This introductory physical chemistry course examines the connections between molecular properties and the behavior of macroscopic chemical systems.

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Del curso dictado por University of Minnesota

Statistical Molecular Thermodynamics

162 calificaciones

This introductory physical chemistry course examines the connections between molecular properties and the behavior of macroscopic chemical systems.

De la lección

Module 6

This module introduces a new state function, entropy, that is in many respects more conceptually challenging than energy. The relationship of entropy to extent of disorder is established, and its governance by the Second Law of Thermodynamics is described. The role of entropy in dictating spontaneity in isolated systems is explored. The statistical underpinnings of entropy are established, including equations relating it to disorder, degeneracy, and probability. We derive the relationship between entropy and the partition function and establish the nature of the constant β in Boltzmann's famous equation for entropy. Finally, we consider the role of entropy in dictating the maximum efficiency that can be achieved by a heat engine based on consideration of the Carnot cycle. Homework problems will provide you the opportunity to demonstrate mastery in the application of the above concepts.

- Dr. Christopher J. CramerDistinguished McKnight and University Teaching Professor of Chemistry and Chemical Physics

Chemistry

It's time to delve into the statistics that underlay entropy and it's

properties. So I'll ask you to remember Boltzmann,

who we talked about early in the course. And you'll recall Boltzmann's famous

equation is S entropy is equal to Boltzmann's constant times the log of w.

Where w was some measure of disorder, possibility, probability.

And so I'll ask you to remember that we discussed ensembles and we had a vision

of near infinite water cooler, filled with bottles.

Where the bottles were systems, and the collection of bottles was called the

ensemble. And I worked with that ensemble in

particular, a particular flavor of ensemble, the NVT ensemble.

Which is called the canonical ensemble. That is where you are specifying a

number, a volume, and a temperature. I want to work, instead, with a somewhat

different ensemble now. Still a collection of bottles, that's the

ensemble. Still, each system is a bottle.

But I'm actually going to control N, V, and E.

Not the temperature, but rather, the internal energy, E.

I'm going to use E here instead of U. That is a micro canonical ensemble, as

oppose to a canonical ensemble. So the difference being that E is a state

variable, not T. Now, even though every system has the

same energy. That doesn't mean that their all in the

same state. Because there can be degeneracy.

So there can be many states that have the same energy.

We indicate that by this capital omega. So that is the degeneracy associated with

a given energy. And in molecular systems and macroscopic

volumes, that number is typically enormous.

Ten to very large powers. And so, I want to actually look at

multinomial statistics. And so you may have heard of binomial

statistics we're actually going to be a little more general than that but I'll

make it clear here hopefully what I'm talking about.

So let's take w, this number to be the number of ways of having a certain number

of systems in a given state call it state 1 and so I'll indicate that by a

subscript 1. The subscript tells you which state are

you in. In another number of systems in state 2,

so that's A sub 2, and so on, there can stage three, four, however many I care

about. And let's the systems be distinguishable

for a moment. So in that case the number of ways I can

do that. Well, first off, how many ways can I

order things? So if I am given a collection of things.

And the total number of things is capital A.

How many ways can I put them in order? Well, of course, I have capital a choices

for the first thing. And I have capital A minus 1 choices for

the second thing so that's a times a minus 1.

Then capital A minus 2 for the third thing and so on.

So the number of ways I can order them is a factorial.

But I'm not interested in only how might I order them But I'm going to put them

into subclasses, that's what the little a's are.

a sub1 systems in a given state, a sub2 systems in a second state.

And I don't actually care about how they're ordered in their sub states.

All I cared about was that there was a certain number there.

And so I end up dividing a factorial by the number of ways I can organize them in

their substates. Because I've overcounted those

possibilities. They're all the same possibility as far

as I'm concerned. So, the number in each individual state

factorial. That's how many ways I could reorder

those. Appears in the denominator.

And so, I'll just write that more simply as a product over all the number of

states I'm considering of the number of of systems in that state, factorial.

And just to show an example. Let's say that I have four total systems

and hence capital a is four and I asked the question well how many ways can I

arrange two of them in state one and two of them in state two and that uses up all

my possibilities that, that is four systems.

So none in state 3, and none in state 4. And I'll just explicitly draw them.

So here are 4 states. And I basically will just label them.

This one's in 1, 1, 2, 2. 1, 2, 2,1.

1, 2, 1, 2, and son on. And if you exhaust all the possibilities,

and convince yourself. You see I've got 6 written here.

And so let's just check that out for a moment.

4 factorial, that's 4 times 3 times 2 times 1 is 24.

And divided by 2 factorial is 2. So, 24 divided by 2 times 2, 24 over 4,

6. Great, this works, okay.

It had to work, that's what the statistics say.

And if you think of some of the other possibilities.

A, a trivial one, of course, is how many ways are there to have all four stakes

the same. Well that's four factorial over four

factorial. That is the minimum possible value for W

it's one. And then if you think about what's likely

to be the maximum possible value, it's where every state is unique, and so I can

label these things in a lot of different ways.

I no longer have anything in the denominator associated with ordering

within groups because there's only one in every group.

So that maximizes w at 4 factorial itself, 24.

And I've just listed a couple of other possibilities here.

And you can verify for yourself that, that's actually the case.

So, let's take a moment, and I'll let you do one example of your own, not one of

these. And, once you're comfortable with that

multinomial calculation, we'll move on. Great that's not very hard is it?

So lets come back to Boltzmann's equation, S equals K log omega.

Sorry k log W. And in a perfectly ordered arrangement W

is equal to one. And in that case, the entropy is equal to

0. And we did see on a prior slide, that in

order to maximize W, the most disordered arrangement, that also clearly will give

the maximum entropy. Now, as to why it's log of W, and not

just W itself for instance. Well, that comes from the addativity of

the entropy, but the product-like character of ways of arranging things.

