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 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

Lets talk about the natural independent variables of thermodynamic state

Â functions. So by now, it should almost be second

Â nature to know that when you see d u you can think to yourself TdS minus PdV.

Â That is an expression of the first and second laws being substituted in.

Â To a replace work and heat in the differential form of the internal energy.

Â And so, if you consider S and V the differential quantities appearing on the

Â right side as independant variables of the internal energy.

Â Notice that the coefficients in this case are simple thermodynamic functions, T and

Â minus P. So if I instead just use calculus to

Â express the differential of U with respect to S and V.

Â I would have gotten dU equals partial U partial s partial dS plus partial

Â informing us, so this looks like beginning to assess a Maxwell relation.

Â Informing us that partial U, partial S at constant V is equal to temperature.

Â Partial U partial V at constant S is equal to minus pressure.

Â And, let me just compare with what might have happened if we had chosen two

Â different variables. What if I had expressed U in terms of V

Â and T? Then, I would have gotten partial U,

Â partial V with respect, holding temperature constant dV.

Â Plus partial U partial T at constant volume, dT.

Â Well, actually this we've seen before. The change in internal energy with

Â respect to the change in temperature at constant volume, that's called the

Â constant volume heat capacity. So I'll, I'll just write that.

Â This other term, I'm going to refer you back to video 8.3, where we actually

Â derived that. We looked at the volume dependence of the

Â internal energy, and we used maximal relations to determine this volume

Â dependence. But notice that its considerably more

Â complex. Right, I've got this prefactor that

Â involves temperature, and pressure, and the derivative.

Â It's not just a trivia quantity like temperature or minus pressure, and so we

Â say that entropy and volume are the natural independent variables of U.

Â Because they allow these differentials appearing in the total differential to

Â have very simple thermodynamic forms. So lets pause for a moment and I'm going

Â to let you consider a differential form. So really, as long as you remember the

Â one differential for internal energy, that derives from the first and second

Â laws. And you remember the definitions of the

Â other thermodynamic state functions. You can work out all of the useful

Â differentials. So, here we have the key one, dU equals

Â TdS minus PdV. Now, let's add the differential of PV to

Â both sides. And so, if I do that, I get the d of U

Â plus PV. So I'll need to add a VdP and a PdV.

Â When I do that, U plus PV that's H. So on the left hand side, I've got dH.

Â And when I look at the collection of terms, here's what came from U, here's

Â the new dPV and minus PdV cancels plus PdV so I'm just left with dH equals TdS

Â plus VdP. That's pretty simple.

Â If I subtract ts from both sides of internal energy.

Â So now, I have a, I need a dts. Well, that'll give me minus TdS minus

Â SdT. U minus TS is A.

Â So, this is the differential of A. And again, I'll have a cancellation the

Â TdS cancels minus TdS. I end up with dA is minus SdT minus PdV.

Â At this stage, you see sort see where this is going.

Â So, last thing we're going to do is with the Gibbs free energy I'm going to add a

Â PV and subtract a TS when I put in all those differentials.

Â Plus VdP, plus PdV, minus TdS minus SdT, I get a cancellation of this term from

Â the internal energy. I get a cancellation of these two terms,

Â one from the internal energy, one new. And the only ones that survive dG on the

Â left hand side, because G is U plus PV mius TS.

Â On the right hand side, the surviving terms are minus SdT plus VDP.

Â And so, deriving all these only required remembering the first and second laws and

Â how they allow to write dU in this fashion.

Â And the definitions of enthalpy, Helm Holts free energy, Gibbs free energy.

Â And now that I have these expressions, I also can just look at and discern the

Â natural independent variables. So if U, dU is TdS minus PdV, the

Â variables are these differentials, S and V.

Â Whoops, I, I don't know, I threw two up there, but sure, let's do two at once.

Â And so, the natural independent variables for the entropy can be expressed just by

Â keeping dS on one side, moving PdV over to the other side and dividing everything

Â by T. And so, you get natural independent

Â variables of internal energy and volume, U and V.

Â For the enthalpy you get S and P. And as we saw in the last slide for the

Â helms hold free energy T and V, and for the Gibbs free energy T and P.

Â Maybe these aren't so surprising right that we use these free energies when

Â these quantities are being held constants.

Â So, that's another way to remember what are the natural independent variables.

Â That gives rise to a series of Maxwell relations.

Â So, insofar, as these quantities are the first derivatives, if you like.

Â The first differentiation partial derivatives of the quantity on the left

Â with respect to the variable that they multiply.

Â Then, when I take the next derivative, when I differentiate with respect to the

Â other variable, so T is partial U partial S.

Â If I now differentiate T with respect to V, the other variable, that must be equal

Â to differentiating this, partial U, partial V, with respect to the entropy

Â variable. So remember, that's what a Maxwell

Â relation does. And if in words that sounded a little

Â confusing, let's sort of just look at it, maybe almost graphically.

Â So, here I have the variables themselves S and V that multiple differentials.

Â What happens to them? They stick around in the numerators of

Â the Maxwell relation. Alright, so I differentiate those they

Â are already first differentials I'm going to create a mixed differential.

Â And what goes in the denominator? Well, I'm going to differentiate with

Â respect to the other variable. So, I'm going to hold the first variable

Â constant. So, if I've got dT here, I'm holding

Â temperature constant when I differentiate S.

Â And here I've got P, I'm going to hold pressure constant when I differentiate V.

Â So, those stayed on the same side of the two terms, if you will.

Â And then, there's a reversal of terms. What am I differentiating with respect

Â to? I've got t for the first one and it's

Â going to show up over here. For the second one, I've got P for the

Â second one, it shows up over here for the first one.

Â So, the green and the sort of hot pink colored arrows, they are the ones that

Â invert unlike brown and blue which stay in their respective sides.

Â So operationally, that's how you develop a Maxwell relation by looking at these

Â you can just jot down the Maxwell relations.

Â Okay, well this is important enough and in principle enough once you've done a

Â few of them. Let's pause and, and do one more.

Â Great, hopefully that was pretty easy. As one does this again and again, it sort

Â of becomes second nature to form those differentials.

Â So, that's the end of the material for this video.

Â Got one more, one more video of new material to take a look at.

Â And in particular, given how often chemists preform experiments at constant

Â pressure and constant temperature. Which is to say sort of open glassware,

Â if you will, on a lab bench. I want to focus a little more carefully

Â on the pressure and temperature dependence of the Gibbs free energy.

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