A three part question, all related
Part 1.
My question is about an ideal gas in a rigid container with two (equal volume) compartments separated by a rigid wall. One compartment has an ideal gas at some P, V, T. The other side is vacuum. If suddenly the wall is removed, what happens to temperature?
Would this be any different if it were a piston, where we pulled the piston to double the volume?
Part 2.
I found a similar question that had already been asked. From John Rennie's answer there it appears that the temperature should not change in both cases, but I am not sure.
What is most surprising to me is that everyone seems to be happily agreeing that this can be understood by $PV=nRT$ with $P$ and $N$ constant, and that somehow this equation implies that if you decrease $V$ then $T$ must increase!
Isn't that just the opposite of what the formula says? $V$ and $T$ are directly proportional, so if $V$ increases, $T$ must increase, at least according to the ideal gas law.
Part 3.
Finally, cooling and air-conditioning seems to work on the principle of expanding volume (going from a thin pipe to a wider one) causing a decrease in pressure, and hence temperature. But $V$ increased, and $P$ decreased, so why did $T$ change?
Best Answer
Temperature change upon expansion into a vacuum is the Joule Thompson Effect. In a real gas the temperature can increase or decrease depending upon the particular gas and conditions. For an ideal gas there is no change in temperature. The Joule-Thompson coefficient is zero. See this reference for derivation and further explanation.
No, the direct proportionality of V and T assumes pressure is constant, but pressure is not constant for expansion into a vacuum.
Three effects need to be considered, expansion against a non-zero pressure, the Joule Thompson effect, and evaporative cooling (latent heat of vaporization) any one of which can cause the temperature to decrease.