This introductory physical chemistry course examines the connections between molecular properties and the behavior of macroscopic chemical systems.
Video 8.3 - Maxwell Relations from A
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En provenance du cours de University of Minnesota
Statistical Molecular Thermodynamics
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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.
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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
relations. So, the various functions we've looked
at, we've explored, to some extent, their interrelationships.
How does enthalpy depend of internal energy?
How does Gibbs free energy depend of enthalpy?
And some of the thermodynamic properties we've talked about are not actually very
conveniently measured. So, if I asked you to measure the
internal energy of a gas, you don't necessarily have a tool.
You can reach into a drawer and stick into the gas and it reads out internal
energy. Not like a thermometer reads out
temperature or a manometer reads out pressure.
And so you might be a little confused, how, how do I go about measuring that?
Well, it's very useful if we have a means to express one of those quantities like
internal energy in terms of other quantities we can measure, like
temperature and pressure, for instance. And so here is a general expression for
instance that dA equals dU minus TdS minus SdT.
And for a reversible process, we would say dU equals TdS minus PdV.
So remember that is the first then second law, I've replaced dell q with TdS.
That's the second law and because it's reversible, I can replace the work with
PdV. And so when I do that and I substitute, I
get TdS minus PdV, minus TdS, minus SdT and so all that's left is that dA is
equal to minus PdV minus SdT. Now, I want to step back for a moment and
consider just differentiating A, considering A to be a function of T and
V. And if I do that, this is now just
differential calculus. I'll get that the differential of A is
the partial derivative of A with respect to V, when I hold T constant times DV.
Now the partial derivative of a with respect to T, while I hold V constant
times dT. So that's again just simple calculus if
you will. But notice that this equation and this
equation, they're the same, except these partial derivatives actually are
quantities here that are measurable quantities.
And so I'll just turn on the arrows there.
You see that the pressure is associated with partial A, partial V.
And the entropy is associated with partial A, partial T.
And in particular, partial A, partial V with temperature held constant is minus
the pressure and partial A, partial T. With the volume held constant, is minus
the entropy. Now, why is that interesting, important?
Well, if, I'm just going to carry those equations to the next slide, so I have a
little bit of room to work with them. So given those two differentials, again
from calculus, we know that the mixed partial derivatives of a function are
equal. Irrespective of which order
differentiation is done in. Which is to say, if I take this partial
derivative of A with respect to V, and now I differentiate it with respect to T.
That must give me the same answer as having taken the first derivative that I
take with respect to T. Followed by taking a derivative with
respect to V, that's the quality of the mixed partial derivatives.
And so If I look at partial A partial V, which is minus the pressure, then it's
partial derivative with respect to T is minus partial P partial T at constant
volume. I do the same thing working with negative
S as the partial derivative of A with respect to T.
I take its derivative with respect to V and I get this quantity.
And by the equality of the mixed partials, I have this relationship, that
the change in pressure with respect to the change in temperature.
When evaluated at a constant volume, will be equal to the change in entropy with
respect to the change in volume, when evaluated at a constant temperature.
So that's, that's pretty important. I'll show you more practically why in a
moment. That is one of many so called Maxwell
relations. That is, relationships that are derived
through looking at the equality of mixed partial derivatives that derive from
state functions. And so here we have James Clerk Maxwell,
he was a Scottish scientist of considerable renown.
Did enormous amount of work in many fields of physics, Maxwell's equations
are some of the most famous equations in in the physical sciences.
And while I do like to digress and talk about history the truth is, Maxwell's
contributions are so many. And so much more in the physics area
perhaps than in the chemistry area, that I'm going to defer to some great
physicist biographer. And let you go look that up on you own.
But I do want to continue looking at how useful these relations that Maxwell
established, can be for measuring thermodynamic quantities.
So here I'll just bring back that relationship that partial P, partial T is
equal to partial S, partial V, holding their respective volume or temperature
constant. And the utility is that given this
Maxwell relation, we can now determine how S, the entropy changes with respect
to the volume, given an equation of state.
Remember what an equation of state is? Is tells you how pressure, volume and
tempature is related. Those are three things I know how to
measure, P, V, and T. I have meters that measure that, I do not
have a meter that measures entropy. I don't know how to do that.
But, now I have a means to do it. So if I move this differential over to
the other side if you will, call it dV. I'll get dS is equal to this derivative
times dV. So if I want to know delta S for a
non-infinitesimal change, I would integrate from an initial volume to a
final volume. What is the change in pressure with
respect to the change in temperature evaluated at that volume, dV?
So, T is being held constant, while I'm integrating over V.
So I would measure this at given volumes of gas with given pressures.
I would simply change the temperature by a degree and look at, how much did the
pressure change? It's enough to tabulate.
So I can get my volume or density, density is just the intensive quantity
related to the volume a mole occupies. I can get the dependence of entropy from
PVT data. So I just do a lot of pressure volume
temperature measurements. And just to do the example for an ideal
gas, actually. Where partial P, partial T, I know that
from the ideal gas equation, right? P is equal to nRT over V, so partial,
partial T of that's pretty easy. It's just NR over V, so when I plug that
in, the nR constants, they come out front.
