# Thermodynamic process in auto-refrigeration

I'm trying to understand why a high-pressure liquid drops in temperature when it's pressure is reduced in a refrigeration cycle. I can't use PV=nRT which only applies to gases, but a lower pressure would mean the liquid would start boiling if it was at a sufficient temperature. From just looking at a phase change diagram, the liquid could change to a gas with a drop in pressure and no temperature change. But according to Wikipedia:

"The saturated liquid refrigerant passes through the expansion valve and undergoes an abrupt decrease of pressure. That process results in the adiabatic flash evaporation and auto-refrigeration of a portion of the liquid (typically, less than half of the liquid flashes)."

If it's an adiabatic process and Q=0 then ΔU=-Q-W and internal energy decreases (Q=0 for the process but the latent heat of vaporization -Q would have to come from the liquid). I don't know how internal energy changes for liquids (which is supposed to be more complicated than gases) but for an ideal monoatomic gas a negative ΔU=(3/2)nRT would result in a decreased temperature. Does something similar happen for liquids and is this why you get a net decrease in temperature of both the liquid and vapor?

Thank you.

In addition to the $PV = nRT$ which you mentioned, substances have phase diagrams, or imperative states if you will. These are the states that substances must conform to given certain environmental conditions, because of the elements they are made of and the resulting bonding structures. They have been charted from experimental data.
For example, according to the phase diagram for water, water at atmospheric pressure and $110^\circ C \space(383K)$ is a gas. If we start with a few grams of ice at $-20^\circ C$, and at atmospheric pressure, and place it in an oven at $110^\circ C$, at the same pressure, it has to abide by the phase diagram and evaporate.
• @Idan I see. In the question you mentioned adiabatic flash evaporation. The water to steam example you mention in the comment is not adiabatic. To obtain steam at $100^\circ C$ from water at the same temperature and pressure (atmospheric), you would need to add energy. This is the latent heat of vapourisation of water (it is listed), which the water absorbs while transforming from a liquid to a gas. – Dlamini Mar 10 '18 at 0:07