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

Loading...

From the course by University of Minnesota

Statistical Molecular Thermodynamics

143 ratings

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

From the lesson

Module 8

This last module rounds out the course with the introduction of new state functions, namely, the Helmholtz and Gibbs free energies. The relevance of these state functions for predicting the direction of chemical processes in isothermal-isochoric and isothermal-isobaric ensembles, respectively, is derived. With the various state functions in hand, and with their respective definitions and knowledge of their so-called natural independent variables, Maxwell relations between different thermochemical properties are determined and employed to determine thermochemical quantities not readily subject to direct measurement (such as internal energy). Armed with a full thermochemical toolbox, we will explain the behavior of an elastomer (a rubber band, in this instance) as a function of temperature. Homework problems will provide you the opportunity to demonstrate mastery in the application of the above concepts. The final exam will offer you a chance to demonstrate your mastery of the entirety of the course material.

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

Chemistry

Okay. Let's look at rubber band thermodynamics.

Â And, you know, when I did the thermite, reaction as a demonstration, that was

Â definitely a kids-don't-try-this-at-home sort of demonstration.

Â But actually, in the course of this lecture, there will be a

Â please-try-this-at-home, aspect. So, if you would like to go and get

Â yourself a rubber band, I have one here late at the end of the lecture there'll

Â be a chance to try something out. And see if what I'm telling you is true

Â or not. But for now let's dive into the

Â thermodynamics of a rubber band. So, when you stretch a rubber band, I'll

Â take my rubber band, I'll stretch it a bit, there is a restoring force, let's

Â call it f. That's present when you stretch the

Â rubber band longer than its equilibrium length, that is, the length when it's

Â just sitting there. And, as long as we don't stretch it too

Â far, that restoring force will be a constant that does not depend of the

Â length l. And to accomplish that stretching, I'm

Â doing work on the rubber band. And so what is my work It is f times dl,

Â force times distance is work, that's classical physics.

Â And that's non-PV work. And so I promised a couple of videos ago

Â I'd show an example of non-PV work. Here's an example of non-PV work.

Â Moving something a force times a distance.

Â There are many kinds of non-PV work. There's chemical work, changing the

Â nature of chemical compounds. There's electrical work, moving electrons

Â against potentials. There's magnetic work.

Â Lots of different sorts of work we can extract from a system.

Â So, in this particular case we're doing non-PV work of stretching the rubber

Â band. And then of course there is PV work as

Â well and that's minus PdV, the way it always is.

Â So, note our sign convention here. If I am stretching the rubber band, dl is

Â positive, and I'm doing work on the system.

Â So, the non-PV work should be positive, and so f must be positive, the force, the

Â restoring force. Because I am increasing the rubber band

Â length and I'm doing work on the system. Now, actually for a rubber band, when I

Â stretch it, it's volume really does not change.

Â It's, it's getting longer, certainly, but it's also getting a little thinner as I

Â stretch it. And it's not a gas, it, it really is

Â occupying pretty much the same volume. And so we can treat dV as being roughly

Â 0. And if I'm doing this isothermally, room

Â temperature, I'm really operating at constant temperature and constant volume.

Â And that suggests that the right thermodynamic variable for this process

Â that I should be considering is the Helmholtz free energy.

Â So let's look at that. A equals U minus TS.

Â And I want to take the differential. So, when I take that differential, I get

Â dA equals dU minus TdS minus SdT. And I'll replace, as always, using the

Â first law. That dU is del q reversible, plus del w

Â reversible. Remaining terms here.

Â Using the second law I replace del q reversible with TdS.

Â And the reversible work in this case, because I don't have a, a PV change, I'm

Â working at constant volume. Is fdl.

Â It's that remaining work. And so the TdS cancels this TdS.

Â I'm left with fdl minus PdV minus SdT, but I'm at constant volume and I'm at

Â constant temperature. And so I get simply fdl.

Â So because dV equals dT equals 0, I have that the change in Helmholtz free energy

Â is equal to the force times the differential of the length change.

Â And so if I write that in another way, I have that the force, if I were to need to

Â quantify it, I would look at the change in the Helmholtz free energy.

Â With respect to a change in length at constant temperature.

Â Well, let me go on and relate that force to entropy.

Â So I have A is equal to U minus ts, and I'll just keep around what I already

Â determined about force. So, let me differentiate A this

Â expression for the Helmholtz free energy, with respect to l length.

Â So, partial A, partial l is equal to partial U, partial l minus T partial S

Â partial l. Well, a rubber band is a so called

Â elastomer and the internal energy U of an elastomer, it's a little bit like an

Â ideal gas. In the sense that, that is the ideal

Â elastomer. It depends only on the temperature T, and

Â not on the length. So, just as U depends only on the

Â temperature and not on the volume, U for a perfect elastomer depends only on the

Â length, sorry, only on the temperature, and not on the length.

Â And a rubber band is close enough to a perfect elastomer that we can make that

Â approximation. In which case we have, since this will be

Â 0, that partial A partial l, which I already know is equal to f is equal to

Â minus T, this remaining term. Minus T partial S partial l, with respect

Â to temperature. Alright, so I'll just rewrite that.

