One mole of ideal monoatomic gas is taken through the cyclic process as shown in T-P diagram. Process $\text{C}\to \text{D}$ is such that PT = constant (take $\mathcal{l}n3=1.0986$)  $\text{B}\to \text{C}$ isobaric process  $\text{A}\to \text{B}$is isothermal process.

Since,  $\text{C}\to \text{D},\text{PT}=\text{constant}$

$\Rightarrow {\text{P}}_{\text{C}}={\text{P}}_{\text{B}}={\text{3P}}_{\text{0}}$

$$\begin{gathered} \therefore Q_{BC}=n{C}_P\Delta T=n\frac{5}{2}R(\frac{4}{3}{T}_0-T_0)=\frac{5nR{T}_0}{6} \\ \text{C}\to \text{D,PT}=\text{constant}\Rightarrow {\text{PV}}^{1/2}=\text{constant } \\ \therefore \Delta {\text{Q}}_{\text{CD}}=\frac{3}{2}\text{nR}({\text{2T}}_{\text{0}}-\frac{\text{4}}{\text{3}}{\text{T}}_{\text{0}})+\frac{\text{nR}({\text{2T}}_{\text{0}}-\frac{\text{4}}{\text{3}}{\text{T}}_{\text{0}})}{1-\frac{1}{2}}=\frac{\text{7}}{\text{3}}{\text{nRT}}_{\text{0}} \\ \text{A}\to \text{B},\Delta Q=nR{T}_0\mathcal{l}n\left(\frac{{\text{P}}_{\text{A}}}{{\text{P}}_{\text{B}}}\right) \\ D\to A \\ \Rightarrow V=\text{constant } \\ \therefore \Delta Q_{D\to A}=n{C}_V\Delta T=n\frac{3}{2}R(T_0-2T_0)=-\frac{3}{2}nR{T}_0 \\ \eta =1-\frac{(\frac{3}{2}nR{T}_0+nR{T}_0\mathcal{l}n3)}{(\frac{7}{3}+\frac{5}{6})nR{T}_0}=0.1793 \end{gathered}$$