Why Does Temperature Remain Constant in Adiabatic Free Expansion?

[Scintillation]

Homework Statement

We start with an ideal gas confined to half a container. Then, without heat flow, the gas flows into the other half as the partition is punctured. Find the new temperature after the expansion.

Homework Equations

Since there is no heat flow, I assume the process is adiabatic.

Relevant equations are:
(T1)(V1)^(y-1)=(T2)(V2)^Y-1

(P1)(V1)^y=(P2)(V2)^y

The Attempt at a Solution

Using my first equation, with T1= To, and V2=2V1, my final Temperature should be To/(2^y-1).

Except, my solution key states that there is NO change in temperature. Using the equation 1/2m(vrms)^2=3/2kt, since the velocity of each individual particle has not changed, then the temperature has not changed either.

But then what does my value of To/(2^y-1) mean? I always have trouble with these equations for adiabatic expansion/pressure. Another question I had (I can discuss this in more information if needed) is I was given the original volume, original and final pressure, as well as the original and final temperatures, and asked to find the final volume. I understand I could have used PV=nRT, but I tried using (P1)(V1)^y=(P2)(V2)^y to see if it would be the same. It was not the same, unfortunately, but I don't understand why. Can anyone help clarify either of these problems?

 

[TSny]

Relevant equations are:
(T1)(V1)^(y-1)=(T2)(V2)^Y-1

(P1)(V1)^y=(P2)(V2)^y

These equations are derived assuming that the adiabatic process is slow (quasi-static). For a slow adiabatic expansion, the gas does work as it expands and the gas cools. But if the gas expands freely into a vacuum, the gas doesn't do any work and the only effect is for the molecules to spread apart. For an ideal gas where there is no interaction between the molecules (other than collision forces) the spreading out does not change the average speed of the molecules.