Chemistry · Thermodynamics · NEET
For a gas expanding or getting compressed against a constant external pressure, w = −pₑₓₜ × ΔV, where ΔV = V_final − V_initial. The pressure used is the EXTERNAL (surroundings) pressure pₑₓₜ, not the pressure of the gas itself. The minus sign comes from the IUPAC convention that work done BY the gas on the surroundings is counted as negative for the system.
When a gas expands, its volume increases, so ΔV is positive. In w = −pₑₓₜ × ΔV, a positive ΔV multiplied by the minus sign gives a negative w. Physically, the gas pushes the surroundings outward, so the gas GIVES energy to the surroundings. Energy leaving the system is written as negative work in the IUPAC convention. So: expansion → w is negative; compression → w is positive.
You use the EXTERNAL pressure (pₑₓₜ), the pressure of the surroundings pushing on the gas. This is the number two most students get wrong. For an irreversible expansion against a fixed outside pressure, only pₑₓₜ matters, even if the gas's own pressure is higher. The gas pressure is used only in the reversible case, where you write w = −nRT ln(V₂/V₁) for isothermal reversible expansion.
They are equal in size but opposite in sign. In the NEET/IUPAC convention, w always means work done ON the gas (the system). If the question asks 'work done BY the gas,' that value is −w. Example: if w (on the gas) = −900 J, then work done by the gas = +900 J. Always read the wording carefully — this is a classic NEET trap.
If pressure is in atm and volume in litres, work comes out in L·atm. Convert using 1 L·atm = 101.3 J. If pressure is in bar and volume in litres, use 1 L·bar = 100 J. If pressure is in N/m² (pascal) and volume in m³, the answer is already in joules. Check the units the question gives you before you convert.
If the volume does not change, ΔV = 0, so w = −pₑₓₜ × 0 = 0. No volume change means the gas does not push its boundary, so no pressure-volume work is done. This is why, at constant volume, the first law simplifies to ΔU = q (all heat goes into internal energy).
Under isothermal condition, a gas at 300 K expands from 0.1 L to 0.25 L against a constant external pressure of 2 bar. The work done by the gas is (Given 1 L·bar = 100 J):
An ideal gas expands isothermally from 10⁻³ m³ to 10⁻² m³ at 300 K against a constant pressure of 10⁵ N·m⁻². The work done on the gas is:
A gas is allowed to expand in a well-insulated container against a constant external pressure of 2.5 atm from an initial volume of 2.50 L to a final volume of 4.50 L. The change in internal energy ΔU of the gas in joules will be:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Work is a path function. The amount of PV work depends on HOW the gas expands — for the same start and end states, a reversible path gives more work than an irreversible path against a fixed pressure. Because it depends on the path, work is not a state function.
Use w = −nRT ln(V₂/V₁) only for a REVERSIBLE ISOTHERMAL process of an ideal gas, where the external pressure is always just slightly less than the gas pressure. Use w = −pₑₓₜΔV for an IRREVERSIBLE process against a constant external pressure. Picking the wrong formula is a common NEET mistake.
In a free expansion, the gas expands into a vacuum, so the external pressure pₑₓₜ = 0. Then w = −pₑₓₜΔV = 0. No work is done even though the volume increases, because the gas pushes against nothing.
NEET follows the IUPAC sign convention, where w is the work done ON the system. When a gas expands it does work ON the surroundings and loses energy, so w must be negative. The minus sign in w = −pₑₓₜΔV builds this convention directly into the formula.
