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PROFESSOR: Last few lectures, and we started to see some of
the consequences and looked at simple changes,
transformations of expansion and mixtures
of liquids and gases.
And saw how in the framework of statistical mechanics, we
could derive the thermodynamic results that you saw before,
based on an empirical framework.
On the thermodynamic framework we've been
working with all term.
But you saw them derived from a microscopic perspective.
Now what I want to do is move to examples that are more
commonly encountered in chemistry.
One that we did last time actually was very common, or
is at least a prototype for something common.
That is we looked at what would be a simple model for a
polymer with different configurations
available to it.
And just tried to understand what the basic thermodynamics
is that it would exhibit.
What the heat capacities would be in various limiting cases.
High and low temperature.
I want to continue that today.
But for an even more common case.
So the model that I want to lay out is indicated on your
notes for today.
And essentially, you could look at it as
double stranded DNA.
Any kind of polymer with a whole variety of
configurations.
Ultimately what's going to matter is what is the set of
energy levels available to the system.
And so here's what it looks like.
The way I've tried to depict it.
I'm imagining something like, covalently bonded chains that
have links between them.
So of course, it's reminiscent of DNA with hydrogen bonding
between the strands.
And the idea in this simple model is that you can have
various numbers of these hydrogen bonds.
So this is the case that would be the most
energetically stable.
That is, all the available hydrogen bonds are formed
between neighboring pairs.
So this would be our lowest energy.
Now let's starts breaking some hydrogen bonds.
And what I'm imagining is they break starting at one end and
working their way down.
So the next highest energy state would be like this.
So now these are still hydrogen bonded.
This is not.
And there's a cost in energy for breaking
that hydrogen bond.
We'll call it epsilon zero.
It's a positive number.
So that this is higher energy than this.
And now we'll go to the next one, which could have two
broken bonds.
Broken hydrogen bonds.
And so on.
And if we imagine that this is very long, there may be many
many, members of the chain.
Then of course this sequence of structures with
corresponding energies might go on for a very long way.
So this is a simple construction that gives us a
set of states with distinct energies.
And so we have a series of configurational energy levels.
They're evenly spaced in this model.
So there's zero, E zero, two E zero.
And so on.
So those are available energy states.
And now, just based on that simple model, we should be
able to figure out all the thermodynamics.
We should should be able to figure out what the
equilibrium energy is at a particular temperature.
What the average energy is.
And so forth.
And we can do it in a straightforward way.
So we'll start with our molecular partition
function, as always.
So it's q configurational.
And it's just a sum over our available states with their
associated Boltzmann factors.
So sum over n of e to the minus En over kT.
Now the way I've modeled this, there are no degeneracies.
Each energy level has just one microscopic state of the
molecule corresponding to it.
So there are no molecular degeneracies.
Turns out we'll be able to put this into a simple form.
So the first thing is, let's approximate that we could take
the sum -- not to some finite level, which would be the case
if there's a finite number of elements in the chain.
But all the way through infinity.
And the idea here is the following.
We'll assume that the chain at least has
a great many members.
Even if it's finite.
And then at some point, this becomes a really small number.
Even if epsilon naught, the energy of one of the bonds, is
less than kT.
Once we multiply it by a very large number n, it becomes
very much greater than kT.
So the underlying assumption here is, well there will not
be a significant number of molecules in the highest
energy state anyway.
In other words, the terms higher than a finite but large
value of n between there and infinity are going to be so
small that those terms will contribute negligibly to the
sum anyway.
And on that basis, we could make this approximation.
And the incentive to make this approximation and carry the
sum to infinity is that then we can put this in what turns
out to be a very simple closed form.
So now if we just write out a few of the individual terms in
the sum, it's one plus e to the minus
epsilon zero over kT.
Plus e to the minus epsilon zero over kT.
I'll write this as squared. e to the minus two
epsilon zero over k t.
And so on.
So this has the form of a sum that looks like one plus x,
plus x squared, and so on.
