# Time to establish saturated vapour pressure above liquid

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?

• This will mostly depend on the circumstances of your experiment such as vapor pressure, empty volume above liquid, surface area, temperature and enthalpy of vaporization. Is the liquid connected to a thermal reservoir? – Alexander Aug 7 '12 at 11:54
• Are you asking about the case where there is air in the container (which becomes dry air when the vapour is magically removed), or the case where the liquid and the vapour are the only things in the container? It makes quite a big difference because in the latter case the low pressure will cause the liquid to start boiling until the vapour is restored, whereas in the former case the air next to the liquid becomes saturated very quickly, but you have to wait for diffusion to transport the vapour away from that layer, so saturating all the air takes a long time. – Nathaniel Aug 7 '12 at 17:30

## 2 Answers

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.

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:

http://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&ved=0CGUQFjAA&url=http%3A%2F%2Fwww.mie.utoronto.ca%2Flabs%2Ftkl%2Fpublications%2FWardFangExpress.pdf&ei=2jshUJjeGMe90QXpiYCQDQ&usg=AFQjCNE1eyAlPck_tx2MaKcCeqkobYHaJw&sig2=hw6URy7aqJAnBWdB_lNccA