JEE Advanced 2021 — Paper 2 — Question 15

1 st process is adiabatic since entropy is constant.

$$W_{1} = \Delta U$$ $$\Delta U = 450R - 2250R = -1800R$$

$W_{1} = -1800R \quad \ldots (1)$In 2 nd process internal energy is constant, it means it is an isothermal process.$W_{2} = -2.303 nRT \log \frac{V_{3}}{V_{2}} \quad \ldots (2)$

$$= -nRT \ln \frac{V_{3}}{V_{2}} \quad \ldots (3)$$

Given, $n = 1$ mole, here temperature is unknown.

$$U = nC_{V}T \quad \text{for process II}$$ $$450R = 1 \times \frac{5}{2}RT$$ $$T = \frac{450 \times 2}{5R} = 180 \, \text{K}$$

Equation (1) = equation (2):

$$W_{1} = W_{2}$$ $$-1800R = -1 \times R \times 180 \ln \frac{V_{3}}{V_{2}}$$ $$\ln \frac{V_{3}}{V_{2}} = \frac{1800}{180} = 10$$