Sign In | Subscribe

Enter your Sign on user name and password.

Forgot password?
  • Follow us on:
Start learning today, and be successful in your academic & professional career. Start Today!
Loading video...
This is a quick preview of the lesson. For full access, please Log In or Sign up.
For more information, please see full course syllabus of AP Chemistry
  • Discussion

  • Study Guides

  • Download Lecture Slides

  • Table of Contents

  • Transcription

  • Related Books & Services

Lecture Comments (21)

1 answer

Last reply by: Professor Hovasapian
Thu Dec 17, 2015 12:55 AM

Post by Tammy T on December 12, 2015

Hello Prof. H!
Is the Pressure P the internal gas pressure exerted on the container or the external Pressure of gas exerted on the container? At some point I thought it is P internal gas, but at some point during the lecture I think it is P external gas. Please clarify this for me. Thank you!

1 answer

Last reply by: Professor Hovasapian
Wed Nov 4, 2015 9:29 PM

Post by John-Paul Kliebert on November 2, 2015

At 38:19 the end result of the example is 3.03 L. I understand everything else up until that point. Could you please tell me how you came to that result?

Thank you.

1 answer

Last reply by: Professor Hovasapian
Fri May 8, 2015 1:11 AM

Post by BRAD POOLE on May 7, 2015

Hey Professor Raffi,

Great lectures for MCAT prep!  I understand mathematically (V = kT) if k is constant and T increases then for k to remain constant V has to increase as well, but conceptually this doesn't seem right.  If the volume of a container increases then the gas particles have more room and decrease their chance of colliding with one another, if there is a decrease in collisions then a decrease in KE and thus a decrease in temperature, Right?  If my understanding is correct then how does an increase of temperature happen when volume increases?  Thanks for the advice!

2 answers

Last reply by: sadia sarwar
Sun Nov 16, 2014 6:47 AM

Post by sadia sarwar on November 15, 2014

how is the final concentration of 100mLof 0.20M NaOH and 50.0mL of 0.20M the same as the initial even though the volume is different?

1 answer

Last reply by: Professor Hovasapian
Sun Nov 2, 2014 3:13 AM

Post by David Gonzalez on November 2, 2014

Hi professor!

We know that gas particles move very quickly in random motions in whatever container they are in. My question is: assuming that no gas particles ever exit or enter the container (and that the temperature remains the same), at what point will the gas particles stop moving? From what I think I understand about the second law of thermodynamics, nothing can have an infinite amount of kinetic energy- not even gas particles, right?

P.S. I know that, according to the KMT, gas particles collide and transfer their energy to each other. But even then, can this energy really be transferred back and forth between the particles forever? Wouldn't this violate the second law of thermodynamics? The question is probably very fundamental, but at the moment, it's torquing my brain.

2 answers

Last reply by: Professor Hovasapian
Sat Apr 5, 2014 1:35 AM

Post by Angela Patrick on April 3, 2014

Not a big deal, but the R value in the quick notes is given as .08306 instead of .08206. Would be nice if you or an educator tech could fix this slight problem. Thanks.

2 answers

Last reply by: Hyun Cho
Thu Dec 19, 2013 10:54 PM

Post by Hyun Cho on December 18, 2013

at 38:22 the equation says (0.467atm x 3.25L)/261K=(0.592atm x 3.03L)/308K, i thought according to the boyles law, p1v1=p2v2. so why doesnt 0.467atm x 3.25 L= 0.592atm x 3.03L

1 answer

Last reply by: Gowrish Vaka
Thu Mar 17, 2016 5:14 PM

Post by KyungYeop Kim on September 8, 2013

Adding to the questions, I just want to say that I've read all the rules of significant figures. But this one seems really arbitrary.

For instance,

0.7x0.13 = 0.091. Textbook key and my teacher would assume that to be true.

But if I apply the rules,

0.7x0.13 = 1 sig fic = 0.09.

But it seems to me ludicrous to do this for each step of every phase of a calculation.

Can I not round off and just go with 0.091, and round off whenever there's to be the final answer, I will never know..

Thanks again.

1 answer

Last reply by: Professor Hovasapian
Sun Sep 8, 2013 9:46 PM

Post by KyungYeop Kim on September 8, 2013

Hi Professor Raffi, I have a very important question to ask regarding significant figures. This is a very long question but I would really appreciate if you could answer....

I know there are rules about significant figures, but I find them very arbitrary in some cases. (which I'm about to show you)

So, I was doing a simple calculation on finding partial pressures of HE and NE.

Firstly I found that the # of moles of He = 0.15 x 2.5 / 0.0821 x 288 K = 0.0159 and the # of moles Ne = 0.42 x 2.5 / 0.0821 x 288 = 0.0444
and total # of moles of He + Ne = 0.0603.

