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Library / Thermodynamics / THE-03

Thermodynamics

Boundary work at constant pressure

Work done by a gas expanding against a constant pressure.

Ref
THE-03
Rev
v1
Field
Thermodynamics
Result
Wb [kJ]

For a quasi-equilibrium process with a moving boundary, V is total volume. Expansion gives positive work by the system. The second form applies at constant p; use absolute or gauge pressure consistently with the surroundings.

A gas that pushes a piston out at constant pressure does work equal to the pressure times the volume change.

Calculator

Solve for
kJ
kPa
105,000
m³
0.00110
m³
0.00110
  • Volume change0.25 m³
  • Work per cubic metre swept200 kJ/m³

What drives the result

Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.

Boundary work (kJ) current value 50 kJ
V₁Initial volumeKey driver
−1,93069.8
V₂Final volume
−19.81,980
pPressure
2.51,250
  • kPa × m³ gives kJ directly.
  • Compression gives negative work by the system.
Chart data as tables

Ranked variables

VariableTypical rangeLowest resultHighest resultSwing (kJ)Local stepChange in result
V₁ Initial volume0.001 to 10 m³−1,93069.82,000+10%−4 %
V₂ Final volume0.001 to 10 m³−19.81,9802,000+10%+14 %
p Pressure10 to 5,000 kPa2.51,2501,247+10%+10 %

Sweep of Final volume

Final volume (m³)Boundary work (kJ)
0.001−19.8
0.003162−19.37
0.01−18
0.03162−13.68
0.10
0.316243.25
1180
3.162612.5
101,980

Sweep one variable

Boundary work (kJ)Hover, touch, or use the arrow keys on the chart
0500100015002000 0.0010.010.1110 Final volume (m³), log scale now 50

Variables and how each one contributes

Wb

Boundary work

Result · kJ

What the equation gives you. Everything else feeds it.

p

Pressure

kPa · typical 10 to 5,000

Constant pressure on the boundary. Directly proportional.

V₁

Initial volume

m³ · typical 0.001 to 10

Initial volume. Lowers the work.

V₂

Final volume

m³ · typical 0.001 to 10

Final volume. Raises the work.

Worked example with the values above

  1. Volume change: 0.35 − 0.1 = 0.25 m³.
  2. Multiply by p = 200 kPa.
  3. Wb = 50 kJ.

Assumptions and limits

  • Quasi-equilibrium process.
  • Constant pressure.

Common mistakes

  • Using the wrong pressure (gauge vs absolute) against ambient.
  • Forgetting the sign.

Statistics and uncertainty

Volume changes come from small differences between large readings, so a ±2% error in V can mean a larger error in ΔV.

Where it is used

  • Piston and cylinder energy balances.
  • Expansion of gases against a constant load.

Maths you need

  • Pressure and volume: kPa and m³
  • Work: force × distance

Related formulas

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