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

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Из курса от партнера 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.

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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

In this lecture, we're going to continue working with ideal diatomic gases.

So just to review that you saw in the lsat lecture, for the energy of an ideal

diatomic gas, there are four contributors.

There is the electronic energy, the translational energy and the vibrational

energy which we dealt with in the last lecture.

And in this lecture I'd like to focus on the remaining term, that is, the

rotational energy, and its contributions to the total.

In order to compute that energy, we can use the quantum mechanical rigid rotator,

Schrodinger equation to compute allowed energy levels, and degeneracies.

And as we saw early on in the course, within that approximation, the energies

are given by the equation shown here. So they depend on a quantum number J, and

the quantum number expression J times J plus 1 can range from 0 when j equals 0,

to increasingly high integer values. and there's a pre-factor involving H bar

squared, and the moment of inertia of the molecule.

And the degeneracy of these levels is two J plus one, where J is the quantum

number. So, if we then construct a partition

function over levels, so we'll include the degeneracy explicitly in the

summation. Then we would write it this way.

It's the sum from j equals 0 to infinity 2j plus 1 e to the minus beta, the energy

expression. It's a little bit more convenient just

from a manipulation from the equation standpoint, and also for conceptual

purposes, to define a rotational temperature much as we defined a

vibrational temperature when working with the vibrational partition function.

And so the goal effectively is to take all of the constants that appear in this

energy expression, other than beta, and to replace them with this rotational

temperature capital theta, and it is then h bar squared over 2 times the moment of

inertia, times Boltsman's constant which of course is lurking here in beta.

Beta, remember is 1 over Boltsman's constant times temperature.

So when we take that approach. The rotational partitional function

becomes sum over J equals zero to infinity, 2 J plus 1, E to the minus

Theta Rot, so the rotational temperature, times J, times J plus 1, divided by

temperature. So that's what's left from beta when we

take the Boltzmann's constant out. A 1 over T term.

Now, if we would like to solve for that expression, unforutnately there's no

closed-form solution for that series, so we can't just look it up and see what it

converges to as we sum out to the last term.

However, just as for the translational partition function If it's the case that

the energy levels are sufficiently closely spaced, we can replace this sum

by an integral. And we can ask the question, what, what

would it take for those energy levels to be very closely spaced?

Well, it requires that the argument of the exponential is very small

essentially. So, as J is increasing, we're not really

changing this argument very much. And a way to make it very small would be

to make the rotational temperature, theta, small relative to the actual

temperature, T. So that's the condition for making this

replacement of the sum by the integral. That the rotational temperature is

considerably smaller than the actual temperature in which we are interested.

So let's just take a look at a few typical diatomic gases and see if this

condition is satisfied under typical conditions for typical gases.

So recalling exactly what is the rotational temperature, it's the quotient

of h bar squared. With 2 times the moment of inertia times

Boltzmann's constant. So the only variable in there that

depends on the nature of the molecule is the moment of inertia.

And remember the moment of inertia gets larger and larger as the masses of the

two atoms involved in the bond gets larger and larger.

So we would expect as the diatomic gets heavier, uses more heavy atoms Then the

rotational temperature should go down. And if we actually carry out the

calculation using actual atomic masses and bond distances, since the moment of

inertia depends on bond distance as well. We get these results and so what you see

is as expected, the lightest possible diatomic, that's molecular hydrogen using

prodium as the nucleus so no neutrons, just a mass of one.

Has a rotational temperature of 85.3 kelvin.

All the others are much, much smaller. Of course if you do have a proton, as you

see here for hydrogen chloride or hydrogen bromide, now we have sort of low

teens down to 12 for the rotational temperature, kelvin.

So those are very cold temperatures obviously.

And as we go to two heavier atoms like Carbon Monoxide, Nitrogen, Nitrous Oxide

and so on, single digit rotational temperatures ultimately by the heavier

dihalogon models Chlorine and Bromine, below one Kelvin.

So certainly if we're interested in a gas at room temperature which is 298 Kelvin,

we have satisfied the condition that rotational temperature is much, much

lower. Then the observed temperature that we're

working with. So that's easily satisfied then.

Really the only gas that occasionally might be of interest would be molecular

hydrogen working at quite cold temperatures.

Well, let's then replace our sum with an integral, and it turns out this is a very

friendly integral, so statisctical mechanics is occasionally very kind to

one. And so if we look at that interval it

might look a little bit imposing, but let's just make some quick substitutions.

let's call j times j plus 1 by a new variable name.

We'll call it x. That means that dx, I have to take the

derivative of the right hand side so if I think of that as j squared plus j that's

pretty easy to take a derivative of. I'll get 2 j plus 1.

All times d j. And so that's what's right here.

And then I'll also put in a variable a which is going to be my rotational

temperature divided by the temperature of the system.

If I now rewrite my integral using all those substitutions, I get integral 0 to

infinity dxe to the minus ax. And probably no one really needs to go to

an integral table to look that up. That's pretty trivial.

