So recalling exactly what is the rotational temperature, it's the quotient

of h bar squared. With 2 times the moment of inertia times

Boltzmann's constant. So the only variable in there that

depends on the nature of the molecule is the moment of inertia.

And remember the moment of inertia gets larger and larger as the masses of the

two atoms involved in the bond gets larger and larger.

So we would expect as the diatomic gets heavier, uses more heavy atoms Then the

rotational temperature should go down. And if we actually carry out the

calculation using actual atomic masses and bond distances, since the moment of

inertia depends on bond distance as well. We get these results and so what you see

is as expected, the lightest possible diatomic, that's molecular hydrogen using

prodium as the nucleus so no neutrons, just a mass of one.

Has a rotational temperature of 85.3 kelvin.

All the others are much, much smaller. Of course if you do have a proton, as you

see here for hydrogen chloride or hydrogen bromide, now we have sort of low

teens down to 12 for the rotational temperature, kelvin.

So those are very cold temperatures obviously.

And as we go to two heavier atoms like Carbon Monoxide, Nitrogen, Nitrous Oxide

and so on, single digit rotational temperatures ultimately by the heavier

dihalogon models Chlorine and Bromine, below one Kelvin.

So certainly if we're interested in a gas at room temperature which is 298 Kelvin,

we have satisfied the condition that rotational temperature is much, much

lower. Then the observed temperature that we're

working with. So that's easily satisfied then.

Really the only gas that occasionally might be of interest would be molecular

hydrogen working at quite cold temperatures.

Well, let's then replace our sum with an integral, and it turns out this is a very

friendly integral, so statisctical mechanics is occasionally very kind to

one. And so if we look at that interval it

might look a little bit imposing, but let's just make some quick substitutions.

let's call j times j plus 1 by a new variable name.

We'll call it x. That means that dx, I have to take the

derivative of the right hand side so if I think of that as j squared plus j that's

pretty easy to take a derivative of. I'll get 2 j plus 1.

All times d j. And so that's what's right here.

And then I'll also put in a variable a which is going to be my rotational

temperature divided by the temperature of the system.

If I now rewrite my integral using all those substitutions, I get integral 0 to

infinity dxe to the minus ax. And probably no one really needs to go to

an integral table to look that up. That's pretty trivial.

And so the solution is minus 1 over a, e to the minus ax evaluated at the limits

on this definite integral. So e to the minus a times infinity gives

0 and e to the minus a times 0 is e to the 0 so that's 1.

So I'll get minus 1 over a and just end up with 1 over a.

What was A? Well A was the rotational temperature

divided by the temperature. So the solution, the whole partition

function, is just the temperature divided by the rotational temperature.

It's a tremendously simple expression. Of course we can then re-expand what the

rotational temperature is. And if we do that, if we put back in the

constants and the moment of inertia, and if we want to express h bar squared in

terms of Planck's constant and the 2 pis, remember h bar is Planck's constant

divided by 2 pi, we get this expression for the rotational partition function.

8 times pi squared times the moment of inertia, so that's the molecularly

dependent quantity appearing in the partition function.

Times Boltzmann's constant times the temperature divided by [UNKNOWN] constant

squared. Well, with that partition function in

hand, we can now compute useful properties of the gas as an ensemble.