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 8

This last module rounds out the course with the introduction of new state functions, namely, the Helmholtz and Gibbs free energies. The relevance of these state functions for predicting the direction of chemical processes in isothermal-isochoric and isothermal-isobaric ensembles, respectively, is derived. With the various state functions in hand, and with their respective definitions and knowledge of their so-called natural independent variables, Maxwell relations between different thermochemical properties are determined and employed to determine thermochemical quantities not readily subject to direct measurement (such as internal energy). Armed with a full thermochemical toolbox, we will explain the behavior of an elastomer (a rubber band, in this instance) as a function of temperature. Homework problems will provide you the opportunity to demonstrate mastery in the application of the above concepts. The final exam will offer you a chance to demonstrate your mastery of the entirety of the course material.

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

Chemistry

With a full complement of state functions in hand, let's take a look at Maxwell

Â relations. So, the various functions we've looked

Â at, we've explored, to some extent, their interrelationships.

Â How does enthalpy depend of internal energy?

Â How does Gibbs free energy depend of enthalpy?

Â And some of the thermodynamic properties we've talked about are not actually very

Â conveniently measured. So, if I asked you to measure the

Â internal energy of a gas, you don't necessarily have a tool.

Â You can reach into a drawer and stick into the gas and it reads out internal

Â energy. Not like a thermometer reads out

Â temperature or a manometer reads out pressure.

Â And so you might be a little confused, how, how do I go about measuring that?

Â Well, it's very useful if we have a means to express one of those quantities like

Â internal energy in terms of other quantities we can measure, like

Â temperature and pressure, for instance. And so here is a general expression for

Â instance that dA equals dU minus TdS minus SdT.

Â And for a reversible process, we would say dU equals TdS minus PdV.

Â So remember that is the first then second law, I've replaced dell q with TdS.

Â That's the second law and because it's reversible, I can replace the work with

Â PdV. And so when I do that and I substitute, I

Â get TdS minus PdV, minus TdS, minus SdT and so all that's left is that dA is

Â equal to minus PdV minus SdT. Now, I want to step back for a moment and

Â consider just differentiating A, considering A to be a function of T and

Â V. And if I do that, this is now just

Â differential calculus. I'll get that the differential of A is

Â the partial derivative of A with respect to V, when I hold T constant times DV.

Â Now the partial derivative of a with respect to T, while I hold V constant

Â times dT. So that's again just simple calculus if

Â you will. But notice that this equation and this

Â equation, they're the same, except these partial derivatives actually are

Â quantities here that are measurable quantities.

Â And so I'll just turn on the arrows there.

Â You see that the pressure is associated with partial A, partial V.

Â And the entropy is associated with partial A, partial T.

Â And in particular, partial A, partial V with temperature held constant is minus

Â the pressure and partial A, partial T. With the volume held constant, is minus

Â the entropy. Now, why is that interesting, important?

Â Well, if, I'm just going to carry those equations to the next slide, so I have a

Â little bit of room to work with them. So given those two differentials, again

Â from calculus, we know that the mixed partial derivatives of a function are

Â equal. Irrespective of which order

Â differentiation is done in. Which is to say, if I take this partial

Â derivative of A with respect to V, and now I differentiate it with respect to T.

Â That must give me the same answer as having taken the first derivative that I

Â take with respect to T. Followed by taking a derivative with

Â respect to V, that's the quality of the mixed partial derivatives.

Â And so If I look at partial A partial V, which is minus the pressure, then it's

Â partial derivative with respect to T is minus partial P partial T at constant

Â volume. I do the same thing working with negative

Â S as the partial derivative of A with respect to T.

Â I take its derivative with respect to V and I get this quantity.

Â And by the equality of the mixed partials, I have this relationship, that

Â the change in pressure with respect to the change in temperature.

Â When evaluated at a constant volume, will be equal to the change in entropy with

Â respect to the change in volume, when evaluated at a constant temperature.

Â So that's, that's pretty important. I'll show you more practically why in a

Â moment. That is one of many so called Maxwell

Â relations. That is, relationships that are derived

Â through looking at the equality of mixed partial derivatives that derive from

Â state functions. And so here we have James Clerk Maxwell,

Â he was a Scottish scientist of considerable renown.

Â Did enormous amount of work in many fields of physics, Maxwell's equations

Â are some of the most famous equations in in the physical sciences.

Â And while I do like to digress and talk about history the truth is, Maxwell's

Â contributions are so many. And so much more in the physics area

Â perhaps than in the chemistry area, that I'm going to defer to some great

Â physicist biographer. And let you go look that up on you own.

Â But I do want to continue looking at how useful these relations that Maxwell

Â established, can be for measuring thermodynamic quantities.

Â So here I'll just bring back that relationship that partial P, partial T is

Â equal to partial S, partial V, holding their respective volume or temperature

Â constant. And the utility is that given this

Â Maxwell relation, we can now determine how S, the entropy changes with respect

Â to the volume, given an equation of state.

Â Remember what an equation of state is? Is tells you how pressure, volume and

Â tempature is related. Those are three things I know how to

Â measure, P, V, and T. I have meters that measure that, I do not

Â have a meter that measures entropy. I don't know how to do that.

Â But, now I have a means to do it. So if I move this differential over to

Â the other side if you will, call it dV. I'll get dS is equal to this derivative

Â times dV. So if I want to know delta S for a

Â non-infinitesimal change, I would integrate from an initial volume to a

Â final volume. What is the change in pressure with

Â respect to the change in temperature evaluated at that volume, dV?

Â So, T is being held constant, while I'm integrating over V.

Â So I would measure this at given volumes of gas with given pressures.

Â I would simply change the temperature by a degree and look at, how much did the

Â pressure change? It's enough to tabulate.

