For more information, please see full course syllabus of AP Chemistry

For more information, please see full course syllabus of AP Chemistry

### Related Articles:

### Kinetic Molecular Theory and Real Gases

- The Kinetic Energy of an Ideal Gas is directly proportional to its Temperature (Kelvin).
- At high pressures gases deviate from ideal behavior because the particles are pushed together much more tightly:
- A volume adjustment must be made.
- A Pressure adjustment must be made.
- The van der Waals equation accounts for these deviations.

### Kinetic Molecular Theory and Real Gases

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro 0:00
- Kinetic Molecular Theory and Real Gases 0:45
- Kinetic Molecular Theory 1
- Kinetic Molecular Theory 2
- Kinetic Molecular Theory 3
- Kinetic Molecular Theory 4
- Equations
- Effusion
- Diffusion
- Example 1
- Example 2
- Example 3

### AP Chemistry Online Prep Course

### Transcription: 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, 6.02x10 ^{23}.*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 6.02x10^{23}.*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 CH _{4}) at 40 degrees Celsius.*1194

*So, calculate the root mean square velocity of atoms in a sample of CH _{4} 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 CH _{4}; 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-meter ^{2} per second^{2}; 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-meters ^{2} per seconds^{2}, over kilograms.*1374

*Well, kilograms cancels with kilogram, leaving us meters ^{2} per seconds^{2}; 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 (O _{2}) 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 rate _{1}, times the molar mass of 1, equals rate_{2}, 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 O _{2}, 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 rate _{1} over rate_{2} = 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 n ^{2} over V^{2}...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 n ^{2} over V^{2}.*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

1 answer

Last reply by: Professor Hovasapian

Thu Dec 17, 2015 12:49 AM

Post by Tammy T on December 11, 2015

Hello Prof. Hovasapian!

I understand how when the Volume of the container shrink (due to applying external pressure), we switch to using Real gas law due to the volume of the particle is significant compared to where it is in.

However, how is it that when the pressure is high and temperature is low, we must account for the real gas behavior? I thought the P in ideal gas law is not P external.

2 answers

Last reply by: Jackson Forrestall

Wed Nov 12, 2014 8:53 PM

Post by Jackson Forrestall on September 27, 2014

Hello Professor Hovasapian! I understand the difference and the importance of knowing and using both the ideal and real gas laws, but, when time is of the essence, especially on the AP exam, would it be more beneficial for me to just use the ideal gas law due to it being quicker? Thank you so much for your lectures as well. They help me so much!

1 answer

Last reply by: Professor Hovasapian

Sat Jul 6, 2013 6:46 PM

Post by KyungYeop Kim on July 4, 2013

Hi Professor Raffi, I have a question about effective nuclear charge. As you go down a group in the periodic table, why is it that the effective nuclear charge decreases? from what I know, is it true that as the attraction decreases down the group, it somehow counterbalances the increase in nuclear charge? I'm confused.

2 answers

Last reply by: KyungYeop Kim

Thu Jul 4, 2013 8:11 PM

Post by Antie Chen on April 30, 2013

Hello Raffi, I am really confused about these equations. In the equation about 8:00, what's the Urms? rms? and what's the difference between these two equation?

And in the equation of real gases, a&b are van der waal constants, are they constant for specific gases, but why a&b are different in the example 3?