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 2

This module begins our acquaintance with gases, and especially the concept of an "equation of state," which expresses a mathematical relationship between the pressure, volume, temperature, and number of particles for a given gas. We will consider the ideal, van der Waals, and virial equations of state, as well as others. The use of equations of state to predict liquid-vapor diagrams for real gases will be discussed, as will the commonality of real gas behaviors when subject to corresponding state conditions. We will finish by examining how interparticle interactions in real gases, which are by definition not present in ideal gases, lead to variations in gas properties and behavior. 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

Let's take a look in this lecture at gas liquid P V diagrams.

Â Let me start with a P V diagram for carbon dioxide.

Â So what is a P V diagram? It is a plot of pressure.

Â In this case in units of atmospheres on the y-axis, and molar volume and units of

Â liters per mole on the x-axis. And shown on the plot itself are so

Â called isotherms. That is it is a plot of p versus v bar at

Â constant temperature. That's what isotherm means, at a constant

Â temperature. So the highest temperature on here in

Â degrees celsius, 47.8 degrees celsius, and then we drop to 36.2, 32 and so on,

Â 22.6, 13.2 degrees C. So, there's a few interesting features on

Â this plot I want to call to your attention.

Â One is this horizontal behavior of some of the lines.

Â So let's think about what you're actually, you might be doing in an

Â experiment. Let's say again that I've got a

Â container, and I've got a piston. And I'm pressing harder and harder with

Â my piston. And so what I would expect to happen If

Â I'm starting over here on the right hand side, at a low pressure.

Â At a low pressure I would expect a larger molar volume.

Â I am able to have ore volume per mole because I'm not pressing on it so hard.

Â So as I press, the molar volume should go down.

Â And, so I'm pressing higher, and higher pressure and the volume's going down,

Â down, down. And then suddenly my pressure state is

Â constant for quite a while, while the molar volume goes down, down, down, down,

Â down. And remember the relationship between

Â molar volume and density is inverse. So, if the molar volume is going down,

Â the density is going up. So this is the situation I described in

Â the last lecture. I'm liquefying my gas.

Â So when you see this kind of behavior, where there's a sudden change in the

Â slope of the line. It's basically a non differentiable point

Â in that line. It remains horizontal for a period of

Â time and then suddenly the relationship between pressure and molar volume shoots

Â up. And that's because liquids are very

Â incompressible so to get a smaller molar volume takes an enormously larger amount

Â of pressure. Now if i raise the temperature from 13.2

Â degrees to 22.6 degrees in this case. Well, I can press to a higher pressure

Â before and a smaller molar volume before I again see that horizontal behavior take

Â place. Over which time period I'll liquefy the

Â gas. And if I were to connect up all of these

Â discontinuity points, that's what this dotted line is.

Â If there were many, many isotherms, the dotted line would follow the,

Â discontinuities. And we call that the Co-existence curve.

Â That is, it marks the edges of a region of pressure and molar volume where the

Â two phases co-exist, both liquid and gas. And the right-hand side tells you what's

Â the molar volume, or the density, of the gas during that compression period.

Â And the left hand side tells you what's the molar volume of the liquid during

Â that period. Of course after you've got a pure phase

Â on one side or the other, the molar volume will again become sensitive to the

Â pressure. But during that period, as one phase

Â transforms to the other, you've got both and they've got their specific molar

Â volumes. Now notice what happens as I trace the

Â coexistence curve up, up, up, up, up. At some point it turns over, that is

Â there is a point where one isotherm just kisses the very top of the coexistence

Â curve. That is called the critical temperature.

Â It is the temperature above which you cannot liquefy a gas anymore.

Â It doesn't matter how much you press on it, it will never be in the liquid phase.

Â So a term to remember the critical temperature.

Â So as the temperature approaches the critical temperature, the curve flattens

Â and you get this point of inflection essentially.

Â And then the T below the critical temperature, you see horizontal regions

Â where gas and liquid co-exist. So remembering these different regions is

Â critical to understanding a P V diagram. So how did non-ideal cubic equations of

Â state do for that. So here is the Van der Waals equations

Â state plotted for carbon dioxide. I'm going to use the a and b parameters

Â Specific to carbon dioxide at various temperatures.

