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>> So, let me remind you, we're going to take a quiz
on Friday during the [inaudible] of class.
They'll be a stack of Scantrons right there at the front
of the lecture hall there.
Pick them up.
Pick up one please and find a seat.
Turns out there's a 116 of you
and there's 240 seats in this lecture hall.
So what that means is we don't need two forms for this exam.
Every one -- we can sit every other seat in this room
and there shouldn't be a problem with that.
Okay? So if you would, please make sure
that you don't sit exactly next to somebody else.
Leave an empty seat between you and the next person next to you.
Okay? We shouldn't have any problem.
There's plenty of seats in here for us to do that.
Okay, it's going to be five short problems keen
on concepts covered in lectures one and two,
homework one and/or discussion study guide one which is posted
to the website on Tuesday morning.
All right?
That's the discussion study guide you're going to be talking
about in discussion all week.
I also posted three quizzes from last year.
I've already posted to the announcements page
so you can get an idea for what these problems are going
to look like.
One of the things you'll notice is that you have to be able
to calculate a factorial using your calculator.
Might be the first time you've had to do that in any class.
I don't know.
But I know when I had to calculate factorial
on my calculator I had to pull out the owner's manual
and figure out how to do it because there's no key
that says factorial on it [laughter].
Okay? So Friday morning would be the wrong time
to try and figure that out.
Look at it before that.
Yes?
>> [Inaudible].
>> I'm sorry.
>> [Inaudible].
>> There is no calculator policy.
Anything that you can carry in here is fine with me.
All right?
Not only that but this quiz is open book, open notes.
But if we -- I escape to imagine you're going to have the luxury
of time to sort through your notes and pause and look
through the book and think about what you might want to --
time is going to be an issue for you on this quiz.
You're going to have to know what you're doing,
find the right equation,
if you don't remember it, boom, knock it out.
Right? There isn't going to be much time
to do problems on this quiz.
You're going to have to know what you're doing.
Okay? So write anything down if you want on a piece of paper.
If you want to have a list of constants or conversion factors,
or equations that have been involved in the stuff
that you've been studying this week,
put that on a piece of paper.
Anything you want.
Yes?
>> Since you emphasized not printing
out the lecture notes should we bring our laptops for like --
to have the lecture slides [laughter]?
>> You are not going to have time to click
through the lecture slides [laughter].
What was on slide 40-?
No. You are not going to have time to do that.
I -- that would be a waste of time for you I think.
All right, any questions on quiz one?
Yes?
>> What if I have a discussion on this Friday?
[Inaudible] you said [inaudible].
>> Yes.
>> Well that's why I posted
that discussion study guide on the website.
>> I cannot get the answer before [inaudible].
>> Well, it turns out you can go to any discussion session.
Right? You know, you might be assigned
to the Friday discussion session but you can go to the one
on Tuesday, Wednesday, Thursday.
There are six of them.
So if you want to go to an earlier one
because you're concerned that -- and you ought to be...
>> [Inaudible].
>> Okay. Other questions?
Okay. We're talking about the Boltzmann Distribution
Law today.
It will be about microstates and configurations
and W. We're going to derive the Boltzmann Distribution Law using
a slightly different approach from the one's that taken
in your book because you can read what's
in your book already.
You don't need to hear me regurgitate that to you.
I'm going to introduce you to partitions,
we're going to do some samples.
Boom! This is a molecule.
All right?
It has evenly spaced vibrational states to zero order.
We're going to assume it's a harmonic.
Vibrations of this molecule are harmonic.
Okay, now it's tedious for me to draw this monstrous thing.
I'm going to draw something smaller that looks like this.
This is three molecules.
We're going to label them A,
B and C. I've condensed these three molecules
into this little notation that I'm showing you here now.
These are the quantum numbers for the vibrational states.
B equals zero.
That's the ground vibrational energy level.
One, two, three.
You remember all this from quantum mechanics.
I'm calling this a three-dimensional array
or three-dimensional lattice because I want to emphasize
that these three molecules are distinguishable
in terms of their position.
In other words I can tell where A is.
I'm not going to confuse it with B. It's this --
because my lattice has A next to B next to C, I know where A,
B and C are and they're not going to move around.
