So now that we have an idea of what the ideal gas law is, and how it works.
Let's do a quick example.
Suppose that you know that a diver's lungs hold about 6 liters of air.
Now that the diver we know that this volume of his lungs.
He goes down into the water to a pressure of 3 atmospheres.
So we're not actually asking about the depth,
we're just going to a particular pressure.
The question could be, how many liters of air
would it take to fill his lungs if that air were brought back up to sea level?
In other words, if we took a picture of him at the bottom of the ocean
and knew how much air was in a particular volume which is his lungs.
We know we brought that air back up to sea level
where the pressure is much less, the air would expand.
The question is to what volume that air expand
if we were brought up to sea level
or equivalently what volume of air do we need
to put in to the diver's gas tanks for example
in order to make sure that this gas can actually fill his lungs?
Which is of course an important question.
So give this a problem a shot, again using the ideal gas law
and using the way that we discussed to derive quantities
based on knowing which things are being held constant.
If you solve this problem, hopefully it look a little something like this.
What we can say, is again, that we have this diver.
He is going underwater and so I'm just representing his lungs as a box here.
His lungs are full of some air.
So we have a volume, down here at the depth,
and we also have some pressure way down here.
What we're going to do is take this air
and bring it back up to the surface where it's going to expand
because it's much less pressure on it.
So now, we only have the atmosphere pressure
and we have a much greater volume.
So let's maybe call this pressure 1 and volume 1.
Maybe instead of atmosphere pressure, we can just keep in general,
call this pressure 2, and volume 2.
So in this problem what you can see
is that obviously both the pressure, and the volume are changing.
Which means the other variables in our ideal gas law which we should write here.
PV equals n times R times the temperature.
The number of molecules that we're talking about.
So this quantity, or this quantity is the same, so this is a constant.
R of course is always a constant
and the temperature we're assuming to be a constant
because were not given in this problem and the temperature is changing in any way.
We're just bringing the gas right back up to the atmosphere pressure
and seeing what happens.
So what we can say and since this entire thing is constant.
The product of pressure times volume, also has to be a constant.
So what we know is that pressure times volume
initially has to be equal to pressure times volume finally.
In other words, there can't be any difference here.
So now this isn't so bad, all we have to do is solve this for the ratio of the volumes,
so let's maybe say, the new volume,
divided by the old volume just by dividing both side of this equation is now equal to.
We have to be careful here, we divided both side by V1
now we're going to divide both side by P2 is equal to P1 over P2.
So always be careful in this step, this is a very easy step to make a mistake
and we might have had V2 over V1, equals P2 over P1.
And that would have been incorrect.
So now that we have this we know the ratio of the pressures,
because this pressure P1, were given in the problem,
is equal to 3 times the atmosphere of pressure or 3 times P2 the pressure up at the top.
So this is the last key to this puzzle,
so we just plug this in, we have 3 times P2 over P2.
So you see, I just re-written P1 as 3 times P2 from this equation here.
Now, the P2 is cancel and we see that the ratio of the volume must be 3.
In other words, the volume of the air,
this volume V2 is 3 times whatever the volume was here.
So the question is, what is the volume there?
So we know that the new volume is equal to 3 times the original volume,
so we have one last step which is what is the original volume.
What we know that this volume of air was in this diver's lungs.
So we already in fact know that V1 is equal to 6 liters.
So this is the other bit of information we're going to need to use.
So let's use this. In this case, the new volume, the volume way up here the atmosphere pressure
is equal to 3 time 6 liters or 18 liters of air.
So in this problem, you can see exactly how you would solve something like the ideal gas law.
Again, taking whatever constants you have in the problem that you can assume to be constant
making sure they are all in their own in your equation,
and then knowing that everything else in your equation
is something that's going to be held at a constant value.
This completes the lecture on the ideal gas law
as well as some basic properties of gases and how they work.
We're going to explore this ideas a little more
and get into some more practical applications of how gases work as well.
And until then thanks for listening.