And so the solution is minus 1 over a, e to the minus ax evaluated at the limits

on this definite integral. So e to the minus a times infinity gives

0 and e to the minus a times 0 is e to the 0 so that's 1.

So I'll get minus 1 over a and just end up with 1 over a.

What was A? Well A was the rotational temperature

divided by the temperature. So the solution, the whole partition

function, is just the temperature divided by the rotational temperature.

It's a tremendously simple expression. Of course we can then re-expand what the

rotational temperature is. And if we do that, if we put back in the

constants and the moment of inertia, and if we want to express h bar squared in

terms of Planck's constant and the 2 pis, remember h bar is Planck's constant

divided by 2 pi, we get this expression for the rotational partition function.

8 times pi squared times the moment of inertia, so that's the molecularly

dependent quantity appearing in the partition function.

Times Boltzmann's constant times the temperature divided by [UNKNOWN] constant

squared. Well, with that partition function in

hand, we can now compute useful properties of the gas as an ensemble.

And so, if you remember the ensemble energy.

Is equal to number of molecules time Boltzmann's constant times t squared

times the patial derivative of the log of the partition function with respect to

temperature. So we'll plug in our partition function

here it is appearing again. It has a very simple dependence on

temperature. When I take the log all of these terms

will break out as separate terms. The only one that depends on T will be

actually log T and the derivative of log T with respect to T is just 1 over T.

So when I carry out this multiplication I'll end up with this very simple Total

rotational energy N K T. If I'm working molar, N would be

avogadro's number and this would R T. Recalling that the heat capacity is

simply the partial derivative, excuse me, the actual derivative of the expectation

value for the rotational energy, and a molar quantity in this case.

I'm looking at molar heat capacity. With respect to t well the derivative to

rt with respect to t is just r. So the molar heat capacity of a diatomic

ideal gas is r. And a way to thik of that if you remember

in the transitional partition function when we worked with it.

We found that there was one half R contribution to each of the translational

degrees of freedom. There's an x, a y and a z mode of

translation, each contributing one half R.

The way to think about a diatomic, so if my hands are the atoms, it can rotate in

two directions - I can rotate like this, or I can rotate end over end, so kind of

like that. So if you think of the two ways that a

diatomic can rotate that are unique, each of those contributes one half R.

And they add up then to R. There is no rotation about its axis, that

doesn't change anything. So that's what is sort of unique about a

linear molecule. So, two degrees of rotational freedom and

they each contribute R over 2, just as each degree of translational freedom

contributes R over 2. Alright.

Well, let's pause for a moment. I'll let you take a look at a self

assessment problem. See if you're picking up on these

concepts and then we'll return. Alright.

The last item I'd like to look at just to make a contrast with the vibrational

partition function. You might remember that it Typical

temperatures, say room temperature. We've found that for most molecular

vibrations, unless they have very low vibrational frequencies.

Almost all the molecules would be found in the ground vibrational state.

And we also know that in the translational levels those are extremely

dense and easily accessed. Well what about the rotational energy

levels. So, if we wanted to compute the fraction

of molecules that you would find in a given rotational level, remember what the

formula is for a fraction in a level. It's degeneracy times "E" to the minus

beta, the energy of that level, divided by the full partition function.

And so, if I plug in, then, for an arbitrary level that the degeneracy is

"2J" plus one And the energy is rotational temperature times j times j

plus one divided by t. So I'll just simplify that and make it

look like this well increasing j you would expect to see the.

The fraction increased because 2j plus one is increasing so the degeneracy

favors more and more occupation of higher levels.

On the other hand the exponential as j is increasing because its e to the minus

this quantity. Is being killed off as j goes to higher

and higher levels So, it's a question of how long does it take before the

exponential dominates because this is a linear increase.

And this is an exponential decrease. And if we just pick an example molecule,

let's do carbon monoxide. And we'll look at it at 300 kelvin.

And so if you recall from a table I showed earlier, the rotational

temperature for carbon monoxide is only 2.8 kelvin.

So we can actually carry out the calculation, plug in the necessary moment

of inertia which would appear in the rotational temperature and here's the

distribution you get. You get that there's less than 1% in the

ground rotational state. It goes up for a while, it peaks at about

the seventh rotational energy level, and that's a little less than 9%.

And then the exponential begins taking over and the population decreases and

decreases. But there's still appreciable population

of the twentieth rotational level and above.

And so, rotation is somewhere between translation and vibration.

As we'd expect, given that the spacing of the energy levels is somewhere between

translational and vibrational energy levels.

That's what contributes in some sense to the heat capacity's ability to put energy

into the rotations. And not necessarily into the

translations, which causes the temperature to go up, so that it takes

more energy into a diatomic than into a monoatomic in order to raise the

temperature by the same amount, because we'll be exciting rotations easily just

as we excite translations easily. Alright, well, that completes our

examination of the rotational partition function itself.

In the next lecture let's put all of these peices together and actually work

with a full ensemble partition function Q for the ideal diatomic gas.

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