In other words these molecules can be labeled
and kept track of.
That's an important point.
Because later on [inaudible].
But for the time being we know where these molecules are
and can call them A, B and C. We're not going to lose them.
So, now imagining a three dimensional array
of these molecules.
[Inaudible] lattice [laughter].
To keep it simple lets say that we just got three molecules; A,
B and C. Now let's add three quantum energy
to these molecules.
All right?
There's three ways to do that but all the energy
in molecule A. We could put all the energy
into molecule C. We could put all the energy
into molecule B. That's one way to configure the energy.
Put all the energy into one of the three molecules.
We could also put two quantum into one of the molecules
and one into the other.
Leaving the third with no energy in it at all
and there are six ways to do that.
Finally, we could just put one quantum of energy into each
of the three molecules and there's only one way to do that.
I think you can see that intuitively.
Now notice I'm not labeling these molecules any more.
Here they're labeled A, B and C but I'm going to get rid
of that because I'm lazy.
But don't be confused.
I've always talked about three molecules here.
Somebody sent me an email that asked me about that.
It's a good question.
A, B and C are still here.
Okay, we're going to call each one of these guys a microstate.
And we've got this short form notation that we could use
to identify not microstates but families
of microstates, configurations.
All right?
These curly brackets are going
to help us identify different configurations
that these microstates can be in.
This first digit inside the curly brackets is going
to be the number of molecules in the ground statement.
The next digit is going to be the number of molecules
in the first excited statement and so on and so forth.
That's what this notation will mean.
So you can see, these guys are all in the same configuration.
How do I know that?
There's two that have the ground state occupied
and one has the excite -- the third, equals three occupied.
Right? That's also true here; two and one.
Also true here, two and one
so they all have this configuration here.
Two, zero, zero, one.
And these guys analogously; one, one, one, zero.
Boom. And of course there can be any number of zero's.
Okay? This guy is zero, three, zero, zero.
So we have a notation that we can use
to identify these different configurations.
That's all this is.
There's no interesting science here.
Okay, now short of figuring out
and drawing every single configuration we need a formula
that allows us to calculate how many microstates there are
for each configuration.
Let's see if we can derive one from the simple model
that we've looked at already.
Because instead of having three molecules we're going
to have 10 to the 23.
Right? So it will not be practical for us to think
about these molecules in the terms that we've been using
so far but [inaudible] about three.
Okay, so as I build up -- let's think about configuration two.
You'll recall configuration two has one molecule --
has two quantum molecules.
[Inaudible].
All right?
As I build up that configuration I have to --
I can put that first V equals two into any one
of these three molecules; A,
B and C which are no longer on here.
Right? It can go into any one of the three molecules.
I put it in A but it could have gone into B
or it could have gone into C. In terms
of where I put the second one, once I've put
that one there I have two possibilities
of where I can put the second one, and the third one then,
once I put that into A and that one
into C the third one can only go into B. The last parcel
of zero goes to the remaining one molecule.
Okay? And so if I want to know how many versions
of this are there, it's three times two times one.
Three times two, times one, that's just three factorial
that means there should be six ways to do this and there are.
Okay? What about configuration one?
Configuration one was three quantum and zero zero.
All right?
Starting with the first, I can put the first one at A,
B or C. This time I relabeled these guys.
Then the next one they can all go into B here.
In this case it can go into A
because I've already put the three [inaudible] into B.
In this case can go into A
because I've already put the three [inaudible] into C. Okay?
And then finally it remains --
it's going to go into the place that's left over.
Here it can only go into C. here it can only go
into C. here it can only go into B. Now,
in principal the counting in this case works exactly the same
as it worked in this case.
Here we got six possibilities, but here I think we can see
that there's only the possibility
of having three microstates that have this configuration.
What changed?
What's the difference between configuration two
and configuration three?
Or configuration one rather.
Well the difference is that this state is degenerate.
V equals zero.
It's go two molecules.
And in other words there's two verbally distinguishable ways
to create that microstate.
But this guy here and this guy here
or this guy here makes no difference.
