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

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From the course by 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.

From the lesson

Module 7

This module is relatively light, so if you've fallen a bit behind, you will possibly have the opportunity to catch up again. We examine the concept of the standard entropy made possible by the Third Law of Thermodynamics. The measurement of Third Law entropies from constant pressure heat capacities is explained and is compared for gases to values computed directly from molecular partition functions. The additivity of standard entropies is exploited to compute entropic changes for general chemical changes. 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

Alright. Let's take advantage of the tools we've

Â got so far this week to do some calculations of entropies, standard

Â entropies. So again a quantity an equation we've

Â seen before, third law entropy values. Entropies at a given temperature are

Â equal to the integral from 0 to T of the heat capacity dT, at constant pressure

Â heat capacity, that is dT over T. And, in this case, because I've made my

Â target temperature T. That's actually the limit on the definite

Â integral, so I'll use a T prime as an integration variable just to be

Â notationally careful. But, I've simplified this a bit compared

Â to some of the earlier videos. I've left out the S at temperature 0.

Â Because we have accepted the third law, namely that S of 0 should be equal to 0,

Â as long as something goes through a perfect crystal.

Â Now, for this expression to hold, there is one last stipulation, and that is that

Â there to be no phase transition of the substance between 0 and the target

Â temperature T. If there is a phase transition, for

Â example, if you go from a solid to a liquid phase, you melt.

Â Then, we have to account for the entropy change associated with that.

Â And because phase transitions are reversible processes, that means that the

Â delta S associated with that phase transition is going to be the amount of

Â heat that goes into the process. Divided by the temperature at which the

Â phase transition takes place. And so, if it takes place at fixed

Â pressure, then, the fixed pressure heat transfer is equal to the change in

Â enthalpy. And so delta S for a transition is equal

Â to delta H for a transition over the temperature at which it occurs.

Â So, if I now think of breaking this integral up into integrals over phases,

Â with different integrals for the different phases.

Â And then also some additional entropy associated with phase transitions, here

Â would be a very general expression. Its as I start at absolute 0 when my

Â substance is a solid. I go up to the temperature of fusion at

Â which point it begins melting integrating the heat capacity.

Â Then I would add the entropy associated with melting.

Â Now I continue my integral from the fusion temperature to the boiling point

Â the vaporization temperature. Now keeping track of the liquid heat

Â capacity. Where here I had the solid heat capacity.

Â Now I need the entropy associated with vaporization.

Â And, finally, I would continue to integrate from the boiling point up to

Â whatever temperature I'm interested in, If there is one above that.

Â Now the gas heat capacity at constant pressure.

Â Now, it's also helpful to take a more careful look at the low temperature

Â behavior of the constant pressure heat capacity.

Â And, if we have a nonmetallic solid, so an insulator, that is.

Â The Debye T cubed law is generally observed to hold.

Â And that says the following. It says that the constant pressure heat

Â capacity, the molar constant pressure heat capacity, is equal to a series of

Â constants, 12 pi to the fourth power over 5 times the universal gas constant.

Â Times the quantity T over capital thetas sub D, that's called the Debye

Â temperature, and that quantity all cubed. So it has units of kelvin, so this

Â becomes a a unitless quantity. And R has the same units as heat

Â capacity, so that all holds. So, as long as the temp is below some

Â limit, this this expression holds. Now Debye actually derived this

Â particular relationship through a analysis of quantized phonon energy

Â levels. And solids, so that would be sort of a

Â solid states physics thing, it's beyond what we want to do in this course.

Â But suffice it to say there is sort of a first principles explanation for why this

Â equation is valid. And, of course, experiment bears it out

Â as well. And, so if we ask, then, what is the very

Â lowest temperature contribution to a third law entropy.

Â We would want to compute the molar entropy.

Â I'll integrate from 0 to a given temperature T, the molar heat capacity dT

Â over T. But now I can replace the molar heat

Â capacity with the Debye T cubed expression.

Â So I'll do that, I'll pull all the constants out front, and I end up

Â integrating from 0 to T, T prime squared, because I had a T divide, T cubed divided

Â by T. So I'm left with T squared dT.

Â So when I do that integral I'll get T cubed over 3.

Â And actually if I look at this resulting expression, it's just this the constant

Â pressure molar heat capacity divided by 3.

Â And so that's very useful. That's says I don't necessarily need to

Â buy a big expensive piece of equipment to do, gokes magnetic, cooling, magnetic

Â refrigeration all the way down to some ridiculously low temperature.

Â Instead, I only have to get to a reasonably low temperature maybe, maybe I

Â go to five Kelvin, I ought be able to use liquid helium for that.

Â I measure the constant pressure heat capacity at that temperature.

Â How much heat does it take to raise my substance by 1 degree Kelvin?

Â And then, I just divide by 3, and that ought to be equal to the molar entropy at

Â that temperature. It's at that stage I can begin doing

Â additional measurements, in order to keep adding the extra entropy, above and

Â beyond that starting point. So, I don't really have to start from

Â zero, that's the utility. So, that just restates what I, what I was

Â commenting on, it's very convenient, because it obviates our need to measure

Â constant pressure heat capacity. All the way down to absolute zero.

