Equipartition Theorem - Example 4
University of North Carolina at Chapel Hill
Equipartition Theorem - Example 3


Kinetic Theory Of Gases


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Video Transcript

welcome to our third example video. Looking at the equity partition, the're in this video. We'll consider our box full of two different gasses A and B except in this case, we're going to have a mon atomic gas for gas A and A die atomic gas for gas be. We'll still have a number of moles and a in a number of moles. And be now, given this difference between Amman Atomic and Die atomic gas, how does that affect what we did before? Where we start with some initial temperature difference and we would like to know what is the final temperature going to be? Well, it turns out that all that really is going to change is that we know a the an initial energy of gas A is going to be equal to three halfs and r t a nod, and the initial energy for B is going to have a different constant in front of it. It will be five halfs n b r t b zero, and then we can do exactly as we did before, Which is to say if I have and e a final equal to n a over n a plus n b multiplied by E. Total. And I know that in the final energy is going to be equal to for gas a at least three halfs n a r T final which will be the same for a and for B then I can set this equal to and a over n a plus. Nb well replied by E total, which is the sum of our initial energies, that is to say, three halfs and a r t a initial plus five halfs and be are TV initial. So notice when we cancel things in this case N A. Cancels our cancels. But then when we multiply the three halfs over, what we're going to get is that TF is equal to N A. I'm ta zero as before, But now we're going to have a five thirds and be times t b zero divided by n a plus and be so effectively, what we've done here is we've waited the B term more heavily. In this case, it means that it's gonna have a greater effect on the final temperature because it has thes extra ways of storing energy over the mon atomic gas. It's kind of an interesting result, but again, a remarkably simple solution. If you've noticed, if we had instead gone with E B, then when we tried to cancel everything out, then we would have had a slightly different equation. We would have had a five Hafs on the left, and when it tried to cancel here, we would have ended up with 3/5 multiplied by N a. T A plus a N b T B, and this would have ended up being the exact same number if you go through and calculated it all out with given quantities.