I have been looking up the basics on evaporation, boiling, humidity, and the dew point, in order to try to understand these everyday concepts with precise physics. I have noticed that "vapor pressure" is often used to describe these phenomena. E.g., evaporation [boiling] occurs when the vapor pressure of a liquid exceeds the vapor pressure [total pressure] of the environment.

But isn't chemical potential the "right" concept here, not pressure? In stat mech, pressure governs a mechanical trade-off of volume between two systems, whereas chemical potential governs a trade-off of particles. And with evaporation, what matters is whether particles tend to escape from a liquid or tend to deposit into it. But vapor pressure is a concrete quantity that is measured in units of pressure and used in calculations.

So why does it work? Is it a "hack" because water is incompressible, so its volume is directly proportional to its particle number? Is it a technically incorrect method of calculating things like dew points, that happens to work in practice? Or something else?

  • $\begingroup$ Vapor pressure, as you note, is a concrete measurable quantity. And it is part of the Gibbs free energy. They link is clear when you look at boiling, but remember G is a state function so the link is always there. $\endgroup$
    – Jon Custer
    Feb 22, 2021 at 14:10

1 Answer 1


You are correct that the chemical potential $\mu=\left(\frac{\partial G}{\partial N}\right)_{T,P}$, being the conjugate thermodynamic variable to matter, is the true arbiter of whether matter will shift phases. If the chemical potential of the gas phase is lower than that of the condensed phase, for example, then evaporation is spontaneous.

But "chemical potential" is notoriously vague, and the vapor pressure (meaning not just any pressure but specifically the equilibrium pressure of the gas phase above the condensed phase) is more easily measurable and intuited, as you note.

The "hack" I'm familiar with to easily associate the two is based on an ideal-gas assumption ($Pv=RT$), which is usually reasonable because the vapor is usually sparse. Since $d\mu=-s\,dT+v\,dP$ (with molar extensive parameters), or $d\mu=v\,dP=RT\,dP/P$ at constant temperature, we can integrate to obtain $$\mu-\mu_0=RT\ln\left(\frac{P}{P_0}\right),$$ which provides a useful relation for the gas phase.

Perhaps one get the same result or a similar one by assuming the bulk modulus $K\to\infty$ for the condensed phase, as you describe, but I'm not familiar with that method. That would give $\mu-\mu_0=v(P-P_0)$ for the condensed phase, as $v$ can be assumed constant. (These relations are used in analyzing the Poynting effect, for instance.)


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