This introductory physical chemistry course examines the connections between molecular properties and the behavior of macroscopic chemical systems.

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From the course by University of Minnesota

Statistical Molecular Thermodynamics

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This introductory physical chemistry course examines the connections between molecular properties and the behavior of macroscopic chemical systems.

From the lesson

Module 4

This module connects specific molecular properties to associated molecular partition functions. In particular, we will derive partition functions for atomic, diatomic, and polyatomic ideal gases, exploring how their quantized energy levels, which depend on their masses, moments of inertia, vibrational frequencies, and electronic states, affect the partition function's value for given choices of temperature, volume, and number of gas particles. We will examine specific examples in order to see how individual molecular properties influence associated partition functions and, through that influence, thermodynamic properties. Homework problems will provide you the opportunity to demonstrate mastery in the application of the above concepts.

- Dr. Christopher J. CramerDistinguished McKnight and University Teaching Professor of Chemistry and Chemical Physics

Chemistry

Welcome, to week four of Statistical Molecular Thermodynamics.

Â In this week, I want to focus on, the partition functions of gases.

Â So, in the last week's material, we explored how to use certain partition

Â functions, that just seemed to come out of nowhere in a sense.

Â And this week, I want to illustrate how those partition functions are derived.

Â Where they come from? How they connect with the various

Â components of the energy of gas molecules, let's dive in.

Â So, remember that for an ideal gas, that is a gas where none of the particles

Â interact with one another. In that case, we can write an ensemble

Â partition function. In terms of a molecular or atomic

Â partition function. So, the ensemble partition function is

Â capital Q. The molecular or atomic partition

Â function is little q. And the relation between them, is that

Â the ensemble partition function, which depends on number of particles, volume,

Â and temperature, that's the canonical ensemble.

Â Can be written as the particle partition function, again I'll say molecule or

Â atom, from now on I'll probably just say molecule, and we'll accept atoms as sort

Â of a subset of molecules. So, the molecular partition function

Â raised to the power of n, where n is the number of molecules, all divided by n

Â factorial. So, notice that the molecular partition

Â function does not depend on n, but n appears on this side that is why the

Â ensemble partition function depends on n. And so, now let's apply this to an ideal

Â gas where the molecules are indistinguishable.

Â And, the number of states greatly exceeds the number of molecules.

Â So, that's the assumption of low pressure, and that number of states being

Â greater than the number of molecules, remember, was critical to being able to

Â use this simple n factorial in the denominator.

Â To account for over-counting of states associated with using the molecular

Â partition function to the nth power. Alright, but now let's look at the

Â molecular partition function. So, let's consider each degree of freedom

Â separately. For an ideal monatomic gas, the only

Â degrees of freedom that are available, are translational, and then there's

Â electronic energy as well. And that's a significant simplification.

Â And so, the total energy, of an atomic, of an atom, that is, in an atomic ideal

Â gas would be its translational energy summed with its electronic energy.

Â And so, the atomic partition function then will be the product of the partition

Â functions from each degree of freedom. So, we've seen that before that when you

Â have an exponential of a sum of energies, that's like a product of exponentials of

Â energies, and each of those exponentials is a partition function.

Â So, we can write for the atom, the atomic partition function which depends on

Â volume and temperature, is equal to the translational partition function times

Â the electronic partition function, product of the two.

Â So, we're going to need to consider each of those two separately in order to

Â construct the total molecular partition function, in this case atomic partition

Â function. So, we have a general form for any

Â partition function that it is a sum over states, e to the minus beta remember beta

Â is 1 over boltzmann's constant times temperature 1 over KT.

Â And the energies of allowed states, so I'll just generically say here the

Â allowed energy levels for translation. And in the very first week of the course,

Â we looked at the allowed energy levels for certain systems that could be solved

Â exactly with quantum mechanics. One of those was the particle in a box.

Â And you should remember that for a particle in a three dimensional box, and

Â in particular let's take a cubic box that has a side length of a.

Â Then the allowed translational energy levels, are equal to h squared.

Â That's [UNKNOWN] constant squared, divided by 8 times the mass of the atom,

Â times the side length squared, all multiplied by a series of quantum

Â numbers. Though the, so the square of the x, the

Â square of the y, and the square of the z quantum numbers, where these numbers nx,

Â ny, nz they can take on any integer values 1, 2, 3, etc.

