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

Aug 6, 2018

Probably not....

#### Explanation:

We consider the equilibrium phase change, say...

$\text{Liquid "rightleftharpoons" Vapour}$

And this is generally a function of temperature....water for instance at a given temperature exerts a equilibrium vapour pressure, at $298 \cdot K$ the so-called ${P}_{\text{saturated vapour pressure}} = 23.8 \cdot m m \cdot H g$. In higher level calculations, where a gas is collected over water, we subtract ${P}_{\text{SVP}}$ 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, ${P}_{\text{SVP}} = 1 \cdot a t m$ AT $100$ ""^@C. Why should this be so?