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Lecture Comments (12)

1 answer

Last reply by: Professor Hovasapian
Wed Aug 3, 2016 5:59 PM

Post by tae Sin on August 3 at 08:30:24 AM

isn't concentration the same as molarity, or is molarity only used for liquids?

1 answer

Last reply by: Professor Hovasapian
Sat Jun 11, 2016 5:42 PM

Post by Neil Kotta on June 11 at 02:10:45 PM

For question 4 where did you get the "n" from? I thought the formula originally was (PV)/T. Thank you so much; you're videos are really good!

1 answer

Last reply by: Professor Hovasapian
Wed Mar 18, 2015 12:27 AM

Post by Rohan Suri on March 17, 2015

For part d on example 2 shouldnt No2 have a greater average kinetic energy because you said that Average Kinetic Energy is KE=1/2mv^2 and m being the mass in kg. Since NO2 has a higher mass shouldnt it have a higher average KE? I thought the 2 equations KE =3/2RT and KE=1/2mv^2 are interchangable.

1 answer

Last reply by: Professor Hovasapian
Thu Jan 8, 2015 2:10 AM

Post by Stephen Donovan on January 7, 2015

Isn't the symbol for density rho?

1 answer

Last reply by: Professor Hovasapian
Sat Nov 15, 2014 10:37 PM

Post by Shih-Kuan Chen on November 12, 2014

Dear Professor,

I am confused when should R= 0.08206 and when R= 8.31, can you please explain to me when I should use which?

1 answer

Last reply by: Professor Hovasapian
Sat Jul 14, 2012 6:19 PM

Post by Ciara Flynn on April 9, 2012

In example 2, part A, why doesn't NO2 have the greater partial pressure, since equimolar quantities only mean there are equal moles of the COMPOUND N2 and the COMPOUND NO2, meaning NO2 will have more atoms per mole? Shouldn't it exert greater pressure than N2 gas? Or is this a case where we're assuming ideal gas behavior, and no effect made by the molecules themselves?

Related Articles:

AP Practice for 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
  • Example 1 0:34
    • Example 1
  • Example 2 6:15
    • Example 2: Part A
    • Example 2: Part B
    • Example 2: Part C
    • Example 2: Part D
    • Example 2: Part E
    • Example 2: Part F
  • Example 3 14:45
    • Example 3
  • Example 4 18:16
    • Example 4
  • Example 5 21:04
    • Example 5

Transcription: AP Practice for Gases

Hello, and welcome back to

Last lesson, we finished off the discussion of gases with real gases and the kinetic-molecular theory.0002

This lesson, we're just going to sort of tie it all together; we're just going to do some practice problems.0008

We're just going to work on the entire range of gas problems.0013

We're not going to do 1,000 problems, but we're going to do a fair sampling, and it will be the type of problems that you will see on the AP exam, and the type of thinking that you're going to have to do, and a little bit of the manipulations.0017

So, let's just go ahead and get started--that is the best way to make sense of anything.0029

The first example that we're going to do is the following (let's see...go ahead and write 1 here)--the problem is as follows.0036

The density of a gas was measured to be 1.25 grams per liter (now, notice: we have grams per liter--it's still a mass per volume; the actual, individual units don't really matter; for gases, grams per liter is pretty standard, simply because gases don't weigh very much).0046

So, we measured this density at 29 degrees Celsius and 1.75 atmospheres.0073

OK, we want to know: What is the molar mass of this gas?0082

A pretty standard problem--it's often how we find the molar mass of a gas.0089

What is the molar mass of the gas?0093

Now, I can guarantee you that this problem, in one form or another, will show up on the AP exam--perhaps even twice; I guarantee it; it always does.0098

So, let's see; let's talk about how we're going to approach this.0106

Well, let's just look at the definition of molar mass: they are asking for a molar mass, so what is molar mass?0110

Let me use red ink here.0115

Molar mass is equal to the mass in grams, over the number of moles; that is it--that is all molar mass is.0118

Let me rearrange this equality; let me write it as mole equals mass over molar mass.0129

Now, let me use a capital M for molar mass, but I'll go ahead and leave mass itself as...0139

And you know this already--you know, when you're finding the number of moles, you take the mass of something and you divide it by the molar mass; you get the number of moles.0145

Well, mol is n in the ideal gas law, so let's go ahead and put--wherever we see n, let's put mass over molar mass; how is that?0152

We write PV=nRT; that means PV=mass, over molar mass, times RT; it equals--and then we drop the V down; we solve for P; we get P=mass, times R, times T, over molar mass, times volume.0163

