One mole of an ideal gas at 900K , undergoes two reversible...

Process -I: Adiabatic reversible process.

(Since entropy is constant)

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

$= \left(450 - 2250\right)R$

$= - 1800R$

Process II: Isothermal reversible process.

(since internal energy is constant and entropy is increased)

Work done:

$$\begin{array}{ll}{W}_{II} = - nRTln\dfrac{{V}_{f}}{{V}_{i}}\end{array}$$

$$\begin{array}{ll}{W}_{II} = - nRTln\left(\dfrac{{V}_{3}}{{V}_{2}}\right)\end{array}$$

${W}_{II} = - \dfrac{900}{5}Rln\left(\dfrac{{V}_{3}}{{V}_{2}}\right)$

\left|{W}_{II}\right| = \dfrac{900}{5}Rln\dfrac{{V}_{3}}{{V}_{2}}$</p><p>$U = \dfrac{5}{2}nRT$</p><p>$450R = \dfrac{5}{2}nRT$</p><p>$nRT = \dfrac{900}{5}R$</p><p>Given</p><p>\left|{W}_{I}\right| = \left|{W}_{II}\right|$

$1800R = \dfrac{900}{5}Rln\dfrac{{V}_{3}}{{V}_{2}}$

$ln\dfrac{{V}_{3}}{{V}_{2}} = 10$