Lecture 11. Midterm I Review.
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Transcript: English(auto-generated)
00:04
So today we're going to be going over some midterm review questions. It's going to be a lot similar to 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 midterm is going to cover chapters 13 and 14.
00:22
And today we're going to go over a good portion of the 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 molecular, statistical mechanics of a particular system. And the other one is going to be covering thermodynamics of gases.
00:42
Basically the equal partition theorem. And there's going to be a 10 point extra credit problem which is extra credit. So we're not going to divulge the details to you. And so here's a summary of all the partition functions in translational, rotational, vibrational states.
01:02
Some things you want to keep in mind is that you 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 canceling out in the way that they should, go back and rerun through your answer because well the numbers are sometimes hard to get ahold of.
01:24
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 contributions from each state.
01:43
And beta again remember is in 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 talk about nl and it's doubly degenerate electronic
02:03
state at 121 wavenumbers and a doubly degenerate ground state at 0 wavenumbers. So the first thing, and this is obviously not going to be asked, well it could be asked on the test, but we have to plot out the partition function as a function of temperature from 0 to 1000k.
02:22
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 MO diagram depicting this system with the two degenerate ground states
02:42
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.
03:01
We'll set e to equal 0 and the contribution from the excited state where the energy is going to be equal to 121 wavenumbers. So when we put those two together this becomes a partition function. And so once we multiply it out, minding our own units,
03:21
this is the resulting equation. As we plot this out as a function of temperature, we can see that at 0 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 0 kelvin. And as we scale the temperature up to 1000 kelvin, we can see how the contribution
03:41
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,
04:03
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 the ground state, if we go back and we look at the wave function, the contribution from the ground state, or not the wave function, the partition function, the contribution
04:22
from the ground state is simply 2. So we leave that up here and we carry out the overall partition function that we derived earlier to get that at 300 kelvin, the population of the ground state is about 64 percent.
04:42
Now the way to get the excited state is we can simply subtract 1 from this quantity and get 36 percent. Or if you want to, kind of like as a challenge to yourself, carry out the same calculation with the contribution from just the excited state in an i. And now we have to go
05:03
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
05:20
which one specifically to use. And this is the one we 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 at 300 kelvin,
05:42
we will 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 respect to beta. And this turns out to be in the simplified version of this.
06:01
So then once we mined 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. Do you guys have any questions so far? I'm kind of blasting through this pretty quickly.
06:21
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 midterm exam question from a couple of years ago. We have three vibrational modes
06:41
at 680 wave numbers, 330 and 973. So the first question we are 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 questions as we get to them. So how do we figure this out? Well, we sum over all the contributions
07:02
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 going to be in wave numbers, which is going to equal 991 wave numbers. Or if you want to carry out the calculation in joules,
07:20
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.
07:42
So for the second part of the equation, we're asked that if a liter of this system is warmed up to 2,000 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?
08:04
This part, okay. Well, we're trying to explain that the system is, yeah, basically at the ground state there is no contribution of translational rotational vibrational energy in the excited state.
08:23
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?
08:42
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 use from problem one.
09:03
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
09:22
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.
09:42
However, if in the question it's denoted 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 yeah,
10:01
we use the vibrational partition function for 680 wave numbers. It's just e to the negative or yeah, e to the negative ei 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 form of it
10:21
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
10:44
in the denominator. So back to the question. So what fraction of these molecules is the 680 wave number vibrational mode excited? So we calculate our term population from using our well used by now,
11:04
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.
11:22
Yeah, it is negative. 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
11:42
of the vibrational mode, we find that the vibrational temperature is at 978 kelvin, which is well below 2,000 kelvin. So we expect the vibrational partition function to be appreciable. It turns out that using this,
12:01
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 guys have any questions
12:21
on this problem? I'm kind of blasting through, so we'll have some time at the end for questions and whatnot. So here's the equal partition equation that we've seen. Now, if you guys remember the discussion section that we went over, I think it was in week three, we exhausted the kt over 2 contribution
12:43
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,
13:01
I think it's the translational energy. We have a quadratic term here, momentum squared over 2m, 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,
13:23
which you guys will remember that r and k are just the same thing, but one refers, 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,
13:41
all we have to do is take the kt over 2, or specifically our entire contribution from kt over 2 from vibrational, rotational, and translational states, and derive it in terms of temperature. And we find that the heat capacity for one quadratic contribution is going
14:00
to be equal to k over 2. And there it is. So if we take the classical Hamiltonian for 3D translation, where we have three 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 3kt over 2, which makes sense
14:22
because there's three quadratic term. 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.
14:42
So as we can see on the graph, this is just the 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 the heat capacity for all monatomic gases, because if it's a single molecule,
15:04
it can't, or not a single molecule, a single atom, it can't, 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
15:22
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 isn't significant,
15:40
but only the rotational is 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.
16:03
So since we have two terms, our contribution is 2kT over 2, or for a mole of these guys, it will be 2RT 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 a linear molecule
16:21
where the rotational states partake, we can approximate the heat capacity of the system for translation or rotational states by 3R over 2 plus 2R over 2, which will be 5R over 2. So this is again denoting that there is no vibrational contribution
16:40
to the heat capacity at these temperatures. So once we take into account rotation, we can see that the heat capacity contribution will be 5 over 2, as we have just shown. Now, for a nonlinear molecule, which can exhibit three orders or three quadratic terms for the rotational contribution,
17:04
all we have to do is simply add another kT over 2 to it to make it equal 3kT over 2. And similarly, again, for a mole of these guys, it's going to be 3RT over 2. And once we take the heat capacity of these guys together for a nonlinear molecule, our total contribution is going
17:23
to be equal to 3R. Again, this is not taking into account vibrational states because the vibrational temperature is too low 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
17:41
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
18:03
or p squared, sorry, momentum, over the reduced mass. So following the rules of the equal partition theorem, we will get 2kT over 2 or RT per mode. But we have to take into account that any system can have for a linear molecule,
18:23
for a diatomic, sorry, 3N minus 5 vibrational modes or 3N 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 5R over 2. And since it's a diatomic system, this contribution ends
18:42
up equaling 6 minus 5, 1R. And so as we can see, all of these things put together, it's 5R plus 3N minus 5. So then we sum all those together and we get 7R over 2.
19:02
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
19:22
in the equal partition theorem, we can solve for CH4 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?
19:43
So let's calculate the vibrational temperature where we have HV over K. We plug in our vibrational mode, mind our units, and we calculate that at 308 Kelvin, we expect this mode to be populated. So because 25 degrees Celsius, which is 298 Kelvin, is pretty close to this, we'll take
20:01
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 want to include it.
20:21
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
20:41
in wave numbers, irrespective of the units, it will contribute. However, if it's something about half, so if you're talking about a system that's a 300 Kelvin, or 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.
21:00
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
21:20
or vibrational partition function, vibrational temperature. So yes, we decide to include it. So we have two degrees of rotational, two rotational degrees of freedom because this is a diatomic species. And we include one vibrational mode because the only way this can vibrate, since it's a diatomic, is amongst itself.
21:42
So once we calculate this out, taking these constraints into account, we get 5R over 2 plus R, which is 7R over 2. But the real value is, it's pretty close, it's 3.4R. Note that if we leave out the vibrational mode, then our contribution becomes 2.5R, which is significantly less than the heat capacity
22:02
if we don't include the vibrational mode. And my remote is having problems. I think this is the end. Do you guys have any questions? So we're kind of blasted through this pretty quickly. Feel free to ask questions in any of the discussion sections
22:22
on anything as far as stuff we've gone over. If you don't feel comfortable with the material, it'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.