That is, I want the total entropy to be equal to entropy for one set of systems.

Call them capital A. And another set of systems call them

capital B but the number of ways I can accomplish those arrangements is the

product of all the ways I can do it over here because for every number of ways I

can do it over here that's true for each one of the ways I can do it over here.

So I have to muliply all of these possibilities times these possibilities.

And so what is the function for which, k log this product is what, well, I guess I

should say what's the function. You need a log function to take the log

of this product and turn that into a sum of logs.

A log of products is a sum of logs. So it does have the correct behavior

then. The S total is equal to SA, this defines

SA Plus SB. Now, I want to look at another way to

express the entropy. And this is known as the degeneracy form.

So, let me take this expression, S equals k log w.

I've just written w out with all its factorials.

And now I've got a log of a quotient. So that's a log of differences.

So it's a log of the numerator minus log of the denominator.

But the denominator is itself a product. And the log of a product is a sum of

logarithms. So when I subtract, I won't subtract log

of the product, I'll subtract sum of the logs.

And what are those logs? The logs of each of the individual little

aj factorial terms. So we need to deal with the log of a

number factorial. And Stirling, a mathematician, worked

out, it's not too hard a proof, you can give it a shot yourself if you're

interested that the log of n factorial, as n grows to a very large number.

Is extremely well approximated by N log N, no factorial minus N.

So once you know what N is it's trivial to work out log of in factorial.

And so if use that approximation, what do I get here.

So when I transform A factorial log of A factorial to N log N minus N.

I'll get an a log a minus a. Similarly, each of these little [UNKNOWN]

factorial's, I'll get an a ln a, little a's, plus sum over all the j's, little

a's, but what is this sum of all the little a's Is capital A, that was the

definition of capital A, the sum of all the little As.

And so those two terms drop out and I'm left with this, Boltzmann's constant

times this expression. So now let's let the number of degenerate

states, j, that are available to the system, be represented by capital omega.

So that's what I'm summing over here, all the degenerate states.

And the population of each degenerate state is going to be equal to some number

N. And there all going to be the same number

because remember that's when W is maximized, when all of the states have

the same population. And in that case then, what is the total

number of systems? Well, it's n times the degeneracy.

So I'm going to, going to swap that in, everywhere I had an a before I'm going to

put n times the degeneracy and everywhere I had a Little a.

I've got an n. So here was a capital A, n times the

degeneracy. Here it was again.

Minus, now my sum, I'll, I'll emphasise that the upper limit is the degeneracy,

the capital omega, of n log n. And so I will expand that out a bit.

So this term is still here and this is the same term.

N's a constant. And I just add it to itself capital omega

times. So I've just multiplied times omega.

It's omega times n log n or n omega log n.

So now I have this. And if I look at that, and I expand it.

I have n omega log this is a product so I get n omega log n.

Plus n omega log omega minus n omega log n.

That third term cancelled the first term. I'm left only with n omega Boltzmann's

constant log omega. Well, that's, like, omega to the n omega

power. Since I have this multiplier out here.

That could be an exponent on the argument to the logarithm.

And what is n omega? It is the total number in the, in the

ensemble, capital A. So it's like log of omega for a system to

the 8th power, and what does that mean? The degeneracy of a system, times all the

systems in the ensemble. Well, the product of all the

degeneracies, is the degeneracy of the ensemble.

And so I've ended up transforming to entropy is k log degeneracy for the

ensemble. This is another way to write entropy.

Which I accidentally sort of misspoke a little earlier in the video.

So, in addition to s equal k log W, you can say s equals k log omega.

And, let's actually consider a specific example maybe to hammer this point home.

So, imagine that I'm going to isothermally expand an ideal gas, and I'm

interested in an entropy change. So, what is the degeneracy available to

that ideal gas? Well, it's going to be some function that

depends on how many molecules I have. It's going to be some function that

depends what is the energy of the ideal gas.

And this is isothermal, so it's a constant energy as I do the expansion.

But, it's, g will depend on E. And then, finally, as I make the volume

bigger, every single one of the molecules is going to have an opportunity to

explore more volume. And so, there's a volume to the nth power

term in the degeneracy. It's an ideal gas.

None of the molecules interact with each other.

They're all getting to explore more volume.

So, product of all of these is the degeneracy and that just emphasizes.

That's how many molecules I have. Okay, so, in that case, I want to work

out delta s. And let's use a molar quantity, so that n

is Avogadro's number. As I go from volume 1 to volume 2.

That's going to be then k log degeneracy for the final state, state 2, minus k log

the degeneracy for state 1. And so I'll just write out those

degeneracies, f of N g of E V2 to the Avogadro's number power.

Same thing in the denominator except V1. F of n will cancel because the number of

particles didn't change. G of e will cancel because the energy

didn't change. So I'm left with log v two over v one to

Avogadro's number power. It's a log that can take that power out

front, Boltzmann's constant over Avogadro's number are I get r log V2 over

V1. So this just emphasizes.

That simplification came because there's no change in number, and it's isothermal.

So, no change in energy. Notice that that entropy change.

R log, V2 over V1. That's exactly what we worked out from

considering, sort of, the classical thermodynamic considerations of energy,

heat. Work transfer for the expansion of an

ideal gas. I showed it to you here in the isothermal

path but remember, it doesn't matter what path we take.

So this is a statistical way to come up with the same result.

Okay well, now that we've examined sort of the fundamental statistics, I want to

come back to entropy in sort of real model systems.

Maybe that's a contradiction in term, a real model system.

But systems that chemists think about changes in and are interested in.

Computing entropy and entropy changes. So that'll come next.

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