Â I'll get rid of the common thing in the middle, and I've got force is equal to

Â minus t partial s, partial l. So, looking at that relationship I now

Â see that the force ought to decrease. It will have a negative value, if the

Â entropy increases with increasing l. So, that would be a positive divided by a

Â positive. Temperature is always positive.

Â So, I'd get a negative quantity leading to a reduced force.

Â On the other hand, if the entropy decreases with stretching, so if the

Â change in S is negative divided by a positive change in length, this quantity

Â will be negative times a negative. The change in force would be positive, so

Â the force would go up. And so that leads to the question, what

Â do we expect to be the change in entropy when we take a rubber band and stretch

Â it? So, what does one expect?

Â Well, you need to think about the molecular nature of a rubber band to

Â answer that question. So, a rubber band is really a collection

Â of tangled and cross linked polymers. And, when I stretch the rubber band, I

Â force those polymers on the left hand side of this picture, which are all sort

Â of tangled and hear are the little cross links.

Â I force them to better align along the axis that I'm stretching.

Â Alright. By pulling the ends apart, those polymer

Â chains have to lengthen, and to do that they have to align along the axis of the

Â stretch. That leads to a greater ordering of those

Â chains and hence it leads to a decrease in entropy.

Â Alright. That means that delta S is negative.

Â Delta l is positive, it's going up. So, the force should be increasing as I

Â stretch the rubber band. As, as I raise the temperature, sorry,

Â that's actually what we're doing of course, we're changing the temperature.

Â So, temperature is going up. Delta s is going down, negative quantity

Â times negative quantity is positive, so the force gets larger and larger at

Â higher temperatures. And that's exactly what we saw in the

Â demonstration video, when we heated the rubber band with our heat source it

Â pulled harder against the weight making the apparent weight decrease on a scale.

Â So, the rubber band increases its restoring force.

Â Okay. Well, hopefully you saw how that all went

Â together, how we were able to use the concept of the helm holds free energy.

Â And these differentials to relate the force to the entropy.

Â Let me, give you one thing to think about for yourself, having to do with the

Â adiabatic stretching of a rubber band as opposed to the isothermal, and then we'll

Â come back. All right.

Â Well, that example we just did, if you stretch your rubber band suddenly, that

Â can correspond to an adiabatic change. Remember, adiabatic means no heat

Â transfer from the surroundings to the system, in this case, our system is our

Â rubber band. Now, we're not insulating our rubber

Â band, but if I do it really really quickly, there's not much time for heat

Â to transfer to the system. And so if I do that, I have dU is equal

Â to dell q plus dell w. It's not reversible anymore, I'm doing it

Â quickly. It's presumably irreversible.

Â But, I have dell q as equal to 0, the adiabatic limit, that leaves del w which

Â is equal to fdl plus the PV work, but we aren't changing the volume so there is

Â none of that. So, I just get that dU is equal to fdl,

Â which is the result you should have just got.

Â Now for a perfect elastomer, U depends only on T.

Â And so, you can define a constant length heat capacity, just like we defined a

Â constant volume heat capacity for a gas. So if you like, C sub l for length in

Â this case, T, is equal to the change in internal energy with respect to the

Â change in temperature at a given length. And if I rearrange this, I have that dU

Â is equal to ClT, dT, right? How much heat does it take to raise the

Â temperature of my ideal rubber band by 1 degree?

Â That's the same sort of heat capacity that we use in gases.

Â So dU equals ClT, dT. But it's also equal to fdl, and so that

Â says that force times distance displacement will be equal to the heat

Â capacity dT. So, let me just rewrite that in terms not

Â of d's, but deltas, so we'll actually make a non infinitesimal change.

Â So, force times delta l is equal to heat capacity times delta T.

Â Well, we know that f is positive, and the constant length heat capacity must also

Â be positive. Because as we add temperature, sorry as

Â we add heat to the system as we increase the temperature, the internal energy goes

Â up. So, it's a, increase divided by an

Â increase, positive over positive, it's a positive quantity.

Â And so that says that if I make delta l positive, delta T must be positive.

Â That is, if I suddenly increase the length of the rubber band, the

Â temperature of the rubber band should go up.

Â So, this is a lovely experiment to try at home.

Â So, take your rubber band, hold it between two fingers.

Â Make sure that your rubber band is not so old that it will snap and hurt you, but

Â pull it suddenly. And then you want to use a sensitive

Â thermometer, and in this case, your upper lip actually works out pretty well.

Â So, the rubber band ought to go from feeling, Mm-hm, sort of cool, actually, a

Â little bit less than room temperature, it seems like, or at least less than body

Â temperature. And I pull it suddenly, and I place it

Â against my lip, and it's noticeably warm. Yes, that is one hot rubber band.

Â and that conforms to the thermodynamics we just derived, that delta T must go up

Â as delta l goes up. So, thermodynamics in action, what, what

Â more could one ask for? Alright, that's our rubber band house for

Â this particular lecture. I want to return to something a little

Â bit more formal for the next video. And in particular we'll look at the

Â natural independent variables of thermodynamic functions.

Â Coursera provides universal access to the worldâ€™s best education,
partnering with top universities and organizations to offer courses online.