Derivation of the Poynting and Kelvin equation (Effect of external pressure on vapor pressure)

Question

Consider a condensible liquid. Using the Clausius-Clapeyron equation one can derive an expression for the vapor pressure $p_{sat}$. This is the pressure at which liquid condensat and vapor coexist. In equilibirum the pressure in vapor $p_v$ and liquid $p_l$ is both $p_{sat}$.

Consider now that an external force $p_{ext}$ acts on the liquid. This pressure can for example be caused by a non-condensible inert gas with pressure $p_{int}$ in the vapor phase or by the surface tension of the liquid. For a spherical droplet of radius $r$ the Laplace pressure is $p_{laplace}=\frac{2\gamma}{r}$.

The external pressure $p_{ext}$ has an effect on the pressure of the vapor $p_v$ in the coexistence phase. The effect of the inert gas is accounted for by the Poynting equation $$ p_v=p_{sat}\exp\left(\frac{v_l}{RT}(p_l-p_{sat})\right) $$ the effect of the surface tension by the Kelvin equation $$ p_v=p_{sat}\exp\left(\frac{2\gamma v_l}{rRT}\right) $$ How can these equations be derived and how are they related?

Answer 1

This answer is mostly based on @Chet Millers work, but gives a bit more detail.

With no external pressure applied in equilibrium the pressure in liquid and vapor are the same $p_l=p_v=p_{sat}$. The vapor pressure curve $p_{sat}(T)$ is implicitley given by the chemical equilibrium condition $$ \mu_l(p_l,T)=\mu_v(p_v,T)\rightarrow\mu_l(p_{sat},T)=\mu_v(p_{sat},T) $$ Assuming that the vapor behaves like an ideal gas we have \begin{equation} \mu_l(p_{sat}(T),T)=\mu_v(p_{sat}(T),T)=\mu_v^0(T)+RT\ln(p_{sat}(T)) \tag{1} \end{equation} Now consider the external pressure $p_{ext}$. The total pressure $p_l$ acting on the liquid is now given by the sum of external pressure $p_{ext}$ and new vapor pressure $p_v$. This new vapor pressure is once again determined by the phase coexistence equilibrium condition $$ \mu_l(p_l,T)=\mu_v(p_v,T)\rightarrow\mu_l(p_v+p_{ext},T)=\mu_v(p_v,T)\tag{2} $$ Now we use that $\frac{\partial\mu(p,T)}{\partial p}=v_l(p,T)$ and therefore $$ \mu_l(p_v+p_{ext},T)=\mu_l(p_{sat}(T),T)+\int_{p_{sat}(T)}^{p_v+p_{ext}}dpv_l(p,T) $$ Assuming that a liquid is mostly incompressible $v_l(p,T)\approx\textrm{const.}$ we recover $$ \mu_l(p_v+p_{ext},T)=\mu_l(p_{sat}(T),T)+v_l(p_v+p_{ext}-p_{sat}(T)) $$ Using eq. 1 $$ \mu_l(p_v+p_{ext},T)=\mu_v^0(T)+RT\ln(p_{sat}(T))+v_l(p_v+p_{ext}-p_{sat}(T)) $$ Plugging this in eq. 2 and using that the vapor behaves like an ideal gas the equilibrium condition becomes $$ \mu_v^0(T)+RT\ln(p_{sat}(T))+v_l(p_v+p_{ext}-p_{sat}(T))=\mu_v^0(T)+RT\ln(p_v) $$ Which is equivalent to $$ p_v=p_{sat}\exp\left(\frac{v_l}{RT}(p_v+p_{ext}-p_{sat})\right) $$ From here we want to discuss the two special cases mentioned in the question

1. Poynting equation: Trivially follows by setting $p_{ext}=p_{inert}$ and using that the total pressure on the liquid $p_l$ is the sum of the vapor pressure $p_v$ and inert gas pressure $p_{inert}$ $$ p_v=p_{sat}\exp\left(\frac{v_l}{RT}(p_l-p_{sat})\right) $$ In first order $p_l-p_{sat}\approx p_{inert}$ which makes it clear that a increase of inert gas pressure also increases the vapor pressure. Because of this inert gas is used in technical applications. It allows to conduct reactions at higher pressures, whilst still staying in vapor-liquid coexistence.

2. Kelvin equation: Assume we have a spherical droplet. Here the external pressure is equal to the laplace pressure $p_{ext}=p_{laplace}=\frac{2\gamma}{r}$. Furthermore note that generally the laplace pressure is much greater than the change in vapor pressure $\frac{2\gamma}{r}>>p_v-p_{sat}$ and therefore $$ p_v=p_{sat}\exp\left(\frac{2\gamma v_l}{rRT}\right) $$ For large, almost plane drops ($r\rightarrow \infty$) the effect of the surface tension on the vapor pressure vanishes as expected.

Answer 2

My approach to this would be different. At temperature T and total pressure P, the chemical potential of the liquid condensible would be $$\mu_L=\mu^0(T)+RT\ln(p_e(T))+v(P-p_e(T))$$where $p_e(T)$ is the equilibrium vapor pressure of the pure condensible species. The chemical potential of the condensible species in the vapor would be $$\mu_V=\mu^0(T)+RT\ln(p(T,P))$$ where p(T,P) is the partial pressure of the condensible species in the gas mixture. For equilibrium, the chemical potentials of the condensible species in the liquid and vapor must match. This leads to: $$p(T,P)=p_e(T)\exp{\left(\frac{v(P-p_e(T))}{RT}\right)}$$The rest is easy.