Two moles of an ideal monoatomic gas is expanded according to the process AB as shown in the figure. Then heat supplied to the gas is 

The process AB is given by, $\mathrm{P}=\frac{{\mathrm{p}}_0}{{\mathrm{v}}_0}\mathrm{V}$

Also, PV = nRT

$$\begin{gathered} \therefore \mathrm{T}=\frac{\mathrm{PV}}{\mathrm{nR}}=\frac{{\mathrm{p}}_0{\mathrm{V}}^2}{{\mathrm{nRv}}_0}. \\ \mathrm{At} \mathrm{V}=2{\mathrm{v}}_0, {\mathrm{T}}_{\mathrm{B}}=\frac{{\mathrm{p}}_0{\left(2{\mathrm{v}}_0\right)}^2}{{\mathrm{nRv}}_0}=\frac{4{\mathrm{p}}_0{\mathrm{v}}_0}{\mathrm{nR}}, \mathrm{and} \mathrm{V}={\mathrm{v}}_0, {\mathrm{T}}_{\mathrm{A}}=\frac{{\mathrm{p}}_0{\mathrm{v}}_0}{\mathrm{nR}} \end{gathered}$$

$\therefore ∆\mathrm{U}=\mathrm{n}.\frac{3\mathrm{R}}{2}({\mathrm{T}}_{\mathrm{B}}-{\mathrm{T}}_{\mathrm{A}})=\mathrm{n}.\frac{3\mathrm{R}}{2}\left(\frac{4{\mathrm{p}}_0{\mathrm{v}}_0}{\mathrm{nR}}-\frac{{\mathrm{p}}_0{\mathrm{v}}_0}{\mathrm{nR}}\right)=\frac{9}{2}{\mathrm{p}}_0{\mathrm{v}}_0$

W = Area under P-V graph $=\frac{1}{2}(2{\mathrm{p}}_0+{\mathrm{p}}_0)(2{\mathrm{v}}_0-{\mathrm{v}}_0)=\frac{3}{2}{\mathrm{p}}_0{\mathrm{v}}_0$

$\therefore$Heat supplied $=∆\mathrm{U}+\mathrm{W}=6{\mathrm{p}}_0{\mathrm{v}}_0$