AP Chemistry Kinetic Molecular Theory and Real Gases

Welcome back to Educator.com, and welcome back to AP Chemistry.0000

Today, we're going to be discussing kinetic-molecular theory and properties of real gases.0003

Up until now, we've been discussing the ideal gas law, PV=nRT; well, the real gases actually behave well at low pressures and high temperatures.0008

When the pressures start to get high (let's say above 3, 4, 5, 6 atmospheres), and the pressures start to drop, volume starts to drop--at that point, the gas behavior starts to deviate from the ideal.0023

Towards the end of this particular lesson, we will discuss about how we adjust for that--how van der Waals adjusted for that.0037

Let's go ahead and get started.0043

We're going to run through the kinetic-molecular theory first.0047

So, again, what we are about to list right here--the four or five axioms of the kinetic-molecular theory--are precisely that: they are axioms; they are observations that we have made, and they are assumptions that we are making.0050

Based on those assumptions, we can start to build a theory, and hopefully have it correspond with what we observe empirically (in other words, what we get from experimentation).0062

Let's just list them out.0073

Let me see: our first axiom: The kinetic energy of an ideal gas (and again, we are talking about ideal gases' behavior--toward the end, we will discuss the deviation from ideal behavior, but the kinetic-molecular theory applies to ideal gases) is directly proportional to the temperature.0076

What that means, in terms of an equation, is: KE (the kinetic energy--the energy of motion of the molecules of the gas) is equal to 3/2 RT.0118

Now, R is the gas constant; however, here it is not equal to .08206 liter-atmosphere per mole-Kelvin.0131

In this particular case, because energy is expressed in Joules, and T here is in Kelvin, this R has to be Joules per Kelvin per mole; so, as it turns out...let's see: 8.31...it's going to be 8.31 Joules per mole-Kelvin.0139

And again, it's just a way of making the units work out; that is all it is; it's the same constant, R, the Rydberg constant; it's just so that it actually works out in terms of the appropriate units.0163

T is the temperature in Kelvin.0172

OK, now, also, what I'll put down here (this is a very, very important relationship--notice, it establishes the relationship between the energy, or the temperature; that is really what is going on here--so temperature is really just a measure of the average kinetic energy of the molecules)...0182

Most people confuse temperature and heat; temperature and heat are not the same thing.0200

Temperature is a measure of the motion, random motion, of the molecules; heat is that energy that actually flows from something hot in the direction of something cold.0204

They are two different things; temperature does not measure heat; it measures kinetic energy--it measures the random motion of the molecules, the vibrations, the bouncing around of each other.0214

OK, let's say there is one other little mathematical thing I want to put down here: the average kinetic energy of a gas particle (this one is not altogether that important, but I might as well just put it down here) equals 1/2 mass, times the average velocity squared.0223

Again, in this particular case, mass has to be in kilograms.0249

When velocity is expressed in meters per second, we get our kinetic energy in Joules.0255

That is why we need mass in kilograms.0261

OK, the second assumption of the kinetic-molecular theory is that the particles are so small (the particles of gas that are bouncing around) compared to the distance between the particles that the volume of the molecules is negligible.0264

In other words, molecules are very, very tiny; and when they're bouncing around in a gas, they are really very far apart from each other; so essentially, you can think of them as volume-less point particles that are just bouncing around.0308

Now, obviously, when we squeeze, when we drop the volume, increase the pressure, lower the temperature--when the volume gets really, really tiny--now the volume of the molecules actually plays an important role.0318

We can't really make this assumption anymore; we have to adjust this; but, for kinetic theory--ideal behavior--we just ignore the volume of the actual particles themselves.0329

They do have volume; they're just not important.0339

Three: the third axiom is: Particles exert no forces on each other.0345

In other words, they are just bouncing around randomly; one has nothing to do with the other; they don't stick together; they don't repel; they just bounce off of each other in perfectly-elastic collisions.0364

Now, that is not the case--we know for a fact that it doesn't happen that way--but for ideal behavior, we can make this assumption.0377

