We've already worked extensively with entropy and the partition function, but now I want to continue our exploration by looking at what happens near 0 Kelvin. And so I'll recall for you that the entropy can be expressed as Boltzmann's constant times the log of the partition function. Plus kT times the partial derivative of the log of the partition function, with respect to temperature, holding number of particles and volume constant. And now, let's actually expand Q, and remember what the partition function is. It's the sum over all states, exponential minus the energy of that state, divided by Boltzmann's constant times temperature. So, I've swapped that in for Q here in the first term, and I've swapped it in for the partial log of Q. I've actually carried out the differentiation, and so I get a 1 over T term, and I get this expression and I take the derivative. We've taken this derivative multiple times because T appears here, I'll pull down the E sub j over k. That'll cancel k over k, I'm left with this E sub j term, and I end up also with the partition function again. So, I'll let you revisit that derivative if you'd like to, but it's one we've taken before in the past. What I want to do here, is explore the question of, how does this expression behave as the temperature goes to 0? Which is to say, is the statistical thermodynamic definition of entropy consistent with the third law of thermodynamics? Well, so let's look at the partition function as temperature goes to 0, the low temperature partition function. And, to make things a little more convenient, let me work with levels, instead of states. Just to remember what that means is, I'm now going to switch my index, over which my sum is running. I'm going to run over energy levels, so all levels that have the same energy. 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. All right, hopefully the sanity check all made sense and now let's take a closer look at this expression for the molar entropy of a diatomic ideal gas. So, this first term contains all of the contribution from translation. And it also, by convention, if you had actually worked out the full computation of entropy. You'd have had a Stirling's approximation term that came from the log of n factorial, and by convention, that gets lumped into the translation piece as well. There's a term deriving from the rotations of the diatomic molecule, two terms that derive from the vibration. There's only one vibration in a diatomic, and finally there's an electronic term. And so what do we need in order to actually compute the entropy? Well, this is nitrogen, nitrogen gas. And so, appearing in this first expression are the masses of the two atoms in the diatomic. And if we were doing 2 nitrogen 14 isotopes, they'd, m1 would be equal to m2. So it's about 28 atomic mass units. A tiny bit different, because the atomic mass unit is defined for carbon 12, but close enough. You can go look up the exact number in kilograms if you want to. The symmetry number appears in the rotational term and its two for a homo-nuclear diatomic. And we need the rotational temperature and if you go look that up after we had it on previous slides as well, its 2.88 Kelvin. In the next two terms the only thing we need to know is the vibrational temperature. And again that's something we've had on prior videos as well as you can find it in tables, 3374 Kelvin. And lastly, you need to know the ground electronics state degeneracy and its 1 for nitrogen, its not a degenerate state, it's just non-degenerate. So, if I take all these values and plug them in, and everything else up here is a constant, Boltzmann's constant, pi, Planck's constant, Avogadro's number. By convention we choose a molar volume at which we're tabulating things, and that's the volume of an ideal gas, say, at room temperature. So, plug all those numbers in, and, you will find that I'll, I'll, just give the numbers by piece, 150.4 Joules per Kelvin per mole, deriving from this first term. Translation, 41.13 from rotation, 1.15 times 10 to the minus third. So, that's really quite small compared to these other two terms. From vibration, none at all from the electronic degeneracy. Because really, there's no disorder associated with the with the electronic state it's in exactly one state, the ground state. And so there's no entropy, because there's no disorder. Similarly, this vibrational temperature is so high, 3,374 Kelvin. They are effectively, all the vibrations are in the ground state. So, again not much disorder there and it only contributes a thousandth of a joule per Kelvin per mole. But add them all together and you get that the standard entropy is 191.5 joules per Kelvin per mole and I should but a bar over this, this really is the molar entropy. And I'll just remind you that if you go back to video 7.3, where I showed an example of using. or at least, tabulated experimental data that would have come from heat capacity measurements. The value that was determined there was 191.6 joules, per kelvin, per mole. So, within 0.1 joule per kelvin per mole, that is quantitative agreement. So, I think this is an amazing result to some extent. It is an incredibly powerful example of just how much insight we gain into the behavior of a macroscopic gas. Because we understand its molecular properties. By plugging those molecular properties into an expression that depended on them, we were able to compute the entropy effectively exactly. And the important, maybe even beyond that, we know where that entropy comes from. We know that more than almost three quarters I'd say comes from the translation. And most of the rest comes from the rotation, and that I think is a fascinating molecular level of insight that comes from statistical molecular thermodynamics. Alright, well that was a single example, I'd like to now expand and do some additional considerations of third-law entropies in general. So we'll get to that next. [SOUND]