Two moles of helium gas undergoes a cyclic process as shown in figure. Assuming the gas to be ideal. ($R=\raisebox{1ex}{$25$}\!\left/ \!\raisebox{-1ex}{$3 J/mol$}\right. K$and l$\ln \left(2\right)=0.693$). Then

As we know in a cyclic process, the change in heat energy or heat supplied to the gas is equal to the net work done by the gas. Here, $AB$ is isobaric process. Hence, work done during this process from $A$ to $B$ is

$$\begin{gathered} W_{AB}=P\left(V_B-V_A\right)=nR\left(T_B-T_A\right) \\ \mathrm{or} W_{AB}=2\times \raisebox{1ex}{$25$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\times \left(400-300\right)=1666.66 J \end{gathered}$$

Work done during isothermal process from $B$ to $C$ is

$$\begin{gathered} W_{BC}=nR{T}_C\ln \left(\raisebox{1ex}{$V_2$}\!\left/ \!\raisebox{-1ex}{$V_1$}\right.\right)=nR{T}_C\ln \left(\raisebox{1ex}{$P_1$}\!\left/ \!\raisebox{-1ex}{$P_2$}\right.\right) \\ =2\times \raisebox{1ex}{$25$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\times 400\ln \left(2\right) \\ =2\times \raisebox{1ex}{$25$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\times 400\times 0.693=4620 J \end{gathered}$$

Work done during isobaric process from $C$ to $D$

$$\begin{gathered} W_{CD}=nR\left(T_D-T_C\right)=2\times \raisebox{1ex}{$25$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\times \left(300-400\right) \\ =-1666.66 J \end{gathered}$$

Work done during isothermal process from $D$ to $A$

$$\begin{gathered} W_{DA}=nR{T}_D\ln \left(\raisebox{1ex}{$P_D$}\!\left/ \!\raisebox{-1ex}{$P_A$}\right.\right) \\ =nR{T}_D\ln \left(2\right) \\ =-2\times \raisebox{1ex}{$25$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\times 300\times 0.693 \\ =-3465 J \end{gathered}$$

Net work done $=W_{AB}+W_{BC}+W_{CD}+W_{DA}$

                        $=1666.66+4620-1666.66-3465=1155 J$

Now from first law of thermodynamics

Here $∆Q=∆U+∆W$  thus we have

$$\begin{gathered} ∆U=0 \\ ∆Q=∆W=1152.3 J \end{gathered}$$

So the heat given to the system is $1152.3 J$ As the gas returns to its original state, there is no change in internal energy.