And because the sum goes to infinity, we can simply write
it as one over one minus x.
Sort of a very simple result.
And now putting this back into x, it's one over one minus e
to the minus epsilon zero over kT.
That's our q zero.
Our molecular partition function.
So this model yields a particularly simple
result for q zero.
And then everything else follows from there.
So then we can write the canonical partition function
capital Qn.
It's just q configurational to the capital Nth power, where
capital N is the total number of these molecules.
So it's just one over one minus e to the minus epsilon
over kT to the Nth power.
And then we can start writing out the results for the
various thermodynamic properties.
So A configurational is minus NkT, log q configurational.
That is, remember, A is minus kT log capital Q. And this is
just going to come straight out.
So it's minus NkT log of little q.
Minus NkT, log of one over one minus e to the minus
epsilon over kT.
Or positive NkT, log of one minus e to the minus
epsilon over kT.
So a pretty simple form for the Helmholtz free energy.
And I'm not going to write out all of the individuals
thermodynamic terms, but I'll write a few of them.
So mu, the chemical potential for these configurations, is
just dA/dN.
With T and V constant.
And so the only capital N dependence, the only
dependence on the number of molecules is multiplicative.
Right here.
So this just gives us kT log of one minus e to the minus
epsilon naught over kT.
And the important point to realize here that we mentioned
last time too, in the examples we considered then, especially
the last one, is that because the only place that N figures
in is as a multiplicative factor.
What this is telling us is that we just have a chemical
potential, of Helmholtz free energy per molecule.
The molecules are all independent.
The total energy doesn't depend on any interaction
between this molecule and its neighbor somewhere else.
So terms like this are simply additive.
So if we work out the chemical potential, it's just one over
N times A.
And I'll just write the result for u configurational.
Because I do want to look at the heat capacity.
So it's Nk T squared, d log of little q configurational / dT.
With N and V held constant.
And it turns out to just be N epsilon zero.
One over E zero over kT minus one.
So now we can look at the heat capacity.
Cv configurational.
So it's du configurational / dT.
With constant N and V. And taking this derivative with
respect to temperature.
Gives us an expression of the following form.
It's Nk epsilon zero over k T squared. e to the E zero over
kT, over e to the E zero over kT minus one squared.
Just taking the derivative in the usual way.
Now, this looks kind of complicated
for the heat capacity.
But let's take a look, just like we did last time for the
simpler case that we treated then, let's take a look at the
high and low temperature limits of what happens to the
heat capacity.
So at high temperature, well we can start by just looking
at the energy.
That has a form that's fairly simple.
So when temperature is large, this is small, and we can
Taylor expand it.
So then we're just going to have one plus epsilon zero
over kT minus one.
Which means the ones will cancel.
And we end up with a fairly simple result.
So u configurational, in that case, is just NkT.
The epsilon zeroes are going to cancel also.
And what that says is that the heat capacity -- and of course
we could take the limit of this, but we can just take the
derivative of this with respect to
temperature more easily.
So Cv configurational.
It's just N times k.
In other words, the heat capacity in the high
temperature limit is a constant.
Last time we treated a simpler case, where there were only
four configurations altogether available, to this sort of
simple polymer model that we drew.
Unlike this present model where we're saying there are
essentially infinite number of configurations and different
energies available.
So many that the highest ones we're never even going to
access, because they'll be much higher than kT.
Here it's different, so now we have --
You know before, when we had a limited number of total
states, remember what happened in the heat capacity at high
temperature.
What was the limiting case?
Finite number of states.
What's the heat capacity at high temperature?
This is going to be on the exam.
What's the heat capacity at high temperature, if there's a
finite number of states available?
STUDENT: Zero.
PROFESSOR: It's zero.
What was the low temperature limit of the heat capacity?
STUDENT: Zero.
PROFESSOR: It was also zero.
Right.
And the idea was for a system with a
finite number of states.
