Area under curve in reversible isothermal is more. So, more work will be done by gas.

${\mathrm{T}}_1={\mathrm{T}}_2$

$\mathrm{\Delta }\mathrm{U}=\mathrm{n}{\mathrm{C}}_{\mathrm{V}}\mathrm{\Delta }\mathrm{T}=0$

In reversible adiabatic expansion, ${\mathrm{T}}_2<{\mathrm{T}}_1$

So $\mathrm{\Delta }\mathrm{T}=-\mathrm{v}\mathrm{e}$ , $\mathrm{\Delta }\mathrm{U}=-\mathrm{v}\mathrm{e}$

In free expansion, ${\mathrm{P}}_{\mathrm{e}\mathrm{x}\mathrm{t}}=0$

So $\mathrm{W}=0$

If carried out isothermally $\left(\mathrm{\Delta }\mathrm{U}=0\right)$

$\mathrm{q}=0$ (adiabatic); from I law

If carried out adiabatically $\left(\mathrm{q}=0\right)$

$\mathrm{\Delta }\mathrm{U}=0$ (isothermal); From I law



During irreversible compression, maximum work is done on the gas (corresponding to shaded area).