Two moles of an ideal monoatomic gas undergoes a process 1 - 2 - 3 as shown in figure.  Then total heat supplied to the gas in the process is

 

 

(4)

$\begin{array}{l}{Q}_{12}=Heat\sup plied\mathrm{in}theprocess1-2\\ =2\times \frac{3R}{2}\times \left(2T_0-T_0\right)=3R{T}_0\\ \mathrm{Pr}ocess2-3isisothermal.Thereforeheat\sup pliedto\\ thegas,Q_{23}=Workdonebythegas\\ =P_2{V}_2\ln \left(\frac{V_3}{V_2}\right)\\ =nR{T}_2\ln \left(\frac{V_3}{V_2}\right)=2.R.2T_0\ln \left(\frac{V_0}{V_0}\right)\\ \therefore Q_{23}=4R{T}_0\ln \left(2\right)\\ \therefore Totalheat\sup pliedtothegas\\ =Q_{12}+Q_{23}=R{T}_0\left(3+\ln 16\right)\end{array}$