The idea was -- so let's say, the way we had it before,
there was one state with energy zero.
And we had three states with some energy epsilon zero.
And that was it.
Those were all the molecular states available.
So we looked at the two cases.
In one case, when the high temperature limit where kT is
much bigger than any of this stuff.
In that limit, the molecules are just equally likely to be
in any of these states.
Because there's much, much more thermal energy than this
energy difference.
And the molecules are constantly getting kicked
around by the available thermal energy.
So if you raise the temperature a little bit more,
it doesn't make any difference.
The molecules already are evenly
distributed among the states.
There's no additional configurational energy.
So du/dT is zero.
It can't increase any more.
So in that case, the high temperature limiting heat
capacity is zero.
And in the other case --
In the low temperature limit, now let's say
kT is really low.
It's much less than epsilon zero.
So now we'll redraw this as zero.
And put the epsilon zero states up here.
And kT is here.
Well when it's like this, the temperature is so low that
there's not nearly enough thermal energy to populate any
of these states.
And if you change the temperature by a little bit,
it's still not enough thermal energy to populate any of
these states.
So once again, du/dT is zero.
You change the temperature and the configurational energy
doesn't change.
So the heat capacity is zero again.
This is why it's so informative to measure heat
capacities.
Because you can learn an awful lot about the intrinsic
structure of the material.
What are the energy levels available to it?
What do they do?
You can learn a tremendous amount about that by making
measurements of the heat capacity.
Well, now we're in a different case.
We have a whole set of evenly spaced energy levels.
Never ends.
So now let's look at the high temperature limit.
But we're never in a limit that's higher than the highest
level, because we're assuming it goes on forever.
So it's up here somewhere.
There's kT.
So what happens?
Well if you raise the temperature, there still are
higher lying levels that can be populated, and that will
get populated.
So du/dT isn't going to be zero in the high temperature
limit, in this case.
But it stops changing at some point.
Because nothing's very different about this than
having kT be, let's say, up here or up here.
So in the high temperature limit, yes, the energy does
change with temperature.
But it changes in the same way at any temperature.
In other words, the energy is just linearly varying with
temperature.
And the heat capacity is a constant. du/dT doesn't change
anymore, once you're in the high temperature limit.
Now without me writing any expression on the board,
what's the low temperature limit of the heat capacity
going to be in this case?
Going to be on the exam.
STUDENT: Zero?
PROFESSOR: Zero, that's still going to be the same, right?
Put kT way down here.
That's just like this, right?
Not near enough energy to populate even though lowest,
you know, the first level above the ground state.
Change the temperature a little bit, still not enough
thermal energy to get up there.
So the energy doesn't change with temperature in that low
temperature limit.
So you can immediately see what's going to happen at low
temperature.
Any questions?
Yeah?
STUDENT: [UNINTELLIGIBLE] will still be n k t, and
then it's just --
[UNINTELLIGIBLE]
PROFESSOR: So, of course, that's a
limiting case here, right?
That happened because in the limit of high temperature,
then this exponent is really small, right?
So then you can Taylor expand it.
In the limit of low temperature,
this is really big.
This is nearly zero.
So e epsilon zero is much bigger than kT in that case.
This is big.
This is negligible, right?
Oh.
Wait a minute.
Something's making me real unhappy.
I can't have this right.
Ah.
Yes.
It needs to be zero, is what it needs to be.
What am I thinking?
Of course.
It's one over a huge number.
It's zero.
OK.
So of course you can't expanded it.
It's just, this is much bigger than this.
This is an enormous number at that point.
So in other words, what it's saying is the configurational
energy is zero, because everything is stuck in the
ground state.
And it stays zero if you vary the temperature.
So you get the zero limiting value for the heat capacity,
and the energy itself is also zero.
Other questions?
OK.
Let me just make a few comments about the entropy.
So I didn't write it out before.
And I'm tempted not to do it again, but I
suppose I'll do it.
So it turns out to be Nk, minus log of one minus e to
the minus epsilon zero over kT.
