# How to calculate concentration of vapor at the surface of a water drop

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?

Thanks.

• 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? – Omega Centauri Jul 1 '11 at 22:08
• 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. – Ron Maimon Sep 3 '11 at 20:27