In this lecture, we're going to wrap up

our study of the physical properties of gasses.

This will conclude the material from Concept Development Study number 14.

So you may want to review that reading as well.

In this lecture, we're going to discuss a few applications of the

Ideal Gas Law which we developed in the previous two lectures.

And we're also going to extend the Ideal Gas

Law to a discussion of mixtures of gasses.

To review very quickly, you'll recall that the Ideal

Gas Law tells us that for any gas regardless

of it's type, type of molecules, that the pressure times the volume is equal to

the number of moles of the gas multiplied

by the temperature multiplied by a constant R.

That constant is given here on the screen.

It is in fact just experimentally measured

to be 0.08206 liter atmospheres per mole Kelvin.

The Ideal Gas Law is a summary of the three experimental

laws that we've produced. Boyle's Law, relating pressure and volume.

Charles's Law, relating volume and temperature.

And Avogardo's Law, relating the volume and the numbers of moles.

Remember that, ideal means that it is independent of the type of gas molecule.

What that means is, we can apply it to just

about any kind of gas that we are interested in.

Here is a particular example. What volume of gas is occupied

by one mole of the gas if we are at room temperature and one atmosphere pressure?

Again the constant is given here on the screen.

Well, let's write this out.

We know that PV is equal to nRT and we are now asking for what

the volume of the gas is, so we can solve this equation for the volume.

It's just the number of moles times RT divided by the pressure.

And we have given here in this equation, or

in this problem, the temperature which is in degrees centigrade 25.

the pressure on atmosphere and the number of moles here.

So you can tell we have n and temperature and pressure.

And it's just a matter of then plugging

those numbers in to get the calculation correct.

So what do we have?

1.00 moles multiplied by R which is 0.08206

liter atmospheres per mole Kelvin

multiplied by the temperature which is 25 degrees centigrade which is 298.15 Kelvin.

We always have to use absolute temperatures there.

And we divide that by the pressure which is given to

be 1.00 atmospheres and this now just becomes a calculator problem.

You can just plug these in. We'll check the units here.

Here is moles per mole.

Here is Kelvin per Kelvin. Here is atmosphere per atmosphere.

The only unit left is liters.

So in fact our answer's going to be in liters.

Plugging the numbers in and calculating them out.

We wind up with a value 24.5 litres is the volume of a single mole of gas.

And it is interesting that because this is the Ideal Gas Law,

whatever gas this is, regardless of the identity of that gas, a single mole of

gas under normal conditions, 25 degrees centigrade and

one atmosphere of pressure will occupy 24.5 liters.

Let's consider then, another application sort of stepping it up one level farther.

In this case, we're actually going to use the Ideal

Gas Law to determine the molar mass of a gas.

So we're going to study now a

hydrocarbon gas, that'll be a gas consisting

entirely of hydrogen and carbon or molecules.

And consists entirely of hydrogen and carbon.

And what we are given here is the density

of the gas at 25 degrees centigrade and one atmosphere.

But that density is not moles per litre but rather grams per litre.

But from the Ideal Gas Law, we can write

that the number of moles divided by the volume

is equal to the pressure divided by R times T.

That's just a rewriting of PV is equal to nRT.

If we then calculate the pressure divided by the temperature and with R,

you'll notice back over here that we do in fact have the temperature.

And we have the pressure.

So we have everything that we need to know to be able

to plug in this equation and find the number of moles per volume.

Let's do that.

The pressure is, what did we say, 1.00 atmospheres.

R is 0.8206 liter atmosphere per mole Kelvin.

And the temperature is, let's see, 298.15K.

And if we plug all those numbers in, let's check the units first.

Atmospheres per atmospheres,

Kelvin per Kelvin and the denominator, we still have units of litre

per mole which will ultimately then give us units of moles per litre.

And if we just plug the numbers into the calculator, we discover

that the number of moles per litre is 0.0409 moles per litre.

But of course that's not the question we were asked.

We were asked what is the molar mass of

this hydrocarbon?

And right now what we have is, instead

the number of moles per litre of the hydrocarbon.

But what we also know of course is that the

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number of moles per litre or the number of moles is equal to

the mass divided by the molar mass. So, if

we simply write this as the mass per

volume times molar mass. Just substituting n

into the equation up above, is equal

to 0.0409 moles per litre. Then let's

see, we have mass per volume is here. That's the number that we've been given.

That's 1.80 grams per mole. So, we get 1.80 grams per litre times one

over the molar mass, is equal to 0.0409 moles per liter.

If we solve

now for the molar mass, that's just taking let's see,

we'll move mass to, molar mass to the other side.

Divide both sides of the equation by 0.0409.

If we do that calculation we will wind up with the molar mass

being 44.0 grams per mole.

The units of litres cancel out.