Equality of chemical potentials in the presence of surface effects. We proved this using entropy in Section 24.1. Obtain the same result from the Gibbs function, as follows. Consider a system consisting of a drop of liquid in the presence of its vapour. Ascribe to the surface of the drop internal energy $U_s$ and entropy $S_s$, but no volume or particles. Define $G \equiv U+$ $p_v V-T S$, where $p_v$ is the pressure of the vapour. Consider a change in which $p_v, T$ are constant but material moves from one phase to the other. By considering $\mathrm{d} U, \mathrm{~d} S$, and $\mathrm{d} V$ show that
$$
\begin{aligned}
\mathrm{d} G= & \left(u_l+p_v v_l-T s_l\right) \mathrm{d} N_l \\
& +\left(u_v+p_v v_v-T s_v\right) \mathrm{d} N_v+\sigma \mathrm{d} A,
\end{aligned}
$$
if the liquid is assumed to be incompressible. Relate $v_l \mathrm{~d} N_l$ to $\mathrm{d} A$ and hence show that $\mathrm{d} G=\mathrm{d} N_l g_l+\mathrm{d} N_v g_v$, where $g_l$ is evaluated at the pressure of the liquid and $g_v$ at the pressure of the vapour. Hence show that $g_v=g_l$ in phase equilibrium.