One mole of an ideal gas at temperature ${\mathrm{T}}_1$ expand according to the law $\left(\frac{\mathrm{P}}{\mathrm{V}}\right) =$ constant. Find the work done when the final temperature become ${\mathrm{T}}_2$:

$\mathrm{W} = {\int }_{{\mathrm{v}}_1}^{{\mathrm{v}}_2}\mathrm{PdV} = {\int }_{{\mathrm{V}}_1}^{{\mathrm{V}}_2}\mathrm{KV} \mathrm{dV}$

                $(\because \frac{\mathrm{P}}{\mathrm{V}} = \mathrm{K} = \mathrm{constant})$

$\therefore \mathrm{W} = \frac{1}{2}\mathrm{k}({{\mathrm{V}}_2}^2-{{\mathrm{V}}_1}^2)$

     PV = RT But p = KV

$\therefore {\mathrm{KV}}^2 = \mathrm{RT}$

or $\mathrm{K}({{\mathrm{V}}_2}^2-{{\mathrm{V}}_1}^2) = \mathrm{R}({\mathrm{T}}_2-{\mathrm{T}}_1)$

$\therefore \mathrm{W} = \frac{\mathrm{R}}{2}({\mathrm{T}}_2-{\mathrm{T}}_1)$