Consider a two$-$phase, vapor$-$liquid equilibrium. In this case, we will assume the liquid is

a spherical droplet with volume $V,$ surface area $A,$ and radius, $r.$

We have an isolated system $($there is no gravity$).$

The vapor phase can be described as in class, but since the droplet has surface tension,

there is an additional work term: $W^L=-P^{L}d{V}^L+d{A}^L$ for a reversible process, where $$

is the surface tension. Thus, $d{U}^L=T^{L}d{S}^L-P^{L}d{V}^L+{}^{L}d{n}^L+d{A}^L.$ This is the form of

the Fundamental Equation for systems where surface tension is important $($usually only for

very small droplets$).$ Note that $A$ in this problem is the surface area of the droplet, not the

Hemholtz free energy.

a$)$ Given that $V=\frac{4}{3}{r}^3$ and $A=4r^2$ for a sphere, find an equation $d{A}^L=f(r)d{V}^L.$

b$)$ Using the equation developed in part $($a$),$ find $T^L-T^{\text{v}\text{a}\text{p}},P^L-P^{\text{v}\text{a}\text{p}},$ and ${}^L-{}^{\text{v}\text{a}\text{p}}.$ What

is different about the pressure term as compared to phase equilibria problems where

surface tension is not included?

Hint: Consider the following relationships at equilibrium: $d{S}^{\text{v}\text{a}\text{p}}$ and $d{S}^L,d{V}^{\text{v}\text{a}\text{p}}$ and $d{V}^L,$

$d{n}^{\text{v}\text{a}\text{p}}$ and $d{n}^L.$