Question

Derive an expression for work done by the gas in an Isothermal process.

Answer 1

Suppose $1$ mole of gas is enclosed in isothermal container. Let $P_1,V_1,T$ be initial pressure, volumes and temperature. Let expand to volume $V_2$ & pressure reduces to $P_2$ & temperature remain constant. Then, work done is given by

$W=\int dW$

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

as $PV=RT$ ($n=$ mole)

$P=\frac{RT}{V}$

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

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

$=RT{\left[InV\right]}_{V_1}^{V_2}$

$=RT[In{V}_2-In{V}_1]$

$W=RTIn\frac{V_2}{V_1}$

$W=2.303RT{log}_{10}\frac{V_2}{V_1}$

for constant temperature

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

So, also

$W=\mathrm{2.303.}RT{log}_{10}\frac{P_1}{P_2}$