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

Loading...

From the course by University of Minnesota

Statistical Molecular Thermodynamics

143 ratings

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

In this lecture, let's continue to work with Maxwell Relations, but for the Gibbs

Â free energy. So, here's our definition of the Gibbs

Â free energy. G is equal to U minus TS plus PV.

Â When I take a full differential of the Gibbs free energy, I get dU minus TdS

Â minus SdT, plus PdV plus VdP. I make my usual first and second Law of

Â Substitution. The dU is equal to TdS minus PdV.

Â In which case, I get dG is equal to minus SdT plus VdP.

Â So this TdS term cancels this minus TdS term, this minus PdV term cancels this

Â plus PdV term, these are the only two terms that survive.

Â We compare now with the formal differential, the simple calculus

Â differential of G with respect to temperature and pressure.

Â And it's partial G partial T dT partial G partial P dP, and once more there's a

Â relationship here that partial G partial T is negative the entropy.

Â And partial G partial P is positive volume.

Â So, here are the coefficients, then, expressed as differentiables or as

Â measurable properties, or knowable properties.

Â Entropy's hard to measure, but we have ways to relate it to other things.

Â And so here is out friend, James Clerk Maxwell again, and really this slide

Â looks very similar to that for the Helmholtz free energy that we worked

Â with. The difference is really only in the

Â letters that are appearing in our equations.

Â So we want to equate cross derivatives, and so here are our initial derivatives.

Â We now differentiate again, with respect to the other variable, and set the two

Â equal to one another. And when that happens, given that this is

Â the derivative of free energy with respect to pressure.

Â I should now take its differential with respect to temperature, I get partial V,

Â partial T. Given that the derivative of G with

Â respect to T is minus S. I take its derivative with respect to

Â pressure, partial S, partial P, that's all negative, these two must be equal to

Â one another. And so this is another of many possible

Â Maxwell relations. Once more, notice that there is one

Â quantity in here, entropy, for which I do not have convenient meter.

Â But the other quantities are things that I can readily go into a laboratory and

Â measure. [COUGH].

Â And so if I choose to exploit this Maxwell relation, I am likely to try to

Â isolate the entropy. And see how the entropy changes with

Â pressure given an equation of state, that is a relationship between P, V and T.

Â And my mechanism will be that I will integrate at constant temperature, this

Â quantity times dP integrated from an initial pressure to a final pressure.

Â And I've moved the negative sign from this side over to the other, because I'll

Â have entropy being positive on the left. So I'm holding temperature constant while

Â I'm integrating over pressure. And I get this pressure dependence of

Â entropy then simply from knowledge of P-V-T data, equation of state data.

Â Once more, it's always convenient to work with an ideal gas.

Â And so for an ideal gas, volume is equal to nRT over P, so taking the derivative

Â with respect to temperature is trivial, I just get nR over P.

Â That means my integral has the constants taken out front, it's dP over P.

Â And I'll end up with minus nR log, final pressure divided by initial pressure, P2

Â over P1, isothermal. Again, I want to typically start from an

Â entropy that I can perhaps get from a partition function.

Â Because I am so low of pressure that my gas behaves ideally.

Â And so I integrate from an ideal pressure as pressure goes to 0 to some final

Â pressure P2, and here is my same example, ethane at 400 Kelvin.

Â So this is exactly the same value that I had on a slide from the last lecture,

Â where obviously as pressure goes to 0, volume goes infinite.

Â Density goes to 0, so it's all the same number, minus, sorry, excuse me, positive

Â 246.45 Joules per mole Kelvin. And as I increase the pressure, the

Â entropy goes down. And so this plot looks rather similar to

Â what was plotted against density, because naturally density does in fact go up as

Â pressure goes up. But the numbers are different, these are

Â values of bar. The qualitatively, entropy decreases.

Â And so I'm getting entropy data from knowledge of volume, temperature, and

Â pressure variation. Now, what about the enthalpy dependence

Â on P? So, when we worked with Helmholtz's free

Â energy, we had a convenient way to measure the internal energy.

Â With Gibb's free energy we have a way to measure the enthalpy.

Â So, if I differentiate G, which is H minus TS, with respect to the pressure,

Â we get partial G partial P. Is equal to the pressure dependence of

Â the enthalpy minus T partial S partial P. Again, I'll use my Maxwell relation to

Â get rid of the entropy thing that I don't really know how to use a meter for.

Â But I do know how to get volume temperature relationships.

Â I already know that partial g partial p is equal to volume, so I can rearrange

Â solving for the pressure dependence of the enthalpy.

Â And it's equal to V minus T partial V, partial T.

Â And here's the sort of data I might derive from using experimental equation

Â of state data. Here's ethane once more at 400 Kelvin.

Â Here's my ideal enthalpy, which I can get from a partition function, 17.87

Â kilojoules per mole, and here's the behavior as I integrate.

Â And notice that this one is maybe, less easy to come up with a simple intuitive

Â explanation for why the enthalpy behaves this way.

Â There's balance of P V and internal energy.