I'm going to get the same microstate when I do that.
so, when I count microstates, even though there's
in principal six ways, just like there was
in the first configuration, three times two times one,
we're going to need to reduce the total number
of configurations by the degeneracy factor.
Okay? And in this case there are two molecules
that have the same energy and so it will turn
out to be the same case that we need to reduce
by dividing by two factorials.
So instead of three factorial or six,
we're going to have three factorial divided
by two factorial.
That's six divided by two.
That's three.
There should only -- three ways
to make this state and there are.
Now it may not be [inaudible] obvious that is has
to work this way and it's rare for me
that anything's [inaudible] obvious
so I work [inaudible] example like well,
this turns out to be true.
It all works this way.
In other words you always need to divide by the number
of states that have the same energy
that if there's multiple states
that the same energy you take the product.
Two factorial times three factorial times four factorial
for example.
We'll do an example like that in just one second.
In this case there's three guys.
Right? In configuration three.
That's three factorial and three factorials divided
by three factorials, one.
there's only one way to do that obviously.
Okay, now you can do more example like this
and convince yourself that this is true as I had
to when I first looked at this formula.
So, in other words we do have a general formula that we can use.
this is it.
that's the number, total number of microstates.
That big W there.
That is the number of molecules.
That's the number of molecules that have no energy,
one quantum of energy, two quantum's of energy and so on.
So, consider system of eight molecules containing four
energy quanta.
Eight molecules?
Yeah. One, two, three, four, five, six, seven, eight.
Should have made it a little wider.
Right? So there's eight molecules there.
Four energy quanta.
Really? One, two, three, four, and these are all zeros.
All right?
What's it -- what's the shorthand notation
for that configuration?
Well there's five guys that have zero.
Two guys that have one.
one guy that has three and all the rest
of these are going to be zeros.
Problem one on quiz one [laughter].
Okay? What's W?
Problem two [laughter].
All right, eight factorial.
One, two, three, four, five.
Here's another one that's got two and one.
what's zero factorial?
>> One.
>> yeah. Okay, now if you can't --
if your calculator will not do that --
of course you can do it long hand.
Right? One times two times three times four times five.
You know, so eight is sort of doable.
But when you get to like 27, you know, you really need to figure
out how your calculator does a factorial.
All right?
You will need to calculate that factorial
on Friday during the quick.
Okay? Boom, 168 ways to generate that configuration.
Pretty useful formula for us [inaudible].
Now, let's look at this a little bit more closely
because there is something hidden here
that is very, very important.
All right?
That we haven't noticed yet.
Right, this is a fundamental fact of nature
that you have never seen before.
Almost certain.
Right? Really?
Related to this?
I think so.
Here it is.
All right, 10 molecules now.
One, two, three, four, five, six, seven, eight, nine, 10.
But all five quanta energy, now let's make it five in this case.
Put all five into that guy.
All right?
How many microstates are there?
Ten molecules.
Nine of them are all in the energy state down here
so I'm going to divide by nine factorial.
There's 10 ways to generate this configuration
and you can see immediately that that has to be true.
Right? Ten ways to do this.
Ten microstates.
What about if I put four in one and one in another?
All right, what's the W going to be for that?
Still 10. I'm just calculating factorial;
10 times nine times 10 factorial is eight factorial.
Right? There's eight guys that have this ground state energy
so I'm going to have eight factorial there, one factorial,
one factorial, one factorial, a bunch of zero factorials.
So, that's 90.
Or how do I know that?
Eight factorial cancels with eight factorial and I'm left
with nine times 10 is 90.
So there's 90 ways to generate that microstate right there.
Or 90 microstates that have that configuration.
You with me so far?
Now, let's work out all of the other possibilities,
which we can readily do.
We understand how to do that now.
All right, and here are the number of microstates.
That's the one we calculated first, then we calculated 90.
There's another 90 and this guy right here is 360,
and this guy right here is 360.
This guy is 840.
All right?
Eight forty has two quanta and one.
One quanta and three and zero and six.
All right?
And then there's the 252.
That's this guy right here.
Right? Those are the only possibilities that exist.
What's profound about this?
Well, that number is bigger than all the rest of these numbers.
Configuration six -- it's just randomly labels six.
