Equipartition Theorem - Example 2
Equipartition Theorem - Example 3
Equipartition Theorem - Example 4
University of North Carolina at Chapel Hill
Equipartition Theorem - Example 1

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Kinetic Theory Of Gases

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Video Transcript

welcome to our first example video. Looking at the equal, patrician, the're, um, in this video, we're going to consider it die atomic gas that is rotating and moving and all the ways we would expect. So it has two axes of rotation, and then it has the three translational directions. Now, given these degrees of freedom, if I want to find out how fast it's rotating around this access, that is, what is Omega Z? We will call it in this case. So Omega's he would be equal to what? Or perhaps more meaningful, E. We could ask. What is the frequency there? Well, remember that Omega is equal to two pi F. But we need a way to solve for Omega. We know that the average amount of energy along that the molecules storing in rotation around that axis is going to be one half our times T. And then we also can write the amount of energy there as rotational kinetic energy, which is one half I omega squared. So what would I be in this case? Well, if the distance between the two atoms is an amount D, and we'll assume that it's a and two or something like that, where we have two of the same atoms that are bonded together. So we have a distance de over to from this axis of rotation, and we have the mass. Which means I then is going to be equal to the sum of the moment of the nurses for each individual, uh, Adam, which is going to be one half de times m when have t squared times M plus one half D squared times M, which leads us to a total of one half m. D squared as our moment of inertia plugging that in. Then the total amount of rotational kinetic energy is going to be 1/4 times m d squared omega squared. If this is the total amount of rotational kinetic energy, it has to be equal to one half rt. Therefore, 1/4 M d squared omega squared is equal to one half r t. We can get rid of a factor of one half here, so we'll be left with one half m d squared. Omega squared is equal to Artie solving for Omega. Then we have omega is equal to two r t divided by m d squared again where D is the bond distance and we can take the square root of this. If we wanted frequency, then remember that frequency is going to be equal to omega over two pi. So a frequency will be won over two pi multiplied by the square root of to RT over m D squared so we can calculate the frequency of rotation for this molecule around that particular axis.