# Consequences of $T_{sat}$ and $P_{sat}$ dependence

I understand that boiling point increases as pressure increases. This means that if you boil water at higher pressure it is going to need much more energy to break the bonds and thus boil at higher temperature. On a thermodynamics text (written by Cengel, Michael ), the text uses refrigerant-134a as an example.

"Consider a sealed can of liquid refrigerant -134a in a room $25^{\circ}$ . If the can has been in the room long enough, the temperature of the refrigerant in the can is also $25^{\circ}$. Now if the lid is opened slowy and some refrigerant is allowed to escape, the pressure in the can will start dropping until it reaches the atmospheric pressure.if you are holding the can, you will notice it's temperature dropping rapidly and even ice forming outside the can. A thermometer inserted in the can reads $-26^{\circ}C$when the pressure drops to 1 atm , which is the saturation temperature of refrigerant 134a at that pressure the temperature of liquid refrigerant will remain at $-26^{\circ}C$until last drop vaporize."

The part I don't get is the Italicized part. How can the temperature drop when the pressure drops? is there an answer which explains it at a molecular level? Does it work for all cases?(If I decrease the pressure of a can with liquid water in it, will it turn cold too?)

Are you familiar with the concept of "heat of vaporization." When you allow the pressure in the can to drop by allowing some gas to escape from the can, the liquid is still at the warmer temperature, and its equilibrium vapor pressure is now higher than the pressure of the gas in the head space. So some liquid will evaporate in order to try to re-establish equilibrium. This evaporation (and associated heat of vaporization) will cause the temperature of the can contents (both liquid and vapor) to drop. This action will continue until equilibrium is re-established at a lower temperature and equilibrium vapor pressure (equal to the outside pressure).

The transformation you are describing occurs very rapidly, we can thus consider it adiabatic. An adiabatic expansion of an ideal gas leads to a rapid decrease in its temperature. You can read it from Laplace's law :

$P^{1-\gamma} T^{\gamma} = K_0$, where $K_0$ is a constant.

As $\gamma > 1$, if $P$ decreases (which is the case here) then $T$ decreases as well.

Temperature comes from thermal agitation. At a microscopic level, you can picture it as gas molecules being less "agitated" on average as you give them more space to expand.

There are many reasons why water would probably behave differently. Firstly, it is a condensed phase, it thus doesn't follow the ideal gas law. Secondly (and more importantly), liquid water exhibits very strtong interactions (mainly hydrogen bonds), making its behavior very different than most other liquids.

• Is this always the case for ideal gases, or does this occur at saturation temperature and pressure only? Apr 6, 2016 at 14:22