CGPET 2021 Solved Paper

Given, number of mole, $n=1$ Initial temperature, $T_i = 400\,\text{K}$ Since, final temperature, $T_f = 2T_i$ ∴ $T_f = 800\,\text{K}$ Difference in temperature, $\Delta T = T_f - T_i = 800\,\text{K} - 400\,\text{K} = 400\,\text{K}$ Now, internal energy, $\Delta U = n C_v \cdot \Delta T$ ⇒ $\Delta U = 1 \cdot \left(\frac{3}{2} R\right) \cdot 400$ [For monoatomic gas, $C_v = \frac{3}{2} R$] ∴ $\Delta U = 600R$.......(i) According to the problem, process is given as $V^2 T = K$ ........(ii) From the equation of ideal gas for 1 mole, $pV = RT \Rightarrow T = \frac{pV}{R}$ ⇒ $V^2 \cdot \left(\frac{pV}{R}\right) = K \ \text{[From Eq.(ii)]}$ ⇒ $V^3 p = K$......(iii) Standard equation of adiatomic process is $pV^x = K$ On comparing from Eq. (iii), we get $x=3$ Since, the expression for molar heat capacity, $C = \frac{R}{\gamma - 1} + \frac{R}{1 - x}$ ⇒ $C = \frac{R}{\left(\frac{5}{2} - 1\right)} + \frac{R}{1 - 3}$ (∵ For monoatomic gas, $\gamma = \frac{5}{2}$) Solving above equation, we get $C = R$ and heat required, $\Delta Q = n \cdot C \cdot \Delta T = 1 \times R \times 400 = 400R$ .......(iv) Thus, work done, $\Delta W = \Delta Q - \Delta U$ From Eqs. (i) and (iv), we get $\Delta W = 400R - 600R = -200R$