Well, it's finally time to close a loop and circle back and make a connection between Boltzmann's constant and beta. Beta factor we introduced early on in the course, and then took on faith was equal to 1 over kt. So before I actually deliver that proof, I'm going to just get your intellectual juices flowing, by letting you to answer a quick question, and then we'll dive in. Alright, well to start off this proof if you will let me begin with the probability form of the entropy expression. So s is equal to minus Boltzmann's constant, sum over j, probability of state j times the log of the probability of state j. And what I want to do is differentiate this expression. Just apply a differential operator. So, I get ds is minus k, and then, I'll be differentiating a product, so I need to use the chain rule. So first, I'll differentiate the probability. So, I get a dp, and it's left multiplying log p. And then, I will multiply. Excuse me. I'll differentiate log p. That'll give me 1 over p. So the p's will cancel. And I'm just left with sum over j, of the differentials of the probabilities. But, the sum over all probability is a constant. It's equal to 1. And so the sum of the individual differentials must be equal to 0. The derivative of constant is 0. And as a result, I'm left with ds is minus k sum over states log probability, d probability. I'll again substitute in for the probability the actual expression, but it's e to the minus Beta energy of that state, divided by the partition function. So, when I plug that in, I get a log of a quotient. So I'll separate that out into two logs. So the first term would be the log of the exponential. That goes away, I'm just left with the argument of the exponential minus beta ej. Second term is q itself, so minus log q all dp. Okay, so expand that out. I have kb times beta. The two negative terms canceled. Sum over j, energy dp plus k, I'll pull out this constant log, q. Oh, look. There's that sum over dp again. Still equal to 0. And so that term goes away. And I'm left with ds is equal to this expression. Kb beta, sum over j, ej, dpj. Let's keep that in mind. And we're running out of space on this slide. So, move on to the next one. So, just reproducing that. And now I want to recall for you, something from last week. It was in video 5.6, and that was that dU, was equal to this expression. And we then expanded this, noting that energy depended on volume, and we got an expression of probability times derivative of energy with respect to volume, dv. Plus here's this term again, same term here, as here. We associated this first term with work, and this second term with heat. Right? And in particular then, this was Delq reversible. So at this stage, if I make this substitution for this sum, as Delq reversible. I get that ds is equal to Boltzmann's constant times beta times Delq reversible. Mind you the definition of ds is, it's Delq reversible divided by t. And as a result, Kb times Boltzmann's constant must be this 1 over t. Ergo, beta has to be one over kT. Quod erat demonstrandum. So we finally accomplished a proof. And this was really how Boltzmann related, Boltzmann did this statistical analysis. Noted the similarity between his equations, and the classical thermodynamic equations. And was able then to establish the value of his constant, and it's relationship to classical thermodynamics. So a, a tour de force piece of work on, on Boltzmann's part. And thus we've established a connection between statistical and classical thermodynamics. So this was a short proof, that's all there is to this video. We're going to go on, and look at one last item in this weeks work. And this is something that is much more typically covered in a classical thermodynamics class. But there's a few take home messages, that are useful to chemists as well. I want to take a look at the Carnot Cycle, and we'll get to that next. [SOUND] [BLANK_AUDIO]