But here's the question. When I do the calculation(0.15 x 2.5 / 0.0821 x 288 K), if I apply the rules of the significant figures, it should be 2 digits, right? But my teacher and the answer key both assume that that doesn't apply.

Fir instance, they just add up 0.0159+0.0444 = 0.0603.

But if I apply the rules before doing it, I get
0.016+0.044 = 0.06. IF I plug it in to some other formulas, I get a very different number;

So my question is: How do we know when to round off and when not to? in that case, is it okay to round off for each calculations, before combining them? or is it a norm to do that after all calculations are complete?

I'm just very confused because I'm losing points and will lose points for things I don't find very useful. If I rounded off before adding them up, I get different numbers than if I rounded off After adding them up.

Is scientific community aware of this issue? what are their rules for this, if you know any?

Thank you.. I would really appreciate it.. best regards.

Related Articles:

Pressure, Gas Laws, & The Ideal Gas Equation

  • Pressure is defined as Force per Area, and its SI unit is the Pascal
  • Ideal Gas Law: PV = nRT. This equation expresses a relationship among the various measureable quantities of a given gas sample. It might better be written as PV/nT = R.
  • In other words, given a sample of gas, its Pressure times the volume it occupies divided by the number of mols of gas times its Kelvin temperature is equal to a Constant.
  • When some mathematical arrangement of parameters is equal to a constant, this is profoundly important.
  • R is the Gas Constant: 0.08306 Liter-Atmosphere/Mol-Kelvin
  • Units used must match the units of R.
  • Standard Temperature and Pressure (STP) is 0 K and 1 atm.

Pressure, Gas Laws, & The Ideal Gas Equation

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
  • Pressure 0:22
    • Pressure Overview
    • Torricelli: Barometer
    • Measuring Gas Pressure in a Container
    • Boyle's Law
    • Example 1
  • Gas Laws 21:18
    • Gas Laws
    • Avogadro's Law
    • Example 2
  • Ideal Gas Equation 38:20
    • Standard Temperature and Pressure (STP)
    • Example 3

Transcription: Pressure, Gas Laws, & The Ideal Gas Equation

Hello, and welcome back to; welcome back to AP Chemistry.0000

Today, we're going to start talking about gases.0005

For the first part of this, we're going to talk about the gas laws; we're going to talk about the ideal gas equation.0009

But, before we do that, we're going to have to talk about this thing called pressure.0015

Let's just jump in and get started.0018

Let's go ahead and give a quick definition of pressure.0023

Pressure (we'll use P) equals force over area; it's basically the force over a unit area.0028

So, if you have the same amount of force, smaller area, your pressure is going to go up.0040

So, let's write this as F over A, just to use some symbols.0047

Now, as far as units are concerned, the unit of force is something called the newton.0053

Area is just...we use meters squared.0062

As you know, area is length, width--some sort of flat representation--so it's going to be some square meter, square centimeter, square foot...something like that.0065

When we use the newton over meter squared--this unit--we call it the pascal.0075

That is the standard unit for measuring pressure.0085

That is what it is now, but of course, there are going to be many, many different types of pressure units over the years; many units have shown up.0091

So, we'll do some conversion factors really quickly; but I wanted to talk about this pascal, and I wanted to talk about this newton unit--what it actually means--so that you have an idea of where these things come from.0099

Let's go back; the unit of force is the newton, and the unit of area is meters squared.0109

We have no problem with the meters squared; you are all pretty familiar with area.0117

Force and newton: let's see what that means.0121

Well, as far as physics is concerned, you know that force equals mass times acceleration.0124

OK, so when the mass is expressed in kilograms and the acceleration is expressed as meters per square second, we get this unit, which is kilogram-meter per square second.0130

This unit is the newton.0143

It should make sense--for Isaac Newton.0148

It is symbolized with an N; so that is where that comes from.0152

Force is in newton-meters; we call it a pascal.0158

Now, let's go ahead and talk about some other units that we will be using.0162

The one that we will be using most often, basically because it's related to this thing called the gas constant (which we will talk about a little bit later in this lesson) is the atmosphere.0169

Let me fix this little m here: so, one atmosphere is equal to 101,325 pascals.0178

More often than not, we will be working with atmospheres; but that doesn't mean that our...the units will be in atmospheres; they will be given or will be asked for; so there is always going to be some kind of a conversion that we're going to make.0191

One atmosphere is also equal to 760 millimeters of mercury (and we'll talk about what that means in just a second).0202

It is equal to 29.92 inches of mercury (that is not a unit that you see too often anymore).0211

It is also equal to 14.7 pounds per square inch (also listed as psi, but I like to do pounds per square inch, so that you see that this is a conversion factor).0220

