
One mole of an ideal gas is taken from $a$ to $b$ along two paths denoted by the solid and the dashed lines…

Solution

So, work $W_s=-4 \times 0.5 \ln \left(\frac{5.5}{0.5}\right)=-2 \ln 11 \mathrm{~L}-\mathrm{atm}$ Dotted line represents irreversible process So, work $ W_d=-\left\{4 \times 1.5+1 \times 1+\frac{2}{3} \times 2.5\right\}=-\left\{6+1+\frac{5}{3}\right\} \mathrm{L}-\mathrm{atm}=-\frac{26}{3} \mathrm{~L}-\mathrm{atm} $ So $\quad \frac{W_d}{W_s}=\frac{26}{3 \times 2 \ln 11} \approx 2$ Energetics Conceptual III
Asked in: JEE Advanced 2010 (Paper 2)