# Chemistry - Change in water vapor with change in pressure and volume

## Solution 1:

If water is introduced in an empty container, the water will slowly evaporate. The pressure $$p$$ will slowly increase from $$0\ \mathrm{Pa}$$ to a maximum which is $$3167\ \mathrm{Pa}$$ at $$25\ \mathrm{^\circ C}$$. This corresponds to a molar concentration of water vapor equal to: $${c = p/RT = \frac{3167\ \mathrm{J\ m^{-3}}}{8.314\ \mathrm{J\ mol^{-1}\ K^{-1}}\times298\ \mathrm K} = 1.278\ \mathrm{mol\ m^{-3}}}$$ Expressed in gram per cubic meter, the concentration of water is: $$c = 23.0\ \mathrm{g\ m^{-3}}$$.

This means that between $$0$$ to $$23\ \mathrm{mg}$$ water can be evaporated in a $$1$$ liter container at $$25\ \mathrm{^\circ C}$$, and that a maximum amount of $$2.3\ \mathrm{mg}$$ water can be evaporated in a $$0.1$$ liter container. As a consequence, if the gas volume of your water vapor is compressed from $$1$$ liter to $$0.1$$ liter without changing the temperature, $$23\ \mathrm{mg} - 2.3\ \mathrm{mg} = 20.7\ \mathrm{mg}$$ water will be condensed as a liquid on the inner wall of the container.

## Solution 2:

There will be no water vapor in the container if you apply "room pressure" on it (~1 atm, applied with for instance a piston) at room temperature. If you want to have coexistence between liquid water and its vapor you can set T or p but the remaining intensive variables (including density) are then fixed.

The reason is the requirement of chemical, thermal and mechanical balance between phases, which leads to Gibbs phase rule:

$$f=c+2-p$$

where f is the number of degrees of freedom (intensive variables such as T or p which you get to set), c is the number of components (here c=1, there is just water), and p is the number of phases (here p=2, because we want coexistence of liquid and vapor).

Therefore f=1. This means you get to set one intensive variable. The other ones depend functionally on the first, describing a coexistence line.

For a single phase, the application of a specific pressure implies that the density is no longer under your control at a fixed temperature. For two phases , at a specific {T, p} set on a liquid-vapor coexistence line the total volume can be changed independently (up to a point) but the density of both liquid and vapor will be constant).

## Solution 3:

Assuming that $$P_0$$ corresponds to the equilibrium vapor pressure of water at $$T_0$$ and the temperature does not change from $$T_0$$ (i.e., equilibrates thermally with the room air), when you try to compress the saturated vapor, the pressure won't change significantly from $$P_0$$, but some of the vapor will condense to liquid water at $$T_0$$ and $$P_0$$.