The magnitude of work done by one mole of a van der Waals gas, during its isothermal reversible expa — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

For a van der Waals gas, P = RT/(V-b) - a/V^2. Work done during isothermal reversible expansion is: W = int P dV = int_{V1}^{V2} [RT/(V-b) - a/V^2] dV = RT ln((V2-b)/(V1-b)) + a(1/V2 - 1/V1).