Â So I can get my volume or density, density is just the intensive quantity

Â related to the volume a mole occupies. I can get the dependence of entropy from

Â PVT data. So I just do a lot of pressure volume

Â temperature measurements. And just to do the example for an ideal

Â gas, actually. Where partial P, partial T, I know that

Â from the ideal gas equation, right? P is equal to nRT over V, so partial,

Â partial T of that's pretty easy. It's just NR over V, so when I plug that

Â in, the nR constants, they come out front.

Â I get the integral from V1 to V2 of dV over V.

Â And I'll get that delta s is equal to nR log V2 over V1.

Â That's actually a result we derived a couple of weeks ago by a different

Â process, by analyzing paths along isothermal, gas expansions, for instance.

Â And you can go look in video 6.2 at how we did that, but here was a much simpler

Â way to do it just taking advantage of a Maxwell Relation.

Â And moreover, if I choose my initial volume to be so large that my gas is

Â very, very dilute. Then it ought to behave as an ideal gas,

Â and I can use the properties of ideal gases to establish what the entropy ought

Â to be at that very large volume. Or alternatively very low density, so as

Â the density goes to 0. So in that case delta s, which is going

Â to be the entropy at the new volume compared to the entropy at almost

Â infinite volume. Is this integral from ideal volume to V2,

Â and I can plot that, effectively. So, here, I have what is the molar

Â entropy of an ideal gas? And given my gas, if it's behaving

Â ideally, I can do that from the partition function.

Â Remember what I'll need for that, I'll need to know its mass.

Â I'll need to know its moments of inertia, I'll need to know its vibrational

Â frequencies. But with those in hand, I can compute, in

Â a spreadsheet, what the entropy must be. And so for ethane for instance at 400

Â Kelvin, I would get 246.45 joules per mole Kelvin, at one bar.

Â And then, using data that I would measure for, ethylene at 400 Kelvin.

Â I could determine, as this integral goes along, what's happening to the entropy.

Â And so it's going down, down, down, down, down as the density is increasing.

Â And certainly, by the way, from concept standpoint, that should seem sensible.

Â The entropy should be being reduced as I am decreasing the volume, increasing the

Â density. There's clearly less disorder as I give

Â my gas less volume to occupy. So, for real gases, where we don't have

Â an analytical equation of state the way we do for the ideal gas.

Â we really have to do these measurements. We would have to look at, how does the

Â pressure vary with temperature? Over a range of different volumes.

Â And I didn't mean to make you go so, I'll go back there for a second.

Â and in this case we would also you could also do that with different densities as

Â opposed to different volumes, that's what's plotted here.

Â Now the internal energy we can play exactly the same sort of game.

Â So if we differentiate A equals U minus TS with respect to V.

Â We get partial A partial V is equal to partial U partial V minus T partial S

Â partial V. It's isothermal, so I don't have to

Â differentiate temperature. It's not varying at all, it's being held

Â constant. I have a Maxwell relation that we've just

Â been working with. Partial P partial T is equal to partial S

Â partial V. And I've already derived, just a little

Â while ago, that partial A partial V is minus the pressure.

Â And so, if I make these substitutions. I get, as I rearrange this, that partial

Â U partial V, the change in internal energy with respect to the change in

Â volume. Is negative P plus T times partial P

Â partial T. And again the important this is, on this

Â side, these are all things that I can measure at given temperatures, pressures,

Â volumes. What's the dependence of the pressure on

Â the temperature? And, tabulate and work with.

Â And so here is, then, the same sort of idea.

Â Not for entropy, but for molar internal energy.

Â So once again, I'll start at a so low a pressure that my gas is behaving ideally.

Â In which case, from its partition function, I should be able to compute its

Â internal energy, 14.55 kilojoules per mole in this case.

Â And then as I increase the pressure up to rather high pressures here, up to 600

Â bar, plus, that's being shown here. The internal energy is going down and

Â it's gotten from integrating this expression over the volume change.

Â Volume would be related to the pressure here, I would have to go look up how

Â they're related. The plot just just happens to be against

Â pressure. And, I could again establish, what then

Â is the internal energy. Which may be useful in trying to figure

Â out how much work I can extract out of a gas, for instance.

Â Okay, well, let me pause for a moment. I'd like for you to have a chance to

Â maybe work this expression for an old friend, the ideal gas.

Â And see if you can work out the relationship there.

Â All right, maybe the last one to look at something we don't have a meter.

Â Let's look at the volume dependence of the Helmholtz free energy.

Â This relationship between minus the pressure and the derivative of the

Â Helmholtz free energy with respect to the volume at constant temperature.

Â So if I were to integrate this, move the dV over to the other side, I'll get that

Â delta a is equal to minus integral from V1 to V1, PdV.

Â And let's use an ideal gas as an example, so I know in that case pressure is nRT

Â over V. So when I make that substitution,

Â temperatures a constant now because we're working at constant temperature.

Â And so we'll pull that out with n and the universal gas constant, we get delta A as

Â minus nRT the integral dV over V. So, that will give us the log of V2 over

Â V1. And I want to compare this to a previous

Â result for an ideal gas at constant temperature where delta S is equal to nR

Â log V2 over V1. And so notice the relationship between

Â these two is that delta A is minus T times delta S.

Â And that's just what we expect actually for an ideal gas.

Â Remember, delta A is equal to delta U minus T delta S, but on what does delta U

Â depend for an ideal gas? It only depends on temperature.

Â And as a result, if we're working at a constant temperature, delta U is 0.

Â So sure enough, delta A ought to be minus T delta S, and here's the proof indeed

Â that it is. Alright, well, that was some useful

Â Maxwell relations that derive from the Helmholtz free energy.

Â In the next lecture, I'd like to look at some additional Maxwell relations.

Â These deriving from the Gibbs free energy.

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