Thermodynamics Ideal gas problem

[Ron Burgundypants]

Homework Statement

An isolated box contains two chambers separated by a thermally insulating but moveable partition. Both chambers contain dilute gas (same kind) at different densities and temperatures. The left chamber contains 1.0 x 10^22 particles at 25 degrees celsius and the right chamber 6.0 x 10 ^21 particles at 15 degrees celsius. What is the equilibrium temp when the wall is removed?

Homework Equations

PV=NkT
[Left Chamber] (mCv(Tf-Ti)) + [Right chamber](mCv(Tf-Ti))
Maybe the density equation?

The Attempt at a Solution

What I understand;
- The volume is constant
- Chamber left loses some energy this is the same as what chamber right gains
- The Heat capacity is the same as the gas is the same in both chambers so we can forget it.

So I've tried all sorts of things but the problem is I don't know get how the number of particles can be exchanged for a mass seeing as we don't know which type of gas it is, otherwise the problem would be easy right? Just solve for Tf. Should I just use a random gas and get the molar mass etc. or is there a better way?

 

[Chestermiller]

As you said, the gas is the same in both chambers, so you don't need to know the molar mass. You are aware that the gases mix after the partition is removed, correct? Do you know how the internal energy of an ideal gas mixture is related to the internal energy of the pure gases comprising the mixture?

 

[Ron Burgundypants]

The internal energy of an ideal gas is the degrees of freedom x NkT right? Not sure how to relate it to the total internal energy though no...

 

[WrongMan]

just write it out with letters and you will see it all cancels out

- Chamber left loses some energy this is the same as what chamber right gains

 

[Ron Burgundypants]

Ok I tried something and got an interesting answer. Might be right

Sum up the internal energies of both chambers i.e 3/2NRT(left) + 3/2 * NRT(right)

So this is the total internal energy - 3/2 NRT (total) - Solve for T now and I have the equilibrium temperature? Looks about right.

Wrongman I'll try your way too and see what happens if its easier. Thanks guys!

 

[Ron Burgundypants]

Aha! Yes brilliant thanks guys.