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.

This is also true for refrigerant fluids going through a restriction valve from the 'condenser' side to the 'evaporator' side. The choice of refrigerant and the choice of refrigerant flow rate resulting in the pressures and temperatures around the restriction valve, is deliberate so as to effect the required refrigeration. As the refrigerant fluid passes through the valve it finds itself at a (lower) pressure, it evaporates to conform to its phase diagram. At a micro level, it is expanding to fill a larger volume. It maintains the same total energy (adiabatic process) but much less energy per unit volume, causing it to cool. If the fluid has to go from liquid to gas, it will need to scavenge energy from itself to allow some of it to evaporate, thereby cooling the fluid.

  • $\begingroup$ Thanks for the reply. My book has the following example of a liquid (water) being heated and expanding to a gas gaining internal energy. "Determine the change in internal energy of 1.00 liter of water (mass 1.00 kg) at 100°C when it is fully boiled from liquid to gas, which results in 1671 liters of steam at 100°C. Assume the process is done at atmospheric pressure." However, if the refrigerant expands adiabatically Q=0, ΔU=-W, and the system loses energy. If the latent heat of vaporization has to come from within the system then energy is lost and temperature goes down (I think). $\endgroup$ – user177470 Mar 9 '18 at 18:21
  • $\begingroup$ @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. $\endgroup$ – Dlamini Mar 10 '18 at 0:07
  • $\begingroup$ Right because the example is not adiabatic the internal energy increases. However, I was trying to use that example to compare it to the adiabatic process in autorefrigeration. If Q=0 then ΔU=-Q-W because the Q for the heat of vaporization would have to come from the liquid. $\endgroup$ – user177470 Mar 10 '18 at 0:50
  • $\begingroup$ @Idan I see. So the 'scavenging' of energy from the fluid itself to achieve evaporation results in a temperature drop in the adiabatic case. This is true. I have added that aspect to the answer. $\endgroup$ – Dlamini Mar 10 '18 at 1:03
  • $\begingroup$ For an adiabatic process do we just say it's insulated and no Q enters the system or as you wrote that "it maintains the same total energy (adiabatic process)." My intuition is that didn't it didn't maintain the same total energy; it lost energy and cooled because it "scavenged" as you wrote. $\endgroup$ – user177470 Mar 10 '18 at 1:18

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