Consider one mole of an ideal gas in a cylindrical container following the path shown in the $P-V$ diagram. The process follows $PV^{1/2} = \text{constant}$.
$(a)$ Find the work done in taking the gas from state $1$ to state $2$.
$(b)$ If $V_2 = 2V_1$, what is the ratio of temperatures $\frac{T_1}{T_2}$?
$(c)$ The internal energy of one mole of gas at temperature $T$ is $\frac{3}{2}RT$. Find the heat supplied to the gas in taking it from state $1$ to state $2$.

(N/A) The work done in a polytropic process $PV^n = K$ is given by $W = \frac{P_1V_1 - P_2V_2}{n-1}$.
Here, $n = 1/2$. So, $W = \frac{P_1V_1 - P_2V_2}{1/2 - 1} = \frac{P_1V_1 - P_2V_2}{-1/2} = 2(P_2V_2 - P_1V_1)$.
Since $P_1V_1^{1/2} = P_2V_2^{1/2}$, we have $P_2 = P_1(V_1/V_2)^{1/2}$.
Thus, $W = 2[P_1(V_1/V_2)^{1/2}V_2 - P_1V_1] = 2P_1V_1[(V_2/V_1)^{1/2} - 1]$.
$(b)$ Using the ideal gas law $PV = RT$ (for $1$ mole), $P = RT/V$.
Substituting into $PV^{1/2} = K$, we get $(RT/V)V^{1/2} = K$, so $TV^{-1/2} = \text{constant}$.
Therefore, $T_1V_1^{-1/2} = T_2V_2^{-1/2} \implies \frac{T_1}{T_2} = (\frac{V_1}{V_2})^{1/2}$.
Given $V_2 = 2V_1$, $\frac{T_1}{T_2} = (\frac{V_1}{2V_1})^{1/2} = \frac{1}{\sqrt{2}}$.
$(c)$ From the first law of thermodynamics, $Q = \Delta U + W$.
$\Delta U = \frac{3}{2}R(T_2 - T_1)$.
Since $T_2 = \sqrt{2}T_1$, $\Delta U = \frac{3}{2}R(\sqrt{2}T_1 - T_1) = \frac{3}{2}RT_1(\sqrt{2} - 1)$.
$W = 2(P_2V_2 - P_1V_1) = 2R(T_2 - T_1) = 2RT_1(\sqrt{2} - 1)$.
$Q = \frac{3}{2}RT_1(\sqrt{2} - 1) + 2RT_1(\sqrt{2} - 1) = \frac{7}{2}RT_1(\sqrt{2} - 1)$.