P-V curve of a diatomic gas is shown in the figure. Find the total heat given to the gas in the process $\mathrm{A} \to \mathrm{B} \to \mathrm{C}$

From first law of thermodynamics

$∆{\mathrm{Q}}_{\mathrm{ABC}} = ∆{\mathrm{U}}_{\mathrm{ABC}} + ∆{\mathrm{W}}_{\mathrm{ABC}}$

$∆{\mathrm{W}}_{\mathrm{ABC}} = ∆{\mathrm{W}}_{\mathrm{AB}}+∆{\mathrm{W}}_{\mathrm{BC}} = 0 + {\mathrm{nRT}}_{\mathrm{B}}\ln \frac{{\mathrm{V}}_{\mathrm{C}}}{{\mathrm{V}}_{\mathrm{B}}}$

                = ${\mathrm{nRT}}_{\mathrm{B}} \ln \frac{2{\mathrm{V}}_0}{{\mathrm{V}}_0}$

               = ${\mathrm{nRT}}_{\mathrm{B}} \ln 2 = 2{\mathrm{P}}_0{\mathrm{V}}_0 \ln 2$

$∆\mathrm{U} = {\mathrm{nC}}_{\mathrm{V}}∆\mathrm{T} = \frac{5}{2}(2{\mathrm{P}}_0{\mathrm{V}}_0-{\mathrm{P}}_0{\mathrm{V}}_0)$

$\Rightarrow ∆{\mathrm{Q}}_{\mathrm{ABC}} = \frac{5}{2}{\mathrm{P}}_0{\mathrm{V}}_0+2{\mathrm{P}}_0{\mathrm{V}}_0 \ln 2$