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.