That is all this is--the conversion factor: one atmosphere, this many pascals; one atmosphere, this many millimeters of Hg.0235

If we are given it, let's say, in pascal, and we want pounds per square inch, we have to pass through atmosphere; we make two conversions.0242

So, atmosphere is sort of our base unit, if you will.0249

Again, one atmosphere just happens to be the pressure at sea level by the atmosphere of this earth on us--at sea level.0255

OK, so let's talk about measurements of pressure and how we actually go about doing this.0263

You have heard of things called barometers; maybe you have heard of something called a manometer; how do we measure the pressure of a gas, and how do we decide what that pressure is?0268

Let's talk about a barometer, a Torricelli barometer, in fact.0276

Well, here is what happens: if I take a little thing of mercury, and it's full of mercury; and then if I take a long tube--very long tube--and I fill it, absolutely to the top, also with mercury; and if I quickly turn it over and drop it in here, so that no air actually manages to get in here, something very interesting happens.0288

Basically, some of the mercury (this is all filled with mercury, and this is all mercury here; I flip it over so nothing spills out--no air actually gets in) actually goes down, and then what happens is, it equalizes.0330

There is a point where it actually just stops going down, and of course, this thing is also full of mercury, and as this empties, this level rises, but at a certain point, it stops.0348

Here is what is going on: the atmosphere is pushing down on that mercury.0360

The weight of mercury between this and this--between this level and this level--this amount of mercury is actually pushing down, because it has weight.0369

It has mass and mass density, and mass times the acceleration of gravity--it has weight; so it's pushing down this way.0381

The air from the atmosphere wants to push this mercury back up the tube; this mercury in the tube wants to go down this way; at some point, the pressure is equalized.0388

That equalization is the atmospheric pressure.0401

This distance right here is measured at 760 millimeters; that is where it comes from.0407

So, when we put this apparatus together: in general, at sea level, this height of mercury will be 760 millimeters.0415

It will be 29.92 inches of mercury.0425

That is where it comes from.0429

This is also called 760 torr; so you will often, also, see pressures given in torr, or a Torricelli.0431

Torr and millimeters of mercury is the same thing, precisely, because of this.0442

That is all that is happening: air pressure is pushing down on the mercury; the mercury is pushing down on the mercury against the air; at some point, they're going to equalize.0447

Everything comes to a stop; nothing moves anymore; this distance is what we call one atmosphere, because it's the atmospheric pressure pushing down on that.0455

We have standardized that as the 1.0466

Now, let's talk about how we measure the pressure of a gas in a container.0469

Let's see if I can draw this out: let's take some container, a regular round-bottom flask, and let's go up, down, and up; OK.0476

There is some gas here; well, as it turns out (let me just go ahead and draw), there is some mercury here and here; this is full of mercury.0501

But notice, these levels are different.0518

If they are equal--so if we have, let's say (let's use red)--if the levels of the mercury were actually equal, what that means is that the gas, which is some gas in here exerting a pressure on the mercury this way, is also equal to the pressure of the mercury this way.0526

Oh, I'm sorry; this is open, of course; it has to be open, because you have the atmosphere pushing down this way.0548

If these are equal, that means that the gas pressure (pushing the mercury this way) and the atmospheric pressure (pushing the mercury this way) are the same.0558

The pressure of the gas is equal to the pressure of the atmosphere.0568

However, now, let me go ahead and erase the red, and go back to what it was that I actually drew, which was the level of mercury here and the level of mercury here, and this is all full of mercury.0573

What that means is that the gas pressure is not only equal to the atmospheric pressure, but it's actually more, because it has actually managed to push the mercury further down here, and this mercury level has gone up here.0588

Now, the gas pressure is equal to the pressure of the atmosphere, which is generally 760, plus this height, h, in millimeters or inches or whatever graduation we happen to be measuring with.0605

That is all that is going on here; you have some gas, which is exerting a pressure on the mercury; you have the atmosphere exerting a pressure on the mercury.0627

Well, if the gas pressure is more, then it's going to drop in this tube; it's going to go up in this tube; and this difference is going to be a measure of the extra difference.0634

It is always going to be at least atmospheric pressure, plus that extra bit; that is how we measure the pressure of the gas.0644

If it were the other way around (let me redraw it; I have my round-bottom flask, so I go up and down and up again; come this way and this way, and come back)--if the level were here, and this were this way (and again, we have something open on top)--so now, we have the atmosphere pushing down; this is all mercury, and now the gas pressure is pushing down; but now the atmospheric pressure is higher than the gas pressure, because this is low on this tube, high on this tube.0651

Now, the pressure of the gas is equal to the pressure of the atmosphere, generally 760, minus h, which is that.0688