For all practical purposes, it behaves this way at low pressure.0383

OK, now the fourth one: The particles are in constant motion (this makes sense), and the collisions (oops, this random stray line all the way across the page--let's get rid of it here) with the walls of the container are the cause of exerted pressure.0388

So, if I have some gas in a closed vessel, and I measure the pressure, the pressure of that gas is coming from the particles--the gas molecules or atoms--bouncing and hitting the walls of the container.0444

That is where pressure comes from; it's just like a ball that hits a wall--just imagine billions and billions and billions of balls hitting a wall--well, that wall is going to feel it!0457

So, these are the assumptions of the kinetic-molecular theory of an ideal gas.0466

Now, from this, we can go ahead and deduce some things.0473

I'm not going to go ahead and derive any of these equations; I'm just going to throw them out, because, again, we're concerned with using the equations--not necessarily where they come from.0477

You can go ahead and follow the derivation in any one of your textbooks--they are either in the appendix or in the actual text itself--so I'll let you look at them if you want to; it certainly helps if you do.0484

If not, no harm--no foul--we're just going to be able to use them; that is what we're going to do.0496

The root is equal to (let's see) velocity and 3R...actually, that's going to be for the whole thing...3KT over m: so, the root mean square speed--just think of it as the average speed at which a particle is moving--is equal to 3 times K times T times m.0502

Now here, K is something called Boltzmann's constant, a very, very, very important constant--probably the single most important constant in physics (at least in my opinion--other people would say that the Planck constant is--they are closely related, in fact).0540

Boltzmann constant: that is 1.38x10^-23 Joules per Kelvin.0560

T is temperature in Kelvin, and m is mass in kilograms.0573

It is the mass of an individual particle in kilograms; that is why we use the small m instead of the capital m.0582

Now, let's go ahead and write this same thing as 3RT/M, where R is, as we said before, the Rydberg constant: 8.31 Joules per mole-Kelvin--and that actually happens to equal Boltzmann's constant times Avogadro's number, 6.02x10²³.0587

So, this is also a really interesting relationship to keep in mind--that R, the Rydberg constant, equals the Boltzmann constant times Avogadro's number.0623

8.31 Joules per mole-Kelvin equals 1.38x10^-23, times 6.02x10²³.0632

This is where you end up getting R from.0640

M, the capital M, is the molar mass.0643

This just gives me average--the root mean square speed of gas particles.0649

OK, so let's throw a couple of other definitions out there.0657

Essentially, what I'm going to be doing is just sort of laying out these things--what they are--defining them, and then, once I have them on the page, I'm going to go ahead and use them to solve some problems.0660

But, I just wanted to lay them out as they are, as opposed to laying one out, doing a problem...I'm going to save the problems until the end.0669

There is something that we call effusion, and effusion is nothing more than the passage of a gas through a tiny opening--that is it.0677

So, if I have a balloon and I poke a little needle hole in it, the gas from the inside of the balloon effuses out.0696

The rate of effusion is just the rate at which it actually comes out.0702

So, it's going to be a certain volume per unit time, like, let's say, 10 milliliters per second.0708

That means 10 milliliters of air are escaping for every second; that is all it is--a rate is just an amount over a unit time.0713

But, you know this already.0721

Well, somebody by the name of Graham discovered that the rate of effusion is inversely proportional to the molar mass of the particular gas.0723

Or, we can write it as: The rate times the molar mass is equal to a constant.0736

Well, for two gases under similar circumstances--for the same temperature and pressure--they are going to equal the same constant.0750

Therefore, what you have is something like this: you have: The rate of the first gas, times its molar mass, equals the rate at which the second gas effuses, times its molar mass.0760

This is how I like to use it; however, it is totally equivalent to (and more often than not, you will see it written like this): The ratio of the rates--rate 1 over rate 2--is equal to the molar mass of 2 over the molar mass of 1.0774

It really doesn't matter: you can write it this way; you can write it this way--it's a totally personal choice; most books will write it this way, because they like to have ratios of the same thing.0794

