Kelvin equation

  • Thread starter fogl
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Hi
I have problems understanding the Kelvin equation:

[tex]\ln {p_v \over p_0}= {2 \sigma V_m \over rR_mT}[/tex]
You can rewrite Kelvin equation in the folowing form as well:
[tex] p_v =p_0*exp({2 \sigma\over r}*{M \over \rho_lR_mT})[/tex]

It is obvious from the above equations that [tex] p_v [/tex] is always bigger than [tex] p_0 [/tex], since [tex] exp({2 \sigma\over r}*{M \over \rho_lR_mT}) [/tex] is always bigger than 0 (all the paramethers are > 0). Equilibrium vapor pressure [tex] p_v [/tex] should therefore (in the case of validity of Kelvinove equation) always be bigger than saturation vapor pressure over the flat interface. I would like to ask in what case the vapor pressure can be bigger than saturation pressure at the same temperature (is that possible at all?). If vapor pressure cannot be bigger than [tex] p_v [/tex], does it mean, that droplet can only evaporate, but it cannot condense and grow? Where have I done wrong in my understanding of the eqation?
 

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kuruman
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The radius of curvature ##r## can be positive or negative. See here for an explanation.
 

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