Work done by gas pistons
Derive pressure-volume work for piston systems and connect work sign to expansion/compression.
Offline pages
Saved on this device only. Connect at least once every 30 days to confirm Premium. Sign-out or a change of account removes saved copies.
On this page
You stopped at:
ContinueA gas does work when its pressure force moves a piston. Pressure supplies the force; the volume change records how far that force acts. To calculate the work, identify both the pressure acting at the moving boundary and the direction of motion. The initial and final volumes alone are insufficient.
At a glance
- Prerequisites: mechanical work and the first-law conventions
- You will learn: derive pressure–volume work; integrate a stated pressure path; distinguish gas output from load work when a piston stores energy
- Key result: for a quasistatic gas-pressure path, Wₒᵤₜ = ∫_Vᵢ^VfP dV, Wₒₙ = -Wₒᵤₜ.
- Common trap: treating every pressure in the apparatus as interchangeable, or using an endpoint pressure throughout a varying-pressure path.
Choose the moving boundary
Choose the gas as the system. Let a plane piston have area A, and take its outward displacement dx as positive. If the gas exerts a uniform boundary pressure Pb on its face, the force it supplies is Fb = PbA. The corresponding gas-volume change is dV = A dx.
The opposing load can include atmospheric pressure and a mechanical load. Write its total resisting force as F_load = P_loadA. For a massless frictionless piston in mechanical balance, Pb = P_load. For a massive accelerating piston, the forces need not balance: part of the gas output can increase piston kinetic energy.
In a quasistatic process the gas remains close to equilibrium, so its thermodynamic pressure P can represent Pb. Quasistatic motion alone does not guarantee reversibility: friction or heat transfer across a finite temperature difference can still make a process irreversible.
Positive Wₒᵤₜ means energy leaves the gas through work. Positive Wₒₙ means energy enters it. With unchanged bulk kinetic and potential energies, the first law is Δ U = Q + Wₒₙ = Q-Wₒᵤₜ. Here U is internal energy, also written as E elsewhere in this course.
From piston displacement to volume work
The mechanical work supplied by the gas is δ Wₒᵤₜ = Fb dx = (PbA) dx = Pb dV. For a quasistatic gas-pressure path this becomes Wₒᵤₜ = ∫_Vᵢ^VfP(V) dV, Wₒₙ = -∫_Vᵢ^VfP(V) dV. The area under a pressure–volume curve represents signed work. At positive absolute pressure, expansion gives positive output; compression reverses the integration limits and gives negative output. Only constant pressure allows Wₒᵤₜ = P(Vf-Vᵢ).
The units explain the numerical scale: Pa m³ = N m = J, 1 kPa L = 1 J. Convert both quantities to a matching pair. Multiplying a pressure in pascals by a volume difference in litres without conversion makes the numerical work a thousand times too large.
The OpenStax pressure–volume work derivation uses a quasistatic gas-pressure curve. In a rapid process there may be no single equilibrium gas pressure along the route. A stated boundary pressure or external-load model can still determine work, but a guessed internal-pressure curve cannot. The familiar expression ∫ Pₑₓₜ dV for gas output requires a piston model that accounts for any energy stored or dissipated between gas and load.
Constant-pressure piston work
Worked example 1
Work during constant-pressure piston expansion
Problem
A gas expands quasistatically at constant pressure 2.5 × 10⁵ Pa from 2.0 × 10⁻³ m³ to 3.2 × 10⁻³ m³ through a massless frictionless piston. Find its work output and the work done on it for the first law.
Show full solution
Find the area under the pressure–volume path
Method
Calculate the signed volume change and multiply by the constant pressure.Reason
The pressure is constant, so the work integral reduces to a rectangle.Working
Answer
Δ V = (3.2-2.0) × 10⁻³ = 1.2 × 10⁻³ m³, Wₒᵤₜ = (2.5 × 10⁵)(1.2 × 10⁻³) = 300 J. The units are Pa m³ = J.
