Calculate constant-pressure work in a gas
Use W = pΔV for a constant-pressure volume change and distinguish work done by a gas from work done on it.
A Level; Cambridge International Physics 9702, 2025–2027 syllabus, Topic 16. These original explanations and calculations are study aids; full timed assessment and human subject-teacher review remain outstanding.
Before you begin
- Use temperature, molecular kinetic energy, pressure, volume and energy-transfer conventions.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
For an expanding gas at constant pressure, what is the sign of work done by the gas?
What you will learn
- Use W = pΔV for constant-pressure volume changes
- Distinguish work done by the gas from work done on the gas
- Interpret expansion and compression signs
Pressure and volume change determine boundary work
For a gas changing volume at constant pressure p, the magnitude of the work is pΔV, where ΔV = Vfinal − Vinitial. With p in pascals and ΔV in cubic metres, the work is in joules because Pa m³ = J.
In an expansion, ΔV is positive and the gas does positive work on its surroundings: Wby = pΔV. The work done on the gas is the opposite sign, Won = −pΔV.
Compression reverses which system does the work
During compression, ΔV is negative. Work done by the gas is negative, while work done on the gas is positive: the surroundings transfer energy to the gas through mechanical work.
Keep the stated convention visible before substituting values. The first-law form in this syllabus uses W for work done on the system, so W = −pΔV when applying that law to a constant-pressure gas-volume change.
Pause and explain
A gas is compressed at constant pressure. What is the sign of work done on the gas?
Worked example
A gas expands by 2.0 × 10⁻³ m³ at constant pressure 1.0 × 10⁵ Pa. Find the work done by the gas and the work done on it.
Show the worked solution
- Use Wby = pΔV.
- Wby = 1.0 × 10⁵ × 2.0 × 10⁻³.
- Reverse the sign for work done on the gas.
Answer Work done by the gas = +200 J; work done on the gas = −200 J.
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
A gas expands by 3.0 × 10⁻³ m³ at constant pressure 2.0 × 10⁵ Pa. Calculate work done by the gas and on the gas.
Give me a hint
Calculate pΔV for work by the gas, then reverse the sign for work on it.
Compare my reasoning
- Wby = pΔV.
- Wby = 2.0 × 10⁵ × 3.0 × 10⁻³ = 600 J.
- Won = −600 J.
The gas does +600 J of work; the work done on the gas is −600 J.
Look for these in your work
- Constant-pressure formula used
- Joule magnitude calculated
- Opposite work convention stated
A gas is compressed by 1.5 × 10⁻³ m³ at constant pressure 4.0 × 10⁵ Pa. Find the work done on the gas.
Give me a hint
For compression ΔV is negative; work done on the gas is −pΔV.
Compare my reasoning
- ΔV = −1.5 × 10⁻³ m³.
- Wby = pΔV = −600 J.
- Won = −Wby = +600 J.
Work done on the gas is +600 J.
Look for these in your work
- Compression sign applied
- Work-by calculated
- Work-on sign correct
Explain why the work done by a gas during expansion is the negative of the work done on the gas.
Give me a hint
Identify the two interacting systems and apply energy-transfer sign conventions.
Compare my reasoning
- During expansion the gas transfers energy mechanically to the surroundings.
- Work done by the gas is positive for ΔV > 0.
- The work done on the gas is the equal-magnitude energy transfer in the opposite direction, so it is negative.
The same boundary energy transfer is positive for the gas as giver and negative for the gas as receiver.
Look for these in your work
- Expansion identified
- Work by sign stated
- Opposite system convention explained
Common mistakes
- Using pressure in kPa without converting if the volume is in m³
- Giving the same sign for work done by and work done on the gas
- Using pΔV without checking that pressure is constant
What can you explain now?
For an expanding gas at constant pressure, what is the sign of work done by the gas?
Compare with the explanation
Positive
Expansion gives positive ΔV and Wby = pΔV.
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Make it stick
How do you distinguish work done by an expanding gas from work done on it?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.