I get the integral from V1 to V2 of dV over V.
And I'll get that delta s is equal to nR log V2 over V1.
That's actually a result we derived a couple of weeks ago by a different
process, by analyzing paths along isothermal, gas expansions, for instance.
And you can go look in video 6.2 at how we did that, but here was a much simpler
way to do it just taking advantage of a Maxwell Relation.
And moreover, if I choose my initial volume to be so large that my gas is
very, very dilute. Then it ought to behave as an ideal gas,
and I can use the properties of ideal gases to establish what the entropy ought
to be at that very large volume. Or alternatively very low density, so as
the density goes to 0. So in that case delta s, which is going
to be the entropy at the new volume compared to the entropy at almost
infinite volume. Is this integral from ideal volume to V2,
and I can plot that, effectively. So, here, I have what is the molar
entropy of an ideal gas? And given my gas, if it's behaving
ideally, I can do that from the partition function.
Remember what I'll need for that, I'll need to know its mass.
I'll need to know its moments of inertia, I'll need to know its vibrational
frequencies. But with those in hand, I can compute, in
a spreadsheet, what the entropy must be. And so for ethane for instance at 400
Kelvin, I would get 246.45 joules per mole Kelvin, at one bar.
And then, using data that I would measure for, ethylene at 400 Kelvin.
I could determine, as this integral goes along, what's happening to the entropy.
And so it's going down, down, down, down, down as the density is increasing.
And certainly, by the way, from concept standpoint, that should seem sensible.
The entropy should be being reduced as I am decreasing the volume, increasing the
density. There's clearly less disorder as I give
my gas less volume to occupy. So, for real gases, where we don't have
an analytical equation of state the way we do for the ideal gas.
we really have to do these measurements. We would have to look at, how does the
pressure vary with temperature? Over a range of different volumes.
And I didn't mean to make you go so, I'll go back there for a second.
and in this case we would also you could also do that with different densities as
opposed to different volumes, that's what's plotted here.
Now the internal energy we can play exactly the same sort of game.
So if we differentiate A equals U minus TS with respect to V.
We get partial A partial V is equal to partial U partial V minus T partial S
partial V. It's isothermal, so I don't have to
differentiate temperature. It's not varying at all, it's being held
constant. I have a Maxwell relation that we've just
been working with. Partial P partial T is equal to partial S
partial V. And I've already derived, just a little
while ago, that partial A partial V is minus the pressure.
And so, if I make these substitutions. I get, as I rearrange this, that partial
U partial V, the change in internal energy with respect to the change in
volume. Is negative P plus T times partial P
partial T. And again the important this is, on this
side, these are all things that I can measure at given temperatures, pressures,
volumes. What's the dependence of the pressure on
the temperature? And, tabulate and work with.
And so here is, then, the same sort of idea.
Not for entropy, but for molar internal energy.
So once again, I'll start at a so low a pressure that my gas is behaving ideally.
In which case, from its partition function, I should be able to compute its
internal energy, 14.55 kilojoules per mole in this case.
And then as I increase the pressure up to rather high pressures here, up to 600
bar, plus, that's being shown here. The internal energy is going down and
it's gotten from integrating this expression over the volume change.
Volume would be related to the pressure here, I would have to go look up how
they're related. The plot just just happens to be against
pressure. And, I could again establish, what then
is the internal energy. Which may be useful in trying to figure
out how much work I can extract out of a gas, for instance.
Okay, well, let me pause for a moment. I'd like for you to have a chance to
maybe work this expression for an old friend, the ideal gas.
And see if you can work out the relationship there.
All right, maybe the last one to look at something we don't have a meter.
Let's look at the volume dependence of the Helmholtz free energy.
This relationship between minus the pressure and the derivative of the
Helmholtz free energy with respect to the volume at constant temperature.
So if I were to integrate this, move the dV over to the other side, I'll get that
delta a is equal to minus integral from V1 to V1, PdV.
And let's use an ideal gas as an example, so I know in that case pressure is nRT
over V. So when I make that substitution,
temperatures a constant now because we're working at constant temperature.
And so we'll pull that out with n and the universal gas constant, we get delta A as
minus nRT the integral dV over V. So, that will give us the log of V2 over
V1. And I want to compare this to a previous
result for an ideal gas at constant temperature where delta S is equal to nR
log V2 over V1. And so notice the relationship between
these two is that delta A is minus T times delta S.
And that's just what we expect actually for an ideal gas.
Remember, delta A is equal to delta U minus T delta S, but on what does delta U
depend for an ideal gas? It only depends on temperature.
And as a result, if we're working at a constant temperature, delta U is 0.
So sure enough, delta A ought to be minus T delta S, and here's the proof indeed
that it is. Alright, well, that was some useful
Maxwell relations that derive from the Helmholtz free energy.
In the next lecture, I'd like to look at some additional Maxwell relations.
These deriving from the Gibbs free energy.
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