Equipartition Theorem - Example 3
Equipartition Theorem - Example 4
University of North Carolina at Chapel Hill
Equipartition Theorem - Example 2


Kinetic Theory Of Gases


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Video Transcript

welcome to our second example video and the equip partition there. Um, in this video, we're going to consider a box with the thin membrane in it that has a gas A and a gas be. And in this case, we will consider both gas is to be mon Atomic gas is we're gonna have t a greater than TB and have some number of moles and a and some number of moles and be given this and that it's Amon atomic gas. Remember that the initial temperatures are related to the thermal energies. E A zero is equal to three halfs n a are t a and we have e B zero is equal to three halfs n b r t b. Now what we would like to find out here is what is the final temperature going to be? So in order to find the final temperature, what we need to remember is that as thes to exchange energy that the total amount of energy e b not plus e a not is going to be constant. We also can remember the relationships that e a final will be equal to n a over n a plus and be and multiplied by E total and E Be final is going to be equal to end. Be divided by n A plus and be times E total. Now, having tracked all of this, you need Capital A's here, having tracked all of this, Then we can ask ourselves, What is the total change in temperature going to be or what is temperature final going to be? Well, we know what the final energies air going to be, and we know how those energies are related to temperature. Therefore, we can say that three halfs n a r t a final is going to be equal to n a divided by n a plus nb times e total Which e total is going to be the some of these two initial energies that we've written out here. So that will be three halfs and a r ta not plus three halfs n b r t b not looking at we what we can cancel here will have an n a that cancels and then to solve for t a final we have. We can also lose our our which shows up in every term. And we have a three halfs in every turn, which means Ta Final, then is simply going to be equal to n a times ta initial plus and be times TB initial all divided by N A plus and B, which is a remarkably simple result, given all the complexity that it took conceptually and mathematically to arrive here.