That is all that is going on here: in order to measure the pressure of a gas, we take the atmosphere as the standard, and we see how much more it has pushed the mercury up this way or how much the mercury level has dropped from the equilibrium value.0704

That is our h, so we either add it or we subtract it to get the measure of the pressure of the gas.0718

This is called a manometer.0725

That is how we measure the pressure of a gas; and again, pressure is equal to the amount of force per unit area--force over area.0730

Generally, it's newton over square meter, which is called a pascal, and then you have, of course, your conversions: through atmosphere, through pounds per square inch, and things like that.0739

Now that we have an idea of what pressure is, let's go ahead and start talking about gases and the behavior of gases, and talk about some gas laws.0748

Let's start off with discussing (let's see...let's go back to blue; oops--that's OK; we can leave it as red) pressure, volume, temperature, and number of moles.0760

All right, so when we talk about a gas, we can talk about four different variables.0778

We can talk about the pressure of a gas, the volume that a gas occupies, the number of moles of gas that are actually present, and we can talk about the temperature.0786

Those are things that we can vary.0796

So, as it turns out, we want to try to find some sort of a relationship among these four variables.0798

Again, that is what you do: you take a look at the variables that are involved in a particular situation, and you see if you can come up with some explicit relation.0804

Well, in order to do something scientifically, you want to deal with two things at a time.0812

You can't really deal with them all at once, when you're running your experiments to try to elucidate what the relationship is, because usually, you're changing one thing, and you're measuring something else.0817

That is the basic scientific method: you're changing one thing; you're measuring something else.0828

Now, if we take...let's say we decide to go with (now I'll go to blue) pressure and volume.0832

As it turns out, somebody by the name of Boyle (so we'll call it Boyle's law--the name is actually not that important, unless you're interested in scientific history--it's the mathematics that is important; it's the concept that is important)...0842

Boyle's law says that the pressure of the gas is equal to some constant (call it K) over V; it is inversely proportional to volume--that is what this relationship expresses.0855

When I say something is inversely proportional to something, that means the one variable on the left is in the numerator; the other variable is actually in the denominator; that is what "inverse" means.0871

Directly proportional means P=K times something, where there is no fraction.0880

At a constant temperature: so, if we keep the temperature constant, and we have a certain amount of gas (also, the number of moles is constant in this case)--if we increase the pressure, the volume goes down; if we increase the volume, the pressure goes down.0887

That is what "inverse relationship" means; another way to write this is PV=K; if I move this over to one side, now you can sort of see the equality a little bit better.0908

If the pressure is raised, in order to retain the equality, the volume has to go down.0919

You know this intuitively: if I take something and I squeeze it (put more pressure on it), the volume goes down.0925

That is what is going on.0932

Now, because the product of the pressure times the volume is equal to a constant, if I make a change in the system (whether changing the pressure, changing the volume, or changing both)...0934

Now, let's go to the pressure 1 times volume 1; I know that that is equal to some constant.0948

Well, if I change those, now I have a different pressure and/or a different volume; I call those P2 and V2; it doesn't matter--this is constant; that is the whole idea.0955

This is what Boyle discovered--that it doesn't matter what the pressure and the volume are: if the pressure changes, the volume is going to change; if the volume changes, the pressure is going to change.0966

But, no matter what, the product of the two always stays constant.0974

Therefore, because P1V1 equals a constant, and P2V2 is equal to the same constant, I can say that P1V1=P2V2.0979

So, when I talk about a change in a system, Boyle's law can be expressed this way: At a constant temperature, if I change pressure and volume, I can set it equal to the new pressure and volume.0990

So, if I have three of these, I can solve for the fourth; that is what this equation is telling me.1001

It began with an inverse relationship of pressure and volume, and it turns into something that can express the change of a system.1007

Let's just do a quick example.1016

Let's consider a sample of gas--a sample of...let's use sulfur dioxide gas at 1.37 liters and 4.6x103 pascal.1022

If the pressure is changed to 1.4x104 pascal (in other words, if I increase the pressure), what is the new volume?1047

OK, well, your intuition should tell you that, if I'm increasing the pressure, just based on this, the volume should go down.1072

So, if we end up with a volume that is going to be less than 1.37, we know we're good.1079

If we did something, and we end up with a volume of 3, 4, 5...something greater than this, that is a check; that tells us that we did something wrong with our arithmetic.1083

So, you can use your intuition to help guide you to decide whether something is correct.1091

Again, we're just going to use Boyle's law, that says in any given system, at a constant temperature, the pressure times the volume in that system is a constant.1096

Therefore, if I change the pressure, the volume has to change in order to retain that equality.1105

That means P1V1=P2V2; or, if you want to use PinitialVinitial, PfinalVfinal, that is fine, too.1110

