- #1
fogl
- 1
- 0
Hi
I have problems understanding the Kelvin equation:
\ln {p_v \over p_0}= {2 \sigma V_m \over rR_mT}
You can rewrite Kelvin equation in the folowing form as well:
p_v =p_0*exp({2 \sigma\over r}*{M \over \rho_lR_mT})
It is obvious from the above equations that p_v is always bigger than p_0, since exp({2 \sigma\over r}*{M \over \rho_lR_mT}) is always bigger than 0 (all the paramethers are > 0). Equilibrium vapor pressure p_v 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 p_v, 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?
I have problems understanding the Kelvin equation:
\ln {p_v \over p_0}= {2 \sigma V_m \over rR_mT}
You can rewrite Kelvin equation in the folowing form as well:
p_v =p_0*exp({2 \sigma\over r}*{M \over \rho_lR_mT})
It is obvious from the above equations that p_v is always bigger than p_0, since exp({2 \sigma\over r}*{M \over \rho_lR_mT}) is always bigger than 0 (all the paramethers are > 0). Equilibrium vapor pressure p_v 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 p_v, 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?