Plus epsilon zero over kT.
Over e to the epsilon zero over kT, minus one.
And this comes from combining the terms for A and u.
It comes from minus A over T. Plus u over T. And what I want
to do is look at its limiting cases also.
In particular, what happens here in the limit of high
temperature.
And what happens turns out to be k log kT over epsilon zero
to the Nth power.
Or put the N over here.
OK.
And what this is telling us about is the number of
available states.
Roughly, how many states are there that are accessible to
the system at some temperature.
So in other words, think of it as -- let's put the N back
there. k log of kT over epsilon zero to the Nth power.
And think of it as k log capital omega.
Where that would be the degeneracy.
Now all the states are in equal energy, but remember for
the whole system, remember we discussed this before.
How you have a very, very narrow distribution of system
energy states at equilibrium.
So you can think of this as the degeneracy of the system
states that are actually going to exist at a particular
temperature.
So if we look at the limiting value of the partition
function, it's just kT over omega.
Or the same thing for capital Q. kT over
epsilon zero, sorry.
To the Nth power.
So you have a very simple expression.
And so again, what this is doing, is it's giving us a
measure of about how many states are available.
And it's particularly informative to look at that
for the molecular partition function.
What it's telling us is, if I look at kT over epsilon.
And these levels, remember, are evenly spaced.
So here's epsilon zero, two epsilon zero, three epsilon
zero, and so on.
It says, you know, if kT is about ten times bigger than
epsilon zero.
So this is ten.
It's telling us, roughly how many states does the system
have thermal access to.
The molecular state.
It has about ten states.
So going over to our picture of the set of structures, you
could have anywhere up to about ten bonds broken.
Now, the individual molecules are going to be in a whole
range of states.
Some of them will have fewer than that, and some of them
will have more than that number of bonds broken.
But on average, it's going to be about that number.
And then, if you look at the whole system, the number of
states available, of course it's astronomical.
Because you have to take each molecule, and say, well it
could be in any one of something on the
order of ten states.
And then whole set of N other molecules can be in the states
they might be in.
Change the first one, and do it again.
So you have this astronomical number of system states.
But remember, like we discussed once before, it'll
turn out that although the individual molecule states
vary considerably with energy, the system states, which are
averaging over some astronomical number of
molecules, where capital N is something
like ten to the 24th.
Once you average over that many individual molecules, you
find that there's very, very little
fluctuation in the energy.
In the system energy.
The individual molecule energies vary considerably.
Realistically the variation of the molecular energies -- that
variation is going to be comparable
to the energy itself.
To the average energy.
So if the average energy is roughly you know, the ten
epsilon zero, you say okay, how much might it vary?
Well, there are going to be some molecules that have only
a couple of bonds broken, and some that might
have 20 bonds broken.
The variation will be on the same order of magnitude as the
average itself.
But then you average over capital N of them.
And then you immediately discover that there's very,
very, very little variation.
And in particular, what happens then, is, you know,
for molecular average energy, epsilon zero, and variance, or
standard deviation about the same magnitude.
System average energy is u, and it's going to
be capital N times.
Well let's say that -- average --
Oh sorry, let me not put the zero There.
It's just average energy.
And the system average energy is just N times
the molecular energy.
But the system variance is going to be on the order of
the square root of N times epsilon.
If you've done statistics, then you've seen
that sort of result.
You do a bunch of samplings a capital N number of samplings,
and the variation looks like the square
root of that number.
So what ends up happening then, if you look at the
relative variation, you might look and then say, well it's
pretty big.
This is ten to the 12, right?
It's still a huge variance.
But let's compare it to the average.
So this is ten to the 24th.
This is ten to the 12th.
It's on the order of ten to the minus 12.
An incredibly tiny fractional variation
in the system energy.
You could never --
There would be no practical way to measure it.
And of course that is consistent with
what you would expect.