The rate of one over the rate of the other equals the square of the molar mass of one over the square of the molar mass of the other...absolutely the same thing.0801

OK, now let's define something called diffusion.0812

Diffusion just means one gas mixing with another.0816

If I open up a vial of pure ammonia, and if I put it on the table, let's say 5 feet away from me, it's going to take a little while, but eventually, I'm going to actually smell the ammonia.0826

Well, the ammonia is mixing with the air--the oxygen and nitrogen--the 70%...air is nitrogen and oxygen; it's mixing with the air, and diffusion is just the extent to which it mixes.0837

The rate of diffusion is how quickly it actually mixes.0851

Effusion and diffusion are actually closely related; you can actually use the same equation for both.0855

Again, we will talk more about this when we actually do a problem; it will make more sense; but I just want to throw out the meanings.0861

So, effusion: how quickly a gas escapes from a tiny opening; diffusion is how a gas actually mixes with another gas--how quickly it penetrates that other gas.0866

OK, now let's go ahead and talk about real gases, as opposed to ideal gases.0879

Well, we know that the ideal gas law is PV=nRT; I am going to rearrange this, and I am going to write it as nRT/V, and here is the reason that I am going to do this.0888

We said that the ideal gas law works for low pressures--low pressures where the gas molecules are flying around; they're really far apart from each other; the individual volumes don't matter; but, if I were to all of a sudden take a volume, where the individual gas particles really fall apart, and they are floating--they don't matter--and I increase the pressure, by increasing the pressure, I reduce the volume.0902

So now, I have dropped it down to something like this, where the same number of particles...now the particles don't have as much space as they did before.0930

So now, the volumes of the individual atoms and molecules start to matter.0944

What ends up happening is that this initial volume...when we use this particular volume, now, because there are so many particles, and the volume of the particles is actually a fair percentage of the total volume--now, the volume available for the particles to move around in is not the same as--is actually lower than--the ideal volume, by assuming that volumes don't matter.0952

So, if I assume I have a point--if I have a particle and another point--there is a certain volume that is available to it; but now, if this point is an actual volume--occupies volume--it's taking away volume, so there is less room for this other particle to move around.0979

We have to make an adjustment to this side by reducing the volume.0993

We write this as P=nRT/V-nb; and I'll talk about what b is in just a second (n is just the number of moles).0997

This is a volume adjustment; it is saying that, as we increase the pressure or decrease the volume or lower the temperature (which also decreases the volume), now the volume of the individual particles matter, and I have to make an adjustment for the volume.1011

There is less volume available for the particles of gas that are initially there to move around in.1024

It is not V; it's less than V--that is why it is V-nb.1030

OK, now, the next adjustment is this: we assumed, in the kinetic-molecular theory for ideal gases, that the particles exert no forces on each other.1034

They have no attractive force to each other.1042

Well, as it turns out, particles do have an attractive force to each other, and in fact, the more polar the particular molecule (like, for example, water molecule)--they're going to stick together.1044

Because they stick together, the collisions that they experience are not elastic, so the pressure that we actually measure is going to be less than the pressure that it would be under ideal conditions, because now you have fewer particles actually bouncing around and hitting the containers, because more of these particles are actually sticking together.1055

There is loss of energy, if you will, in some sense; the total energy of the system is conserved, but individually, there is sort of a loss; so the pressure that we measure is actually going to be less than the pressure...I'll say that the pressure observed is going to be less than ideal pressure.1078

That is why I put the "obs" here.1095

So, this factor is going to be a, times n over V squared; this is the pressure adjustment.1097

This is the volume adjustment; this is the pressure adjustment.1107

Now, I'm going to rearrange this again, and I'm going to write it as P+a times (n over V)², times V-nb, equals nRT.1110

This is PV=nRT, but because of real gas behavior, I have made adjustments to the pressure and the volume for real gas behavior.1126

Now, this a and b are called van der Waals constants, and we have calculated different constants for different gases.1135