Â So, given those allowed energy levels we simply substitute those in to the

Â expression for the partition function. So, I could write this is a single sum e

Â to the minus beta that epsilon. And now I'll just write in what that

Â epsilon is this is just explicit, so here's beta and then here's all the other

Â terms that are written up there for the allowed energy levels.

Â So, let's continue to work with that, we'll just move it to the next slide.

Â And I'm going to note that since I have a sum appearing in the argument of my

Â exponential, I can convert that into a product of exponentials.

Â So, we've done this a lot, this is the beauty of exponential you keep

Â transforming between sums and products. So, here is a sum over the energy levels

Â for n sub x, the quantum number in the x direction.

Â Here we're summing over the quantum number in the y direction.

Â And now we're summing over the quantum number in the z direction, alright?

Â So, the form of all these arguments of the exponentials is the same, it just

Â involves a different quantum number. But since only the indexes is different,

Â each one of those sums is identical. So, I can think of it as the

Â translational partition function is the cube of this sum.

Â So, a sum over some index n e to the minus beta h squared n squared all

Â divided by 8ma squared. Notice that the volume appears in the

Â partition function for translation, because there is this length unit that's

Â being used to define the particle in a box energy levels.

Â So, we have to pick a certain box in which we will find the energy levels, and

Â by picking a side length, that's equivalent to picking a volume.

Â And we'll see that in a little more detail in just a moment.

Â But first, let's actually take a more careful look at this sum.

Â So, It's relatively easy to write down, there's a few imposing Greek letters and

Â Roman letters, but otherwise not, not so complicated.

Â However, in order to evaluate that sum, well there's no closed form that we can

Â just jot down an answer for what that sums to as n goes from one to infinity.

Â However, the translational energy levels, you'll recall, are very closely spaced

Â one to another. And calculus as a field was basically

Â born, when people tried to figure out how to do sums over densely spaced well in

Â this case I'll call them levels. So, given that kind of dense spacing that

Â sum can be transformed to an integral. Alright, that is a continuous function

Â instead of a sum. So, if I do that I would write my

Â partition function, and this is equivalent to moving from a quantum

Â mechanical system to a classical system. In a classical system of course you would

Â view all translational energies are possible.

Â Its a little odd actually to think that only certain translational energies are

Â possible. That's quantum mechanics for you its

Â always a little bit odd. But they're so dense that its very nearly

Â classical, and we can make this approximation.

Â So, then what I want to evaluate is the cube of not the sum, but the integral

Â from 1 to infinity. The index is n, so I'll go over an

Â infinite test mode dn, e to the minus beta h squared n squared over 8ma squared

Â right the, the same argument but now inside an integral instead of a sum.

Â Well, it turns out that that integral, this one here on the left, that's

Â actually not any easier to solve than the sum itself.

Â However, if all we do is make a pretty small change, let's change the bottom

Â index on the integral, this is a definite integral, let's change it from 1 to 0.

Â In that case if you look in a integral table you will discover that integrals

Â from 0 to infinity of the form dne to the minus alpha m squared, and that's called

Â a gaussian function. So, gaussians appear in many places in

Â science, and it turns out that integral has a nice analytic solution.

Â It is the square root of pi over 4 Alpha. Right, so, nice and straightforward, easy

Â to write down. So, in our case, that that integral table

Â alpha is equal to everything that multiplies n squared.

Â That is, h squared divided by 8ma squared k t, and so when we plug in for that

Â integral. We get 8 pi ma squared k t all divided by

Â 4 h squared to the 1/2 power. And so just in case I, I went a little

Â fast here on something, I did transform along the way.

Â Here's beta, remember beta's 1 over k t. So, I just write k t down the

Â denominator, I'm going to find it a little more convenient to look at the k

Â and the t. And so if this was alpha, I need to have

Â alpha in the denominator of this expression.

Â So, I take this whole thing and everything that was in the denominator

Â here will go up to the numerator, and sure enough there's the 8 and the m and

Â the a squared and the k t and the pie sticks around from this.

Â Meanwhile this h squared went into the denominator, and here's the 4, alright.

Â So, just fatefully plugging in the appropriate values given our integral.

