(a) For diagram the process $(A \rightarrow B),(d V=0)$
So, $d W=\int p d V=\int p \times 0=0$
Hence, work done $d W=0$
By first law of thermodynamics
$$
\begin{aligned}
d Q &=(d U+d W)=(d U+0) \\
d Q &=d U \\
d Q &=n C_V d T=n C_V\left(T_B-T_A\right) \\
d Q &=\frac{3}{2} R\left(T_B-T_A\right) \quad(\because \mathrm{n}=1 \text { given }) \\
d U &=d Q=\frac{3}{2}\left(R T_B-R T_A\right) \\
&=\frac{3}{2}\left(p_B V_B-p_A V_A\right) \quad\left[\because C_V=\frac{3}{2} R\right]
\end{aligned}
$$
Heat exchanged (to system), (as $V_A=V_B$ )
$$
\begin{aligned}
d Q_1 &=d U=\frac{3}{2}\left(p_B V_B-p_A V_A\right) \\
&=\frac{3}{2}\left(p_B-p_A\right) \mathrm{V}_A
\end{aligned}
$$
(b) For diagram the process $(B \rightarrow C)$,
$$
\begin{aligned}
&(p=\text { constant }) \\
&d Q_2=d U+d W=C_V d T+p_B d V \\
&\quad=\frac{3}{2} R\left(T_C-T_B\right)+p_B\left(V_C-V_B\right)
\end{aligned}
$$
$$
\begin{aligned}
d Q_2 &=\frac{3}{2}\left(p_C V_C-p_B V_B\right)+p_B\left(V_C-V_B\right) \\
&=\frac{5}{2} p_B\left(V_C-V_B\right)
\end{aligned}
$$
$$
\begin{aligned}
d Q_2(\text { Heat exchanged })=& \frac{5}{2} p_B\left(V_C-V_A\right) \\
&\left(\because p_B=p_C \text { and } V_B=V_A\right)
\end{aligned}
$$
(c) For diagram the process $(C \rightarrow D)$.
Because $C D$ is adiabatic change, so $d Q_3$ (Heat exchanged $=0$
Hence no exchange of heat.
(d) For diagram the process $(D \rightarrow A), \Delta p=0$ Involves compression of gas volume from $V_D$ to $V_A$ at constant pressure $p_A$. Hence heat exchanged can be calculated by similar way as $B C_1$.
(Heat exchange) $d Q_4=\frac{5}{2} p_A\left(V_A-V_D\right)$