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

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

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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

All right, let's wrap up our study of entropy by consideration of something

that's important to power and engines. And that's the Carnot cycle.

So prior to focusing on what Carnot discussed, let's take one moment and

review. The expansion in our toolbox that's

occurred as a result of bringing in entropy.

So our ability to do calculations of thermodynamic properties includes

knowledge derived from the following expressions.

So we've got dell q reversible, the reversible heat change, is equal to TdS.

I've just rearranged the definition of dS.

We know that reversible work for an ideal gas is minus PdV.

And finally we've got just the first law of thermodynamics, dU is equal to del q

plus del w, reversible. So, if I rearrange these various

expressions that all involve similar terms I can write, and I'll let you do

the arrangement if you want, but you'll see it.

dS is equal to dU over T plus P over T, dV.

And that just consists of swapping in these various expressions and rearranging

them. Now, in an ideal gas remember, let's work

with molar quantities just to make life a little simpler.

That dU bar is equal to CV bar, dT, and PV bar is equal to RT.

So again I'll make some substitutions and I'll get that dS bar the molar entropy

change is heat capacity times dT over T plus R times dV over V.

And so if Cv, that is if the heat capacity is just a constant independent

of T, I can integrate this exact differential expression on both sides and

I end up with delta S is equal to heat capacity times log T2 over T1 plus R log

V2 over V1. As I go from a state one to a state two

with their respective temperatures and volumes.

And for an idea gas, of course, I can relate the molar volume to a pressure.

So I could equally well write this, it's equal to Cp log t two over T1 minus R log

P2 over P1. Notice I switched to a pressure here and

I have pressure as an expression here. Remember that for a spontaneous

irreversible process delta s is greater than zero.

Alright, that's another key feature that we've got in our toolbox.

So these are equations we're going to have occasion to work with again in the

future. But now let's, let's move on to consider

something out of the history of thermodynamics.

Let me introduce you to Sardi Carnot. And Carnot was a young French military

officer engineer, was related to powerful members of the French aristocracy.

And at quite a young age, at age 28, he wrote a tredis that was entitled

Reflections on the Motive Power of Fire. Which is a wonderful title I think for

anything having to do with engines. And it was a monograph on heat engines

and in particular the factors that effected their efficiency.

And so later on that was read by people like Clausius who we already saw in the

context of entropy in the first and second laws And also, Lord Kelvin, who

we'll see again in a moment as they developed the second law in a more firm,

mathematical footing. Sadly Carnot perished in a cholera

epidemic at a very young age in 1832, he died at the age of 36, but his

contributions are, are very long lasting. And so let me describe for you an ideal

gas Carnot engine. So this is an engine that does work based

on a heat differential. And so in the Carnot cycle, the power

cycle, here's what happens. At point one, here I am at point one, I

have my gas in contact with a cold reservoir.

And I, isothermally compress the gas with a piston to here at a point four.

And I let the heat flow into the cold reservoir.

At point four, I disconnect my gas out of contact, I adiabatically, no heat flow

out of the gas at this point, continue to compress it to 0.3.

The temperature will go up because I'm compressing it and not allowing heat to

go away. When I get to 0.3, I'm at a higher

temperature. In fact it's a higher temperature of

another bath, a hydrogen temperature bath.

I put it in contact with that bath, and I now let it Isothermally expand.

So now it's doing work. It's expanding against external pressure

and it's absorbing heat from the hot bath as it's doing that.

At point 2 on this cycle, I remove it from contact with the hot bath.

I continue to let it expand adiabatically doing work.

It's dropping in temperature because it's adiabatic and when it's finally dropped

enough that it's all the way back to 0.1, that completes the cycle.

If you asked how much work did I do in total, well, it was the area under this

curve. That's the expansion.

And then the area under this curve, that's an expansion.

Minus the area under this curve, because I was doing work to compress the gas so

I've got to put that work back in. I get net work and the area under this

curve. So if you think about that just

geometrically this area plus this area minus this area minus this area.

That's the area inside this cycle. Alright, so if I can maximize this area

I'm getting more work out of my engine. Alright, so I'm, I want to continue

working with that in a moment. I want to pause for a, a second and let

you consider a problem that's related to this, and then we'll come back to the

Carnot cycle. Okay we're back with the Carnot cycle

then. Let's consider the work that is done

here. There are two isothermal steps.

We've already worked out in detail what the work associated with an isothermal

expansion is. I'll just recapitulate it here.

It's minus n, R, the temperature at which you're working T sub j, log, final

volume, divided by initial volume. So that's for the reversible isothermal

steps. What about the adiabatic steps?

Again, if you go back and you go back and look at the relevant material in week

five. You find that for the adiabatic steps

it's n times the molar heat capacity at constant volume, final temperature minus

initial temperature. So if I add those together, well, notice

that Tf minus Ti. For this adiabatic step, final is the

cold temperature and initial is the hot temperature.

So I get Tc - Th. For this adiabatic step, final is the hot

temperature and initial is the cold temperature.

So I get Th - Tc, those two cancel one another, so all that's left is the work

associated with the two isothermal steps. So minus n r, hot temperature, log V3

over V2, plus cold temperature, log V4 over V1.

