Consider the example system shown in Figure 17.8. Model the two fluids as ideal gases, and suppose the rubber has a constant surface tension $\sigma$ and heat capacity $C$.
(i) Show that changes in the total entropy and energy are governed by
$$
\begin{aligned}
\mathrm{d} S= & \frac{C \mathrm{~d} T}{T}+\frac{N_1 k_{\mathrm{B}} \mathrm{d} V_1}{V_1}+\frac{C_1 \mathrm{~d} U_1}{U_1} \\
& +\frac{N_2 k_{\mathrm{B}} \mathrm{d} V_2}{V_2}+\frac{C_2 \mathrm{~d} U_2}{U_2}, \\
\mathrm{~d} U= & \mathrm{d} U_1+\mathrm{d} U_2+C \mathrm{~d} T+\sigma \mathrm{d} A,
\end{aligned}
$$
where $C_1, C_2$ are the heat capacities of the two gases, $T$ is the temperature of the rubber, $A=4 \pi r^2, V_1=(4 / 3) \pi r^3$, and $V_2=V-V_1$. Hence show that the maximum entropy occurs when
$$
\left(p_1-p_2\right) \mathrm{d} V_1=\sigma \mathrm{d} A
$$
where $p_1, p_2$ are the pressures in the fluids.
(ii) Now suppose the chamber is in thermal contact with a reservoir at fixed temperature. Show that, in an isothermal change, the free energy is governed by
$$
\mathrm{d} F=\sigma \mathrm{d} A-p_1 \mathrm{~d} V_1-p_2 \mathrm{~d} V_2,
$$
and hence obtain the equilibrium condition again.