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A gas at 4 bar pushes a piston out against 1 bar until it, too, is at 1 bar. Which pressure goes into the work formula? Many students average the two, or use the gas pressure, or reach for nRT ln(V2/V1). Only one choice is right. This post explains why irreversible work uses the outside pressure, works through one-step, two-step and reversible versions of the same expansion, and finishes with an irreversible adiabatic case.
Where is this in the syllabus?
The NCERT Class 11 Chemistry unit Thermodynamics (2026-27 reprint) derives w = −p_ex(Vf − Vi) for a single-step change against a constant external pressure, and states that for an isothermal irreversible change q = −w = p_ex(Vf − Vi). The JEE (Main) 2026 syllabus lists the concept of work in chemical thermodynamics and "reversible and irreversible processes" in physics; the NEET (UG) 2026 syllabus has the same physics line. Check your year's syllabus.
This post uses the chemistry sign: w is work done on the gas, so expansion gives negative w. In the NCERT Physics chapter on thermodynamics, W is work done by the gas, so the same numbers appear with a plus sign. Our post on thermodynamics sign conventions compares the two.
Why the external pressure?
Work is force times distance, and the force that the piston works against is the outside pressure times the piston area. NCERT writes it as w = −p_ex · A · l = −p_ex ΔV. During a sudden expansion the gas is not in equilibrium. Its pressure is not uniform, so "the gas pressure" has no single value to use. The external pressure is the one you know.
Only in a reversible change, where p_ex differs from the gas pressure by an infinitesimal amount at every stage, can you replace p_ex with p = nRT/V and integrate to get w = −nRT ln(Vf/Vi).
Worked example 1: one step
One mole of an ideal gas at 300 K expands isothermally from 4.0 bar to 1.0 bar against a constant external pressure of 1.0 bar.
Volumes: V1 = nRT/p1 = (8.314 × 300)/(4.0 × 10^5) = 6.24 × 10^−3 m³ = 6.24 L. V2 = 24.94 L.
w = −p_ext ΔV = −(1.0 × 10^5 Pa)(18.71 × 10^−3 m³) = −1871 J ≈ −1.87 kJ.
Isothermal ideal gas, so ΔU = 0 and q = +1.87 kJ.
A unit shortcut: 1 L bar = 100 J. So 1.0 bar × 18.71 L = 18.71 L bar = 1871 J.
Worked example 2: two steps
Same start and end, but first against 2.0 bar (to V = 12.47 L), then against 1.0 bar.
Step 1: w = −2.0 bar × (12.47 − 6.24) L = −12.47 L bar = −1247 J.
Step 2: w = −1.0 bar × (24.94 − 12.47) L = −1247 J.
Total: −2494 J ≈ −2.49 kJ.
Worked example 3: the reversible limit
w = −nRT ln(V2/V1) = −8.314 × 300 × ln 4 = −3458 J ≈ −3.46 kJ.
Put the three side by side: the gas does 1.87 kJ, 2.49 kJ and 3.46 kJ of work. On a p–V graph, each answer is the area under the external-pressure line. More steps make the staircase hug the isotherm, and the reversible path covers the whole area under it. NCERT's Fig. 5.5 draws the same idea for compression, with single-step, multi-step and reversible paths shown as shaded areas.
What changes for compression?
Run the one-step version backwards: compress from 24.94 L to 6.24 L against a constant 4.0 bar. w = −4.0 bar × (6.24 − 24.94) L = +74.8 L bar = +7.48 kJ. The reversible compression needs only +3.46 kJ. A sudden compression needs more work done on the gas than the reversible one, while a sudden expansion gets less work out of it.
Worked example 4: irreversible and adiabatic
One mole of a monatomic ideal gas (Cv = 3R/2) at 300 K and 4.0 bar expands adiabatically against a constant 1.0 bar until its pressure is 1.0 bar. Find T2.
With q = 0, ΔU = w:
n Cv (T2 − T1) = −p_ext (V2 − V1) = −p_ext (RT2/p2 − RT1/p1).
With p_ext = p2 = 1.0 bar and p1 = 4.0 bar: 1.5(T2 − 300) = −(T2 − 75).
2.5 T2 = 525, so T2 = 210 K, and w = 1.5 × 8.314 × (210 − 300) = −1122 J.
The reversible adiabatic formula TV^(γ−1) = constant would give 300 × (1/4)^0.4 = 172 K. That formula needs a reversible path, so here it is wrong. The irreversible gas cools less because it does less work.
Common mistakes
Using the gas pressure, or an average, in w = −pΔV for a sudden change.
Using nRT ln(V2/V1) whenever T is constant. It needs a reversible path too.
Treating free expansion as special maths. With p_ext = 0, w = 0, as NCERT's Problem 5.2 shows.
Mixing L bar and L atm. 1 L atm is 101.3 J, not 100 J.
A practice routine
Redo examples 1 to 3 with three steps (against 3, 2 and then 1 bar) and check the answer falls between 2.49 and 3.46 kJ.
Sketch each path on a p–V graph and shade the work.
Solve two adiabatic irreversible expansions with a diatomic gas (Cv = 5R/2).
Write the first-law line before every calculation.
If the staircase picture is hard to imagine, Fermi Simulation Lab has interactive simulations for ideas that are easier to see than read.
Today, rework example 4 for a diatomic gas and compare your T2 with the reversible value.