Now, let me switch some things around here; I'm going to bring the V and put it under the m, under the mass, and I'm going to write mass over V, volume.0190

I'm going to separate it out from everything else, and I'm going to write RT over (no, wait...PV over mass...oh, I'm sorry; I made a little mistake here); I drop down the m; I drop down the V; I actually want to drop...I'm looking for molar mass, so that is what I want to solve.0201

I want to bring the M up here, drop the V and the P down here...there we go: it looked like there was something wrong.0229

And then, I'm going to combine the mass and the mass over V; I'm going to write that as a unit and write RT/P (so this is the same thing--we just played with some variables).0237

Well, what is mass over volume?0247

You know that it is density; so, there you go--the molar mass is equal to the density (which--I used a Greek small letter delta), RT, over P.0250

That is how you find the molar mass.0266

If you had it the other way--if you had the molar mass and you wanted the density--you just rearrange it; that is what is nice about this--you just rearrange your variables, based on the units that they are made of, and you can come up with all kinds of different variations of the particular equation at hand.0268

That is often how you do it in laboratories.0282

So, let's go ahead and put our values in.0285

We have the density, which is 1.25 grams per liter; we have R, which is .08206 liter-atmosphere per mole-Kelvin, and then we have the temperature, which is 29 degrees Celsius, which is 302 Kelvin (again, we always work in Kelvin).0287

Then, our pressure, it says, is 1.75 atmospheres.0312

OK, that is it: so we get: atmosphere cancels atmosphere; Kelvin cancels Kelvin; liter cancels liter; what we'll end up with is g on top, mole on the bottom--grams per mole--molar mass; our units work out.0317

Our final answer is 17.7 grams per mole; that is it.0333

We took the PV=nRT; we solved the molar mass thing for moles; we put it in; we rearranged, and we get the molar mass equals the density, times R, times T, divided by P.0339

This equation right here will show up on the exam, in various forms; you will see it in the multiple choice, and more often than not, chances are pretty good (I would say 80 to 85%) that you will see it in the free response section or the essay section (the part that asks you for some descriptive--where you are not really doing anything mathematical, but you're describing what is going on).0350

So, it will more than likely show up.0370

OK, so let's go to another example; that is what we are here to do.0375

2: we have: Equimolar quantities of nitrogen gas and nitrogen dioxide gas are put into a closed vessel at constant temperature (so, closed vessel--fixed volume; constant temperature; we're good).0382

This is going to be an example of a qualitative problem that you see in the free response section, but they are not going to ask you to do any calculations, but they are going to ask you to reason things out.0419

So, this is a typical, typical example of what you will see on there.0428

Part A (and it comes in multiple parts): A: Which gas has the larger partial pressure?0432

These kind of questions want to test your qualitative understanding of what is going on.0445

In other words, you might be able to solve a problem mathematically, because you have sort of seen it a thousand times, and you know how to fiddle with it; but if you don't actually know what is going on, you won't be able to answer these questions.0449

So, these questions are probably the hardest, in the sense that they are testing whether you can reason things out, based on what you know chemically.0459

So, you have to know the chemistry; it is not just the math.0467

The larger partial pressure: OK, well, all right; we know that the partial pressure is equal to the number of moles, times R, times T, over V; it's just the ideal gas law for a mixture of gases.0471

We know that the total pressure is equal to the sum of the partial pressures; each partial pressure is equal to the number of moles of that particular gas, times RT, over V.0487

Well, b (a and b are just the nitrogen and O2)--that equals the number of moles of b, times RT, over V.0496

It is equimolar--that means the same number of moles.0504

That means the number of moles of nitrogen gas equals the number of moles of nitrogen dioxide gas.0508

Therefore, the partial pressures are equal.0513

The partial pressure of nitrogen gas is equal to the partial pressure of nitrogen dioxide gas.0516

You have reasoned it out; there it is.0523

Part B: here it is interesting: Which has the greater density?0526

Which has the greater density: well, in the problem that we just did, we said that the molar mass is equal to the density, times RT, all over P; let's just solve for the density now, instead of the molar mass.0538

So, let's go ahead and convert this to: Density equals P times M, over RT; in other words, P, R, T...those are fixed.0553

Molar mass; if the molar mass is higher, the density is higher--that is what this equation tells me; I just rearrange an equation.0565

Well, of these two--the nitrogen gas and the nitrogen dioxide gas--which has a higher molar mass? Nitrogen dioxide.0572

