For a reversible isothermal expansion of an ideal gas, the work done in terms of the initial and final volumes is which expression?

• A. W = n R T ln(Vf/Vi)
• B. W = n R T (Vf − Vi)
• C. W = P ΔV
• D. W = n R T ln(Vi/Vf)

Answer: (C)

For an isothermal (temperature constant) reversible expansion of an ideal gas, the pressure changes as the volume changes, following P = nRT / V. The work done by the gas is the integral W = ∫ from Vi to Vf of P dV. Substituting P gives W = ∫ Vi^Vf (nRT / V) dV = nRT ln(Vf/Vi). This is the exact expression because pressure is not constant during the process; you can’t use W = P ΔV unless the pressure remains constant throughout. Since Vf > Vi in an expansion, ln(Vf/Vi) is positive, so the work done by the gas is positive. The form nRT ln(Vi/Vf) would give the negative of the correct result.