The P-V diagram of 2 gm of helium gas for a certain process A$\to$B is shown in the figure. what is the heat given to the gas during the process $\mathrm{A}\to \mathrm{B}$

Change in internal energy from $\mathrm{A}\to \mathrm{B}$ is

$\begin{array}{l}\mathrm{\Delta U}=\frac{\mathrm{f}}{2}\mathrm{\mu R\Delta T}=\frac{\mathrm{f}}{2}\left({\mathrm{P}}_{\mathrm{f}}{\mathrm{V}}_{\mathrm{f}}-{\mathrm{P}}_{\mathrm{i}}{\mathrm{P}}_{\mathrm{i}}\right)\\ =\frac{3}{2}\left(2{\mathrm{P}}_0\times 2{\mathrm{V}}_0-{\mathrm{P}}_0\times {\mathrm{V}}_0\right)=\frac{9}{2}{\mathrm{P}}_0{\mathrm{V}}_0\end{array}$

Work done in process $\mathrm{A}\to \mathrm{B}$ is equal to the area covered by the graph with volume axis, i.e.,

$\begin{array}{l} {\mathrm{W}}_{\mathrm{A}\to \mathrm{B}}=\frac{1}{2}\left({\mathrm{P}}_0+2{\mathrm{P}}_0\right)\times \left(2{\mathrm{V}}_0-{\mathrm{V}}_0\right)=\frac{3}{2}{\mathrm{P}}_0{\mathrm{V}}_0\\ \mathrm{Hence}, \mathrm{Q}=\mathrm{\Delta U}+\mathrm{W}=\frac{9}{2}{\mathrm{P}}_0{\mathrm{V}}_0+\frac{3}{2}{\mathrm{P}}_0{\mathrm{V}}_0=6{\mathrm{P}}_0{\mathrm{V}}_0\end{array}$