If you say let's measure the configurational energy of a
bunch of molecules in a liquid solution, or molecules in a
gas floating around.
And it's a mole of them, that total average energy is not
going to fluctuate significantly, even though
individual molecules that you pick out of that whole system
might have quite widely varying energies.
And that's the point.
So it's an incredibly small variation.
Now you can derive this.
I didn't derive this, of course.
I asserted that this is the case.
And it's probably familiar to some of you if you've seen
some statistics.
But the way you would derive it is, you know in addition to
calculating the average energy, the average of E, you
can also calculate the average of the energy squared.
And you can calculate the standard deviation that way.
You calculate the average of E squared.
And then you minus the average of E, the quantity squared.
Take the square root.
That's your root mean square.
And that's what leads to this result.
So it's pretty straight forward to calculate it also.
Alright.
So any questions about just the extent of variation of the
individual molecule energies or the system energies?
Okay.
Then, now what I'd like to do is look at another kind of
energy that's going to turn out to have exactly the same
set of levels that we just derived
from this simple model.
And some of you may have seen this before.
Many of you may not have.
And for right now, I'll just assert it.
But it will illustrate why this is so useful.
It turns out that if I've got molecular vibrations -- you
know, there's nitrogen and oxygen in the air.
If I look at those nitrogen vibrational energy levels.
Or oxygen energy levels.
They look like this.
Evenly spaced, non degenerate energy levels.
So the model that we've constructed here, based on
this simple two-chain polymer, actually gives us a set of
energy levels that maps directly onto the vibrational
energy levels of a molecule.
So all the results that we've just seen apply, not just for
conformations of a polymer, but for
vibrations of a molecule.
So everywhere where it says configurational, you can just
write in vibrational.
And you'll still be right. because of course, what
matters is what are the states that are available to the
system, and what are their energies?
After that, the formalism of statistical mechanics takes
over, and calculates partition functions and
thermodynamic functions.
The input into that is states.
And their energies.
Well, we have the exact same set of states and energies.
So we immediately arrive at a super important case and its
results for molecular vibrations.
And not only molecular vibrations, but vibrations of
a crystal lattice.
Acoustic vibrations of a glass.
Or even a liquid.
So there's an enormously wide ranging set of results that
we've derived by starting at this simple picture.
And when you take quantum mechanics, you'll see that
indeed you do get this set of evenly spaced energy levels.
You may well have seen the results before, and you'll see
it derived there.
OK.
Given that, then I want to talk a little bit further
about the heat capacity.
So we've seen the limiting cases for the heat capacity.
Namely, the low temperature limit is zero.
That's the case whenever you have quantized levels.
You can always get kT lower by far than the
lowest excited state.
So that everything is stuck in the ground state.
How low that is depends on how far apart
the states are spaced.
But there's got to be some temperature somewhere that's
down there.
So we've seen the heat capacity limiting cases.
Let me just sort of draw them again.
So here's kT much less than epsilon zero.
And up here, kT is much bigger than epsilon zero.
Now I'm just going to sketch the full temperature
dependence.
In other words, I'm going to connect those two limits that
we've seen.
So here's what that looks like.
And I'll write it as vibrational.
So we know it's got to be zero at the low temperature limit.
And we know that it's this constant value in the high
temperature limit.
It's Nk.
And in some way, it smoothly connects.
And if you look at the actual scale, here -- so here's kT
over epsilon zero.
Here's the limit of low temperature.
Actually, you don't have to be very high above epsilon zero
to be in in this high temperature limit.
So if you look at the plot that's on your notes, this is
already -- it's not quite leveling off precisely.
But it's already pretty close.
When kT is just two times epsilon zero.
It's already almost at the limiting case.
Now this has tremendous significance.
So a long time before quantum mechanics was developed,
people made measurements of the heat
capacities of materials.
And they were familiar with the fact that when you got to
ordinary temperature -- room temperature, the heat capacity
was temperature independent.
And that was understandable for reasons
we'll discuss shortly.
