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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University of Minnesota

122 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.

- 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.

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

Â diatomic gas, there are four contributors.

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

Â energy which we dealt with in the last lecture.

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

Â rotational energy, and its contributions to the total.

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

Â Schrodinger equation to compute allowed energy levels, and degeneracies.

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

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

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

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

Â squared, and the moment of inertia of the molecule.

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

Â number. So, if we then construct a partition

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

Â summation. Then we would write it this way.

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

Â expression. It's a little bit more convenient just

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

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

Â vibrational temperature when working with the vibrational partition function.

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

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

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

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

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

Â So when we take that approach. The rotational partitional function

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

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

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

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

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

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

Â converges to as we sum out to the last term.

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

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

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

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

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

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

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

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

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

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

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

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

Â condition is satisfied under typical conditions for typical gases.

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

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

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

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

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

Â two atoms involved in the bond gets larger and larger.

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

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

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

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

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

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

Â Has a rotational temperature of 85.3 kelvin.

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

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

Â teens down to 12 for the rotational temperature, kelvin.

Â So those are very cold temperatures obviously.

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

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

Â dihalogon models Chlorine and Bromine, below one Kelvin.

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

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

Â lower. Then the observed temperature that we're

Â working with. So that's easily satisfied then.

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

Â hydrogen working at quite cold temperatures.

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

Â friendly integral, so statisctical mechanics is occasionally very kind to

Â one. And so if we look at that interval it

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

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

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

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

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

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

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

Â temperature divided by the temperature of the system.

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

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

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

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

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

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

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

Â What was A? Well A was the rotational temperature

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

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

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

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

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

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

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

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

Â dependent quantity appearing in the partition function.

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

Â squared. Well, with that partition function in

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

Â And so, if you remember the ensemble energy.

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

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

Â temperature. So we'll plug in our partition function

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

Â temperature. When I take the log all of these terms

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

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

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

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

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

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

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

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

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

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

Â in the transitional partition function when we worked with it.

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

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

Â translation, each contributing one half R.

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

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

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

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

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

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

Â linear molecule. So, two degrees of rotational freedom and

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

Â contributes R over 2. Alright.

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

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

Â concepts and then we'll return. Alright.

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

Â partition function. You might remember that it Typical

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

Â vibrations, unless they have very low vibrational frequencies.

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

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

Â dense and easily accessed. Well what about the rotational energy

Â levels. So, if we wanted to compute the fraction

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

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

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

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

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

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

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

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

Â favors more and more occupation of higher levels.

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

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

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

Â exponential dominates because this is a linear increase.

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

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

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

Â temperature for carbon monoxide is only 2.8 kelvin.

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

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

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

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

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

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

Â decreases. But there's still appreciable population

Â of the twentieth rotational level and above.

Â And so, rotation is somewhere between translation and vibration.

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

Â translational and vibrational energy levels.

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

Â into the rotations. And not necessarily into the

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

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

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

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

Â examination of the rotational partition function itself.

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

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

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