Â I will plot these isotherms on a P V diagram.

Â And what I see is that at the higher temperatures, that looks pretty similar

Â to what we saw in the last P V diagram. And sure enough, as I lower the

Â temperature, I start to see this flattening effect.

Â And it looks as though I'm getting to some kind of a point of inflection.

Â And then I saw something a little odd. It actually rolls over, drops for a

Â little while and then goes up again, as, as they all should at this side, right,

Â takes a lot of pressure to compress a liquid.

Â So, this is reasonable behavior and really, why is that?

Â So I mentioned in an earlier lecture, if I were to rearrange the van der Waals

Â equation. In order to express it with molar volume

Â taken to positive powers instead of appearing in denominators.

Â This would be the correct expression. It's v bar cubed minus quantity b plus rt

Â over p times v bar squared. Plus a over pv bar minus a constant, ab

Â divided by p equals 0. So that is a cubic equation.

Â And we know what cubic equations look like, they, well, they look kind of like

Â this actually. And it can rise, dip and rise again.

Â And under certain circumstances, maybe there is a point of inflection.

Â So, it, it's understandable why a cubic equation might be capable of reproducing

Â this physical behavior. And it turns out, I'm not going to show

Â you the equations. But the Redlich-Kwong and the

Â Peng-Robinson can similarly be rewritten as cubic equations in the molar volume.

Â And they show similar behavior in terms of being able to match pv isotherms.

Â So, let me take a little bit closer look at the process and remind you of what

Â happened in a real pv diagram. In a real PV diagram, if I start at a

Â lower pressure, and a large molar volume. As I increase pressure, the molar volume

Â drops and drops as I'm compressing. And then I would reach some point at

Â which in principle, the pressure remained constant while the volume continued to

Â drop. And then finally, I would have

Â transformed gas, at this point, to liquid, at this point and I would rise

Â substantially in pressure. What we see with the Cubic equation of

Â state is, so called van der Waals loops. These are unrealistic, why are they

Â unrealistic? Well, think about what that slope of that

Â line there in the isotherm means. The slope would be the change in

Â pressure, divided by the change in molar volume.

Â And in this case, it is positive, right? It's actually saying that as I increase

Â the pressure, I increase the molar volume.

Â That's what it takes to have a positive slope.

Â And that doesn't make any sense at all, as I increase the pressure on something,

Â I ought to decrease its molar volume. So that's a flaw in the equation of

Â state. I'll also note, I,I called out these

Â points A and D and said, here's where we would see the transformation from gas to

Â liquid. And one might ask, why there, I could

Â gave drawn this line down here or I could have drawn it up here?

Â So you can make thermodynamic arguments that I don't actually want to detour off

Â to derive. But something called the Maxwell equal

Â area construction. If you place the line where the area

Â above the line is equal to below the line, that's a pretty good estimate of

Â where the coexistence curve will be passing through.

Â Now, one of the points I want to make about this cubic equation of state, is

Â that because it's a cubic equation. At least at a temperature like this,

Â there are three roots to this polynomial cubic equation.

Â And they're real roots. But as the temperature goes closer and

Â closer to the critical temperature. The positions of the roots, so if I look

Â at root zero would be for instance here, and here, and here.

Â Right, they cross an axis, that cubic equation crosses an axis.

Â But if I get to the point where instead of going up, down, I go to a point of

Â inflection. That's equivalent to all of the roots

Â converging on a single triply degenerate root.

Â And that happens at the critical temperature.

Â Which is to say that the cubic equation at that point is molar volume minus the

Â critical molar volume. So that's the molar volume at the

Â critical temperature, cubed is equal to zero.

Â Right, it's a triply degenerate root. V is Vc, V bar is V bar c that is.

Â So, let me just expand out that cubic equation.

Â That very simple one. V bar minus v bar c cubed.

Â So all I did was cube that expression. And so I'll get v bar cube minus 3 v bar

Â c, v bar squared plus and so on. And the nice thing about that is let me

Â just compare here. I've got v cubed compares to v cubed.