Six doesn't mean anything.
It's favored by more than two to one compared with any
of these other configurations.
Okay? Favored by more than two to one.
It's special.
That's not super special.
I mean it's only special by a factor of two or so.
All right?
Now I'm just plotting W versus the configuration number,
which remember is just arbitrary.
Don't worry about --
the configuration number doesn't mean anything.
But I'm just pointing out here's that configuration 840.
All right, it's special.
Now, how special is it?
well, we've only got 10 molecules here.
Right? What if we had 10 to the 23rd?
right? We've only got 10 here.
What if we had -- what we're dealing
with in chemistry is not 10 molecules very often.
Once in a while.
Most of the time we're dealing with 10 to the 20,
10 to the 23rd, 10 to the 24th.
Right? Ten to the 23rd is a mole.
Right? We're dealing with fractions of a mole or a mole.
All right?
So it's that level that we want to be able to understand here.
We want to be able to understand the statistics
of large molecules.
So let's make the system a little simpler.
Let's make it a two state system.
Heads and tails.
Right? Only two states for the molecule now.
Excited or not excited?
Let's see.
Now our W, here's the number of coin flips.
Right, because that's just a way to think about molecules.
Okay? Nothing has changed.
I'm just going to devaluate the system multiple times.
That's the number of heads, that's the number of tails,
that's the number of microstates.
For example, if I flip four times I could get heads,
heads, tails, tails.
I could get heads, tails, heads, tails.
I could get tails, heads, heads tails.
I think this notation is now obvious to you.
All right?
If I want to know I'm going to get two of each, all right,
that's a configuration.
How many microstates are there for that configuration?
Boom, six.
How do I know that?
Four factorial, four flips, two of each,
six possible ways to do it.
Boom, there they are.
Here are all the different ways that you could that.
All heads, all tails.
What's the probability of that?
One, two, three, four, five, six, seven, eights, nine, 10,
11, 12, 13, 14, 15, 16.
One part in 16.
You got one change in 16
of getting all heads flipped [inaudible].
Okay? You with me on this?
All right, here's the [inaudible].
On this axis I'm plotting the number of microstates divided
by the total maximum number of microstates.
Right? In other words the maximum number of microstates
for different numbers of coin flips
or different numbers of molecules.
All right?
Here's six coin flips, 10, 20, 100, 1,000.
Okay, I'm -- of course the number of microstates goes way
up as the number of molecules goes up.
But let's normalize by the maximum number of microstates
that can exist for each one of these systems as it gets larger.
Because we want to see what happens
to the shape of this distribution.
All right?
What happens is very interesting.
Here's six coin flips.
Here are the data points that I just showed you.
All right?
Plotted on here.
All right?
And now if you increase that 10, 20, 100, 1,000;
look what happens to this distribution.
All right, it becomes needle sharp.
Right? It becomes almost infinitely narrow.
It's not infinitely narrow but it is darn narrow.
That's only a 1,000 molecules.
We're not dealing with a 1,000 molecules very often.
We're dealing with 10 to the 23rd.
what is that distribution going to look like for 10 to the 23rd?
it's just going to be a needle!
It's just going to be a delta function
at this preferred configuration.
What is on this axis here?
These are just different configuration for the system.
Right? This axis is just different configurations.
This axis is the normalized number of microstates.
All right?
What am I saying here?
I'm saying there is one configuration
or a small ensemble of configurations that dominate
as the system gets bigger.
That is a very surprising conclusion.
Right? Are you telling me that I can't have four heads?
No, you can't have four heads when you've got 10
to the 23rd coin flips.
In other words you can have 10 to the 23rd heads.
Now, that's not going to happen.
Well it's sort of intuitively obvious.
Isn't it? all right, but it's less obvious
that you can't have any other possibility either.
In other words there's really only one possibility it's going
to be exactly probable it turns out to be half heads
and half tails in case of coin flips.
All right?
But more subtle if it's multiple states available
to the molecule.
More subtle.
So we can -- the main point here is that we need to --
we really need to understand this configuration right here.
What is that?
what are its properties?
If we can understand that configuration that is sufficient
for us to understand how this chemical system is going
to function.
