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>> So today we're going to be going
over some mid-term review questions.
It's going to be a lot on how the discussion sections work.
We're going to go over some questions
that you guys have all seen before so
and this is the review.
So the mid-term is going to cover chapters 13 and 14
and today we're going to go over a good portion of Chapter 13
and yeah, so let's get started.
So there's going to be two problems on this exam.
One of them is going
to be covering the statistical mechanics of a particular system
and the other one is going to be covering thermodynamics
of gases, basically the Equipartition Theorem
and there's going to be a ten-point extra credit problem
which it's extra credit so we're not going
to divulge the details to you.
And so here's the summary
of all the partition functions translational,
rotational, vibrational states.
Some things you'll want to keep in mind is that you'll want
to remember that the mass that you include
in the translational partition function is in kilograms
and also for the most part just remember to mind your units.
If the units aren't matching up and cancelling out in the way
that they should go back and rerun through your answer
because the numbers are sometimes hard to get a hold of.
The other thing you guys might want to remember
and maybe go back and look at is the symmetry in Lecture 8
where we go over how to calculate the symmetry
for any particular system
and for the electronic partition function there's really no
formula, you just add up the degeneracies
and the contribution from each state.
And beta, again remember is just 1 over KT,
so anything that you see that's 1
over KT you can just translate into beta.
So for the first example let's take a look at this problem.
We've seen this before I think in Discussion Section 1.
Where we're talk about NO
and it's doubly degenerate electronic state
at 121 wave numbers and a doubly degenerate ground state
at zero wave function or zero wave numbers.
So the first thing and this is obviously not going to be well,
it could be asked on the test but we have to plot
out the partition function as a function of temperature
from zero to 1000 K. The next part
of the question asks the term populations.
What is the population in the excited state
and what is the population in the ground state
and the electronic contribution
to the internal energy at 300 Kelvin?
So right here we can see the simple NO diagram depicting the
system with the two degenerate ground states
and the two degenerate excited states.
And so the partition function for the levels
as seen here is the degeneracy
and the contribution from the energy.
So as we can see in this partition function we're going
to have a contribution from the ground state
which is doubly degenerate.
We'll set E to equal zero and the contribution
from the excited state where the energy is going
to be equal to 121 wave numbers.
So when we put those two together it becomes a
partition function.
And so once we multiple it out, minding our own units,
this is the resulting equation.
As we plot this out as a function
of temperature we can see
that at zero Kelvin the only contribution
that we're getting is from the ground state
as intuitively followed by knowing
that the excited state is not going
to be populated at zero Kelvin.
And as we scale the temperature
up to 1000 Kelvin we can see how the contribution
from the second state contributes
to the overall partition function.
So in order to calculate the term populations we will use
this formula which you guys have all seen where we take
into account the number of particles or number of molecules
in the particular state in question, in this case, state i,
over all of the contributing partition function elements.
So in this case since we want to calculate the population
of ground state if we go back and we look
at the wave function the contribution
from the ground state, or not
to wave function the partition function, the contribution
from the ground state is simply two.
So we leave that up here and we carry
out the overall partition function that we derived earlier
to get that, sorry, to get that at 300 Kelvin the population
of the ground state is about 64 percent.
Now the way to get the excited state is we can simply subtract
one from this quantity and get 36 percent or if you want to,
kind of like a challenge to yourself,
carry out the same calculation with the contribution
from just the excited state in Ni.
And now we have to go over the electronic contribution
of the molar energy.
So which equations will we end up using
and this will be the equation sheet that you guys will see
on Friday and so the idea is that we throw a bunch of these
on and if you studied well enough you should know
which one specifically to use
and this is the one we'll want to use.
Notice how it looks like the average energy per molecule
except there's no brackets on the E
and there's an N depicting the actual number per mole.
So as we use this equation of 300 Kelvin we'll denote
that the partition function contribution
of 300 Kelvin is 3.119 and then we plug
in our partition function to find the derivative
with the respect to beta and this turns out to be
in the simplified version of this, so then once we mind all
of our units and we use this equation we will carry
out the long unit kind of approximation there
and the end result is going to be 519 joules per mole.
You guys have any questions so far?
I'm kind of blasting through this pretty quickly
and remember any questions that you guys have feel free
to ask them in discussion sections.
I have one today and two tomorrow.
Even if you're not technically registered in them feel free
to attend and so here's another mid-term exam question
from a couple of years ago.
We have three vibrational modes
at 680 wave numbers, 330 and 973.
So the first question we're asked is
if the molecule is cooled
to 4 Kelvin how much vibrational energy does it retain?
And we'll get to the second two parts
of the question as we get to them.
So how do we figure this out?
Well, we sum over all the contributions
of the vibrational wave function, vibrational modes,
I'm sorry and so we plug them in as wave numbers
and our end result is of course known to be in wave numbers
which is the Ni equals 991 wave numbers or if you want to carry
out the calculation in joules you can convert all
of these wave numbers to joules using HC
and then we'll get our answer in joules.
