# 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 $$P_0$$).

• "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?" Can you explain this assumption? You also need to indicate whether T is maintained constant during compression or allowed to rise. Look at a phase diagram of H2O to figure out how much pressure is needed to liquify the vapor at a given temp. Jul 31 '20 at 11:49
• since the change in volume etc. happens in the open air and the effective external environment temperature won't change significantly we can assume the initial and final temperature of container to be effectively equal.. so yes the T is effectively constant. Jul 31 '20 at 12:05
• I am not a chem student so I might get things wrong (phase diagram suggestion).. can you help me assuming the temp as reasoned above doesn't change much.. Jul 31 '20 at 12:12
• I think the more interesting would be “ehat would happen if the chamber were insulated?” Jul 31 '20 at 15:46

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.

• This neglects the small volume occupied by the liquid water(20 cc). Jul 31 '20 at 15:52

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 $$P_0$$.

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).

• I should add a caveat: on the coexistence line at a specific T, p set on the line, the total volume can be changed independently (up to a point) but the density of both liquid and vapor will be constants. Jul 31 '20 at 14:51