What the properties are.
We only have to understand that guy.
We don't have to understand this one or, this one or,
this one or, this one, or any
of the other 10 configurations that may exist here.
We only have to understand one or a small ensemble of them.
Right? All the properties of the other systems are going to be
so rare that they're not going to contribute to the behavior
that we observe in chemistry for macroscopic amounts of stuff.
Imagine if that's 10 to the 23rd and not 1,000.
Boom.
What does the N stand for in Cornhuskers Stadium?
Who knows the answer?
>> Nebraska.
>> [Laughter] [Inaudible] from like last year [laughter]?
>> Knowledge is the correct answer [laughter].
Football players [laughter].
Okay. Knowledge of this highly preferred configuration equates
to knowledge of the system as a whole.
We only have to know about that guy.
We only have to know about him.
all the rest of these guys we don't need to know about.
Okay, this gets sharper.
That's the whole point.
All right, now your book derives the Boltzmann distribution law
by finding the W where -- the maximum W. Right?
Now I think you can see that at the maximum W by golly,
that's the one we want.
That's the preferred configuration right there.
So if you find the maximum of any one
of these curves you're going to locate that point
and that's what you want to do.
Right? But there's another approach that one can take
and that is that the derivative here is zero.
Isn't it? the derivate with respect to configuration of W
or W max at this point is zero.
In other words if I draw a tangent there I'm just going
to get a horizontal line.
That's another way to derive the Boltzmann distribution law.
That's what I'm about to do.
Today find the configuration for which W does not change
when the configuration
of the system is infinitesimally altered.
In other words, if I shift the configuration left or right
of that peak point slightly, all right, and W doesn't change
because the derivative there is zero,
why I've located the peak of that distribution.
I've located that magic configuration.
That's the whole point of the Boltzmann distribution law.
Consider -- okay, so we're going to do this [inaudible].
Ready? All right.
Consider an isolated, in other words,
number of molecules is constant.
Q; what's Q?
Q is the total number of energy quanta in the system.
So this -- when I say isolated I mean we're not going
to change the number of particles.
We're not going to change the amount of energy.
We're going to leave that constant.
Macroscopic.
Large number of molecules.
[Inaudible] and oscillators
that share a large number of energy quanta.
Large number of oscillators, large amount of energy.
Neither one of those things is going to change.
Which configuration is preferred?
How do -- in other words where is the peak in that curve
that I just showed you?
Where is it?
Right? Let's consider just three states.
There could be thousands.
Let's consider just three.
Let's label them L, M and N, randomly.
If we want to know how many microstates there are we use our
standard formula.
That's the total number of molecules and if this is L,
M and N there must be an A, B, C, D and so on.
All the way to L, M and N, and then O, P,
Q, R, S, T and so forth.
Okay? Now what we're going to do --
you know, this is the configuration
that we're starting with.
Let's shift if slightly.
All right, and see if we can find the configuration
that doesn't change W. that's the strategy here
in this derivation.
All right, we're going to shift it infinitesimally
and we're going to maintain the energy constant while we
do that.
how are we going to possibly -- well, here's what we do.
Energy on this axis here.
Here's our states; M, N, L. right?
Here's the energy difference between M and L. the energy
of M minus the energy of L, that's the difference between M
and L. here's the energy different between M and N. Just
that difference right there.
Right? Graphically this is our situation.
There's a bunch of states below this
and a bunch of states above this.
Okay? So now what we're going to do is we're going to shift.
Right? We've got some number at state N, some number at state M,
some number at state L. now we're going to shift.
Boom! We're going to move Q guys from M into N and P guys
into L. we're going to choose P and Q
so that the total energy here doesn't change.
All right?
We're going to choose P and Q
so that we maintain the same total energy.
Right? But we're going to move some guys from here to here
and some guys from here to here and I think you can see
because we're going -- we're moving some of the lower energy.
We're moving some to higher energy.
It's reasonable that you could maintain the same total energy.
Right? Yes.
>> [Inaudible].
>> No, they're not supposed to be evenly spaced.
But thank you for asking that question [laughter].
Okay, so what concerns energy?
Well Q times that difference M minus L. M minus N rather.
