# Complicated liquid to vapor conversion problem

I had a question about the volume of vapor produced from an amount of liquid after increasing the pressure inside a container containing the liquid, then rapidly decreasing it back to the original pressure.

Say you have an amount of water, say 1mL (or another liquid, not sure what descriptors you would need to know, possibly boiling point?) in a closed container, let's say 2L. you pressurize the container (initial pressure 1atm/14.69psi) to a pressure, lets say 100 psi. Now if you immediately decrease the pressure back to 1atm (instantly in less than 1 second), causing the liquid to vaporize, how could you calculate the volume of the now vaporized water in the bottle? in other words, if all of the water in the bottle is not vaporized, how could you calculate the volume that was? If you included other factors such as temperature of the air, water, exact amount of time spent depressurizing the container, physical properties of the liquid, etc. I also know the easiest way to solve this would be to remove the vapor then re-measure the amount of liquid remaining in the bottle, but that shortcut isn't allowed in this problem.

Could someone create a blueprint equation of this problem and solve it using made up amounts? I was trying to figure it out using PV=nRT but I haven't taken chem in a while and couldn't go much further than that. I appreciate any help!

I doubt anyone will sit around and solve your problems for you but we will try and aid you in your search.

You can check out:

1. Phase Diagrams for a pure 1 component-2 phase system.
2. The Clausius-Clayperon Equation can be used.

I'm not sure what you are trying to accomplish by "quickly" lowering the pressure. For Thermodynamics to be applicable you need to system to reach equilibrium.

Borek
Mentor
Just depressurizing the container is not a reason for the water to evaporate. And the volume is limited by the container, you are always going to have 2 L of the water vapor.