Â Here's a term in v bar squared. So, I can relate 3 times the critical

Â molar volume to b plus r t over p. And I can relate 3 v bar c squared to a

Â over p. And, evidently, v bar c cubed is ab over

Â p. So, if I have this relationship between

Â the two equations. And I do the algebra required to actually

Â solve for the critical volume, the critical pressure, and the critical

Â temperature. I get these expressions.

Â So I'll let you do the algebra for yourself if you like to.

Â There's a whole lot of, you know, pushing terms around.

Â But the bottom line is the critical volume, critical molar volume, is 3 times

Â B. The critical pressure is a over 27 b

Â squared. And the critical temperature is 8a over

Â 27 times b times the universal gas constant.

Â And that actually is where these parameters in the Van der Waals equations

Â of state come from. So they're not just plucked out of thin

Â air. They're not just necessarily played

Â around with until they seem to fit some set of experimental data.

Â Instead, you do experiments required to get at the the critical properties.

Â The critical pressure, the critical temperature, the critical volume.

Â And once you have those in hand, you select the a and b parameters that best

Â reproduce the experimental data. And so if we just want to take a look at

Â a few of those data, I've, I've tabulated some of them here.

Â Helium, neon, argon, krypton hydrogen. So these are, these are the noble gases

Â increasing in size and mass. And then molecular hydrogen and carbon

Â dioxide. And so these are the critical

Â temperatures. Notice that helium's critical temperature

Â is only 5.2 roughly degrees Kelvin. So above five degrees Kelvin.

Â That's only five degrees above absolute zero.

Â You can not liquefy helium. So if you hear someone refer to liquid

Â helium, it is a very cold substance. It's typical about four degrees Kelvin

Â actually. neon has a higher critical temperature,

Â 44. Argon, 151, krypton is actually starting

Â to get closer to room temperature. Molecular hydrogen, again very cold, the

Â critical temperature. Carbon dioxide, 304.14 degrees, sorry,

Â don't say degrees because it's Kelvin, Kelvin, so that's above room temperature

Â in fact. And one can also look at the crit,

Â critical pressures expressed in bar or atmosphere.

Â The relevant molar volumes. They're a little bit less interesting.

Â And then over here is the compressibility at the critical temperature.

Â So remember the compressibility z. That was pv bar over rt.

Â And now we're in the specific instance where critical pressure, critical volume,

Â molar volume, and critical temperature are the relevant p v bar and t.

Â And notice something kind of interesting here.

Â Although there's enormous variation in these numbers, the critical temperature,

Â the critical pressure, the critical volume.

Â The compressibility actually spans a fairly small range.

Â Looks like the smallest one is about 0.274.

Â The largest one is 0.304. That's only about a 10% range in

Â compressibility. Why might that be?

Â Well, actually. Now that we know how to relate the Van

Â der Waals parameters to the critical properties.

Â Let's plug that in. That is, here's r.

Â R is always r. Let's just take out the 1 over r.

Â And we'll replace the critical pressure with its expression using Van der Waal's

Â parameters. We'll replace the critical molar volume

Â with this expression. And finally we'll replace the critical

Â temperature with its expression, 1 over that because the critical temperature is

Â in the denominator. When you do that, let, let's just see

Â here, I gotta b squared down in the denominator, I've got two b's up here, so

Â that's be squared. They all cancel out.

Â I've got an a in the numerator, I've got an a in the denominator, they all cancel

Â out. After that it's a little bit of

Â arithmetic with numbers. Oh, actually the 27 cancels out the 27

Â and the r cancels out the r. I'm left with three divided by eight.

Â I do not even need a calculator to do that.

Â Three divided by eight, 0.375. Not factorial, but an exclamation point.

Â That's amazing. It says that for every gas at least

Â within the Van Der Waals equation of state approximation.

Â The critical compressibility is a constant at 0.375.

Â Now in fact, if we refer back to this column, it's not 0.375 but it's not that

Â far. So there's an apparent correspondence

Â between different real gases. That's independent of the Van der Waals

Â equation of state. Right?