Note incidentally that the fact that the work for the two adiabatic steps cancel,

that means that the area under this curve from three to four That's the work.

Must be equal to the area under these two under this curve from 2 to 1.

So if you were just glancing at this geometric figure.

I'm not sure you'd appreciate that those 2 areas are equal.

But, from thermodynamics, we know they must be.

The path, the Pv path that will be followed will guarantee that.

And so, the heat is equal and opposite to the work on the isothermal steps.

And because they're isothermal, the internal energy is not changing.

So, whatever work we do, either on the system or we extract from the system, is

balanced by the heat, which must flow out of the system when we're doing work on

it. In which must flow into the system when

it's doing work. The adiabatic steps, of course, the heat

transfer is 0, that's what adiabatic means.

So, what's the efficiency of our engine? Well, the efficiency is actually defined

as, how much work did I get out of the heat.

That I took out of the hot reservoir. That's really my energy source.

It's that hot reservoir that I can transform heat energy into work.

So efficiency then is minus w and minus because the work I'm getting out is

negative work by definition. But I want my efficiency to be positive.

So minus w divided by the reversible heat transferred when in contact with the hot

bath. And so here is the question Carnot is

interested in what's the maximum possible efficiency for an engine?

Right if I have an engine I can measure it's efficiency, measure how much heat is

flowing in and how much work I'm getting out of it But should I spend a lot of

time trying to improve it or might I have actually designed the best engine I can

within the context of the heat I'm drawing.

Well the maximum efficiency will occur for a reversible engine cycle.

Right, because the maximum work we can get out of expansion of an ideal gas is

along the reversible path. So in that case, delta U is going to be

W, the work, plus the reversible heat, in and out.

And it's going to be equal to zero because it's a full cycle.

Alright? And so that's all the components of going

around the full cycle. And because internal energy is a state

function, it's equal to 0. So that means that negative w, just

rearranging this equation. I'll move w over on this side.

Now it's negative w. It's equal to the heat in plus the heat

out. And those have different signs.

So, the maximum efficiency, then, if I replace negative w here with this

expression. The maximum efficiency is: the heat in,

plus the heat out, divided by the heat in.

From our knowledge of entropy, though, we can go a little beyond this.

We know again it's a, it's a full cycle. We're doing a circular path, if you will.

So the change in entropy, because it's a reversible path, is equal to zero.

So, del q reversible over t, that's dS associated with one path.

And del q invisible over t the entry point associated with the other path must

be equal to zero. And if I rearrange this then I can

express the reversible heat being dumped into the cold reservoir In terms of the

reversible heat coming out of the hot reservoir, and the differ by a factor of

Tc over Th, and of course they're different signs.

So, again, I can make a substitution, so here is my maximum efficiency.

I will now express this. By substituting in for cold minus q

reversible h over times tc over th. But look, in that case, here's q

reversible h. This will be a q reversible h.

Here's q reversible h in the denominator. Actually the amount of heat transferred

drops out. All that's left are the temperatures And

you get that the maximum efficiency available to your engine is dictated only

by the two temperatures involved, one minus the cold temperature divided by the

hot temperature. That's a pretty profound result.

So, it actually also has lead to another statement of the second law.

This one was expressed by Lord Kelvin. So Lord Kelvin began as William Thomson

and he did so many critical and important experiments in physics and thermodynamics

that he was elevated to the peerage in England and became a Lord and he took the

name Kelvin, which is a river near where he grew up.

and his expression of the second law then was that no net work can be obtained from

an isothermal process. So, what would an isothermal process mean

in terms of maximum efficiency? Isothermal means Tc is equal to h.

It's, it's all one temperature. And so the efficiency is zero.

You get no work out of an isothermal process.

So Lord Kelvin had tremendous chin whiskers amongst other things, but this

was a tremendous observation and boy we could spend a lot of time about all the

fascinating things Kelvin determined. But gotta move on with the Carnot cycle I

think. So here's our statement of the second law

from Lord Kelvin, no network can be obtained from an isothermal process.

But now let's go back to the 19th century What was the practical takeaway here?

They had steam engines mostly. They were interested in making those

steam engines better. So how do you improve the efficiency of

your steam engine? Use superheated steam.

Right? That increases Th, the denominator in

this expression. And as a result you get one minus

whatever your cold temperature is divided by a large number.

In principle, if you could get this number up infinitely high, you would hit

an efficiency of one, that would be fabulous.

Practice of course, it's kind of hard to get infinitely hot steam, but you can

make it hotter. And so people began experimenting with

super heated steam. And I want to emphasize that this above

analysis, that applies for any engine that converts heat to work.

because you can always establish a equilibrium between that engine, and one

that's done with an ideal gas. that's, I'll again emphasize that's why

ideal gasses are useful. They give us general insights.

Okay well that's the Carnot cycle. That's engines, good practical stuff.

Here's something that has an engine. Not necessarily a steam engine but could

be. Actually that looks like it might be a

steam engine now that I look at it. It's pulling us down the tracks further

and further. Now that we've made a lot of progress, we

need to review week six and talk about entropy, what were the really high and

important points. And then we will have an opportunity to

explore new laws of thermodynamics. Well, there's only one left actually, it

will be the third. But we will do the review next and I'll

see you then.

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