I'm reading a paper that examines the evaporation rates of water. In the final formula, it has the following constant:

$c_s - c_\infty $ where $c_s$ is the concentration of the vapor at the sphere surface and $c_\infty$ is the concentration of the vapor at infinity.

I'm fairly confident in how to derive $c_\infty$:

1) Calculate water vapor pressure $p_s = 610.78 e^{\frac{17.2694 T}{T+238.3}}$

2) Actual vapor pressure is then: $ p=(Relative Humidity)p_s$

3) Using ideal gas law gives concentration at infinity: $c_\infty = \frac{(Molar Mass of Water)p}{RT}$

This generally looks like it gives me answers consistent with the paper's values. But, if I screwed up, please let me know.

My problem is with $c_s$. It's weird, because I recognize how easy it should be to get this, but just can't. I've asked some of the other grad students here and they basically all are befuddled. I think this is the physicist's version of a "tip of my tongue" experience. So, can anyone give me a link or method by which I can get this?


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    $\begingroup$ Wouldn't Cinfinity be given (as by a measurement), rather than calculated from first principles. Is there a correction involving surface tension, and the radius of the droplet? $\endgroup$ – Omega Centauri Jul 1 '11 at 22:08
  • $\begingroup$ The concentration at the surface of the bubble is 100% relative humidity, because the surface is in equilibrium with the air in immediate contact with it. The concentration at infinity is whatever the environmental relative humidity is. The profile is determined by diffusion, assuming still air. $\endgroup$ – Ron Maimon Sep 3 '11 at 20:27

So, it turns out that the answer is incredibly simple. Which is what I suspected given the mechanism: Question seems super easy, but can't seem to get answer of the top of my head. So I go searching for an answer, but because the solution is so obvious, no book has it explicitly. Ask people, again, no one seems to have it in mind, but can use complicated formulas based on it using already derived numbers. Ladeeda.

All the paper is looking for is the difference between actual vapor pressure at the drop as opposed to vapor pressure in the room. So all one needs to do is do steps 1 and 3 above and skip 2. Then multiply the result by (1-H) where H is the relative humidity. Good Times!

  • $\begingroup$ What you are saying here can be summarized more simply: the relative humidity is 100% at the surface of the bubble. That's what "skip step 2" means. This is correct, and it is exactly analogous to the "no slip boundary condition" in diffusion of momentum (viscosity), except here you are dealing with diffusion of water vapor. At the interface, there is local equilibrium, assuming the continuum approximation doesn't break down. $\endgroup$ – Ron Maimon Sep 3 '11 at 20:29

Very close to the liquid surface, you assume that the vapor and liquid are in diffusive equilibrium (equal chemical potentials).

If the droplet is very very large, you can start with the ordinary saturation vapor pressure and figure it out as in pballjew's answer.

If the droplet is small enough, you need to use the Kelvin equation which takes into account the way that the water's surface tension alters the diffusive equilibrium between vapor and solid.


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