One mole of an ideal gas at 900   K , undergoes two reversible processes, I followed by II , as shown…

Process -I: Adiabatic reversible process.

(Since entropy is constant)

${\mathrm{W}}_{\mathrm{I}}=\mathrm{\Delta U}$

$=\left(450-2250\right)\mathrm{R}$

$=-1800 \mathrm{R}$

Process II: Isothermal reversible process.

(since internal energy is constant and entropy is increased)

Work done:

 $\begin{array}{c}{\mathrm{W}}_{\mathrm{II}}=-\mathrm{nRTln}\frac{{\mathrm{V}}_{\mathrm{f}}}{{\mathrm{V}}_{\mathrm{i}}}\end{array}$

$\begin{array}{c}{\mathrm{W}}_{\mathrm{II}}=-\mathrm{nRTln}\left(\frac{{\mathrm{V}}_3}{{\mathrm{V}}_2}\right)\end{array}$

${\mathrm{W}}_{\mathrm{II}}=-\frac{900}{5}\mathrm{Rln}\left(\frac{{\mathrm{V}}_3}{{\mathrm{V}}_2}\right)$

$\left|{\mathrm{W}}_{\mathrm{II}}\right|=\frac{900}{5}\mathrm{Rln}\frac{{\mathrm{V}}_3}{{\mathrm{V}}_2}$

$\mathrm{U}=\frac{5}{2}\mathrm{nRT}$

$450 \mathrm{R}=\frac{5}{2}\mathrm{nRT}$

$\mathrm{nRT}=\frac{900}{5}\mathrm{R}$

Given

$\left|{\mathrm{W}}_{\mathrm{I}}\right|=\left|{\mathrm{W}}_{\mathrm{II}}\right|$

$1800\mathrm{R}=\frac{900}{5}\mathrm{R} \ln \frac{{\mathrm{V}}_3}{{\mathrm{V}}_2}$

$\ln \frac{{\mathrm{V}}_3}{{\mathrm{V}}_2}=10$