As given that $p V^{1 / 2}=$ Constant $=K$
$$
\begin{aligned}
&p_1 V_1^{(1 / 2)}=p_2 V_2^{(1 / 2)}=K \\
&p=\frac{K}{\sqrt{V}}
\end{aligned}
$$
(a) Work done for the process 1 to 2 , Work done
$$
\begin{aligned}
&=\int_{\eta_1}^{v_2} p d V=\int_{\eta_1}^{v_2} \frac{d V}{\sqrt{V}}=K\left[\frac{\sqrt{V}}{1 / 2}\right]_{\eta_1}^{v_2} \\
&=2 K\left(\sqrt{V_2}-\sqrt{V_1}\right)
\end{aligned}
$$
Work done from $V_1$ to $V_2$ i.e.
Work done $=2 p_1 V_1^{1 / 2}\left(\sqrt{V_2}-\sqrt{V_1}\right)$
(b) From ideal gas equation,
$$
\begin{gathered}
p V=n R T \Rightarrow T=\frac{p V}{n R}=\frac{p \sqrt{V} \sqrt{V}}{n R} \\
\text { or } T=\frac{K \sqrt{V}}{n R} \quad \text { (As, } p \sqrt{V}=K \text { ) }
\end{gathered}
$$
So, $T_1=\frac{K \sqrt{V_1}}{n R}$ and $T_2=\frac{K \sqrt{V_2}}{n R}$
$$
\begin{aligned}
&\frac{T_1}{T_2}=\frac{n R}{\frac{K \sqrt{V_2}}{n R}}=\sqrt{\frac{V_1}{V_2}}=\sqrt{\frac{V_1}{2 V_1}}=\frac{1}{\sqrt{2}} \quad\left(\because V_2=2 V_1\right) \\
&\frac{T_1}{T_2}=\frac{1}{\sqrt{2}} \quad \text {...(ii) }
\end{aligned}
$$
so required ratio is $1: \sqrt{2}$
(c) As given that, internal energy of the gas $(U)=\left(\frac{3}{2}\right) R T$
$$
\begin{aligned}
&\Delta U=U_2-U_1=\frac{3}{2} R\left(T_2-T_1\right) \\
&\because T_2=\sqrt{2} T_1 \text { from equation (ii) } \\
&=\frac{3}{2} R T_1(\sqrt{V}-1)
\end{aligned}
$$
Fromeq. (i),
$$
\begin{aligned}
d W &=2 p_1 V_1^{1 / 2}\left(\sqrt{V_2}-\sqrt{V_1}\right) \\
&=2 p_1 V_1^{1 / 2}\left(\sqrt{2} \times \sqrt{V_1}-\sqrt{V_1}\right) \quad\left(\because V_2=2 V_1 \text { given }\right) \\
d W &=2 p_1 V_1(\sqrt{2}-1)=2 R T_1(\sqrt{2}-1) n \because P_1 V_1=n R T \\
\Rightarrow d Q &=d U+d W \\
\Rightarrow d Q &=\frac{3}{2} R T_1(\sqrt{2}-1)+2 R T_1(\sqrt{2}-1) \\
&=(\sqrt{2}-1) R T_1(2+3 / 2) \\
d Q &=\left(\frac{7}{2}\right) R T_1(\sqrt{2}-1)
\end{aligned}
$$
That is the amount of heat supplied.