You can see them in any chapter on gases in a chemistry book.1154

They have them for (most of them list maybe 10 or 12) the most common gases--methane, oxygen, nitrogen, hydrogen...things like that.1157

That is all it is; these are just numbers that you put in there; n is just the number of moles of particles that you are dealing with, and this is a better representation of how gases behave at high pressures and low temperatures--in other words, small volumes, because volume matters now.1165

That is all this is; this is just PV=nRT, adjusted for real gas behavior.1183

OK, let's go ahead and jump into our examples, and I think a lot of this will start to make sense.1189

So, our first example is going to be: Calculate the root mean square velocity of atoms in a sample of methane gas (which is CH₄) at 40 degrees Celsius.1194

So, calculate the root mean square velocity of atoms in a sample of CH₄ at 40 degrees Celsius.1227

Basically, how far is your average atom flying around at?1231

OK, well, let's (let me see--we know what we're going to deal with...so...) just use our equation: root mean square speed is equal to 3 times R times T, over the molar mass; square root.1236

Well, T (temperature) is in Kelvin; 40 degrees Celsius becomes 313 Kelvin; we have to make sure to work in Kelvin.1253

Also, remember molar mass: molar mass has to be in kilograms per mole--not grams per mole, so we know that oxygen is 16 grams per mole, but oxygen is going to be .016 kilograms per mole; that is what is really important here.1264

These problems are not difficult; the difficulty is going to be remembering to work in the appropriate unit, so that we actually get our answer in meters per second.1280

OK, so methane is CH₄; C is 12; there are 4 H's--that is 16; so, 16 grams per mole becomes 0.016 kilograms per mole.1287

And now, when we put these numbers in, we get 3, and we said R is 8.31, not .08206, so 8.31 Joules per mole-Kelvin, and the temperature is going to be 313 Kelvin, and we have 0.016 kilograms per mole; all of this under the square root sign.1304

Now, when we do the mathematics, as far as the numbers, we're going to get 699 meters per second.1335

Now, I want to show you where the meters per second comes from.1341

Kelvin cancels with Kelvin; mole cancels with mole; now, what we end up with, as far as units, is Joules over kilograms.1344

All right, here we go: J over kg; well, the Joule is kilogram-meter² per second²; that is the unit--force times distance, a newton times a meter, gives me a Joule.1355

So, the unit of a Joule is a derived unit; it's kilograms-meters² per seconds², over kilograms.1374

Well, kilograms cancels with kilogram, leaving us meters² per seconds²; then, when I take the square root of that, I get meters per second.1382

So, the units work out--very, very important.1391

R has to be 8.31 Joules per mole-Kelvin; temperature has to be Kelvin; molar mass has to be in kilograms per mole--very, very important.1394

OK, let's move on to a second example here.1403

We have...let's see...the problem says: The effusion rate of an unknown gas is found to be (we can measure this) 32.50 milliliters per minute.1407

In other words, 32.50 milliliters are leaking out of a hole every minute.1438

That is all that is; it's a rate--an amount per time.1444

Now, under identical conditions, the effusion rate of oxygen gas (O₂) is found to be 31.50 milliliters per minute.1448

Is the gas (the unknown gas) methane, carbon monoxide, nitrogen monoxide, carbon dioxide, or nitrogen dioxide?1482

Well, let's use what we know.1500

We know that Graham's law says that the rate₁, times the molar mass of 1, equals rate₂, times the molar mass of 2.1502

Well, we know the rate of the unknown gas--we measured it--that is 32.50 milliliters per minute (I'll go ahead and leave the unit off--it's not that important).1514

I don't know what its molar mass is; I would like to know that, because, when I know that, I can just compare it to the molar masses of my choices and pick the right one.1526

Well, I know the rate for the second one is 31.50 milliliters per minute--that is the oxygen--and its molar mass is going to be...it's O₂, so it's not 16; it's 32--32 grams per mole.1534

In this case, we can use the 32 grams per mole; it doesn't have to be in kilograms per mole, because, again, the ratios cancel out.1550

