Would you expect a dynamic equilibrium in a liquid in an open container?

1 Answer
Aug 6, 2018

Probably not....

Explanation:

We consider the equilibrium phase change, say...

"Liquid "rightleftharpoons" Vapour"

And this is generally a function of temperature....water for instance at a given temperature exerts a equilibrium vapour pressure, at 298*K the so-called P_"saturated vapour pressure"=23.8*mm*Hg. In higher level calculations, where a gas is collected over water, we subtract P_"SVP" from the measured pressure.

Now in an open container, the liquid water (or other liquid) will slowly evaporate, and the container will empty eventually. In a closed container, there is dynamic equilibrium between the vapour phase, and the liquid phase, and the rate of evaporation is equal to the rate of condensation.

Of course for water, P_"SVP"=1*atm AT 100 ""^@C. Why should this be so?