(a) $\rightarrow \mathrm{s}$ (b) $\rightarrow \mathrm{p}, \mathrm{r}$, (c) $\rightarrow \mathrm{r}$ (d) $\rightarrow \mathrm{q}, \mathrm{s}$
In process $\mathbf{J} \rightarrow \mathbf{K}: V$ is constant whereas $P$ is decreasing, Therefore, $T$ should also decrease.
$$
\therefore \quad W=0, \Delta U=-\text { ve and } Q < 0
$$
In process $\mathbf{K} \rightarrow \mathbf{L}: P$ is constant while $V$ is increasing. Therefore, temperature should also increase.
$$
\therefore \quad W>0, \Delta U>0 \text { and } Q>0
$$

In process $\mathbf{L} \rightarrow \mathbf{M}$ : This is inverse of process $J \rightarrow K$.
$$
\therefore \quad W=0, \Delta U>0 \text { and } Q>0
$$
In process $\mathbf{M} \rightarrow \mathbf{J}: V$ is decreasing. Therefore, $W < 0$
$$
\begin{array}{lcl}
& (P \mathrm{~V})_J & < (P V)_M \\
\therefore & T_J & < T_M \text { or } \quad \Delta U < 0 \\
\text { Therefore, } & Q & < 0 .
\end{array}
$$
No, solution is required.