Â So, in order then to get the translational partition function that was

Â simply the cube of this integral. And so, I'll cube this expression, and

Â when I do that I'll get the whole thing to the 3/2 power instead of the 1/2

Â power. And I'll take an 8 divided by a 4, and

Â I'll just replace that with a 2. And the last thing I'll do, is I'll

Â notice, here I had a squared, all to the 1/2 power, so that's just a.

Â So, when I cube it I'll get a cubed. And what is a cubed?

Â a cubed is the volume of the box we were solving, the particle in a box equation

Â for. So, I'll just pull that out to really

Â emphasize, here's where the volume dependence comes in to the translational

Â partition function. So, it's this expression, which depends

Â on the mass of the atom, boltzmann's constant, temperature, plung's constant,

Â and volume. So, just keep in mind then, that the

Â reason there is a volume is that the side of the box dictates the allowed energy

Â levels. The actual choice of this volume is part

Â of a so called standard-state convention. So, we would get different values for the

Â translational partition function if we chose different volumes for the particle

Â in a box. And so, thermodynamicists just get

Â together every few years in a meeting somewhere on earth, I suppose.

Â And they decide that listen, I'll report all my values for a certain size box, if

Â you do the same size box for your values. And that way we'll always be comparing

Â apples to apples. And so that is a standard state

Â convention to choose a particular size for your box.

Â And so just to to put a number on the partition function here, let's consider

Â this is not a single atom anymore. But this'll work fine for a molecule as

Â well. Let's consider methane as a molecule.

Â So methane, if you look up the masses of the atoms, add them together, it's 1.993

Â times 10 to the minus 26 kilograms. And we'll adopt a standard state volume

Â of 24.47 cubic decimeters. That's 24.47 liters, and why that number

Â looks a little bit odd? Well, that's how big a box an ideal gas

Â would occupy at room temperature and one atmosphere of pressure.

Â So, that's a convenient thing to choose. So, I will use room temperature 298.15

Â kelvin. And under those conditions if you plug

Â these values into this expression, and you're welcome to check my math at your

Â leisure, but I'll tell you what the answer is.

Â You get that the translational partition function is equal to 1.519 times 10 to

Â the 30th power. And I'll carry along the units to

Â emphasize I did have to make a standard state volume choice.

Â So, I'll carry along decimeters cubed, but usually we just sort of remember that

Â that's there. So, let me remind you, what does the

Â translational partition function measure? What does any partition function measure?

Â It's a quantitative measure of the number of quote, accessible, unquote, states.

Â So, in essence this says for the molecule methane at room temperature and about 1

Â atmosphere pressure, so its occupying the volume that an ideal gas would occupy.

Â There are roughly 1 and a half times 10 to the 30th accessible states.

Â Well how many states is that per molecule?

Â Well if you work out how many moles of methane would be there, it's about one if

Â we're treating it as an ideal gas because it's in 24.47 liters.

Â So, that would be Avogadro's number, 6.02 time 10 to the 23rd.

Â So, there's something on the order of 10 to the 6th to 10 to the 7th, as many

Â accessible states as there are molecules. And remember, we had a discussion last

Â week about what would be the likelihood of finding two molecules in the same

Â state. And you can see that that likelihood

Â clearly is, is reasonably small given how many more levels the are than there are

Â molecules, alright. Well that completes what I'd like to sort

Â of illustrate with the translational partition function.

Â I'd like to give you a chance to get a better feel for the density of the levels

Â by considering the impact of that approximation we made.

Â So, when we moved from setting this lower limit from one to zero, what really

Â happened? So, I'll let you address that, and play

Â with it a bit, and then we'll come back. Well so hopefully you convinced yourself

Â that that missing little piece of integral we tacked on was equal to 1, and

Â 1 is rather small piece of something times 10 to the 30th power.

Â So, it wasn't much of an approximating to change the limit on the intergral.

Â And you've gained an appreciation then for just how little each level is

Â contributing, and how many we must be summing over in order to add up to such a

Â large value for the transitional partition function.

Â And that was just a one of the two partition functions we need to consider

Â for the ideal monatomic gas. Next time, we'll look at the electronic

Â partition function, and putting the two together in order to get the ensemble

Â partition function.

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