The equilibrium between the liquid and the gas is determined by the condition,
\mu _\ell \Bigl(T,p(r)\Bigr) = \mu _g \Bigl(T,p_0 (r)\Bigr) thus μ_\ell (T, p_∞) = μ_g (T, p_∞) .
We assume that the effect is small and do a first-order series expansion of the terms on both sides of the first equation around μ_\ell (T, p_∞) and μ_g (T, p_∞) ,
μ_\ell (T, p_∞) + \frac{\partial \mu _\ell }{\partial p} \Bigl(p(r) – p_\infty \Bigr) = μ_g (T, p_∞) + \frac{\partial \mu _g }{\partial p_0} \Bigl(p_0(r) – p_\infty \Bigr) .
thus taking into account the second equation and the Laplace formula,
\frac{\partial \mu _\ell }{\partial p} \biggl(p_0 (r) + \frac{2\gamma }{r}- p_\infty \biggr) = \frac{\partial \mu _g }{\partial p_0} \Bigl(p_0(r) – p_\infty \Bigr) .
The Schwarz theorem applied to the Gibbs free energy G(T, p,N) yields,
\frac{\partial }{\partial p} \biggl(\frac{\partial G}{\partial N}\biggr)=\frac{\partial }{\partial N} \biggl(\frac{\partial G}{\partial p}\biggr)
which gives the following Maxwell relations for the liquid and the gas,
\frac{\partial \mu _\ell }{\partial p} = \frac{\partial V _\ell }{\partial N _\ell} = ν_\ell and \frac{\partial \mu _g }{\partial p_0} = \frac{\partial V _g }{\partial N _g} = ν_g = \frac{R T }{ p_\infty} .
The last partial derivative is evaluated at p_0 = p_∞. Therefore,
\nu _\ell \biggl(p_0(r) +\frac{2\gamma }{r} -p_\infty \biggr) = \frac{R T}{P_\infty} \Bigl(p_0 (r)- p_\infty \Bigr) .
which implies that the vapour pressure is given by,
p_0 (r) = p_\infty +\frac{2\gamma }{r}\Biggl(\frac{\frac{p_\infty \nu _\ell }{R T} }{1- \frac{P_\infty \nu _\ell }{R T} }\Biggr) .
In the limit where P_\infty \nu _\ell \ll R T , the vapour pressure becomes,
p_0 (r) = p_\infty +\frac{2\gamma }{r} \frac{P_\infty \nu _\ell }{R T} .