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}$$\begin{array}{rrrr} & \therefore & p V^2 & =\text{ constant } \\ & \text{\text{ or } } & \left(\frac{n R T}{V}\right) V^2=\text{ constant } \\ & \therefore & T \propto \frac{1}{V} \end{array}$$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 Tis 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.