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

Welcome to the second week of material on molecular statistical thermodynamics and

Â in think lectures I want to take a look at the ideal gas equation of state.

Â So, a question that arises is, why study gases?

Â Well, gases have a number of advantages when it comes to learning about

Â thermodynamics. Gases are certainly the simplest phase.

Â They're very dilute in character over many ranges of pressure and temperature.

Â And as a result, we can change those variables by large amounts, and look at

Â how the behavior of the system changes. And that allows us to investigate key

Â principles that underlie thermodynamics. And it's really the imperical study of

Â gases that led to the hypothesis and ultimate development of many of the key

Â laws of thermodynamics. And so of course in these pictures we see

Â some of the uses that we put gasses to, today.

Â Obviously we change temperature of gas in hot air balloons in order to make them

Â buoyant in the gas of the atmosphere. You're looking at, in the upper right, a

Â geothermal plant and the steam that's being produced by it which can be used to

Â generate power. And then in the lower right are a number

Â of cylinders of compressed gases which may have different industrial utilities

Â as. Fuels, as chemical reagents as whatever.

Â So, probably one of the most most well known formula, in all of chemistry is the

Â ideal gas equation of state. And so before actually tossing up the

Â formula, which many of you will have seen before, let's talk about what is an

Â equation of state? So an equation of state is something that

Â details the mathematical relationship between a set of physical observables.

Â And in this case, the observables in question are the pressure, the volume,

Â and the temperature of a gas. And the ideal gas law says that a gas is

Â defined as ideal if it obeys the following relationship between those

Â variables. Pressure times volume is equal to the

Â number of moles n times a constant, capital R which is called the universal

Â gas constant times the temperature. So we'll talk a little bit more about the

Â universal gas constant in not too long. For the moment, we'll just accept that

Â it's something, for instance, that allows the units to work out.

Â Temperature, obviously, is in degrees, Kelvin for example.

Â And pressure and volume have their own units.

Â A somewhat more convenient way to write this equation, for our purposes, although

Â many of you may have. Memorized PV equals nRT at some point in

Â the past. Is to replace V with V over n.

Â And when we do that, we represent that with a bar over the V.

Â A V with a bar on it is the molar volume, the amount of volume occupied by one

Â mole. And that allows us to express a couple of

Â definitions that will be useful. And that's the difference between an

Â extensive variable and an intensive variable.

Â So what's an extensive variable? Well, it's something that actually

Â depends on the size of the system, the number of particles in the system.

Â And so, for example, the volume depends on the number of particles.

Â the mass depends on the number of particles.

Â So if we're talking about a gas, if there are two moles, it will weigh twice as

Â much as one mole. And finally the energy is an extensive

Â variable. Intensive variables by contrast, do not

Â depend on the size of the system. So, the temperature is simply a property

Â of the entire system, irrespective of whether it is large or small.

Â So, to the pressure is an extensive variable.

Â And the density, that is, the moles per liter, for example, is a constant that

Â would describe a homogeneous system. So let's take a moment here and I'll give

Â you a chance to take a look at one or two variables and access for yourself whether

Â you think they would be extensive or intensive variables within a given

Â system. So, the ideal gas equation of state can

Â be expressed in a number of different ways.

Â And each one of those different ways of expressing it has come to be associated

Â with different ideal gas so called laws. Let's take a look at a few of those laws.

Â So I've just reproduced here the equation of state again in its two forms.

Â PV equals nRT or replacing V and n with V bar.

Â And so let's now consider the situation where the temperature is held constant.

Â Well, if the temperature is constant, given that R is a constant, then the

Â product of P and V bar must also be a constant for any two allowed values of P

Â and V bar. And that's known as Boyle's law.

Â So P1 times V bar 1 would be equal to P2 times V bar 2.

Â If we hold a different variable constant. In this case, the pressure.

Â In that case, the ratio of the volume, the molar volume and the temperature must

Â be constant. So expressed as an equation, that's V1

Â bar over T1 must be equation to V2 bar over T2.

Â Or put perhaps a little bit more more simply, if I were to double the

Â temperature at constant pressure I would have to double the volume in order to

Â keep the same pressure. And that law is known as Charles's law.

Â Last if I were to hold the molar volume constant or that's equivalent to holding

Â the volume and the number of moles constant.

Â Then it must be the ratio of pressure and temperature which is constant.