Â It goes down for a while, and then it flattens, and then it actually seems to

Â be rising again at very, very high pressures.

Â And for a real gas where we don't have a simple equation of state, like the ideal

Â gas equation of state. Then we really have to look up these

Â sorts of volume, temperature, pressure relationships, have to have done the

Â experiments. Of course, once they're available,

Â they're available for all time. Well, let me take a moment here,

Â actually, and let you work with this equation.

Â For a, equation of state that looks similar to some we worked with back when

Â we did consider real gases. Okay, let's now look at the pressure

Â dependence of the Gibbs free energy. So I know that's the volume, I can

Â integrate delta G from an initial to a final pressure, VdP.

Â And as always, it's good to do the ideal gas first, because it ought to conform to

Â our now, you know, familiarity with ideal gas properties.

Â So I have that V is equal to nRT over P. And as a result, when I pull out the

Â constants, this is isothermal, so t is a constant.

Â I get the integral from P1 to P2, dP over P, that gives me the logarithm at

Â constant temperature. And I want to compare this to a prior

Â result for the ideal gas, namely, delta S.

Â So, delta S is nR log V2 over V1. But, for an ideal gas, what's V2 over V1?

Â Well, P time V is a constant for an ideal gas at a given temperature.

Â And so V2 over V1 is equal to P1 over P2, because there inversely related,but I

Â want to put P2 over P1. So I'll just change the sign of the

Â logarithm when I click the fraction. So when I get the delta S is minus nR log

Â P2 over P1. So the relationship between delta S and

Â delta G is again, they are related by multiplication by minus T.

Â So we get delta G is equal to delta H minus T delta S, and since it's also

Â equal to minus T delta S, that implies that delta H must be 0.

Â And once more our familiarity with an ideal gas says, naturally, because we

Â worked out that the enthalpy of an ideal gas depends only on the temperature.

Â Since the temperature is being held constant, delta H must be 0, and delta G

Â must be equal to minus T delta S. Well, that finishes up what I wanted to

Â do with Maxwell relationships. What I want to look at next, it's

Â actually a very practical example of applying all the thermodynamic principles

Â that we have developed and armed ourselves with the tools.

Â So I think we began the course talking about building a lot of tools, so that we

Â could eventually build a house. Well, I think we're ready for to look at

Â a house of sorts, and in particular I want to look at.

Â Well, it doesn't look much like a house but a rubber band.

Â So I would like to take a look at and analyse the behavior of a rubber band.

Â And we will do that from a standpoint of thermodynamics.

Â So first we'll look at a demonstration video of the behavior of a rubber band as

Â temperature changes. And then we'll work out the

Â thermodynamics behind it. Here's a nice practical test of your

Â ability to explain something using thermodynamics.

Â Here, I have a rubber band, and through its tension, it's supporting some of the

Â mass of a weight that otherwise rests on this scale.

Â We can quantify the tension at any instant by comparing the reading on the

Â scale to the unsupported mass of the weight.

Â Here, I have an incandescent lamp which can serve as a heat source.

Â Now, for the practical question. If I heat up the rubber band, will it

Â contract, and pull harder on the weight, thereby reducing its apparent mass?

Â Or will it lengthen, and increase the apparent mass resting on the scale?

Â What do you think, and why do you think it?

Â I'll let you ponder for a moment. Well, if you're just not sure, or even if

Â you are sure. I think we have to do the experiment, no?

Â So, let's heat up this rubber band. [SOUND].

Â Look at the scale. Do you see how the apparent weight is

Â decreasing? Which is to say the tension in the rubber

Â band is increasing. Thus, the rubber band is contracting as

Â it gets hotter. Is that what you predicted?

Â Let me offer a general explanation for this behavior.

Â A rubber band is composed of a number of long polymer chains.

Â Each of the single bonds between two carbon atoms in those chains, can, in

Â principle, rotate. So that the chain is locally either

Â straight or bent. There are many ways to rotate so that the

Â chain bends. But there's only one way to rotate so

Â that the chain is straight, and maximally extended.

Â Thus, entropy favors shorter, bent chains, and there is much less disorder

Â when chains are straight. So, when are chains more straight?

Â When the rubber band is stretched, so the chains must achieve longer lengths.

Â When we heat the rubber band by increasing the temperature, we favor the

Â free energy of structures having more entropy.

Â Which is to say that at equilibrium, we favor shorter chains over longer and the

Â rubber band contracts as a result. You can try a different experiment at

Â home. Take a rubber band, ideally a somewhat

Â larger one, and touch it to your upper lip.

Â It ought to be room temperature and you won't feel much of anything at all.

Â But now search it to much greater lengths suddenly and touch it to your upper lip

Â again immediately. It should feel warmer than room

Â temperature. A more detailed analysis is required to

Â explain this behavior. Although the entropy change we've just

Â discussed, continues to be part of the analysis, and should time permit, we'll

Â explore that process in more detail.

Â Coursera provides universal access to the worldâ€™s best education,
partnering with top universities and organizations to offer courses online.