Question

Derive the expression for the work obtained in an isothermal reversible expansion of an ideal gas.

Answer 1

Solution:-

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-

$dW=-PdV$

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

$W=-{\int }_{V_1}^{V_2}PdV$

For an ideal gas,

$PV=nRT$

$\Rightarrow P=\frac{nRT}{V}$

Therefore,

$W=-{\int }_{V_1}^{V_2}\frac{nRT}{V}dV$

For isothermal expansion, $T$ is constant.

Therefore,

$W=-nRT{\int }_{V_1}^{V_2}\frac{dV}{V}$

$\Rightarrow W=-nRT{\left[\ln V\right]}_{V_1}^{V_2}$

$\Rightarrow W=-nRT(\ln V_2-\ln V_1)$

$\Rightarrow W=-nRT\ln \left(\frac{V_2}{V_1}\right)$

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

At constant temperature,

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

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

Therefore,

$W=-2.303nRT\log \left(\frac{P_1}{P_2}\right).....\left(2\right)$

Equation $\left(1\right)\&\left(2\right)$ are the expression for the work obtained in an isothermal reversible expansion of an ideal gas.