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 5

This module is the most extensive in the course, so you may want to set aside a little extra time this week to address all of the material. We will encounter the First Law of Thermodynamics and discuss the nature of internal energy, heat, and work. Especially, we will focus on internal energy as a state function and heat and work as path functions. We will examine how gases can do (or have done on them) pressure-volume (PV) work and how the nature of gas expansion (or compression) affects that work as well as possible heat transfer between the gas and its surroundings. We will examine the molecular level details of pressure that permit its derivation from the partition function. Finally, we will consider another state function, enthalpy, its associated constant pressure heat capacity, and their utilities in the context of making predictions of standard thermochemistries of reaction or phase change. 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 I want to look at Adiabatic Processes.

Â So I'll remind you where we finished up last time.

Â That is, we'd considered three different paths to get from state one to state two.

Â Because state one and state two have the same temperature, the internal energy of

Â an ideal gas, along these various pressure volume paths is unchanged.

Â But heat and work were path dependent. So, let's pause for a moment, I'd like

Â you to actually work with an example just to have some numbers to play with.

Â And then, I'll put some numbers on these paths.

Â And then we will move on and talk about adiabatic processes.

Â All right, well, you've had a chance to work with some numbers.

Â I'm now going to talk about some numbers with the example that we've done.

Â And I'll let you either try to reproduce some of them, or just accept, accept that

Â they are what they are and move on. But, maybe I'll do the simplest one here.

Â So, if I were actually to put some numbers on the process itself.

Â If I were to say that the initial state has a pressure of 4 bar, a volume of a

Â half a cubic decimeter, that's a half a liter, and the final state is 2 bar and 1

Â liter. And it's good to do a sanity check.

Â Do those numbers work out? We said this is isothermal and it's an

Â ideal gas. So in that case P times V should be a

Â constant. Should be equal to R times the

Â temperatures times the number of, of molecules, times number of moles, excuse

Â me, but in any case a constant. So it's 4 times a half, equal to 2 times

Â 1.,Check, they're both equal to 2. Okay, so that works out fine.

Â Those are acceptable initial and final states.

Â Lets then do the case where we have 0.1 mole, that'll just, it's a number to work

Â with. what happens along the DE path?

Â That's the easiest one, in a sense. Because, if you remember, what's the work

Â involved along this first path? It is a constant pressure, P1.

Â What is that pressure? 4, the volume change delta V is from one

Â half to 1 so that's a half a liter. So 4 times a half is 2.

Â So its work done on the surroundings, its surroundings, it's a negative.

Â So its negative 2 liter bars. If I transform that to joules, and I use

Â the fact that I've got 0.1 moles, instead of using molar quantities, I get minus

Â 200 joules. And because the change in energy was 0,

Â the heat along the path must be 200 joules the opposite of minus 200 joules.

Â because summing the heat in the work gives 0.

Â Alright, so those are numbers you could actually plug in relevant logs of volumes

Â and get these numbers. You can work out these numbers as well

Â and I will leave that to your, your own interest if you want to.

Â But now lets move on, I want to look at this Adiabatic Expansion process.

Â So, adiabatic, I'll remind you, what's it mean?

Â It means q equals 0, heat transfer is 0. So, in that case, the change, the

Â differential change in energy which is equal, is equal only to delta w.

Â And in fact, given that energy i=as a state function and we've now taken q out

Â of the equation. Work becomes a state function as well,

Â because it's just equal to the change in energy.

Â So if you like, dU is equal to dw. Whenever either one of these inexact

Â differentials is 0 along an entire process.

Â Then the other becomes an exact differential, so that's sort of

Â mathematical but okay it's worth keeping in mind.

Â So for an ideal gas reversible expansion, it's going to be the case dw equals dU

Â equals CvdT. So constant value heat capacity dT and we

Â also know that dw is minus pdV, that's the definition of the differential work.

Â Its an ideal gas so the pressure is minus nRT over v.

Â So if I equate these things, I get that this heat capacity times dT, is equal to

Â this dependence on volume dV. Integrate both of those quantities.

Â The one from initial to final temperature, the other from initial to

Â final volume. And so I get, that this integral is equal

Â to minus R log V2 over V1. For a monatomic ideal gas, I also have,

Â from having derived partition functions and figuring out what the molar heat

Â capacity is, it's 3R over 2. So now, let me solve this other integral,

Â this integral over heat capacity. So I pull 3R over 2 out, all that's left

Â is integral T1, T2, dT over T. That's the log of T2 over T1.

Â So I have this relationship, 3R over 2 log T2 two over T1 is equal to minus R

Â log V2 over V1. Well, let me simplify that a little bit.

Â So in particular, if I've got a R on this side and an R on this side, I cast them

Â out. I'm left with the three halves on this

Â side. So three halves times the logarithm,

Â that's like the logarithm of something to the three halves power.

