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

165 calificaciones

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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

I think we're ready to now consider spontaneity and the Second Law of

Thermodynamics. And to do that I want to construct a

special system, one in which that we're going to look at heat flow.

So energy as heat flows spontaneously from a region of higher temperature to a

region of lower temperature. And sometimes this statement is known as

the Zeroth Law of Thermodynamics. And I think I mentioned earlier in the

course that this is not a postulate, you can actually prove this in a

statistical-mechanical way. But it's a bit beyond what we'd like to

do. For now, let's just accept it.

It certainly squares with our experience. The Zeroth Law of Thermodynamics, heat

will flow from high temperature to low temperature.

So now, imagine you can construct the following system.

It is an isolated system, which is to say, it's perfectly insulated from its

surroundings and it consists of two boxes.

Those two boxes are rigid, so there is a hard wall surrounding each box.

But the wall that joins them is able to conduct heat.

So, heat can flow from one side to the other, although the volume of the

individual boxes themselves does not change.

So, the fact that it's isolated, that adjective means no exchange of energy

with the surroundings, whether heat, work, whatever, it's isolated.

The rigid aspect means that there cannot be any work done.

Because dV is equal to zero for both of these two compartments.

So The First Law tells us that the sum of the internal energies in the two

compartments UA plus UB must be a constant.

Right? My system is isolated and I don't have

any work to do, so UA plus UB are constant.

VA and VB are also constant. That's just by definition of being rigid.

So let's imagine that I'm actually working with ideal gasses.

That'll make life a little bit more simple.

It's not a requirement that you work with ideal gasses.

The nice thing is that any other system you might imagine can always be imagined

being brought into contact with these ideal gasses in a very similar sort of

arrangement. And, hence, it would have to follow the

same behavior because of The First Law that energy is conserved as the ideal

gases. That's what makes the ideal gases

wonderful to work with. So, although, they are in equilibrium

when we bring them together, they're not in equilibrium with each other.

They're at different temperatures perhaps.

But in any case the entropy of the total system is defined as the entropy of the

left side, I'll call it a, and the entropy of the left side b.

So what I want to do is I want to determine the energy and the entropy.

So, given that I've got no changes in volume, that means that dU is just

going to be equal to del q. It's also equal to, since this is

reversible heat, it's equal to TdS, where T is the temperature on an individual

side, either A or B. And this depends on there being no work,

this, this dV equals zero situation. So the total entropy at change dS is

going to be the sum of the two individual entropy changes.

dS in box A plus dS in box B. Right?

And so I'll just express that then. I'll move the temperature over on the

other side here. dU of A divided by T at A plus dUB over

tB. Now, in so far, as UA plus UB must be a

constant, d of UA plus UB must be zero because the derivative of a constant is

zero. Which means dUA must be the negative of

dUB, that is any energy change in A must have come from B.

That's, doesn't seem like rocket science. So given that, I can just replace this

dUA by minus dUB over TA, so here's the minus over TA and I'll just bring the dUB

out front. So there's a constant dUB here, so

constant, that's dUB, times 1 over TB minus 1 over TA.

Well, let's just consider the possible situations.

There's, there's three possibilities. Either the temperature in box B started

higher than that in A or it started lower than that in A.

Or, I did bring them together when their two temperatures were exactly equal.

Well, what's going to happen under those circumstances?

If the temperature in box B is greater than A, heat will flow out of B into A

and that will make the internal energy B go down.

It's less than zero. So if this is a negative number and B is

larger than A, then, I've got 1 over a big number, which is a small number,

minus 1 over not as big a number. So a, a bigger thing that's being

subtracted, this also is negative. Negative times negative.

I get the change in entropy to be positive.

By contrast, if A is greater than B in temperature, then the flow of heat will

be to increase the internal energy in U. And so again if I think about that, this

is now a positive quantity. Since A is bigger than B, this inverse is

smaller than this inverse. So this ends up being positive.

So once again, dS is greater than zero. Doesn't matter which direction the heat

flows, the entropy change is greater than zero.

Of course, if TA equals TB, there's no net energy flow, so I get zero DU.

I'll also get this term is equal to zero, so zero times zero is certainly zero.

There's no change in entropy at all. Okay, so, the spontaneity and entropy

then, the feature I'd like you to notice about this system is that the spontaneous

flow of energy as heat from a body at a higher temperature to a body at a lower

temperature is governed by the condition, dS is greater than zero.

The heat stops flowing when they're at the same temperature.

And dS becomes zero thereafter. So, in an isolated system, the energy

remains constant. So any spontaneous process must be due to

an increase in the entropy. Unlike energy, entropy is not conserved.

So that's a key feature of entropy. It increases whenever a spontaneous

process takes place. And the entropy of an isolated system

will continue to increase until the entire system is in equilibrium.

And at that point the entropy remains constant.

At that point it will have its maximal entropy.

Continues increasing until it can't increase even more because it's at

equilibrium. So we say then that dS is greater than 0

for a spontaneous process in an isolated system.

And the sorts of things that can happen at equilibrium are reversible processes.

And for those, dS is equal to 0. So a way to think about this is that in

an isolated system, if you just start it out of equilibrium.

That was when I brought my two gases into contact with each other, at different

temperatures for instance, that spontaneous processes will occur.

So this is a plot of entropy against time.

So over time spontaneity, spontaneity, things are happening, things are

happening, entropy is increasing, increasing, increasing.