Now, we just go ahead and plug things in.1120

The initial pressure is 4.6x103, so we have 4.6x103 pascal, times the initial volume, which is 1.37 liters.1123

That is equal to...well, the new pressure is 1.4x104 pascal, and our new volume is just going to be V2...a standard algebra problem.1139

When you do this, it's going to be this times this divided by that; your new volume is going to be 0.45 liters, which is definitely less than the 1.37 liters; so, this is a viable answer.1151

Sure enough, our volume, when I increase the pressure, is going to drop to 0.45 liters--a standard application of Boyle's law.1164

Constant temperature: again, when the problem doesn't say anything--doesn't mention temperature specifically--that means it's constant.1172

When it doesn't mention the number of moles, that means it's constant.1179

OK, now I do want to talk a little bit about the units in which we work.1184

Notice, in this case: I have pascal and liter, pascal and liter; well, pascal can cancel with pascal.1189

In general, when doing problems like is the issue with the gas laws.1195

In a minute, we're going to be talking about the ideal gas law, and there is a constant, and that constant requires that you work in a specific type of unit.1203

So, in terms of the gas laws, it's always best to work this way: volume (let's do red)--you should always work in liters; pressure--generally work in atmospheres; and temperature--you want to work in something called Kelvin, which...maybe if you remember from a past experience...or if not, we'll talk about it in just a minute.1212

It's basically Celsius + 273; that gives you the measurement in Kelvin.1240

Now, notice here: I didn't actually convert this to atmospheres.1246

It's not just because pascal cancels with pascal; the reason is this (it's actually mathematical): I can convert the pascal to atmospheres (which is the unit that we should be working in); however, I don't need to do that, because, if I were to convert this to atmospheres, it would just be a conversion factor on both ends.1250

It means I would be multiplying or dividing this by something.1269

Multiplication and division don't change the value of an equality.1273

However, if I were to do something like the following: if I were to take 5 over 9, and then if I were to do 5+273 over 9+273, these numbers are not the same; so, when I work in temperature, I absolutely have to work in Kelvin.1280

Do not rely on degrees Celsius cancelling with degrees Celsius; sometimes that will work; sometimes it won't; but in general, it won't.1300

The reason is this plus sign: addition and subtraction are different than multiplication and division.1309

These are not linear.1316

5 over 9 is not the same as 5+273 over 9+273, but 5 over 9 is the same as 5x13 over 9x13.1321

These cancel; these don't cancel.1335

Therefore, when you are working in temperature, always work in Kelvin.1339

In general, always work in liter; always work in atmosphere; always work in Kelvin; but, at the very least, always work in Kelvin.1343

The temperature has to be in Kelvin; otherwise, none of these things matter.1352

Let's go on.1356

Now, we said (let me see) we had P, V, T, and n, and we did pressure-volume; that was Boyle's law.1361

Now, somebody by the name of Charles decided to work with volume and temperature, and what he discovered was the following.1370

Volume is directly proportional to temperature; it means, as he raised the temperature, the volume of the gas expanded.1381

Or, as he changed the volume of the gas, the temperature expanded accordingly.1391

So, V=KT; what he discovered was the following: this is just a linear equation--temperature on the x-axis, volume on the y-axis.1397

Let's put the 0 mark here, and let's go in terms of hundreds; so -100, -200, -300, 100, 200, 300, 400; I'll just put 100 here, and I'll put -100 here.1412

And then volume: what he discovered was the following--that is what this equation says.1428

It says that, as the temperature went up, there is a linear relationship; the volume went up, linearly.1444

This K is just the slope of that line.1450

Another way to write this is K=V/T.1453

Again, because V/T is a constant, if there is a change, the quotient stays a constant.1457

Therefore, V2/T2...therefore, we can write it as V1/T1=V2/T2.1468

If I start the system at a certain volume and temperature, and I make a change, the quotient is equal to the original quotient of volume and temperature.1476

That is what this says.1485

Now, when he drew this graph, you realize that if he extrapolated this graph all the way down to 0 volume, 0 temperature, on the Celsius scale he hit -273.2 degrees Celsius.1486

This is what we call 0 Kelvin.1503

So, degrees Celsius + 273 equals the Kelvin temperature.1507

The degrees Celsius -273, plus 273, gives me 0 Kelvin.1514

That is where this actually comes from.1519

If I keep dropping the temperature and dropping the temperature and dropping the temperature, the volume of the gas drops to 0.1522

This is a fundamental limit in the physical sciences: in terms of speed, you have an upper limit of the speed of light (3x1010 centimeters per second); in terms of temperature, you can keep dropping something down lower and lower and lower, making it colder, but this -273.2 is a limit that you will never reach.1528

