What is an isobaric process?

A thermodynamic process in which pressure stays constant. Charles' law describes the volume-temperature relationship during an isobaric process.

How is work calculated in an isobaric process?

W = P × ΔV. Because pressure is constant, the work done by the gas is simply pressure times the change in volume.

Cp is the molar heat capacity at constant pressure. For an ideal diatomic gas it is about 29.1 J/(mol·K); for monatomic gases it is about 20.8 J/(mol·K).

How does it relate to Charles' law?

Charles' law (V₁/T₁ = V₂/T₂) is the gas-law form of an isobaric process: volume scales with absolute temperature when P is constant.

Are isobaric processes common in real life?

Yes. Many atmospheric processes, open piston engines on the power stroke, and laboratory experiments with open containers approximate isobaric behaviour.

Can work done in an isobaric process be negative?

Yes. If the gas is compressed (ΔV negative), the work done by the gas is negative, meaning the surroundings do work on the gas instead. The formula W = P × ΔV handles both directions automatically.

Isobaric Process Calculator

Compute work W = PΔV, heat Q = nCpΔT, and ΔU = Q − W for a constant-pressure process.

Isobaric Process Calculator

Constant P. W = PΔV, Q = nCpΔT, ΔU = Q − W (Cp ≈ 29.1 J/mol·K).

P (constant)

n (mol)

V₁

V₂

T₁

T₂

Work done by gas (W)

1013 J

W = PΔV = 1013 J

Heat (Q)

4365 J

Q = nCpΔT = 4365 J

ΔU (internal energy)

3352 J

ΔU = Q − W = 3352 J

The Isobaric Process and Charles' Law

An isobaric process is a thermodynamic process at constant pressure. The state variables that change are volume and temperature, and they do so in lockstep according to Charles' law: V₁/T₁ = V₂/T₂. Charles' law is, in fact, simply the gas-law statement of any isobaric process applied to an ideal gas.

Work, Heat, and Internal Energy

Because pressure is constant, the work done by the gas on its surroundings is W = P × ΔV, straightforward to evaluate. The heat input required is Q = n × Cp × ΔT, where Cp is the molar heat capacity at constant pressure. The change in internal energy follows from the first law of thermodynamics: ΔU = Q − W.

A useful relation for ideal gases is Cp − Cv = R, so heat in an isobaric process is always larger than the corresponding heat in an isochoric process at the same ΔT, because some of the energy goes into pushing the surroundings back.

Worked Example: Heating Gas in a Cylinder with a Free Piston

2 mol of an ideal diatomic gas (Cp = 29.1 J/(mol·K)) sits in a cylinder fitted with a free piston, so the pressure stays fixed at P = 100 kPa. Heating raises the temperature from T₁ = 300 K to T₂ = 450 K. By the ideal gas law, V₁ = nRT₁/P = (2 × 8.314 × 300) / 100,000 = 0.0499 m³, and V₂ = nRT₂/P = (2 × 8.314 × 450) / 100,000 = 0.0748 m³, so ΔV = 0.0249 m³. Work done by the gas: W = P × ΔV = 100,000 × 0.0249 = 2495 J. Heat supplied: Q = n × Cp × ΔT = 2 × 29.1 × 150 = 8730 J. Change in internal energy: ΔU = Q − W = 8730 − 2495 = 6235 J. As a check, ΔU also equals n × Cv × ΔT with Cv = Cp − R = 20.79 J/(mol·K), giving 2 × 20.79 × 150 = 6236 J, matching within rounding.

Real-World Examples

The power stroke of a steam engine: steam at constant pressure pushes a piston.
Boiling water in an open pot: vapour forms at atmospheric pressure.
Atmospheric expansions of air parcels that change altitude slowly enough to equilibrate pressure.