Column I contains a list of processes involving expansion of an...

(A) : In case of free expansion under adiabatic conditions, change in internal energy $\Delta U=0$. $\therefore$ Internal energy and temperature will remain constant. (B) $p \propto \frac{1}{v^2}$$\begin{array}{rrrr} & \therefore & p V^2 & =\text{ constant } \\ & \text{ or } & \left(\frac{n R T}{V}\right) V^2=\text{ constant } \\ & \therefore & T \propto \frac{1}{V} \end{array}$If volume is doubled, temperature will decrease as per Eq. (ii). Further, molar heat capacity in process $p V^x=$ constant is $C=C_V+\frac{R}{1-x}$ From Eq. (i), $x=2 \therefore \quad C=\frac{3}{2} R+\frac{R}{1-2}=+\frac{R}{2}$ Since molar heat capacity is positive, according to $Q=n C \Delta T, Q$ will be negative it $\Delta T$ is negative. Or gas loses heat if temperature is decreasing. (C) :\begin{aligned} p & \propto \frac{1}{V^{1 / 3}} \\ P V^{4 / 3} & =\text{ constant } \\ \left(\frac{n R T}{V}\right) V^{4 / 3} & =\text{ constant } \\ T & \propto \frac{1}{V^{1 / 3}} \end{aligned}$$Further, with increase in volume temperature will decrease. Here,$$\begin{array}{lc} \text{ Here, } & x=4 / 3 \\ \therefore & C=\frac{3}{2} R+\frac{R}{1-4 / 3}=-1.5 R \end{array}$$As molar heat capacity is negative,$Q$will be positive, if$\Delta T$is negative or gas gains heat with decrease in temperature. (D)$T \propto p V$In expansion from$V_1$or$2 V_1$, product of$p V$is increasing. Therefore, temperature will increase or$\Delta U=+$ve. Further, in expansion work done is also positive. Hence,$Q=W+\Delta U=+$ ve or gas gains heat.