Â This is known as Amonton's law. And if we were to write it as an equation

Â it would be P1 over T1 equals P2 over T2. Then again if we put that in a somewhat

Â more, typical conversational form, we'd say for a fixed volume, if I double the

Â temperature I double the pressure. And last of all if I have a constant

Â pressure and temperature and I do consider variable number of moles, then

Â it must be the case that the ratio of the volume and the number of moles must be a

Â constant. That's Avogadro's Law.

Â And so given a constant pressure in temperature, if you double the number of

Â moles you must double the volume. And as an equation that's V 1 over n 1 is

Â equal to V 2 over n 2. Note that's not molar volume anymore.

Â We've actually got number of moles in the expression.

Â So let's investigate a little bit more closely temperature as a concept almost

Â as well as the universal gas constant. So what it temperature?

Â It turns out that the operational definition of temperature is the

Â following- Temperature if, if you rearrange the ideal gas equation.

Â Temperature would be pressure times the molar volume divided by the gas constant.

Â Now that's the ideal gas equation, and so we can ask ourselves, when do gasses

Â behave ideally? And the answer to that question as we

Â will see with some data in not too long is when gases are extremely dilute, they

Â have their most ideal behavior and in the limit of infinite dilution they are

Â indeed ideal. And when is a gas at it's most dilute

Â when it has it's lowest pressure. So it's expanded to fill an enourmous

Â volume. So we can say then that the temperature

Â is equal to the limit as pressure goes to zero, pressure times molar volume divided

Â by the universal gas constant. And so using Amonton's law, we can

Â extrapolate the pressure verses the temperature for a sealed, fixed volume

Â all the way back to pressure equals zero in order to define the zero of

Â temperature. And if pressure actually goes to zero

Â given that we fixed volume we would take temperature to zero.

Â And you can select another point which you will arbitrarily define to have a

Â certain temperature. And really the selection is based on how

Â much heat do you want to be associated with a degree?

Â So, many of you probably know that the celsius scale was chosen, so that there

Â were 100 degrees between the freezing point of water and the boiling point of

Â water, at, atmospheric pressure. So, given that you might like that scale

Â of degree and given that you know where zero is, by having done this

Â extrapolation from Amonton's law. We have a pretty decent estimate of what

Â we might like to assign a certain temperature to be.

Â And the temperature that's chosen is the so called triple point of water.

Â So we'll have more to say about triple points in not too long.

Â But for now, I'll just note. A triple point is a point where all three

Â phases of matter of a substance, a pure substance, are present.

Â That is the solid phase,the liquid phase, and the gas phase.

Â And for a pure substance, it turns out that can only happen at a unique

Â temperature. Can also only happen at a unique pressure

Â and volume, but for now we'll just note the temperature.

Â And for water, the triple point is defined to be 273.16 degrees Kelvin.

Â And the reason that's important is, that allows us to actually assign a value for

Â R, the universal gas constant. So, let's just see exactly how that's

Â done. First, I will note that the relationship

Â between the Celsius scale and the Kelvin scale they both have the same size

Â degrees, they differ by 273.15 degrees. So you ned to add 273.15 to a Kelvin

Â scale in order to get a celcius scale. And if you think that must be a typo, why

Â do you add 273.15 when the triple point is 273.16?

Â No, that's right. maybe someone made a 0.01 degree mistake

Â somewhere way back in history that's been replaced.

Â But these are the correct SI numbers for the 21st century.

Â So let me show you the the measurements that might be used in order to advance

Â these concepts I've just been describing. So what I've got here, is I have a plot

Â of pressure times molar volume, so that's the numerator in this expression up here,

Â versus pressure. As I take the pressure from atmospheric,

Â that's going to be one, one atmosphere all the way back to zero.

Â Now, remember that if the gas were to be ideal, P V over R ought to be a constant,

Â alright? And that's what shown here.

Â There is this ideal gas equation, so P V is a constant since R is a constant.

Â And there's just this flat line. It doesn't matter what the pressure is,

Â it ought to have the same value for pressure times molar volume.

Â However, real gases do not behave ideally until, as I noted earlier, they get very

Â dilute. Pressure has to go toward zero.

Â And sure enough, if I look at the behavior of PV bar, as the pressure is

Â going down down down down down toward zero.

Â They're all converging to the same point. So these are all 273.15 kelvin

Â measurements. So I'm holding my gas in contact with

Â water at it's triple point to maintain the temperature.

