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## AP®︎/College Chemistry

### Unit 8: Lesson 1

Ideal gas equation- The ideal gas law (PV = nRT)
- Worked example: Using the ideal gas law to calculate number of moles
- Worked example: Using the ideal gas law to calculate a change in volume
- Gas mixtures and partial pressures
- Dalton's law of partial pressure
- Worked example: Calculating partial pressures
- Worked example: Vapor pressure and the ideal gas law
- The Maxwell–Boltzmann distribution
- Non-ideal behavior of gases

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# Worked example: Using the ideal gas law to calculate number of moles

AP.Chem:

SAP‑7 (EU)

, SAP‑7.A (LO)

, SAP‑7.A.1 (EK)

The ideal gas law relates four macroscopic properties of ideal gases (pressure, volume, number of moles, and temperature). If we know the values of three of these properties, we can use the ideal gas law to solve for the fourth. In this video, we'll use the ideal gas law to solve for the number of moles (and ultimately molecules) in a sample of gas. Created by Sal Khan.

## Video transcript

- [Instructor] We're told an
athlete takes a deep breath, inhaling 1.85 liters of air at 21 degrees Celsius and 754 millimeters of mercury. How many moles of air are in the breath? How many molecules? So pause this video, and see if you can figure
this out on your own. All right, now let's work
through this together. So let's think about
what they are giving us and what we need to figure out. So, they are giving us a
volume, right over here. They are also giving us a
temperature, right over here. They're also giving us, I'm trying to use all of my colors here, they're giving us a pressure. And they want us to figure
out the number of moles. I'm gonna use a green color here. So they want to know, so we often use the lowercase letter, n, to represent the number of moles. And so, do we know something
that connects pressure, temperature, volume,
and the number of moles? Well, you might be thinking
of the Ideal Gas Law, which tells us that pressure times volume is equal to the number of moles, n, times the ideal gas constant, R, times temperature, T. And so we know everything
here except for n, so we can solve for n. I know what some of you are saying, "Wait, do we know R?" Well, R is a constant. And it's going to be dependent
on which units we use, and we'll figure out
which version of R we use. But that's why I gave you
this little table here, that you might see on a formula sheet, if you were taking
something like an AP exam. So we actually do know what R is. So, we just need to solve for n. So, to solve for n, you just
divide both sides by RT, and so you are going to get that n is equal to pressure times the volume over R times T, R times T. And so this is going to be equal to what? Well, our pressure is 754
millimeters of mercury. Now, over here, where
they give us the ideal gas or the different versions
of the ideal gas constants, you don't see any of them that deal with millimeters of mercury. But they do tell us that each millimeter of mercury is equal to a Torr. If you get very, very, very precise, they are slightly different. But for the purposes of a
first-year chemistry class, you can view a millimeter
of mercury as being a Torr. So, you can view the
pressure here as 754 Torr. So, let me write that down. So, this is 754 Torr. And then we're going to
multiply that times the volume. And here, they give the volume in liters in several of these, and we're probably going
to be using this one, this version of the ideal gas constant, that has liters, Torr, moles, and Kelvin. And so let's multiply times the volume, so times 1.85 liters. And then that is going to be divided by the ideal gas constant. I'll use this version
because it's using all of the units that I already have. I know what you're thinking, "Wait, the temperature's
given in degrees Celsius." But it's easy to convert from
degrees Celsius to Kelvin. You just have to add 273 to whatever you have in degrees
Celsius to get to Kelvin, because none of these are
given in degrees Celsius. And so, I will use this
ideal gas constant. So this is going to be 62.36 liter Torr liter Torr, per mole Kelvin. Mole to the negative one
is just one over mole, so I could write it like this. Kelvin to the negative one
is just one over Kelvin. And then, I'm gonna multiply
that times the temperature. So times, what is 21 degrees
Celsius in terms of Kelvin? Well, I add 273 to that, so
that's going to be 294 Kelvin. And we can validate that
the units all work out. This liter cancels out with this liter. This Torr cancels out with that Torr. This Kelvin cancels out with this Kelvin. And so, we're going to be
left with some calculation. And, it's going to be
one over one over moles, or it's essentially going to simplify to just being a certain number of moles. And so, let's get our calculator out to figure out the number
of moles in that breath. So n, I keep using
slightly different colors, so n is going to be equal to 754 times 1.85 divided by 62.36 and then, also divided by, divided by 294, is equal to this thing. And let's see how many
significant digits we have. We have three here, three
here, three here, four here. So, when we're multiplying and dividing, we just want to use the fewest
amount that I'm dealing with. So I wanna go to three
significant figures. So 0.0, one, two, three
significant figures, so 0.0761. This is going to be 0.0761. And I could say approximately
'cause I am rounding. But that's three
significant figures there. So, that's the number of
moles of air in the breath. Now, the next question is
how many molecules is that? Well, we know that each
mole has roughly 6.022 times 10 to the 23rd molecules in it, so we just have to
multiply this times 6.022 times 10 to the 23rd. So, we could write it this way. We could write 0.0761 moles, I'll write mole, times 6.022 times 10 to the 23rd molecules, molecules per mole. Now these are going to cancel out, and I'm just going to
be left with molecules. And I can just take the
number that I had before 'cause it's nice to be
able to retain precision until you have to think about
your significant figures. And so, but once again, because we did this whole calculation, we're going to wanna round everything to three significant figures. So, let's just multiply this times 6.022. EE means times 10 to the, times 10 to the 23rd, is equal to that. And, if I round to three
significant figures, because my whole calculation, that was my limiting significant figures, I have 4.58 times 10 to the 22nd. So, this is 4.58 times 10 to the 22nd molecules. Squeeze that in there, and we're done.