Q times that difference minus P times that difference has
to be equal to zero, so we're going to choose P and Q
so that they satisfy this equality right here.
Now, so before, here's our number of microstates.
We've already said they're given by this equation.
Before the shift, all right, this equation is true.
After the shift I've added some number of molecules
to L. I've removed molecules from M, moved them to P,
moved them P to L and Q to M. all right?
And got larger by an amount Q, and so this is the shift here.
Describing these three equations right here.
All right?
And what I'm saying is that we want to consider the case
where the before equals the after.
That we want to find the microstate
where after we do this infinitesimal shift W
doesn't change.
That's the whole point.
All right, so if the rate of change of W
with configuration is zero that means that these guys have
to be equal to one another.
All right?
In other words, right, the peak of this curve.
Right? So W before equals W after if and only
if we are right there.
If we're here and we shift
in either direction it's not going to be true.
They'll be changes to the configuration.
Right? It only doesn't change configuration up there.
Okay. So, that has to equal that.
and if you work out the math we're going to skip some steps.
Canceling or rearranging terms.
You don't have to do this on quiz one.
Okay? Take my word for it.
lots of cancellations occur here
and when you're done this equality holds,
but a more useful version of this is
if you just take the natural logarithm
of both sides here you get this, right, P times log of NL
over NM equals Q times log NM over NN.
Doesn't look super useful.
Okay, and substituting from the expressions
for energy defined earlier we have, all right,
I've plugged this is for Q, I plugged this in for P
and I get these guys right here after actually I cross multiply.
And then you've got this equation right here
that looks completely useless but you have to realize is
that M, N -- L, M, and N are just random letters.
I could have chosen any three states
and done this [inaudible].
Okay? And so it turns out to be the case
that these two things can only be equal to one another,
you know, if I -- [Inaudible], is that the right word?
Permutated these states.
Right? If it's the case that I could have used A, B and C or X,
Y and Z and gotten the same result that implies
that this has to be a constant.
Right? Or this has to be a constant.
And that may not be intuitively obvious
because it's really not intuitively obvious to me
but it turns out to be the case.
This can only be true if the left and right sides
of the equation are a constant.
They do not depend on exactly which states are involved,
which states we chose.
We chose L, M and N. Let's call this constant beta.
In other words, I'm going to take one of these guys
and just set them equal to a constant.
We're going to call that constant beta.
Okay, so now I can rearrange this.
I can multiply by that energy difference.
Boom, it moves over to the right hand side.
So I've got log on N of N over N of I.
I chose this N and I now to label these states.
Since it's completely general expression it doesn't matter
which two states you choose.
Right? Subject to the constraint
that that's the energy difference
between the two states right there.
That's the only constraint.
Okay, now if we choose one of these guys
to be the ground state, let's say that guy, I.
right? So instead of N sub I, maybe that's zero.
Let's assume that the energy of the ground state is zero.
In other words that there's no zero point energy.
Okay? So instead of having E sub I minus E sub N,
we're just going to have minus beta times E sub N
because E sub I is equal to zero.
It's the ground state.
All right?
So we get this expression here and if we just take E
to the both sides, take the exponential
of both sides rather, we get this expression right here.
This is the famous Boltzmann distribution law.
What it says is the ratio of the populations of these two states;
state N which is elevated by energy E above the ground state.
Right? The energy of the state is E higher
than this ground state.
All right, we can calculate the population
of that excited energy level using this equation right here
where beta turns out to be equal to one over KT,
where K is Boltzmann's constant.
Is that on the next slide?
No. beta is one over Boltzmann's constant times temperature.
One over KT.
Leonard Nash who is my hero says,
for any isolated macroscopic assembly of species,
with any kind of spacing between the states,
any kind of state spacing,
the predominant configuration had been fully defined
by explicit functional relations between the energy
and the population of each quantum level.
In other words this equation hold true,
not for every configuration, right,
for the only one that matters.
Right? The one that dominates the number
of configurations that are out there.
Remember we have a large system it's dominated
by one configuration or a very small number of them.
All right?
What we're saying is that this equation here characterizes
that preferred configuration.
That preferred configuration is very important.