We keep trying to reach it, and some very, very interesting things happen as you get closer and closer; but that is where this comes from.1547

It is basically just an extrapolation of Charles's law for the direct relationship between volume and temperature.1556

Now, we have P1V1=P2V2; we have V1/T1=V2/T2; let's see if we can't do one more.1562

As it turns out, we can; this is going to be Avogadro's law.1579

What Avogadro did is: he worked with volume and the number of moles; so again, we have pressure, volume, temperature, number of moles; he worked with volume and number of moles.1587

He found that volume is directly proportional to the number of moles.1597

If I pump more gas into something, well, the volume is going to expand.1603

You know this from when you blow up a balloon; you blow up a balloon--you're pumping more and more moles into a certain volume; the volume is going to increase--the balloon gets bigger.1608

Let's rearrange this: it becomes K=V/n.1618

So, V1/n1=V2/n2.1624

Now, if pressure is constant, temperature is constant, and I'm just dealing with volume and the number of moles, I have this relationship.1627

Now, let's put them all together.1637

I have Volume=K1 times P (volume and pressure are directly proportional--it doesn't matter which I put where--this is just a proportionality constant, so I could put P=KV, or I could put V=KP; it's the same thing).1640

I'm sorry, I'm getting this wrong--pressure and volume is going to be...volume is K1 over P, because they are inversely proportional.1664

Volume is directly proportional to temperature, and volume is directly proportional to the number of moles.1678

When I put these together, I get Volume=Tn/P; rearrange this; I'll bring the P over to the left: PV=KnT.1688

Now, what are we going to do with this K?1714

We need to find that proportionality constant; well, I'm just going to tell you what it is.1717

It is something called R, the Rydberg constant, and it is 0.08 (you know what, let me do it down here, so I have some room)...1722

Again, at this point we're not necessarily concerned too much with where the constant comes from; we just want to be able to use it in our calculations--we want to be able to understand it.1735

Those of you who go on to study chemistry, physical chemistry, thermodynamics and kinetics--it's usually a third-year college course--you will actually talk about where the R comes from.1745

It also shows up in the study of spectroscopy.1754

0.08206; and the unit is liter-atmosphere per mole-Kelvin.1759

This is why, when we deal with this gas law, we deal in liters, atmospheres, and Kelvin.1769

That is why we specifically chose those things, because the R is expressed in those units.1776

So now, we write PV=nRT; that is usually how most people see it: this is called the ideal gas law--the ideal gas equation.1781

This expresses a relationship; this is an equation of state.1796

What that means is that this talks about, at a given moment, under certain conditions, all of the variables that we can measure for that gas satisfy this law.1801

The pressure times the volume is equal to the number of moles, times the temperature, times this constant--for an ideal gas.1814

Now, we say "ideal" because real gases don't really behave like this.1822

However, for low pressures (basically, for pressures that are 1 to 2 atmospheres and below--let's just say less than 2 atmospheres), and high temperatures with respect to the Kelvin scale (which--we certainly qualify: 23 degrees Celsius is like room temperature; it's 298 Kelvin--that's a really, really high Kelvin temperature)--at low pressures and high temperatures, gases actually behave like this--almost exactly like this.1826

This is a great model for how gases behave at lower pressures and higher temperatures.1871

Let's just dive in and do some examples here.1882

That is going to be the only way to really make sense of any of this: to use these in problems.1887

So, we have our PV=nRT; that is our standard equation--that is actually related to the ones that we did before--this P1V1=P2V2; and you will see how in just a moment, when we do this particular example.1891

Let me do this in blue.1908

This is Example 2.1912

We have diborane, which is B2H6; it has a pressure of 355 torr, which is millimeters of mercury, at -12 degrees Celsius (so C), and a volume of 3.25 liters.1917

I have a sample of diborane gas sitting in 3.25 liters, at -12 degrees Celsius and a pressure of 355 Torricelli.1949

If we raise T to 35 degrees Celsius and raise the pressure to 450 torr, what will be the new volume?1959

I have a sample of gas at a certain volume, at a certain temperature, at a certain pressure; I make a change to the system--I increase the temperature 47 degrees Celsius, and I raise the pressure from 355 torr to 450 torr--so I raise it by 95 torr.1992

Increase the pressure; increase the temperature; I want to know what the new volume is.2012

This is a changing system.2016

Since there is a change in the system, I know that I am going to have a certain set of values initial, and a certain set of values final.2020

Now, let me start with the ideal gas law, and I will show you one in just a minute.2028

Let me rearrange this: PV/nT=R (that is my constant); well, P1V1/n1T1=P2V2/n2T2.2040

Those people who have done some work in gases--you have probably learned them in two different ways.2058

