AQA · A-Level · Physics
Course version: 7408 · Version from Sept 2016
Estimation of work done in terms of area below the graph.
Full specification wording
The p–V diagram (A-level only): 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.
Checked against the AQA 7408 specification (opens in a new tab) on .
- Point 2 of 2
- 4 key terms
Study notes
From the checked teaching notes for this specification point.
The idea
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.
What earns the marks
- 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.
Common mistakes and how to put them right
Using the area under a straight line joining the endpoints rather than under the actual curved path.
Put it right: Estimate the area beneath the drawn path.
Forgetting to convert volume from cm³ to m³ before multiplying by pressure.
Put it right: Convert to m³ so the product is in joules.
Treating the area as always positive regardless of direction.
Put it right: Relate the sign to expansion or compression.
Exam tips
- 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.
Try the lesson
The start of the The p–V diagram (A-level only): Estimation of work done in terms of area below the graph. lesson, made from checked teaching notes for this specification point.
Have a go first
On a p–V diagram, what does the area below the graph represent when the gas expands?
Listen
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. A strong answer earns marks for these points. 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 common slip: Using the area under a straight line joining the endpoints rather than under the actual curved path. Instead, estimate the area beneath the drawn path. Exam tip: Show the subdivision clearly, such as drawing strips or writing the counted squares, so the estimation method is visible.
Key words
Say what each word means before you turn the card.
Work as the area under a p–V graph
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Graphical integration and estimation techniques
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Sign conventions for expansion and compression
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Consistency of units in thermodynamic calculations
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See every definition
- Work as the area under a p–V 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.
- Graphical integration and estimation techniques
- The area is estimated by splitting the region between the path and the volume axis into strips or simple shapes, counting grid squares, or using the trapezium rule.
- Sign conventions for expansion and compression
- 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.
- Consistency of units in thermodynamic calculations
- Units are handled consistently, for example pressure in Pa and volume in m³ giving work in J.
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