Â So let me think of doing that. And then I'll have a log of something

Â equals minus a log of something. Let me make the minus a log of something

Â the log of the, the inverse. So instead of V2 over V1, I'll have V1

Â over V2. Now, I have a log of something equals a

Â log of something, exponentiate both sides, and I finally have this simpler

Â form. T2 over T1 to the three halves power, is

Â equal to V1 over V2. Let me think about what that means for a

Â second. If V2 is larger than V1, this number is

Â smaller than one, and in order for this quantity to be smaller than one.

Â Raising it to a three house power won't change its sign.

Â It must be the case the T2, the final temperature is smaller than one.

Â That's says, smaller than T1 that is. That says that as this gas is expanding

Â adiabatically, it must be cooling. So that's quite important, that gives us

Â a way to control temperature. If we insulate a system and expand a gas

Â within that system, not allowing heat to flow, the temperature of that gas will

Â drop. And so lets take a moment.

Â I'm going to let you work with that concept.

Â Do a little self assessment. And then we'll come back and explore it

Â in more detail. Alright, so you had a chance to see the

Â difference between different kinds of ideal gases when it comes to adiabatic

Â expansion. I also want to compare the adiabatic

Â behavior to isothermal behavior. And so we looked at some gas laws in

Â demonstrations earlier in the course. And Boyle's law, if you'll recall, for an

Â isothermal process, is that the pressure times the volume at a given state, one.

Â Is going to be equal to the pressure times the volume at a given state two.

Â So if I double the volume, I have to half the pressure, if it's isothermal.

Â On the other hand, for the adiabatic process that we just derived for an, for

Â a monatomic gas. Where this relationship prevails If its

Â an ideal gas I can replace the temperature by PV.

Â So T2 is P2 times V2 and T1 is P1 times V1.

Â So I've got some expressions involving V1 and V2 on both sides.

Â I rearrange a bit and I get something that's clearly different from Boyle's

Â law. It's not P1V1 equals P2V2.

Â It's P1V1 to the five thirds power, equals P2V2 to the five thirds power.

Â So if you wi, if you like then, it's less compression power.

Â With nowhere to dump the heat, the temperature has to rise, and that is a

Â consequence of not allowing heat transfer.

Â And because the temperature is rising, the pressure is going to go up, it's

Â harder for me to compress that gas. So if we did not in fact have a monatomic

Â gas. But had say a diatomic gas that would

Â have had a different heat capacity, I would get a different adiabatic

Â expression. And maybe you'll have a chance to see

Â that again later. But, that takes care of what I wanted to

Â do with adiabatic processes. I want to digress briefly as we're

Â looking at PV processes, to think more deeply about pressure actually and in

Â fact to look at the microscopic origin of pressure.

Â And we touched on this very briefly when we were looking at equations of state and

Â their relationship to partition functions.

Â But I'd like to look at it again with a, a bit more care.

Â prior to do that, doing that though let's insert another demonstration into the

Â course here and in fact I'd like to look at the properties of a vortex tube.

Â So after we've had a chance to do that, we'll return to the microscopic origin of

Â pressure. This is a Vortex tube.

Â It has no moving parts, but it has an inlet nozzle to which we can attach a

Â compressed gas tank. For this demonstration, we'll be using

Â compressed air and two outlet nozzles. One of which, has a centrally blocked

Â orifice, and another which has an orifice only at the center.

Â We've already seen how we can use temperature differentials to induce a gas

Â to do work. Remember the Sterling engine?

Â So if we contemplate the reverse, we ought to be able to do work to induce a

Â termperature differential. We certainly know that that is true,

Â because that's precisely what air conditioners are designed to do.

Â But vortex tubes are not the most efficient example of an air conditioner,

Â and they require compressed air, or some other gas or fluid.

Â So they're not as commonly encountered. Nevertheless, let's see what happens when

Â we measure the temperatures of the gas exiting the oppostite ends of the vortex

Â tube. Which will measure with these two thermal

Â couples. I'm going to turn on the computer, which

Â is connected to the thermal couples. And it's recording the temperature.

Â There's a blue line and a red line, and you'll notice they're very close to one

Â another. Now let's turn on the air.

Â [SOUND]. Look at these temperatures.

Â The gas exiting the hot end of the tube is at 28.5 degrees Celsius, while the gas

Â at the cold end of the tube is down to 2.1 degrees Celsius.

Â Without going into all the details, the air injected into the vortex tubes spins

Â very rapidly. And it does so with little exchange of

Â molecules, from the center of the tube to the outside.

Â Since the radius of the rotation is larger for the outer air mass then the

Â inner, this leads to a temperature differential.

Â The shapes of the nozzles caused the warmer outer air to exit one end, while

Â the cooler inner air must exit the other. The effect seems almost magical when

Â first encountered, but of course, it's entirely allowed by the laws of

Â thermodynamics. With work being done by the expanding

Â gas, dedicated to the creation of a temperature differential.

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