It is OK, as long as the units are the same.1557

That is it; so we have that the square root of the molar mass is going to end up being 5.48, and then, when we square both sides, we get a molar mass of about 30 grams per mole, and when we compare it, it looks like nitrogen and oxygen--nitrogen is 14; oxygen is 16; that is 30.1561

So, our gas is nitrogen monoxide.1584

That is it: Graham's law; rate is related this way; you could put it in that other form, where you have rate₁ over rate₂ = molar mass of 2 over molar mass of 1.1588

I like it this way, because everything is consistent--1, 1, 2, 2--but it is your choice.1599

Let's do another problem here.1607

This one says: Calculate the pressure exerted by 0.600 moles of nitrogen gas in a 2.0 liter vessel at 35 degrees Celsius, using a) the ideal gas law, PV=nRT, and b) van der Waals equation (the van der Waals equation was that adjusted one, the P-a over n/V squared--that one).1614

We want to compare them to see what we are looking at--to see how closely, actually, ideal behavior and non-ideal behavior is for this particular situation.1664

Well, 35 degrees Celsius--that is a pretty high temperature, in terms of Kelvin.1673

2 liters volume is a reasonable volume; and it is .60 moles.1678

Well, let's just sort of see what happens.1683

OK, so a) PV=nRT: we rearrange; we get nRT/V, and we just plug the values in.1686

We get 0.600 (is our number of moles); R is .08206.1695

And remember, when you are dealing with the ideal gas law, you have to use the .08206 for R; when you are dealing with issues of Joules and things like that, root mean square speed...that is when you have to use the 8.31 Joules per mole-Kelvin.1701

The problem itself--if you just stop and take a look at what units you want--they will tell you which R you are actually going to use.1719

Then, 308 Kelvin should be for 35 degrees Celsius, and our volume is 2.0 liters.1726

We end up with 7.58 atmospheres; that is pretty high pressure.1735

OK, now let's use pressure + a, times (let me write it as n² over V²...no, you know what...I'm just going to leave it as (n/V)²).1743

Notice this n/V, by the way: it's the number of moles over a volume--it's a concentration.1759

The pressure is actually contingent on the concentration, which makes sense; it has to do with: the more molecules you have, the more heavily concentrated, in a given volume--the pressure is going to increase.1764

So, just recognize that; that is why I didn't do n² over V².1774

I left it that way so that you could see that it is a concentration term.1779

...V-nb=nRT.1782

OK, and then, when we rearrange, we get P=(nRT/(V-nb))-(a(n/V)²).1788

Now, the pressure adjustment constant (van der Waals constant) for nitrogen gas is 1.39, and (the units are irrelevant; you can certainly look up the units if you want--they are not altogether important) B_N2 is equal to...the b, the volume adjustment parameter, is 0.0391.1807

When we put all of these values in here, we end up with 7.67 for this first term, and we end up with 0.1251 for this second term, which gives us a total pressure of 7.56 atmospheres.1833

So notice, under the ideal gas law, we have 7.58 atmospheres; using the real gas behavior, we have 7.56 atmospheres.1853

7.56 and 7.58 are virtually the same, so under these conditions, we are welcome to go ahead and use the ideal gas law.1862

Which--just by looking at this--35 degrees Celsius is a pretty high temperature; 2 liters--it's a pretty big volume, actually, for .6 moles of gas, so you can pretty much say to yourself, "You know what? I'm just going to go ahead and use the ideal gas law; I don't want to use the van der Waals equation--it's not important."1870

This confirms that that is the case.1889

If the conditions were different, you might get a lot of deviation.1892

OK, so we talked about the kinetic-molecular theory; we have talked a little bit about root mean square speed, average kinetic energy of a gas sample...we have talked about effusion and diffusion and done some problems.1896

In the next lesson, we're going to actually sort of tie it all together and do some regular gas problems.1908

Thank you for joining us at Educator.com, and we'll see you next time--goodbye!1914