Change in water vapor with change in pressure and volume

Given a closed (air tight) container (Volume $V\_0$) held at the room temperature ($T\_0$) and pressure ($P\_0$). The container contains 100% saturated water vapor with no air (at $T\_0$ and $P\_0$).


Now assuming the container is compressible (say a balloon whose volume can be changed by applying uniform external force/pressure how does the water vapor inside the container change when we apply external pressure on its surface ($P\_1$) and its volume changes ($V\_1$).


Assuming the Ideal Gas Law I assume none of the water vapor would condensate with any such change in external pressure and volume or i am missing something?


**Edit 1**: the container contains no air and no liquid water either. Just water vapor at 100% saturation (whatever its value is at initial room temperature and pressure i.e. $T\_0$ and $P\_0$).




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.


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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 $P\_0$.


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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).


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