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

The pressure of a real gas obeys $\left(P + \frac{a}{V^2}\right)(V - b) = RT \implies P = \frac{RT}{V - b} - \frac{a}{V^2}$. Work done by the gas is $\int_{V_1}^{V_2} P dV = \int_{V_1}^{V_2} \left( \frac{RT}{V - b} - \frac{a}{V^2} \right) dV = RT \ln \left( \frac{V_2 - b}{V_1 - b} \right) + a \left( \frac{1}{V_2} - \frac{1}{V_1} \right)$.