Video 4.1 - Ideal Monatomic Gas: qtrans

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Statistical Molecular Thermodynamics

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University of Minnesota

Statistical Molecular Thermodynamics

163 ratings

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.

Video 4.1 - Ideal Monatomic Gas: qtrans15:22

Video 4.2 - Ideal Monatomic Gas: Q8:56

Video 4.3 - Ideal Monatomic Gas: Properties17:09

Meet the Instructors

Dr. Christopher J. Cramer
Distinguished McKnight and University Teaching Professor of Chemistry and Chemical Physics
Chemistry

Welcome, to week four of Statistical Molecular Thermodynamics.

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

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

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

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

Where they come from? How they connect with the various

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

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

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

partition function. In terms of a molecular or atomic

partition function. So, the ensemble partition function is

capital Q. The molecular or atomic partition

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

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

and temperature, that's the canonical ensemble.

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

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

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

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

factorial. So, notice that the molecular partition

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

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

gas where the molecules are indistinguishable.

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

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

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

use this simple n factorial in the denominator.

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

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

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

separately. For an ideal monatomic gas, the only

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

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

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

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

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

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

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

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

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

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

the electronic partition function, product of the two.

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

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

function. So, we have a general form for any

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

translational partition function is the cube of this sum.

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

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

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

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

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

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

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

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

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

Roman letters, but otherwise not, not so complicated.

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

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

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

one to another. And calculus as a field was basically

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

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

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

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

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

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

view all translational energies are possible.

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

possible. That's quantum mechanics for you its

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

classical, and we can make this approximation.

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

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

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

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

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

actually not any easier to solve than the sum itself.

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

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

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

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

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

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

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

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

alpha is equal to everything that multiplies n squared.

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

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

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

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

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

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

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

alpha in the denominator of this expression.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

levels. The actual choice of this volume is part

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

translational partition function if we chose different volumes for the particle

in a box. And so, thermodynamicists just get

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

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

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

apples to apples. And so that is a standard state

convention to choose a particular size for your box.

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

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

well. Let's consider methane as a molecule.

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

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

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

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

would occupy at room temperature and one atmosphere of pressure.

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

kelvin. And under those conditions if you plug

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

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

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

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

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

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

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

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

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

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

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

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

Well how many states is that per molecule?

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

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

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

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

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

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

state. And you can see that that likelihood

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

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

of illustrate with the translational partition function.

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

by considering the impact of that approximation we made.

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

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

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

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

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

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

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

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

large value for the transitional partition function.

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

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

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

partition function.

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