One mole of oxygen undergoes a cyclic process in which volume of the gas change 10 times within the cycle, as shown in the figure. Process: AB and CD are adiabatic. Find the efficiency of the process, in percent. Take (10)^0.4 =2.5.

$\begin{array}{l}\mathrm{n}=1,\mathrm{\gamma }=\frac{7}{5}\\ \Rightarrow \mathrm{\gamma }-1=\frac{2}{5}\\ {\mathrm{Q}}_{\mathrm{AB}}={\mathrm{Q}}_{\mathrm{CD}}=0\end{array}$

Let ${\mathrm{T}}_{\mathrm{A}}={\mathrm{T}}_1\text{ and }{\mathrm{T}}_{\mathrm{D}}={\mathrm{T}}_2\text{ with }{\mathrm{T}}_2>{\mathrm{T}}_1$

For the process AB, we have

$\begin{array}{l} {\mathrm{TV}}^{\mathrm{\gamma }-1}=\text{ constant }\\ \Rightarrow {\mathrm{TV}}^{\frac{2}{5}}=\text{ constant }\\ \Rightarrow {\mathrm{T}}_{\mathrm{A}}{\left(10{\mathrm{V}}_0\right)}^{\frac{2}{5}}={\mathrm{T}}_{\mathrm{B}}{\left({\mathrm{V}}_0\right)}^{\frac{2}{5}}\\ \Rightarrow {\mathrm{T}}_{\mathrm{B}}={\mathrm{T}}_1(10{)}^{0.4}\approx 2.5{\mathrm{T}}_1\end{array}$

Similarly, ${\mathrm{T}}_{\mathrm{C}}=2.5{\mathrm{T}}_2$

Now, ${\mathrm{Q}}_{\mathrm{BC}}={\mathrm{C}}_{\mathrm{V}}\mathrm{\Delta T}$

$\begin{array}{l}\Rightarrow {\mathrm{Q}}_{\mathrm{BC}}=\left(\frac{5}{2}\mathrm{R}\right)\left({\mathrm{T}}_{\mathrm{C}}-{\mathrm{T}}_{\mathrm{B}}\right)=(6.25\mathrm{R})\left({\mathrm{T}}_2-{\mathrm{T}}_1\right)\text{ and }\\ {\mathrm{Q}}_{\mathrm{DA}}=-2.5\mathrm{R}\left({\mathrm{T}}_2-{\mathrm{T}}_1\right)\end{array}$

So, ${\mathrm{W}}_{\text{net }}={\mathrm{Q}}_{\text{net }}=3.75\mathrm{R}\left({\mathrm{T}}_2-{\mathrm{T}}_1\right) \{\because \mathrm{\Delta U}=0\}$

$\Rightarrow \mathrm{\eta }=\frac{{\mathrm{W}}_{\text{net }}}{\text{ heat absorbed }}\times 100=\frac{3.75}{6.25}\times 100=60\mathrm{\%}$