Change in water vapor with change in pressure and volume
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.