Basically, in that case, the heat capacity -- it's not just
Nk for one vibrational mode.
Of course, that's what I describe for a single
vibrational mode of, say, a molecule.
If you have a crystal lattice with N atoms in it.
Let's say it's an atomic crystal.
Each atom has three degrees of freedom.
It can move in each of three independent directions.
And there are N of those atoms.
There are 3N total degrees of freedom.
That's how many vibrations the lattice has.
So in fact, you have to multiply this by 3N.
So what would be seen is, you'd have 3R, the gas
constant, per mole, in other words.
And if you looked at Cv over 3R, then this
value would be one.
Very useful to see that.
To figure that out.
And people understood it.
What people didn't understand is this stuff.
Why did it do that?
That was a great mystery.
Why did the heat capacity go to zero at low temperature?
And the reason they didn't understand it is because the
model for vibration was really simple.
The lattice is a bunch of masses and springs.
I know the vibrational energy.
And I can calculate it.
That is, I know the classical vibrational energy.
Remember, the reason it goes to zero is this.
It's all because of the fact that the energy levels are
discrete, quantized levels, with gaps in between them.
If I've got classical mechanics, then that means
it's vibrational energy.
The energy just gets bigger if the amplitude get bigger.
It's not discrete, it's continuous.
There are always energies in there.
In that case, there's never a situation like this.
Where kT is lower than the first available excited level,
and everything's in the ground state.
That never happens in classical mechanics, because
there are always levels there.
But it does happen in quantum mechanics.
And Einstein recognized that this was a way to explain this
low temperature limiting heat capacity.
So actually if you look at early development of quantum
mechanics, really it was all predicated
on statistical mechanics.
And the idea that, well, that you could then do the
statistical mechanics with quantized levels, just the way
we've done it.
And what you immediately discover is, gee,
it all makes sense.
You go to low temperature, everything's stunk down here.
And suddenly you've got zero heat capacity, because you
changed the temperature a little bit, and everything is
still stuck back here.
So that's the vibrational heat capacity of a solid.
That's what it looks like.
And at moderate temperature, this doesn't have to be very
much bigger than epsilon zero.
You're already in, essentially, the limit.
The high temperature limit.
And now for molecules, that limit isn't usually reached at
room temperature.
Here's a kind of calibration. kT at room temperature is
about equal to 200 wave numbers.
So molecular vibrations, you know, you've taken IR spectra
they're typically on the order of 1,000 wave numbers or so.
kT isn't bigger than the vibrational energy.
But a crystal lattice, you know the vibrations are the
acoustics vibrations.
And those are much lower in frequency.
If you have an atomic crystal, it just has the sound
vibrations at all the different
wavelengths that are available.
They're never very high.
So it's easy to get into the high temperature
limit, in that case.
Where you basically see a temperature
independent heat capacity.
OK.
By the way, this was actually used commonly to determine the
molecular weights of molecular crystals.
Because you've got a factor of the number of moles in there.
If you ask how big is the heat capacity?
Well it does depend on how many moles of
material you have.
Because it depends on how many atoms.
That control how many modes there are in the crystal.
How many vibrational modes.
Because each atom has 3N degrees of freedom.
So if you just weigh the whole crystal, and then you measure
the heat capacity, you know how many moles there are, and
now you know the weight, you can figure out
the molecular weight.
OK.
So that's the low temperature limit.
And now I want to talk a little bit further about the
high temperature limit.
And in particular, I want to talk about both the heat
capacity and the energy that we've seen.
Namely, this thing.
Of course it's really the same result for the energy and the
heat capacity.
They're obviously intimately connected.
So these are the high temperature limits.
Now it turns out that that high temperature limit doesn't
just obtain for vibrations.
But it's also the case for molecular rotations,
translations.
All these low energy degrees of freedom.
So let's see why.
It has a name, that result.
It's called the equipartition of energy.
Sometimes it's called the classical equipartition of
energy theorem.
