Consider a $P - V$ diagram in which the path followed by one mole of perfect gas in a cylindrical container is shown in the figure.
$(a)$ Find the work done when the gas is taken from state $1$ to state $2$.
$(b)$ What is the ratio of temperature $\frac{T_1}{T_2}$ if $V_2 = 2V_1$?
$(c)$ Given the internal energy for one mole of gas at temperature $T$ is $\frac{3}{2}RT$,find the heat supplied to the gas when it is taken from state $1$ to $2$,with $V_2 = 2V_1$.

(N/A) From the graph,the process follows $PV^{\frac{1}{2}} = K$ (constant).
$(a)$ The work done $W$ for the process from state $1$ to $2$ is given by:
$W = \int_{V_1}^{V_2} P \, dV = \int_{V_1}^{V_2} \frac{K}{\sqrt{V}} \, dV = K \left[ \frac{V^{\frac{1}{2}}}{\frac{1}{2}} \right]_{V_1}^{V_2} = 2K(\sqrt{V_2} - \sqrt{V_1})$.
Since $K = P_1 V_1^{\frac{1}{2}}$,we have $W = 2P_1 V_1^{\frac{1}{2}}(\sqrt{V_2} - \sqrt{V_1}) = 2P_1 V_1 (\sqrt{\frac{V_2}{V_1}} - 1)$.
Given $V_2 = 2V_1$,$W = 2P_1 V_1 (\sqrt{2} - 1)$.
$(b)$ From the ideal gas equation $PV = nRT$,and $P = \frac{K}{\sqrt{V}}$,we get $T = \frac{PV}{nR} = \frac{K\sqrt{V}}{nR}$.
Thus,$\frac{T_1}{T_2} = \sqrt{\frac{V_1}{V_2}} = \sqrt{\frac{V_1}{2V_1}} = \frac{1}{\sqrt{2}}$.
$(c)$ Heat supplied $Q = \Delta U + W$.
$\Delta U = nC_v \Delta T = 1 \cdot \frac{3}{2}R(T_2 - T_1) = \frac{3}{2}(P_2 V_2 - P_1 V_1)$.
Since $P_2 V_2^{\frac{1}{2}} = P_1 V_1^{\frac{1}{2}}$,$P_2 = P_1 \sqrt{\frac{V_1}{V_2}} = \frac{P_1}{\sqrt{2}}$.
$P_2 V_2 = \frac{P_1}{\sqrt{2}} (2V_1) = P_1 V_1 \sqrt{2}$.
$\Delta U = \frac{3}{2} P_1 V_1 (\sqrt{2} - 1)$.
$Q = \frac{3}{2} P_1 V_1 (\sqrt{2} - 1) + 2 P_1 V_1 (\sqrt{2} - 1) = \frac{7}{2} P_1 V_1 (\sqrt{2} - 1)$.