Our -- any chemical system is dominated by it.
but there are all these other ones that we're really going
to ignore, and we ignore them in all thermodynamics.
At least all that I know of.
Okay, now we also need a different form
of the Boltzmann distribution law and you can look
at these slides later,
and there's a little derivation of that here.
Total number of molecules is equal to the number of molecules
in the ground state, number of molecules
in the [inaudible] state, number of molecules
in the second excited state.
I can write a summation formula in terms of that.
Whatever it is.
These are summation over all of the states that are occupied
and so we can write one of these Boltzmann distribution laws
for each one of these terms.
Look at that.
Isn't that nice?
All right, N2.
If that's N2 and that's got to be E2, yeah,
sure enough there it is.
Right, N2, E2, N zero, yes.
Let me just click through this.
So solving for N zero we have this equation right here,
and I can then plug this N zero into this N zero right here.
Well, let me do that backwards.
Put solve for N zero here, plug that into this guy.
I get this equation right here.
This is your equation for 2.68.
you don't actually have to do this derivation yourself.
I'm just showing you where this [inaudible] came from.
It's not totally obvious where this thing came from [laughter].
All right, you can just derive from the simplest version
of the Boltzmann distribution law.
All right, what matters is how we use this equation.
This ting in the denominator looks innocuous but it turns
out to be super important.
right? It's something called the partition function,
and we're going to keep coming back to this.
As innocuous as this seems this is the central idea
that we keep returning to in statistical mechanics.
What is that thing?
It doesn't look like anything in particular, does it?
It's the summation of this exponential where we're going
to put a bunch of different energies
in here and sum them up.
I mean how important could it possibly be?
Okay, so this is the molecular partition function
for that particular case that we're talking
about where we've got evenly spaced states in a ladder.
[Inaudible] partition function for that piece.
The molecular partition function,
we usually call it Q. There is two versions it turns out.
There's a version that contains the degenerate [Inaudible] the
degeneracy is the number of molecules
that have the same [inaudible] in a particular state.
Right? So the ground state has to have --
happens to have four molecules in it.
No, let me see that a different way.
If there are four ground state energies
that all have the same energy of zero,
the degeneracy would be four.
We'll do an example.
Okay, this is your equation 13.9.
two versions of this -- I never learn anything until we start
to do exams [laughter].
So let's do some examples.
So [inaudible] start to make sense.
What are the relative populations of the states
of a two level system when the temperature is infinite?
[Inaudible] we got two states, right,
temperature is getting very high.
What's the [inaudible] population of those two states?
[Inaudible] got a lot of molecules in it.
Right? This has to be a case that we have all the time.
Here's our equation.
All right?
For the populations of state I.
two states.
So let's call one state two and one state one.
Let's write one of these equations for state two and one
of these equations for state one.
All right?
What's the difference between them?
I've got E2 here for state two.
I've got E1 here for state one.
Otherwise the denominator is the same station.
It's identical.
Okay? So if I cancel terms the --
I'm asking about two states
in the same system so big N is the same.
There's only one system here that has N molecules in it.
Right? This thing in the denominator,
this partition function, that's the same
between these two terms.
Right? The only thing that's different is this exponential
so I can just pull that out, everything else cancels
so I've got this exponential for energy two corresponding to N2
and this exponential for energy one corresponding to state one,
and so I can simplify that.
All right?
And here's what the resulting exponential looks like.
Okay, and remember data is one over KT
where K is Boltzmann's constant.
All right?
And so if I want to calculate this I just plug in one
over KT but my T is infinite.
That's what that symbol is supposed to represent.
Okay? And so if the denominator becomes very large I'm going
to have E to the zero.
E to the zero is one.
So what we conclude is that the population
of these two states will be equal if the thermal energy
of the system is way high then either
of the energies spacing between them.
I think that's already counter intuitive.
Right? Wouldn't you expect [inaudible] state
to be preferentially occupied by temperature?
I mean in my simple way of thinking
about it that's what makes sense to me but that's wrong.
Right? If the energy, the thermal energy of the system,
right, which is dictated by the temperature,
if that is high compared to the energy difference
between these two states, all right,
they will be equally occupied.