And finally, I hit a maximum value of S, the point at which the system is at

equilibrium. And at that stage the entropy does not

change any longer, dS becomes zero. So what about a general situation, a

non-isolated system? So isolated systems are convenient to

think about, but we're often interested in systems that can exchange heat, work,

what have you, with the surroundings. Well, in that case the change in entropy

comes from two sources. So there's the entropy produced just by

an irreversible process. So all these spontaneous irreversible

processes that are taking place, that's creating entropy.

And then, there's entropy because of heat exchanged.

So remember the definition of entropy would be the reversible heat change,

divided by the temperature, that defines dS.

So that's an exchange between the system and the surroundings that becomes

possible once it's no longer an isolated system.

And so we could write that as dS equals dS produced.

So that is always non-negative and associated with spontaneous processes

plus dS exchanged. And that can be positive, or negative, or

zero. It depends on the sign of the heat

transfer which way that might go. So, could also write dS produced over

delta q over T. So, for the reversible process, D S, I've

got a system at equilibrium. So it won't create any produced entropy,

spontaneous entropy. It can, exchange heat with the

surroundings through a reversible process.

Del q reversible over T that's, defines dS under reversible conditions.

Under irreversible conditions, I've got some entropy that comes from the

spontaneous process, plus an irreversible heat exchange with surroundings, delta q

irreversible over T. Because this quantity is always greater

than zero, that means that the net dS is greater than the irreversible heat

exchange with the surroundings divided by T.

And so here, then, is a statement of The Second Law.

It says there is a thermodynamic function of a system, called the entropy S, such

that for any change in the thermodynamic state of the system dS is greater than or

equal to del q over T. Or if I express it not as the

infinitesimal change but as a total change, delta S is greater than or equal

to the integral of del q over T. The equality sign applies if the change

is carried out reversibly. An inequality sign applies if the change

is carried out irreversibly at any stage, so it's a bit like the, the spoonful of

wine and the spoonful of sewage. If you have a barrel of s- sewage and you

put in a spoonful of wine think of sewage as an irreversible process.

You put in a spoonful of wine its still sewage.

If you have a barrel of wine, think of that as a reversible process, and you put

in a spoonful of sewage, now it's sewage. So it doesn't take much irreversibility

to poison the process. It introduces the greater than sign, the

inequality sign. Let's look at that a little more, the

entropy of irreversible process. So, here, I have that, statement of The

Second Law delta S greater than or equal to del Q over T.

Let me consider the following. I've got two states of a system.

I am going to initially isolate my system from the surroundings.

I'm going to perform some sort of irreversible step to get me to State 2.

Then I'm going to put my system back in contact with the surroundings and I will

reversibly transform from State 2 to State 1.

Entropy is a state function and so delta S for the entire process must be zero.

So the, the path integral that is a circular path of dS is zero because of

the state function properties. And so, what is delta S?

Well, it's the integral from State 1 to State 2 of the irreversible heat change

over t. Plus the integral from State 2 to State 1

of the reversible heat exchanged over T. And that is because of the

irreversibility, so there is this greater than sign, I've got an irreversible step.

Mind you what was the irreversible heat exchanged in that irreversible step, it's

zero. I had the system isolated.

What is it in the reversible step? Well, this defines entropy, so this is

just dS, the integral from 2 to 1 of dS is S1 minus S2.

And as a result, I get zero greater than zero plus S1 minus S2.

And that implies, by just adding S2 to both sides, S2 must be greater than S1.

[SOUND]. So, that irreversible step, which

corresponds to looking at S2 compared to S1, increased the entropy.

So we talked about how spontaneous processes increase the entropy.

This is another proof, if you will, that a spontaneous and irreversible process

increases the entropy. All right.

Well, let's take a moment. I want you to have a chance to think a

little more about this and maybe answer a question, and then we'll come back.

Okay. Well I want to close by at least

introducing one small element of historical context and introduce to you

Rudolph Clausius. And so Clausius is shown here and he was

the one who first really expressed The Second Law in detail, in 1865, in a paper

he published on entropy. And the statement that could be used

would, he actually formulated the first two laws of thermodynamics in a common

statement, which is the energy of the universe is constant.

You can think of that as the first law of thermodynamics.

The entropy is tending to a maximum. So that's for the whole universe, and so

in a way you could think of the whole universe is an isolated system.

If there were a part of the system that we hadn't considered, we wouldn't have

the whole universe. We should just add that back in.

And we know that for an isolated system, entropy increases until it hits a

maximum. So some people refer to that as the heat

death of the universe, which hopefully is a long way in the future.

But, The Second Law guarantees it's out there somewhere.

And then, the mathematical expressions for this are delta U is equal to q plus w

and delta S is greater than or equal to the integral of del q over T.

So an enormous amount of power in these two relatively simple equations.

And also, I might add an enormous amount of gravitas.

I, I wish I could sit here looking as, as stern and knowledgeable as Clausius does.

But I'll just settle for what I've got. So after this we will continue.

I want to look next, we've, we really discussed entropy in the way Clausius

discussed entropy to some extent. It is a phenomenal, logical,

observational [SOUND] thing that was deduced from the properties of gases and

studying heat transfers in gases. Classical thermodynamics.

But I want to [UNKNOWN] bring us back to our focus in this course, which is on the

molecular underpinnings of those thermodynamics.

So we're going to take a look at statistical entropy next and see

ultimately how that relates to our molecular picture.

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