(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.