Convert to the first-law convention
Working
Answer
The gas outputs 300 J, so Wₒₙ = -300 J. Expansion gives positive output and negative work on the gas. Shrinking the volume change to zero at this pressure makes both work values zero. This calculation alone does not fix heat transfer or internal-energy change.
Guided practice 2
A straight pressure path is not constant pressure
Problem
A gas expands quasistatically through a massless frictionless piston. Its pressure–volume path is a straight line from 300 kPa at 2.0 L to 100 kPa at 5.0 L. Find work output and work on the gas. Explain why neither endpoint pressure should multiply the whole volume change.
Calculate the signed area
Hints
Hint 1
Read the whole path
A straight-line path has average pressure equal to the mean of its endpoint pressures. This shortcut is specific to a linear path.Show full solution
Integrate the varying pressure
Method
Use the trapezoidal area, or integrate the stated straight line.Reason
Pressure varies linearly with volume, so its volume-average is the mean of the endpoint pressures.Working
Answer
Wₒᵤₜ = ((300 + 100)/2)(5.0-2.0) = 600 J, using kPa L = J. Therefore Wₒₙ = -600 J.
Quick check
The output lies between 100(3) = 300 J and 300(3) = 900 J, as a positive pressure between those limits should. Reversing the same pressure path changes the work signs.
When the piston itself gains energy
For a horizontal frictionless piston, work–energy accounting gives W_(gas → piston) = W_(piston → load) + Δ Kₚᵢₛₜₒₙ. The left side is work output from the gas; the first term on the right is work delivered to the load. They agree when the piston has no net kinetic-energy change. If the piston accelerates, substituting the load pressure for gas-side boundary pressure omits the stored kinetic energy. A different model must also account for piston potential-energy changes or friction where present.
Check your understanding 1
A horizontal frictionless piston of mass 2.0 kg and area 0.010 m² starts at rest. Throughout an outward displacement of 0.020 m, the uniform gas-side boundary pressure is 120 kPa and the total opposing load pressure is 100 kPa. These pressures are supplied boundary data, not a claim that the gas follows a quasistatic equilibrium path. Find the gas work output, work delivered to the load, piston kinetic-energy increase and final speed. Explain why using 100 kPa for the gas work loses part of the energy.
Show answer
The gas-volume change is Δ V = AΔ x = (0.010)(0.020) = 2.0 × 10⁻⁴ m³. The two transfers are W_(gas → piston) = (120000)Δ V = 24 J, W_(piston → load) = (100000)Δ V = 20 J. Thus Δ Kₚᵢₛₜₒₙ = 24-20 = 4.0 J and vf = square root of (2Δ Kₚᵢₛₜₒₙ/m) = 2.0 m/s. As a force check, the net force is (120000-100000)(0.010) = 200 N, whose work over 0.020 m is also 4.0 J.
Choosing the gas excludes piston kinetic energy from its internal energy; that energy has already left the gas as work. Choosing the gas and piston together would instead include the piston’s energy in the combined-system balance. Check that your explanation distinguishes the gas boundary from the load, and accounts for all 24 J as 20 + 4.
Practice set
Check your understanding 2
A gas is compressed quasistatically through a massless frictionless piston at constant pressure 1.8 × 10⁵ Pa from 4.0 L to 2.0 L. Find its work output.
Show hint
Show answer
Check your understanding 3
For that same compression, find the work to use in Δ U = Q + Wₒₙ.
Show hint
Show answer
Check your understanding 4
Along a quasistatic piston process, P(V) = kV with constant k > 0. Write the work output from Vᵢ to Vf as an integral and evaluate it. State the units of k.
Show hint
Show answer
Before you move on
Pressure supplies a force; piston motion converts it into work. Use the signed integral of the appropriate boundary pressure, and convert to Wₒₙ before using the work-input first law. A quasistatic gas-pressure curve supplies a path for that integral. If the piston stores energy, distinguish gas output from the work that reaches the load.
Spotted an error? Report a correction
Syllabus and review details
No official syllabus alignment is listed for this lesson.