Question

Derive an expression for maximum work obtainable during isothermal reversible expansion of an ideal gas from initial volume \( V_1 \) to final volume \( V_2 \).

Hint:

During isothermal processes for an ideal gas, the product of pressure and volume remains constant, reflecting Boyle's Law: ( PV = constant ).

Answer 1

Derivation of the expression for maximum work during isothermal reversible expansion of an ideal gas: 
The work done in an infinitesimal expansion of an ideal gas at constant temperature is given by: \[ dW = - P dV \] Using the ideal gas equation, \[ P = \frac{nRT}{V} \] Substituting this into the expression for work, \[ dW = - \frac{nRT}{V} dV \] Integrating from the initial volume \( V_1 \) to the final volume \( V_2 \): \[ W = - nRT \int_{V_1}^{V_2} \frac{dV}{V} \] Since the integral of \( \frac{1}{V} \) is \( \ln V \), we get: \[ W = - nRT \ln \left( \frac{V_2}{V_1} \right) \] This equation represents the maximum work obtainable during the isothermal reversible expansion of an ideal gas.