Â I've not really invoked any parameters. I canceled them all out, and that is

Â known as a, a correspondence property, and that's really quite an interesting

Â thing, I would say. So let's pause for a moment, and I'll

Â give you a chance to recapitulate what you've picked up about Van der Waals'

Â parameters, and then we'll pick back up. Alright, so you've come to appreciate how

Â attractive and repulsive forces play a role in the Van der Waals equation of

Â state. Let's next spend a little of time looking

Â at the Law of Corresponding States that I've provided some fore shadowing of in

Â the last slide. On this lecture But prior to doing that,

Â since I've shown you the P V diagram for carbon dioxide, let's take a little time

Â off. And we'll do a demonstration of carbon

Â dioxide under various pressures. We'll look at the triple point, and

Â explore some other features of this interesting substance.

Â In this course we won't have time to examine so called phase diagrams in

Â detail. But in discussing gasses, we've had a

Â chance to at least discuss vapor liquid coexistence curve and critical

Â temperatures. And in one the very first lectures, I

Â mentioned the wide spread use of super critical carbon dioxide in the dry

Â cleaning industry. As a green alternative to Chlorocarbon

Â solvents that used to be used. So, in this demonstration I'd like to

Â illustrate properties of carbon dioxide. Some of which are relevant to our

Â discussion of gas and liquid properties. To begin let's add some dry ice to this

Â plastic bottle. If we now seal the bottle tightly and

Â warm it a bit. The dry ice will sublime, increasing the

Â amount of gaseous CO2 and given the fixed volume, also increasing the pressure.

Â [SOUND] As the pressure grows, we eventually come to the point, [SOUND]

Â where some of the gas liquefies. At this stage, we have solid, liquid, and

Â gaseous CO2 all in equilibrium with one another.

Â And this is called the triple point, because all three phases are present

Â simultaneously. It occurs at a precise temperature and

Â pressure. Neither of which is particularly extreme

Â for CO2. About five atmospheres pressure and a

Â temperature of minus 56.6 degrees celsius.

Â Can you see the liquid and solid phases? And of course we can't see the gas, but

Â it is there in the apparent void volume. [SOUND] In the next demonstration, I'd

Â like to illustrate the critical temperature for C02.

Â Which you'll recall is the temperature beyond which the gas cannot be compressed

Â into a liquid at any pressure. And indeed, we stop calling the substance

Â a gas, and call it a supercritical fluid. For CO2, the critical point occurs at

Â about 73 atmospheres and 31.1 degrees celsius.

Â So here I have a sealed tube containing carbon dioxide at high pressure.

Â As room temperature is a bit below 31.1 degrees celsius.

Â There's clearly a liquid phase and you can see that by observation of the

Â meniscus. That is, the boundary between the liquid

Â and the gaseous phases. That we see more clearly, because of the

Â change in indexive refraction across the phase boundary.

Â Now let's gently heat the tube with this economical substitute for a heat gun,

Â namely, a hair dryer. Keep an eye on the meniscus.

Â Do you see how it's starting to disappear?

Â As I warm the tube past the critical temperature, the carbon dioxide becomes

Â supercritical. And I can no longer discern two phases.

Â Now an interesting thing about the critical point is that as it is

Â approached from above in temperature. One often observes the phenomenon known

Â as opalesence. The condensation of gas to liquid that

Â occurs everywhere in the tube leads to very small suspended drops that scatter

Â like beautifully. Until a new liquid phase is formed.

Â So lets watch carefully as the tube cools.

Â There, do you see the opalescence beginning.

Â And as it fades, do you see the liquid phase increasing in volume, and the

Â meniscus rising up the tube? As supercritical fluids go, carbon

Â dioxide is relatively easy to work with, because the pressures and temperatures

Â involved are not especially extreme. And the characteristics of CO2 as a

Â solvent, are useful for a surprising number of things.

Â In addition to dry cleaning, super critical CO2 extraction, is now one of

Â the two most widely used methods to decaffeinate coffee and tea.

Â Indeed new Me Tea Company calls it super critical carbon dioxide process an

Â Organic use of effervescents and further calls the process, chemical free.

Â Such chemophobia can only make one shake one's head.

Â But now, you've seen a supercritical fluid for yourself and we'll look more at

Â critical constants in the lectures.

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