Â And so let me just illustrate sort of a, a practical demonstration of the, the,

Â measurement or computation. Depending on how you want to thing of the

Â arithmetic step of the total entropy for, for nitrogen.

Â So, this is the molar entropy of nitrogen over the temperature range.

Â Absolute zero here at the origin. All the way up to 400 Kelvin.

Â And if we think of this as occurring in various processes, we would start from 0

Â to 10 Kelvin. And then 10 Kelvin to 35.61 Kelvin.

Â And I don't know why the experimentalists went to exactly, oh, I do know why they

Â went to exactly that temperature. So [INAUDIBLE] let's look at the numbers

Â first. So, at the very lowest range here, 0 to

Â 10 Kelvin we pick up 2.05 Joules per Kelvin per mole worth of entropy.

Â And I'll stop saying the unit here for now on, but that's the unit we're

Â tabulating. Over the next 25.61 degrees, we pick up

Â about 25.79 more. At that stage, the solid nitrogen

Â undergoes a phase transition to another solid form.

Â So, that's not that unusual, solids can often have a different phases.

Â They're still solids, but they're just oriented differently in crystals, for

Â example. And so, there is an entropy change

Â associated with that solid, solid phase transition.

Â And it adds 6.43 to the entropy. In some sense, you could think of that as

Â the driving force for the transition. As you are increasing the temperature,

Â entropy is playing more of a role and the more disordered phase becomes the favored

Â phase. We'll see more about that when we think

Â about other thermodynamic quantities next week.

Â But in any case at that stage I'm at 35.61 Kelvin.

Â That phase is stable for another 28 degrees or so, and as I continue to raise

Â the temperature, another 23.4 entropy these units, increase is observed.

Â And remember, what your really doing of course is, you're measuring the heat

Â capacity degree by degree. And then you're just plugging that in to

Â the integral over each of those degree steps.

Â What was the temperature for that step? And what was the heat capacity for that

Â step in adding them up. So, that would be sort of a numeric way

Â to solve that integral, is a way to think about it.

Â Okay, at 63.15 Kelvin, we have a solid to liquid phase transition, so this is the

Â temperature at which nitrogen melts. As it melts, the entropy increases by

Â 11.2. Notice that that's larger than the solid

Â solid, which seems sensible. There's, it seems like there's more

Â disorder in a liquid than there is in a solid, so we get more of a positive

Â increase in entropy. At that stage, the liquid, nitrogen is

Â only liquid over a relatively small range.

Â And, that is 63 to 77 kelvin and that increases the entropy another 11.46.

Â And so I've, I've just been indicating on this curve this is the total entropy,

Â molar entropy as a function of temperature.

Â And so far we've transitioned these two solid phases, we're now transitioning

Â this liquid phase. And the vertical lines then that are

Â shown here, they're a little to small to label conveniently.

Â But those are the phase transitions. So, that's the entropy without changing

Â the temperature that goes in, in order to change the phase.

Â Well, then we get to a really big phase transition, so this really tall vertical

Â line as the liquid goes to a gas. And certainly a gas has vastly more

Â disorder, as the individual molecules of the gas spread out to occupy a much

Â larger volume than is true for the liquid.

Â And so we pick up 72 joules per kelvin per mole associated with that transition.

Â After that point, we have a long range here, I've, I'm showing a curve that goes

Â to 400 but I'm actually stopping at room temperature, 298.15 in the table here.

Â So we pick up another 39 joules per kelvin per mole.

Â And here's the increase if you like until we get to about here, here's room

Â temperature. And finally in tabulations, one often

Â sees a correction for non-ideality. And that is simply to take note of when

Â you do the measurements you tend to try to do it at very low gas pressures.

Â So that things are behaving ideally, and then you correct for non-ideality as you

Â go to actual, say, one bar of pressure. So, the real gas compared to the ideal

Â gas. In nitrogen's case, at this temperature,

Â that's not a very big correction, 0.02, pretty small compared to all these

Â others. But we add all that together and you get

Â a value of 191.61 joules per Kelvin per mole.

Â And so that is a standard entropy, and you can look this up in tables of

Â collected physical quantities. And I've just mentioned to you that, by

Â convention, if you want to know the real gas that there is a correction for

Â non-ideality at one bar of pressure. Okay, so, having gone through that

Â example let me give you a chance to address a question related to it and then

Â we'll come back. All right, well what we've done, to date,

Â has really been classical thermodynamics. And, all the derivations of

Â differentials, and thinking about how to use heat capacity to measure entropy,

Â those are available from classical thermodynamics.

Â But I do want to pull us back to the real focus of this course, and that is the

Â molecular underpinnings. So, next, we'll look at the relationship

Â between entropy and the partition function as it relates to the third law

Â in more detail [SOUND].

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