Step 4: Net work done by gas = W_AB + W_BC + W_CA = 400 + 0 - 200 = 200 J. Alternatively, net work is the area enclosed by the cycle on the p-V diagram: area = (2.0 x 10^5 - 1.0 x 10^5) x (3.0 x 10^-3 - 1.0 x 10^-3) = 1.0 x 10^5 x 2.0 x 10^-3 = 200 J.
- 5.Step 5: State final conclusion: The net work done by the gas per cycle is 200 J.