Therefore, that implies that NO2 has the higher density.0579

This equation tells me so.0585

This is the kind of stuff that you have to do.0589

Much of the multiple choice section is also going to be stuff just like this: reasonably simple--you can reason it out in a number of steps--no real math involved (or if there is math, it's simple numbers--2, 4, 6, 8); but there it is.0593

It's qualitative understanding; that is what is important.0605

Any of your professors will tell you that, if they had a choice between you understanding things qualitatively or quantitatively, your qualitative understanding is actually a little bit more important.0611

Math will always come; qualitative understanding doesn't always come; or, if it does, it comes a lot slower.0620

OK, C: let's see: Which has a greater concentration?--interesting.0628

Which has a greater concentration: Well, what is concentration?0638

Concentration--the unit of concentration is defined as (when you see :=, it means "is defined as") number of moles per liter.0649

Well, they have the same number of moles--their equimolar amounts were put in--and they are in the same volume flask, so concentration of 1, of nitrogen gas, equals the concentration of NO2 gas--nice!0661

OK, let's see what is next.0678

Which has the greater average kinetic energy?0682

OK, which has greater average kinetic energy: in other words, on average, which collection of atoms or molecules are actually moving faster?0687

Well, kinetic energy--remember what we said?--3/2 RT is average kinetic energy.0702

It is directly proportional to temperature.0716

Temperature is constant; therefore, the average kinetic energy of the two gas samples is the same.0721

That is kind of extraordinary: different gases, different molar masses--their average kinetic energy is actually the same.0727

It is a function of temperature; that is it; it's not a function of anything else.0734

OK, E: Which has the greater average molecular speed?0740

Energy and speed: not necessarily the same here.0745

Which has the greatest average molecular speed: well, Urms is equal to 3RT over M; so now is where that molar mass thing comes in.0748

So, nitrogen--the higher the molar mass on average, the slower the molecule.0779

Nitrogen dioxide is heavier than nitrogen gas; therefore, nitrogen gas is lighter; therefore, nitrogen gas is faster.0786

So, that implies that nitrogen gas is faster--moving around faster.0792

OK, now, F: Which gas will show greater deviation from ideal behavior?0803

This one is a bit challenging.0825

Think about these two molecules: nitrogen gas, N2, is nonpolar; NO2 is a bent molecule--it is polar.0830

Because it is polar, there is greater charge distribution, which means that the individual molecules of NO2 are going to attract each other more.0847

Because they attract each other more, the effect that they have on the pressure, the volume, the relationship that we had to adjust in the van der Waals equation, is greater.0856

So, NO2 will demonstrate greater deviation from ideal behavior, precisely because it is more likely to experience that deviation because of the pressure adjustment--the volume adjustment.0864

That's a typical problem that you will see.0881

OK, let's see: number 3: A mixture of neon and argon gas at a total pressure (PT) of 1.4 atmospheres--if there are twice as many moles of neon as argon (so we have twice as much neon as we have argon gas), what is the partial pressure of argon gas?0884

So, we have a mixture of neon and argon gases; the total pressure in there is 1.4 atmospheres; if there are twice as many moles of neon as argon, what is the partial pressure of the argon?0940

OK, this is a mole fraction/partial pressure kind of thing.0951

We said that the mole fraction...remember, we said it is the number of moles of the part, over the total number of moles of the whole, which is also equal to, as far as pressures are concerned, the partial pressure of argon, over the total pressure.0956

Well, let me rearrange this.0973

The partial pressure of argon is equal to the mole fraction of argon, times the total pressure; I know the total pressure--it's 1.4--I should be able to figure out the mole fraction here, and that should give me my answer.0976

OK, well, let's see if we can't figure this out.0988

Let me use this version of it here; yes, I think I will use this version of it here, instead.0996

The mole fraction is also equal to the moles of argon, over the total number of moles, and it says that the moles of argon, let's say, is x.1004

OK, so the moles of argon is x; well, the moles of neon--it says that I have twice as many, so it's 2x.1022

Therefore, the mole fraction of argon is going to equal x over the total (which is x + 2x); or, if I wanted to do just 1 over 1+2, that is fine.1029

That is equal to the partial pressure of argon, divided by 1.4.1041

OK, well, x over (x + 2x) is equal to x over 3x; the three x's cancel; that leaves me with one-third, equals the partial pressure of argon over 1.4; I multiply 1.4 times one-third, and I get that the partial pressure of argon is equal to 0.47 atmospheres--which makes sense.1049

