Question

Derive an expression of the mechanical work done in an isothermal reversible expansion of an ideal gas.

Answer 1

Let us consider n moles of ideal gas enclosed in a cylinder fitted with a weightless and frictionless piston The work of expansion for a small change of volume dV against the external pressure P is given by

${\delta }_w=-PdV$

$\therefore$ Total work done when the gas expands from initial volume $V_1$ to final $V_2$ will be

$W=-{\int }_{v_1}^{v_2}PdV$

For ab ideal gas, $PV=nRT$i.e

$p=\frac{n}{V}RT$

Hence

$W=-{\int }_{v_1}^{v_2}\frac{n}{V}RTdV$

For isothermal expansion , T = constant so that

$w=-nRT{\int }_{v_1}^{v_2}\frac{1}{V}dV$

$=-nRT{I}_n\frac{V_2}{V_1}$

$=-2.303nRT\log \frac{V_2}{V_1}$......(1)

But at constant temperature

$P_1{V}_1=P_2{V}_2$

or $\frac{V_2}{V_1}=\frac{P_1}{P_2}$

$\therefore \left(1\right)becomes,$

$w=-2.303nRT\frac{P_1}{P_2}$.....(2)