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Thought experiment - a liquid is in a closed container in equilibrium with its vapour, and then suddenly all the vapour is pumped away. Switch off the pump so that instantaneosuly there is no vapour present. How long does it take for the equilibrium vapour pressure to re-establish from the time the pumping is stopped? |
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One can make an estimation. Consider the first part, liquid and its vapour in equilibrium. Let the vapour concentration be $n_0$ and temperature $T$ such that vapour can be treated as an ideal gas (that is there isn't much vapour). Then the flow of molecules flying from vapour to liquid is $$F(n_0) = \frac{1}{4} n_0 \langle v\rangle = \frac{1}{4} n_0 \sqrt{\frac{8kT}{\pi m}}$$ There is an equal but opposite flux from liquid to vapour. One may assume that this flux will remain the same if the vapour was removed. Let $n(t)$ be the constantly increasing vapour concentration after the removing. Then it's behaviour is governed by the differential equation $$V \frac{dn}{dt} = \left(F(n_0)-F(n) \right) S $$ where $V$ is the volume available for the vapour and $S$ is the surface area of evaporation. This equation implies that the already evaporated vapour is in equilibrium, which I guess shouldn’t be very wrong in the case in hand. Once again, the argument is not applicable near the critical point since there vapour is almost like liquid. |
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This is a very tough problem because any evaporation flux is always computed when the liquid and the vapor are in a near equilibrium state. In your problem, one needs to use the Boltzmann's equation and solve the full kinetic problem. There has been many attemps to find a simple formula for the flux, the most commonly used can be found in this paper: |
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