It's twice as much; there are 3 total; so therefore, it's going to be one-third of that 1.4.1077

Again, the mathematics sort of bears it out, based on reasoning it out.1085

Mole fraction--just plug it in; everything will fall out.1089

Let's see what we have: let's see, #4: we have: 2 moles of hydrogen gas, plus 1 mole of oxygen gas, gives 2 moles of water gas at a constant pressure and volume.1097

If Ti is the initial temperature before the reaction takes place (meaning here on the left), what is the final temperature after the reaction takes place (in other words, here on the right)?1125

Before and after: those are the important words here.1156

Be very careful; read the question very carefully.1159

Well, there is a change in the system here, right?--so we're going to use the P1V1/T1n1= P2V2/T2n2.1162

But, they say that pressure and volume are constant, so they basically drop off.1175

That leaves me with 1/T1n1= 1/T2n2.1180

Another way of rewriting this is just T1n1=T2n2.1188

Again, it is important to be able to see where all of this comes from.1196

The equation doesn't just drop out of the sky; you can rearrange it; you can fiddle with it to get what it is that you need.1199

Now, you just sort of put them all in.1204

We solve this for T2; we want our final temperature--that is T2.1207

That is equal to the initial temperature, Ti, times n1 over n2--the total number of moles to begin with, the total number of moles that you end with.1211

Well, Ti...the total number of moles that you begin with is 2+1: 3 moles.1224

The total number of moles that you end with is 2; so our answer is...that is it; that is the expression.1233

The initial temperature, times 3, divided by 2, will give me the final temperature--and it's nothing more than an application of the ideal gas law, with pressure and volume completely ignored.1240

All it is is the temperature and the number of moles.1250

Before the reaction, the number of moles is 3 moles of gas particles; after the reaction, the number of particles is 2 moles; that is what we have to watch out for.1253

Let's see: we'll do a quick multiple choice one here.1267

Which of the following conditions would be most likely to cause deviation from ideal behavior in a gas?1272

We have 1) Low pressure, 2) Low volume, third choice is Low temperature, fourth choice is High temperature, and our (oops, can't have these random lines here; let me put temp there) 5)--we have High pressure.1306

OK, so our choices are: 1 only; 2 and 3 only (let's see); 2, 3, 5; 1 and 4; and the final choice is 5 only.1347

Well, which of the following conditions would be most likely to cause deviation from ideal behavior?1380

So, deviation from ideal behavior has to do with high pressure; high pressure or low temperature causes the volume to be small.1386

A small volume, a high pressure, low temperature: these things tend to induce deviant behavior.1396

1 only (low pressure): no, that is not it.1406

B: 2 and 3 only--well, low volume: yes, that will do it; low temperature: yes, low temperature will induce a low volume; that will do it.1414

High temperature: no; but there is also high pressure--high pressure will also cause a low volume, so this says "2 and 3 only," but 2 and 3 and 5 also work, so it isn't B.1425

We look at C (2, 3, and 5): yes, that works.1437

Let's check our others, just to be on the safe side.1441

1 and 4: 1--no (low pressure); 4 (high temperature)--no, that is definitely not it.1443

5 only: well, 5 works (high pressure)--yes, that is certainly viable--but it's not that one only; it's also 2 and 3 (low volume, low temperature, high pressure).1450

The idea is the "low volume" part.1460

Low volume comes from high pressure; low volume comes from low temperature.1463

So, that is what you have to keep in mind; so deviant behavior comes when you have low volumes.1468

OK, so that gives you a sampling of the type of problems that you are going to see on the AP exam.1475

Some of them are from a multiple choice section; some of them are from what they call the essay section; some from the free response section.1479

This is pretty typical of the type of thing that is going on.1487

They are not altogether difficult mathematically; it's just that they require a qualitative understanding.1489

It is very, very important that you know the chemistry.1494

Being able to do the math is nice, but, as you will discover as we go on in the course, a lot of the mathematics tends to be very, very long; it's not a one- or two-step thing.1498

If you understand the chemistry, the math will follow; but just being able to do the math doesn't mean that you will be able to reason out the chemistry, because the math can lead you astray.1506

As you see, if you miss some of the units, you are going to be all over the place; numbers are not going to make sense.1515

Qualitative understanding is what we seek in science.1520

Math will always be there; qualitative understanding will not always be there.1523

OK, thank you for joining us here at; we will look forward to seeing you next time--goodbye.1528