So the end result in joules
for this particular question is going to be 1.97 times 10
to the negative 20 joules.
So for the second part of the equation we're asked
that if a liter of this system is warmed
up to 200 Kelvin what fraction of these molecules is
in the 600 sorry, oh never mind.
I thought you raised your hand, sorry.
Yeah, what's up?
>> What's the relevance in mentioning the 4 degree Kelvin
on the first part of the problem?
>> This part, oh, okay.
Well we're trying to explain
that the system is basically at the ground state.
There is no contribution of translational, rotational
or vibrational energy in the excited state.
>> Okay.
>> So yeah, this is where we were.
So for what fraction of these molecules is the 680 wave number
state vibration excited?
So what part of the-- how many of the molecules are
in the excited 680 vibrational wave number mode?
So which equations do we use for this?
Now there's this one right here
which is the partition function contribution
for vibrational partition function and the calculation
for which is blocked by that little thing up there
and this is the equation that we used from Problem 1
where we denoted to calculate the population
of each excited state or each state in question.
Now if we use the vibrational partition function and plug it
into the overall partition function
in the denominator do we use the vibrational partition function
from simply 680 wave numbers or do we use it from all three.
Well the answer for this one specifically is we use the 680
wave number vibrational mode
because that's specifically what we're asked about.
However, if in the question it's noted that 330
and the 973 wave number modes are equal
to zero then we include them but I'll get to that in a second.
It's actually down here.
So in order to calculate this we use the vibrational partition
function or we use the vibrational partition function
for 680 wave numbers.
It's just E to the negative or E to the negative E i over KT
or beta E if you're used to that
over the vibrational partition function contribution
from 680 wave numbers.
So this is a simplified formula
because if we take the denominator and carry
out the simplification this ends up up here
and this is the resulting part down there.
Now if you're asked specifically
where the vibrational contribution
from 330 wave numbers and 973 wave numbers is equal
to zero then we include them in the denominator,
so back to the question.
So what fraction of these molecules is this 680 wave
number vibrational mode excited?
So we calculate our term population
from using our well-used
by now degeneracy times the contribution
over the vibrational partition function.
The end result after simplification looks like this
and so once we carry out our calculations
of course minding our units we will have negative.489
for the exponent up here.
Once we carry that over and calculate it we notice
that the contribution is.237.
Does this actually make sense?
So if we carry out the calculation for temperature
of the vibrational mode we find
that the vibrational temperature is at 978 Kelvin
which is well below 2000 Kelvin
so we expect the vibrational partition function
to be appreciable.
It turns out that using this,
the overall partition function is equal
to 2.58 specifically in this example.
Now, hold on a second.
Okay and that covers the entirety of this question.
I thought there was a Part C. Do you guy have any questions
on this problem?
I'm kind of blasting through this so we'll have some time
at the end for questions and what not.
So here's the equipartitioning equation that we've seen.
Now if you guys remember the discussion section we went over,
I think it was in Week 3 we exhausted the KT
over 2 contribution for any quadratic terms
in the Hamiltonians for the translational, rotational
and vibrational energy.
So if we were to think about this in the Hamiltonian operator
for instance, in this case I think it's the translational
energy, we have a quadratic term here, momentum squared over 2 M
and the potential quadratic term half KX squared,
so for any quadratic term which can fit in the form
of AP squared or BX squared we will get a contribution of KT
over 2 but for a mole of these guys we use RT over 2
which you guys all remember that R and K are just the same thing
but R refers to a mole of these things while K refers
to one specific molecule.
So in order to get the heat constant which is the amount
of energy a molecule can store per unit temperature all we have
to do is take the KT over 2
or specifically our entire contribution from KT over 2
from vibrational, rotational, translational states
and derive it in terms of temperature and we find
that the heat capacity for one quadratic contribution is going
to be equal to K over 2 and there it is.
So if we take the classical Hamiltonian for 3D translation
where we have 3 quadratic terms for momentum in the X,
the Y and the Z we'll notice
that our equal partition contribution is going
to be equal to 3 KT over 2 which makes sense
because they're three quadratic terms and so if we take
into account our complete energy and derive the heat capacity
for the energy we'll see that it is equal to 3R over 2
for a mole of a system.
So as we can see on the graph this is just a translational
energy contribution that we see and we're going to go
over the other contributions that can happen here and here
as well when we include the rotational
and the vibrational wave functions.
So this also turns out to be heat capacity
for all monatomic gases because if it's a single molecule
or not a single molecule a single atom,
rotation does nothing and vibration does nothing
because it has nothing to vibrate off of so for molecules
with more than one atom vibration
and rotation can also contribute
but vibrational contribution doesn't turn
on until the vibrational temperature approaches
that of the temperature of the system in question.