And what it says is one half kT per degree of freedom
equals energy in the high T limit.
And in particular, for translation, so E
translational is 3/2.
Well, NkT for N atoms.
Or molecules.
E rotational is, now it depends how many rotational
degrees of freedom there are.
If I've got a linear molecule, there are only two.
It can rotate this way, and it can rotate this way.
If I've got a non linear molecule, parts all over the
place, it has three unique axes.
It can also spin this way.
And that's a rotational degree of freedom.
Of course the linear molecule wasn't like that.
Nothing is moving when you do that.
So it's either NkT, linear.
Or 3/2 NkT.
Non linear.
Great.
And then E vibrational is NkT per vibrational mode.
That's because vibrational energy is potential and
kinetic energy, and it's 1/2 kT each.
Why does all that stuff happen?
Turns out, you can see why pretty easily.
All those degrees of freedom.
Those classical degrees of freedom.
If you say, let's write out an expression for the energy.
Classical.
You know it's 1/2 m v squared.
Or it's 1/2 k x squared.
Or for rotational energy it's 1/2 I omega squared.
You see a functional form emerging here?
That looks similar in all these cases?
You know, 1/2 times some constant, times
some variable squared.
Well, now let's look at our expressions for the energy.
Average energy.
We're going to sum over all the energies of epsilon i, e
to the minus Ei over kT.
Over sum over i, e to the minus Ei over kT.
That's just how we originally derived the average of energy.
In other words, it's the probability of each state
times the energy of that state.
Summed up over all the states.
Well, now we have an expression for the energy.
It's one of these things.
It's something times a variable squared.
So, we'll use some general functional form, a y squared.
And now, let's assume we're in the high temperature limit.
And we've seen what that means.
It means kT is big, compared to the
separation between the energies.
When that's the case, there are lots of these
terms in the sum.
We can convert them to integrals.
We can forget about the fact that the
energies are discrete.
We can say look, they're so close
together, compared to kT.
That we can turn the sums into integrals.
So then we have integrals instead of sums.
And here's our energy.
It's a y squared.
Whatever that is.
It could be this, it could be this, it could be this.
And it's not going to matter. e to the minus a y squared
over kT over, integral over e to the minus a y squared over
kT. dy, dy.
OK.
And here's what's going to happen.
If you do this integral by parts.
This one.
What ends up happening is it gives you this integral times
a certain number.
This is going to come out of the integral.
And so it will turn out.
And it's straightforward to do it.
It's in the notes.
It goes through.
But I'll just write the result here.
The result is that you get, in this high temperature limit
where you've gone to the integral form, you get 1/2 kT.
And it will always be the case.
All you need to know is the form of the energy.
As long as that's the case --
Remember y is just a variable of integration here.
That's not going to be preserved.
This comes out because of what happens here.
So what that's telling you is whenever you have an energy of
this form, and you're in the high temperature limit, then
you get this classical
equipartition of energy result.
So, for translation, of course there are three separate
degrees of freedom.
For velocity in the x, y, or z direction.
For rotation, if it's a linear molecule, let's say there are
two separate degrees of freedom.
You have to keep track of how many degrees of
freedom there are.
But that's all you have to do.
That's enormously powerful.
It means that without doing anything, I know the average
translational energy of the molecules in this room.
Because of course I'm certainly in the high
temperature limit.
With respect to translational energy levels, they're really
closely spaced.
Same with rotations.
Now at room temperature vibrations, forget it.
I have essentially no vibrational energy.
I'm at the low temperature limit for the molecular
vibrations of nitrogen or oxygen.
Those are high in frequency.
So much higher than 200 wavenumbers.
But for each molecule, I have 3/2 kT of
translational energy.
Linear molecules, I have kT of rotational energy.
Without doing any work at all.
And since that applies to most molecules at room temperature,
it's an incredibly useful, very, very general result.
OK.
Next time we'll do a little bit of chemistry, and look at
phase transformations.