Â And I extrapolate back all the gases are giving the same value.

Â They're all converging. And what they're converging to is that P

Â times molar volume is equal to 22.414, and the units here are liter atmospheres

Â per mole. Well given that I've defined the triple

Â point of water to be 273.15, and given that I've determined the pressure times

Â the molar volume for this limit, all that's left is R, and so we get R then

Â as, just from solving this equation. 22.414 here and 273.15 here.

Â So that's just rearranging that equation then to place R on the left hand side of

Â the equals sign and if we want to associate some units with that, common

Â ones are shown here. So, within the units I've already been

Â talking about, liter atmospheres per mole for PV bar, R has the value 0.082058

Â liter atmospheres per mole per kelvin. Note that a, a feature of kelvin if

Â you're careful is you don't say degrees kelvin.

Â Kelvin is just the name of the unit. All the other temperature scales you're

Â supposed to say degrees. So you would say degrees Celsius, you

Â don't say degrees kelvin, you just say kelvin.

Â there is a slight difference between the atmosphere as a unit of pressure and the

Â bar which is the SI unit of pressure. It's not very large.

Â It's only about a percent or so, but it does give rise to a percent variation in

Â the constant and so it's 0.083145 leader bar per mole per kelvin.

Â And it's also sometimes convenient to work with the volume not in liters, but

Â perhaps cubic centimeters. And then of course that is a change in

Â order of magnitude because it takes a thousand cubic centimeters to make a

Â liter. And so now it's 83.145 is the numerical

Â value. And finally in, in SI units, it turns out

Â that pressure times volume has an energy is, is an energy unit.

Â And so we can convert to Joules, in which case it's 8.3145 Joules.

Â So, you can look up any of these conversion factors you care to, but that

Â is the numerical values that we'll use whenever we need to plug in the gas

Â constant. Well that brings us to the end of what we

Â need to know, about the ideal gas equation of state.

Â Next we're going to take a look at equations of state that apply not only at

Â infinite delusion, that describe the behavior of gases.

Â But actually over a broader range of pressure or molar volume, or temperature.

Â And therefore there would be non ideal gas equations of state.

Â Before we go to that lecture, we'll also appause or spend some time in any case.

Â Taking a look at a demonstration which will illustrate a few of the gas laws

Â that I called out as being derivable from the ideal gas law.

Â Let's begin with a very simple demonstration to illustrate Amonton's

Â law, which says that the pressure of a gas at a fixed volume is linearly related

Â to temperature. Here, we have a steel sphere, filled with

Â gas and connected to a pressure gauge. And here we have a bath of ice water and

Â one of boiling water. We know that ice water has a temperature

Â of about 273 kelvin and boiling water 373 kelvin.

Â Since an increase of 100 out of 300 kelvin is about a third.

Â We should see the pressure increase by about one third as we transfer the sphere

Â from the ice water to the boiling water. [SOUND].

Â Looking at the pressure gauge, we see that it is indeed increasing.

Â And since we started from 13 pounds per square inch, we ought to get to about 17

Â pounds per square inch. And we're just about there.

Â Niw, if we plunge the sphere back in the ice water the pressure drops again.

Â Did you know that this was how absolute zero was first estimated?

Â If we were to graph our pressure temperature relationship.

Â We could estrapulate our line backwards to zero pressure.

Â And this could define absolute zero. Of course Amotone's law is a special case

Â of the more general ideal gas law. PV equals nRT, where the special

Â situation is that we are holding volume, and amount of gas constant.

Â We'll be exploring the ideal gas law more in upcoming lectures and exercises.

Â Our next demonstration is of Boyle's law. In this apparatus, we have a piston

Â connected to a pressure gauge. As I press the plunger, at constant room

Â temperature, the pressure increases proportional to the volume reduction.

Â Our final demonstration involves a test tube, containing dry ice, that is solid

Â carbon dioxide, capped at the top with a piston.

Â If we immerse the test tube, in a bath warmer than the dry ice, in this case

Â room air, the gas expands and raises the piston in the syringe.

Â A valve at the side of the syringe, permits the expanding gas to escape so

Â that this demonstration cycles. Thus, gas expansion, with increasing

Â temperature, can be employed to do work. In this case, to raise a piston.

Â We'll spend some time looking carefully into just how much work can be done and

Â how most efficiently to extract that work for different combinations of

Â temperatures, volumes, and pressures. /b

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