That's an important thing to understand lots
of different kinds of physical chemistry that we're going
to be getting to later on.
Let's do another example.
A certain atom has a threefold degenerate ground level.
The nondegenerate electronically excited level
at 3,500 wave numbers and a threefold degenerate level
at 4,700 wave numbers.
Calculate the partition function of these electronic states
at 1,900 degrees Kelvin.
Problem three or four.
Here's the situation.
What do these -- what do these words mean?
A threefold degenerate ground level?
Boom! Energy zero.
An electronically excited energy level, 3,500 wave numbers.
Boom! Thirty-five wave numbers.
A threefold degenerate level at 4,700 wave numbers.
Boom. Right?
Calculate the partition function of these electronic states
at 1,900 degree Kelvin.
Calculate the partition function.
What is the partition function?
It's this guy.
This is the denominator of that expression we were looking
at earlier.
I'm using the version where there's two partition functions.
I'm using the version that has the degeneracy in it
because we've been told4 about the degeneracy.
Here's the partition function.
Here's the version with the degeneracy.
Here's the version without it.
We need this one because we have degenerate states in the system.
Okay, so what is the summation mean?
It means that we're going to have to add up --
we've got three states,
we've got three terms in our summation.
Gerund state, first excited state, second excited state.
Let's figure out what those terms are.
The degeneracy of the ground state is three.
Boom! Okay, so that's three times the exponential minus beta
times zero because the energy of the ground state is zero.
Plus none degenerate [inaudible] times 3,500 [inaudible] plus
triply degenerate through the minus state
of 4,700 weave numbers.
Three terms because there's three states.
Three energies.
Oh well seven states.
Three, one, three.
Right? Okay, now one of the things that we're going to have
to do constantly is move between different energy units
in this class.
Wave numbers, jewels, EV.
All right?
The way that I do this is clumsy but reliable.
All right?
I remember two constants.
The conversion from wave numbers to EV that's 855.5.
Wave numbers per EV.
And, the conversion from Jewels to EV is 1.602 times 10
to the minus 19 jewels per EV.
If you remember those two conversion factors I submit
to you that you can obtain any unit that happens to be a jewel,
EV or a wave number [laughter].
For example, 3,500 wave numbers.
What is that in jewels?
Why do I care?
Because Boltzmann's constant is almost always 1.3 at 1/10
to the minus 23 jewels per Kelvin.
Boltzmann's constant is in jewels.
So the other energies, if they're in jewels, very helpful.
Because then you have to know Boltzmann's constant
and different units.
Always use 1.38 times 10 to the minus 23
for Boltzmann's constant
and change the units of the other stuff.
So 3,500 wave numbers.
What is that 3,500 wave number divided by --
you know, go back to your chem one training and make sure
that you cancel units and get the right ones.
All right?
Because we're going to be moving
between energy units all the time,
and this is the way I do it.
Right? Divide by 865.5.
wave numbers cancel so that product is the energy of EV.
That is multiplied by 1.602 times 10
to the minus 23 jewels per EV.
Now I get energy in jewels and then I divide by KT.
And I get 2.649; 2.649 is that number right there.
There will be a minus sign in front of that right there.
Right? And so these three terms then are going to be three
because that's just Z to the zero times three.
That's three times one.
That's three.
That guy turns out to be .07069.
that guys turns out to be .0855.
add them up I get 3.156.
Does that make sense?
What is the partition function?
It's the number of thermally accessible states.
How much thermal energy is there in this system?
What is KT?
KT turns out to be 1,300 wave numbers if you convert
that number 1,900 Kelvin,
convert it into units of wave numbers.
It's 1,321 wave numbers.
All right?
And so what this calculation tells us is
that there should be a little more
than three thermally accessible states in this system.
Well there are.
These three states are thermally accessible.
They're the ground states for goodness sakes.
All right?
And there's a tiny probability
that these other states will be occupied.
That's what this number indicates;
3.1 thermally accessible states.
Yeah, because there's not enough thermal energy
to significantly populate these excited states.
Thermal energy's too low.
Okay, so we'll review this on Friday. ------------------------------20cc1f12e5ee--