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

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来自 University of Minnesota 的课程

分子热力学统计

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

从本节课中

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

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

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

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

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

Substitution. The dU is equal to TdS minus PdV.

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

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

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

We compare now with the formal differential, the simple calculus

differential of G with respect to temperature and pressure.

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

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

And partial G partial P is positive volume.

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

measurable properties, or knowable properties.

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

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

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

with. The difference is really only in the

letters that are appearing in our equations.

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

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

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

the derivative of free energy with respect to pressure.

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

partial T. Given that the derivative of G with

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

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

one another. And so this is another of many possible

Maxwell relations. Once more, notice that there is one

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

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

measure. [COUGH].

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

isolate the entropy. And see how the entropy changes with

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

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

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

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

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

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

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

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

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

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

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

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

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

entropy that I can perhaps get from a partition function.

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

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

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

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

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

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

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

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

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

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

values of bar. The qualitatively, entropy decreases.

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

pressure variation. Now, what about the enthalpy dependence

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

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

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

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

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

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

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

But I do know how to get volume temperature relationships.

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

solving for the pressure dependence of the enthalpy.

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

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

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

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

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

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

explanation for why the enthalpy behaves this way.

There's balance of P V and internal energy.

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

be rising again at very, very high pressures.

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

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

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

experiments. Of course, once they're available,

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

actually, and let you work with this equation.

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

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

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

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

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

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

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

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

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

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

result for the ideal gas, namely, delta S.

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

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

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

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

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

P2 over P1. So the relationship between delta S and

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

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

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

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

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

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

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

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

actually a very practical example of applying all the thermodynamic principles

that we have developed and armed ourselves with the tools.

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

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

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

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

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

And we will do that from a standpoint of thermodynamics.

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

temperature changes. And then we'll work out the

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

ability to explain something using thermodynamics.

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

mass of a weight that otherwise rests on this scale.

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

scale to the unsupported mass of the weight.

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

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

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

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

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

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

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

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

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

decreasing? Which is to say the tension in the rubber

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

it gets hotter. Is that what you predicted?

Let me offer a general explanation for this behavior.

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

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

principle, rotate. So that the chain is locally either

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

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

that the chain is straight, and maximally extended.

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

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

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

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

free energy of structures having more entropy.

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

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

home. Take a rubber band, ideally a somewhat

larger one, and touch it to your upper lip.

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

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

again immediately. It should feel warmer than room

temperature. A more detailed analysis is required to

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

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

explore that process in more detail.