So for a linear molecule
where the temperature is significantly less
than the vibrational temperature where we assume
that the vibrational contribution is insignificant
but only rotational significant
because of the temperature being higher than that
of the vibrational, rotational temperature, sorry,
we take into account
for specifically a linear molecule two quadratic terms,
one for rotation in the X axis and one for rotation in the Y
because rotation in the Z axis does not change the system.
So since we have two terms our contribution is 2 K T over 2
or for a mole of these guys it will be 2 R T over 2.
And again the heat capacity for when we calculate it
out is just R over 2 or K over 2.
So once we have the linear molecule
where the rotational states partake we can approximate the
heat capacity of the system for translational, rotational states
by 3 R over 2 plus 2 R over 2 which will be 5 R over 2.
So this is again denoting
that there is no vibrational contribution
to the heat capacity at these temperatures.
So once we take into an account rotation we can see
that the heat capacity contribution will be 5 over 2
as we have just shown.
Now for a known linear molecule which can exhibit three orders
or three quadratic terms for the rotational contribution all we
have to do is simply add another KT over 2 to it
to make it equal 3 KT over 2 and similarly again for a mole
of these guys it's going to be 3 RT over 2
and once we take the heat capacity of these guys together
for a known linear molecule our total contribution is going
to be equal to 3 R. Again this is not taking
into account vibrational states or, yeah, vibrational states
because the vibrational temperature is too
or the temperature in the room is too low in comparison
to the vibrational temperature.
So this is the contribution from the translational
and this is the contribution from the rotational functions.
So what about cases where the temperature
in the room is high enough
to have the vibrational states contribute?
So we go back to the equation for the vibrational Hamiltonian
which only really contains two terms.
Again, half KX and pi squared or P squared sorry,
momentum over the reduced mass.
So following the rules
of the equipartition theorem we will get 2 KT over 2
or RT per mode but we have to take into account
that any system can have for a linear molecule,
for a diatomic sorry, 3 M minus 5 vibrational modes
or 3 M minus 6 if it's not a diatomic.
So this takes into account the diatomic vibrational
contribution which will equal out our contribution to be 5 R
over 2 and since this is a diatomic system this
contribution ends up equaling 6 minus 5 one R. And so
as we can see all of these things put together,
it's 5 R plus 3 M minus 5.
So then we sum all those together and we get 7 R over 2
and we can see the full heat capacity contribution
from translational, rotational and vibrational states.
So let's use this to solve for the constant volume,
molar heat capacity of specifically we're going to go
over one of these i2 but using the same rules and principles
in the equipartition theorem we can solve for CH 4 and C6H6.
So the vibrational mode of iodine is 214 wave numbers
and will this vibration contribute
to our constant volume heat capacity at 25 degrees Celsius?
So let's calculate the vibrational temperature
where we have HV over K. We plug in the R vibrational mode,
mind our units and we calculate
that at 308 Kelvin we expect this mode to be calculated.
So because 25 degrees Celsius
which is 298 Kelvin is pretty close to this we'll take
into account the contributions
from the vibrational partition function.
So a quick approximation
to determine whether vibrational modes will contribute is
if we take this equation where the vibrational temperature,
if it's less than or equal to 2 times the temperature
in question you'll want to include it.
If you guys have played around with some of these functions,
I've played around with them in my discussion sections
where somebody has asked why or when do we take
into account the vibrations?
Play around with the math.
I've noticed somewhat of a one-to-one relationship,
meaning if the temperature is close to that of the quantity
and wave numbers irrespective of the units it will contribute.
However, if it's something about half, like so if you're talking
about a system that's at 300 Kelvin, oh, sorry 150 Kelvin
and the vibrational mode is something
like 300 it won't contribute very much like something
to a thousandth of a significant figure.
But again play around with these things.
Get comfortable with them.
It's going to make it a lot easier for you.
But this is a simple rule of thumb
that makes it easier to follow.
So yeah, so once we calculate our own inequality
for this we notice that the temperature is going
to be much greater than the vibrational wave function
or vibrational partition function, vibration temperature,
so yes we decided to include it.
So we have two degrees of rotational
or two rotational degrees of freedom
because it's a diatomic species
and we included one vibrational mode
because the only way this can vibrate
since it's a diatomic is amongst itself.
So once we calculate this out taking these constraints
into account we get 5 R over 2 plus R which is 7 R over 2.
But the real value is it's pretty close.
It's 3.4 R. Notes that if we leave
out the vibrational mode then our contribution becomes 2.5 R
which is significantly less than the heat capacity
if we don't include the vibrational mode
and my remote is having problems.
Do you guys have any questions?
So we kind of blasted through this pretty quickly.
Feel free to ask questions in any of the discussion sections
on anything as far as stuff we've gone over.
If you don't feel comfortable with the material that's fine,
just ask us questions.
We'll be able to help you.
I have office hours on Friday an hour before the exam
at 10 o'clock at Natural Sciences 2, Room 1115. ------------------------------b886dd189c25--