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

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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.

From the lesson

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 Î² 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

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

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

Â review. The expansion in our toolbox that's

Â occurred as a result of bringing in entropy.

Â So our ability to do calculations of thermodynamic properties includes

Â knowledge derived from the following expressions.

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

Â I've just rearranged the definition of dS.

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

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

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

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

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

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

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

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

Â with molar quantities just to make life a little simpler.

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

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

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

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

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

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

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

Â with their respective temperatures and volumes.

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

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

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

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

Â irreversible process delta s is greater than zero.

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

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

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

Â something out of the history of thermodynamics.

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

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

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

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

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

Â and in particular the factors that effected their efficiency.

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

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

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

Â mathematical footing. Sadly Carnot perished in a cholera

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

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

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

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

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

Â have my gas in contact with a cold reservoir.

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

Â And I let the heat flow into the cold reservoir.

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

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

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

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

Â temperature. In fact it's a higher temperature of

Â another bath, a hydrogen temperature bath.

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

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

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

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

Â I continue to let it expand adiabatically doing work.

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

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

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

Â curve. That's the expansion.

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

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

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

Â curve. So if you think about that just

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

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

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

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

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

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

Â then. Let's consider the work that is done

Â here. There are two isothermal steps.

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

Â expansion is. I'll just recapitulate it here.

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

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

Â steps. What about the adiabatic steps?

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

Â five. You find that for the adiabatic steps

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

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

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

Â cold temperature and initial is the hot temperature.

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

Â temperature and initial is the cold temperature.

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

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

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

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

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

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

Â So if you were just glancing at this geometric figure.

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

Â But, from thermodynamics, we know they must be.

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

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

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

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

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

Â it. In which must flow into the system when

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

Â transfer is 0, that's what adiabatic means.

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

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

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

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

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

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

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

Â bath. And so here is the question Carnot is

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

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

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

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

Â within the context of the heat I'm drawing.

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

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

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

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

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

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

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

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

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

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

Â out. And those have different signs.

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

Â expression. The maximum efficiency is: the heat in,

Â plus the heat out, divided by the heat in.

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

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

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

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

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

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

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

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

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

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

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

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

Â reversible h. This will be a q reversible h.

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

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

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

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

Â hot temperature. That's a pretty profound result.

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

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

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

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

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

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

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

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

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

Â You get no work out of an isothermal process.

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

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

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

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

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

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

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

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

Â your steam engine? Use superheated steam.

Â Right? That increases Th, the denominator in

Â this expression. And as a result you get one minus

Â whatever your cold temperature is divided by a large number.

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

Â an efficiency of one, that would be fabulous.

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

Â make it hotter. And so people began experimenting with

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

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

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

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

Â ideal gasses are useful. They give us general insights.

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

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

Â be. Actually that looks like it might be a

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

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

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

Â important points. And then we will have an opportunity to

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

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

Â see you then.

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