[maakii]

In the Joule Thomson effect, the change in temperature for an ideal gas is 0.

However, work is done by the ideal gas (p1v1 - p2v2), and since q =0 then the temperature of the gas must drop to supply the energy to do work. unless the work is 0 - which does not seem right - the temperature must change.

So why is the temperature change 0?

[enahs]

A couple of things.

The Joule Thomson effect does not apply to an ideal gas, but real gases.

Second, the temperature does change.
http://en.wikipedia.org/wiki/Joule-Thomson_effect

[maakii]

But for ideal gases, the joule thomson coefficient is 0.

"It should be noted that ?JT is always equal to zero for ideal gases (i.e., they will neither heat nor cool upon being expanded at constant enthalpy)."

So if you wanted to theoretically use an ideal gas in the experiment, its temperature wouldn't change.

[enahs]

I guess that is true, if you wanted to theoretically use an ideal gas...why I have no idea.

Why is it zero then? What is the definition of an ideal gas?

In an ideal gas we assume the molecules are infinitely small points in space that do not interact with each other or the containers of the walls. If you use this assumption, then how could the temperature of an ideal gas change due to them interacting more or less? The answer is, it can not; and thus no temperature change.

[Borek]

In an ideal gas we assume the molecules are infinitely small points in space that do not interact with each other or the containers of the walls.

Do not interact? You mean they are going through the walls and leave the container? 

[enahs]

In an ideal gas we assume the molecules are infinitely small points in space that do not interact with each other or the containers of the walls.

Do not interact? You mean they are going through the walls and leave the container? 

Ok, perfectly elastic collision  .
Same thing in the real world!

[maakii]

Hmm using the angle of intermolecular forces, it does make sense that there shouldn't be any temperature change..

but then where would the work done come from, if not from the gas? and if the gas does work its internal energy and thus temperature should decrease...

maybe the gas doesn't do any work? would that be a possible answer?