The degeneracy, which expresses how many levels have the same energy g, e to the

minus the energy of that level over kT. And let me in fact then pull out from

this expression e to the minus ground state energy, E0 over kT.

So, I'll take that out as a constant, it's just some number, it's multiplying

this sum and so, it's the degeneracy of a given state.

E to the minus the difference between that state, and the ground state energy.

Right? And so here I have e to the minus E0 over

kT. Here I have E to the minus and minus is

plus E0 over kT so I've sort of multiplied by 1.

And turned my 1 into something inside and outside the summation.

But the utility of that is, as the temperature goes to 0, the argument of

this exponential is divided by something going to 0.

So, it's becoming e to the minus a very, very large number since I'm dividing by

something going to zero. So it's going to 0 itself so for every

level other than n all those terms go to 0.

And I will be left with in that case, simply the limit as T goes to 0 of the

partition function is the degeneracy of the ground state times e to the minus E0

over k T. So, given that that's the low temperature

behavior of the partition function, let me then return to my general expression,

here. Apply that same sort of analysis to all

these sums over exponentials. And I'll end up with then, this sum is

replaced by g0, e to the minus E0. This sum is also replaced by that, of

course, it was being multiplied times the state energy, so the only state that

survives is the ground state, so that E0 sticks around.

And this is again, just Q re-expressed. And so now if I look at these in general,

I have a log of a product. So I will get k log degeneracy.

K log of an exponential. So that just becomes the argument of the

exponential. So I'll get E0, negative E0 over kT, by

multiplying by Boltzmann's constant. So that's just minus E0 over T.

And now if I go and look at what survived over here.

G0, g0, that cancels. E to the minus E0 over kT, in numerator

and denominator. That cancels.

The only thing left is E0 over T. So, these two terms are equal and

opposite to one another. Finally, all that is left is k log

degeneracy of the ground-state. Alright, and, therefore, as the

temperature goes to 0, the entropy determined from the partition function,

also goes to 0, to within the ground-state degeneracy.

So, let's really do this. Let's use all these tools to do a first

principles computation of entropy, a third law entropy.

All right, and we're going to do it for, well I'll say what we'll do it for in a

moment. First, let's just recapitulate the

relevant equations. Here is the entropy expressed in terms of

the partition function. And lets use an ideal gas partition

function which allows us to express the total partition function as a molecular

partition function to the nth power over n factorial.

And particular lets do a diatomic ideal gas and so you can go look this equation

up again if you want to in video 4.6. But it has a translational component to

the partition function, a rotational, a vibrational, and an electronic.

And what we need to do, and this is a wonderful math exercise, is take this q,

this little q, plug it in here to the nth power over N factorial, in order to get

capital Q. Take its partial derivative with respect

to T and then also just multiply the log of it by Boltzmann constant, and so I

have here a great math exercise. I'm not actually going to do it term by

term here. It's a lovely thing to sit down at a

table with and try to verify what I'm about to show you, but it's just

straightforward differentiation. Here's the final expression, just barely

fits on the slide. So, I'm going to use N is equal to

Avogadro's number, so that I have the molar entropy.

And I'm going to divide by R, because all these terms multiply R.

So, just for simplicity's sake, I'll put the R over here on the left hand side.

And so I get this term log of things that look like they might be associated with

the translational partition function. Something that involves the rotational

temperature, the vibrational temperature, another term in vibrational temperature.

The ground state degeneracy of the electronic state, and you also see

lurking in here some e's, so that really the number e.

And if you think about that, if I take the log of e to the five halves, that

just gives five halves and this is in units of R.

So, somewhere in here is a five halves R contribution, and you might remember that

entropy has say a half an R from translation in all directions.

And you get an R out of Sterling's approximation, so as I said, I'll let you

verify this expression as a good side exercise if you'd like.

but really what I'd like to do in terms of a little self assessment here is

looking at this expression, try to do a little bit of a sanity check of when

might you expect entropy to be large. Relative to of one molecule relative to

another, who would have more or less entropy so let me give you a moment to

think about that and then we'll come back.