You have learned it as P1V1/T1=P2V2/T2, and you also learned the ideal gas equation.2061

Well, they are just variations of the same thing; the ideal gas equation expresses the state of a gas at that moment--no change; the changing system--I use the ideal gas situation in two different situations: an initial and a final.2067

That is when I set them equal to each other; they are related by the gas constant, R.2082

That is what is constant.2086

Now, I change the pressure; I change the volume, or change the temperature (volume is also going to change); however, in this case, notice: there is no mention of the number of moles of gas.2088

It's just the gas; because there is no mention of it, we don't need it; we can just drop it out of the equation.2099

Therefore, this becomes P1V1/T1=P2V2/T2.2105

Now, I just plug in the numbers that I have and see what I get.2113

Well, the initial pressure is 355 torr; now, let's get in the habit of actually working in atmospheres; so I'm going to go ahead and convert these to atmospheres.2116

P1 equals 355 torr, times one atmosphere, is equal to 760 torr.2129

A torr is just a millimeter of mercury.2140

That is equal to 0.467 atm.2143

Well, the volume 1 is equal to 3.25 liters; that is already in liters.2147

Temperature 1, which is -12 degrees Celsius--I need to add 273 to that, because I want--need--to work in Kelvin.2154

That is the most important one; I could have left the torr alone, because, again, it's just a conversion factor--it's multiplication.2162

Multiplication cancels; addition does not cancel.2168

I have to work in Kelvin.2172

That is that; well, the second pressure, P2, is going to be 450 torr, divided by 760; that is equal to 0.592 atmospheres.2176

The volume--that is what I want, so that is my question mark.2196

My temperature: my final temperature is 35 degrees Celsius, which is equal to 308 Kelvin--again, that plus 273.2204

That is it--now I just put it in--just plug them in.2213

I have 6 variables; I have 5 of them; I'm looking for the sixth.2218

I do 0.467 atm times 3.25 L, divided by 261 K, is equal to 0.592 atm times the volume that I want, over 308 K.2224

Well, of course, atm and atm cancel; K cancels with K; I am left with my liters, which is what I want.2249

I end up with a volume of 3.03 liters; so, sure enough, 3.25 drops to 3.03 liters.2256

I raise the temperature and raise the pressure; if I raise the temperature, you would expect the volume to increase; when I raise the pressure, you expect the volume to decrease; in this case, the pressure increase outweighed the temperature increase, in terms of the effect.2267

The effect was that the volume actually went down.2287

Let's see here: what shall we do next?2295

Before I discuss the next problem, I want to talk about something called STP, Standard Temperature and Pressure.2301

Standard Temperature and Pressure: let's do standard temperature; it's going to be 0 degrees Celsius, which is equivalent to 273 Kelvin.2316

Standard pressure is 1 atmosphere, which is just 1 atmosphere.2336

Now, watch this: this is not actually necessary to do, but historically, we do this, just for the sake of doing it.2342

The volume--if I set one mole of gas...well, let me actually do the math first, so you see where this is coming from.2349

So, if I set PV=nRT, V=nRT/P.2360

Well, if I take one mole of gas, 0.08206, and if I take a temperature of 273 Kelvin, which is standard temperature, and if I divide it by the pressure of one atmosphere, I end up with 22.4 liters.2368

What this means: so, Kelvin cancels with one other Kelvin; atmosphere cancels with the atmosphere; mole cancels with mole; what I am left with is 22.4 liters.2391

So, one mole of gas, at standard temperature and pressure, occupies a volume of 22.4 liters.2403

Any gas--the identity does not matter: argon gas, CO2 gas, diborane gas, phosgene gas--it doesn't matter: any gas--one mole of it at standard temperature and pressure (273 Kelvin, 1 atmosphere) occupies 22.4 liters.2410

You can use that as a conversion factor to help you with some of your problems.2427

For the most part, you can ignore it, because if you just use the ideal gas equation, you will always end up with the right answer.2430

There is no need to learn something else; but again, it is one of those things that people throw out.2436

I figure, well, there it is.2440

OK, let's do another example.2444

A sample of propane gas (propane is C3H8, 3 carbons, 8 hydrogens), having a volume of 2.70 liters at 25 degrees Celsius and 1.75 atm, was mixed with O2 gas--mixed with oxygen gas--having a volume of 40.0 liters at 31 degrees Celsius and 1.25 atm.2450

The mixture is ignited to form CO2 gas plus H2O gas.2516

The question is: how much CO2 gas is formed at 2.6 atmospheres and 130 degrees Celsius (which is pretty standard conditions for once you ignite something).2533

So, I have a sample of propane gas; it is sitting in 2.7 liters; temperature is 25 degrees Celsius; and the pressure on it is 1.75 atmospheres--that is in one vial.2557

