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

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?

• I haven't been able to follow what you did, but, if I understand correctly, your original objective was to determine the change in the Clausius Clapeyron equation when there is a non-condensible gas also present in the gas phase. Do I understand correctly? Commented Mar 16, 2020 at 18:56
• Yes! You're right. I'm very interested in understanding this @ChetMiller. My starting point is that the pressure the liquid is experiencing is the sum of the vapor pressure and the non-condensible gas pressure, instead of just the vapor pressure as in the normal case
– user224659
Commented Mar 16, 2020 at 20:21
• My reasoning here is in the style of Lifshitz and Landau Statistical Physics. Sadly Landau never discusses the effect of a non-condensible gas, so I'm trying to expand his argument. But I'm open for other ways to takle the question.
– user224659
Commented Mar 16, 2020 at 20:23
• chemistry.stackexchange.com/questions/54400/… I read this answer of yours by the way, but it doesn't exactly help me
– user224659
Commented Mar 16, 2020 at 20:38
• Not clear what you are asking. If you want someone to check your calculation, that is off topic here. If you are attempting to answer your own question then you need to have a clear question in the Question Edit Box and the answer should be posted in an Answer Box, so that the Question and your Answer can be voted/commented on separately . The question also needs to be on topic so you will need to avoid presenting is as a "Homework" style of question. Commented Mar 16, 2020 at 22:24

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 $$$$\mu_l(p_{sat}(T),T)=\mu_v(p_{sat}(T),T)=\mu_v^0(T)+RT\ln(p_{sat}(T)) \tag{1}$$$$ Now consider the external pressure $$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.

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.

• Okay, feels like it is necessary to assume that the chemical potential for a liquid is of the form $\mu(T,P)=\mu^0(T)+RT\ln(P)$, which is only exact for the ideal gas. I was trying to get around using this approximation. But when accepting this, then your answer makes perfect sense. Thank you @ChetMiller.
– user224659
Commented Mar 17, 2020 at 10:56
• For future reference: What Chet Miller is doing to get to eq. 1 in his answer is use $\frac{\partial\mu(P,T)}{\partial P}=v(P,T)$ and assume incompressibility therefore recovering $\mu(P_2,T)=\mu(P_1,T)+\int_{P_1}^{P_2}dP\frac{\partial\mu(P,T)}{\partial P}=\mu(P_1,T)+\int_{P_1}^{P_2}dPv(P,T)\approx\mu(P_1,T)+v(P_2-P_1)$ and then assuming that the chemical potential of the liquid is of the form of an ideal gas.
– user224659
Commented Mar 17, 2020 at 11:02
• The whole idea of the Poynting correction is that the chemical potential of a pure liquid at a pressure above the equilibrium vapor pressure is equal to that of the saturated vapor at the same temperature (and equilibrium vapor pressure) plus the integral of $v_LdP$, where $v_L$ is the density of the liquid, from the equilibrium vapor pressure to the total pressure. Commented Mar 17, 2020 at 11:17
• just for clarification: In your notation $P=p+p_{inert}$ right? Where $p_{inert}$ is the pressure of the non condensible inert gas.
– user224659
Commented Mar 17, 2020 at 13:18
• Yes. Correct. Ok. Commented Mar 17, 2020 at 13:37