In another vial, I have 40 liters of gas, oxygen gas; the temperature is 31 degrees Celsius, 1.25 atmospheres.2569

I mix them together, and raise the pressure to 2.6 atmospheres, 130 degrees Celsius; I want to know how much CO2 is formed under these circumstances.2576

Well, this is chemistry; all chemical problems begin with an equation, in general.2589

A lot of kids end up just sort of starting to just jump into the math without thinking about what is going on; let's go ahead and see what is happening.2595

We have propane gas; we're going to mix it (oops, I'm sorry) with oxygen; we're going to do the balancing as we go, because that is part of the process.2603

We're going to create CO2 gas, and we're going to create H2O.2615

Let's go ahead and balance this equation first.2619

I have 3 carbons; I want to put a 3 over here.2621

I have 8 hydrogens, so I'm going to put a 4 over here.2624

4 oxygens, 6 oxygens, is 10 oxygens; so, I put a 5 over here; now I'm balanced.2627

1 mole of propane reacts with 5 moles of oxygen to produce 3 moles of carbon dioxide, 4 moles of water.2634

Well, given these gaseous conditions, I'm going to use the ideal gas law to find out the number of moles of each.2643

This is a limiting reactant problem; I need to find out which is the limiting reactant in order to find out how much CO2 is going to form.2651

So, let's go ahead and get started!2658

C3H8: the number of moles is equal to PV/RT; I have just rearranged the ideal gas law.2661

I stick the values in here; I have 1.75 (I'm going to skip the units--I hope you will forgive me), 2.7 liters; gas constant is 0.08206, and I am at 25 degrees Celsius, which is 298 Kelvin.2671

Number of moles is 0.193 moles of propane.2693

OK, good; now, let's see what else I have.2701

I have to find out how much oxygen gas I have.2708

Well, again, the number of moles is equal to the pressure times the volume, over the gas constant times temperature.2711

Now, the oxygen is at 1.25 atmospheres; the volume is 40 liters (actually, I said that I was going to skip the units, so let me be consistent here--let me just leave the 40.0 there); and 0.08206 is the gas constant2718

And then, I have 304 Kelvin, because it's at 35 degrees (what was the temperature of the oxygen?--31 degrees Celsius); OK, we end up with 2 moles of O2.2737

Now, I need to find what the limiting reactant is.2754

I take 0.193 moles of C3H8; the mole ratio is 5 moles of O2 per every 1 mole of C3H8.2757

That means I need 0.965 moles of oxygen to react with the 0.193 moles of the propane that I have.2778

Do I have .965 moles of O2?--yes, I have 2 moles of O2.2790

That means that C3H8 is limiting.2794

It's going to run out first.2796

Well, now I do the reaction: 0.193 moles of C3H8 times, now, the mole ratio of moles of C3H8...the mole ratio of that and CO2, which is what I am looking for, is 3:1; that is just straight out of the equation.2800

That means I produce 0.579 moles of CO2.2829

Well, if I have that many moles of CO2, I go back to my ideal gas equation, PV=nRT.2835

Now, I'm looking for the volume that this is going to occupy--just the CO2.2844

Volume=nRT/P; the number of moles that I have created is 0.579; R is 0.08206; my temperature now is 403 Kelvin; and my pressure is 2.60 atmospheres.2854

I end up with 7.36 liters of CO2 gas.2881

Now, mind you, what I have calculated here is the volume of CO2 gas produced, not the total volume of gas produced, because you know that water vapor is also produced; but I didn't ask about the water vapor.2888

I used the ideal gas law to find the number of moles of reactants: a limiting reactant problem--that is all this is.2900

The ideal gas law is just a way...when you are dealing with solids and liquids, with solids, you deal in molar mass; when you are dealing with liquids, you deal in molarity, moles per liter, concentration; when you are dealing with gases, you use the ideal gas law.2907

You are still just dealing in moles; so these are just different techniques in order to handle the different states of matter: solid, liquid, and gas.2920

Look at it that way: the same underlying principles are involved.2928

We still want to know how much of something we are producing from how much of something we are given.2931

That is all that is going on here; that is why we study gases.2936

So again, 7.36 liters of CO2 are produced; it says nothing about the oxygen.2940

This is just the volume of CO2, if I were able to actually contain the CO2.2946

OK, so talked about the gas laws; we talked about the ideal gas law; we talked about pressure; and we talked a little bit about the molar volume of a gas at standard temperature and pressure.2952

Hopefully, this gives you a little bit of a sense of the power of working with gases and working with the ideal gas equation in all of its various manifestations.2964

Thank you for joining us here at

We'll see you next time for the discussion of Dalton's law of partial pressures; goodbye.2977