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## Ideal gas equation

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# Ideal gas equation example 2

## Video transcript

Let's do some more problems
that involve the ideal gas equation. Let's say I have a gas in a
container and the current pressure is 3 atmospheres. And let's say that the volume of
the container is 9 liters. Now, what will the pressure
become if my volume goes from 9 liters to 3 liters? So from the first ideal gas
equation video, you can kind of have the intuition, that
you have a bunch of-- and we're holding-- and
this is important. We're holding the temperature
constant, and that's an important thing to realize. So in our very original
intuition behind the ideal gas equation we said, look, if we
have a certain number of particles with a certain amount
of kinetic energy, and they're exerting a certain
pressure on their container, and if we were to make the
container smaller, we have the same number of particles. n doesn't change. The average kinetic energy
doesn't change, so they're just going to bump into
the walls more. So that when we make the volume
smaller, when the volume goes down, the pressure
should go up. So let's see if we can calculate
the exact number. So we can take our ideal gas
equation: pressure times volume is equal to nRT. Now, do the number of particles
change when I did this situation when I
shrunk the volume? No! We have the same number
of particles. I'm just shrinking the
container, so n is n, R doesn't change, that's a
constant, and then the temperature doesn't change. So my old pressure times volume
is going to be equal to nRT, and my new pressure times
volume-- so let me call this P1 and V1. That's V2. V2 is this, and we're trying
to figure out P2. P2 is what? Well, we know that P1 times V1
is equal to nRT, and we also know that since temperature and
the number of moles of our gas stay constant, that P2
times V2 is equal to nRT. And since they both equal the
same thing, we can say that the pressure times the volume,
as long as the temperature is held constant, will
be a constant. So P1 times V1 is going
to equal P2 times V2. So what was P1? P1, our initial pressure,
was 3 atmospheres. So 3 atmospheres times 9 liters
is equal to our new pressure times 3 liters. And if we divide both sides of
the equation by 3, we get 3 liters cancel out, we're left
with 9 atmospheres. And that should make sense. When you decrease the volume
by 2/3 or when you make the volume 1/3 of your original
volume, then your pressure increases by a factor
of three. So this when by times 3, and
this went by times 1/3. That's a useful thing
to know in general. If temperature is held constant,
then pressure times volume are going to
be a constant. Now, you can take that
even further. If we look at PV equals nRT,
the two things that we know don't change in the vast
majority of exercises we do is the number of molecules we're
dealing with, and obviously, R isn't going to change. So if we divide both sides of
this by T, we get PV over T is equal to nR, or you could say
it's equal to a constant. This is going to be a constant
number for any system where we're not changing the number of
molecules in the container. So if initially we start with
pressure one, volume one, and some temperature one
that's going to be equal to this constant. And if we change any of them, if
we go back to pressure two, volume two, temperature two,
they should still be equal to this constant, so they
equal each other. So for example, let's say I
start off with a pressure of 1 atmosphere. and I have a volume of-- I'll
switch units here just to do things differently--
2 meters cubed. And let's say our temperature
is 27 degrees Celsius. Well, and I just wrote Celsius
because I want you to always remember you have to convert to
Kelvin, so 27 degrees plus 273 will get us exactly
to 300 Kelvin. Let's figure out what the new
temperature is going to be. Let's say our new pressure
is 2 atmospheres. The pressure has increased. Let's say we make the container smaller, so 1 meter cubed. So the container has been
decreased by half and the pressure is doubled by half. Let me make the container
even smaller. Actually, no. Let me make the pressure
even larger. Let me make the pressure
into 5 atmospheres. Now we want to know what the
second temperature is, and we set up our equation. And so we have 2/300 atmosphere
meters cubed per Kelvin is equal to 5/T2, our new
temperature, and then we have 1,500 is equal to 2 T2. Divide both sides by 2. You have T2 is equal to 750
degrees Kelvin, which makes sense, right? We increased the pressure so
much and we decreased the volume at the same time
that the temperature just had to go up. Or if you thought of it the
other way, maybe we increased the temperature and that's what
drove the pressure to be so much higher, especially since
we decreased the volume. I guess the best way to think
about is this pressure went up so much, it went up by factor
of five, it went from 1 atmosphere to 5 atmospheres,
because on one level we shrunk the volume by a factor of 1/2,
so that should have doubled the pressure, so that should
have gotten us to two atmospheres. And then we made the temperature
a lot higher, so we were also bouncing
into the container. We made the temperature 750
degrees Kelvin, so more than double the temperature, and then
that's what got us to 5 atmospheres. Now, one other thing that you'll
probably hear about is the notion of what happens
at standard temperature and pressure. Let me delete all of the
stuff over here. Standard temperature
and pressure. Let me delete all this stuff
that I don't need. Standard temperature
and pressure. And I'm bringing it up because
even though it's called standard temperature and
pressure, and sometimes called STP, unfortunately for the
world, they haven't really standardized what the standard
pressure and temperature are. I went to Wikipedia and
I looked it up. And the one that you'll probably
see in most physics classes and most standardized
tests is standard temperature is 0 degrees celsius,
which is, of course, 273 degrees Kelvin. And standard pressure
is 1 atmosphere. And here on Wikipedia, they
wrote it as 101.325 kilopascals, or a little more
than 101,000 pascals. And of course, a pascal is a
newton per square meter. In all of this stuff, the units
are really the hardest part to get a hold of. But let's say that we assume
that these are all different standard temperatures and
pressures based on different standard-making bodies. So they can't really agree
with each other. But let's say we took this as
the definition of standard temperature and pressure. So we're assuming that
temperature is equal to 0 degrees Celsius, which is equal
to 273 degrees Kelvin. And pressure, we're assuming,
is 1 atmosphere, which could also be written as 101.325
or 3/8 kilopascals. So my question is if I have
an ideal gas at standard temperature and pressure, how
many moles of that do I have in 1 liter? No, let me say that
the other way. How many liters will
1 mole take up? So let me say that a
little bit more. So n is equal to 1 mole. So I want to figure out
what my volume is. So if I have 1 mole of a gas,
I have 6.02 times 10 to 23 molecules of that gas. It's at standard pressure, 1
atmosphere, and at standard temperature, 273 degrees, what
is the volume of that gas? So let's apply PV
is equal to nRT. Pressure is 1 atmosphere, but
remember we're dealing with atmospheres. 1 atmosphere times volume--
that's what we're solving for. I'll do that and purple-- is
equal to 1 mole times R times temperature, times 273. Now this is in Kelvin;
this is in moles. We want our volume in liters. So which version of
R should we use? Well, we're dealing
with atmospheres. We want our volume in liters,
and of course, we have moles in Kelvin, so we'll use
this version, 0.082. So this is 1, so we can ignore
the 1 there, the 1 there. So the volume is equal to
0.082 times 273 degrees Kelvin, and that is
0.082 times 273 is equal to 22.4 liters. So if I have any ideal gas,
and all gases don't behave ideally ideal, but if I have
an ideal gas and it's at standard temperature, which is
at 0 degrees Celsius, or the freezing point of water, which
is also 273 degrees Kelvin, and I have a mole of it, and
it's at standard pressure, 1 atmosphere, that gas should take
up exactly 22.4 liters. And if you wanted to know how
many meters cubed it's going to take up., well, you could
just say 22.4 liters times-- now, how many meters cubed are
there-- so for every 1 meter cubed, you have 1,000 liters. I know that seems like
a lot, but it's true. Just think about how big
a meter cubed is. So this would be equal to
0.0224 meters cubed. If you have something at 1
atmosphere, a mole of it, and at 0 degrees Celsius. Anyway, this is actually
a useful number to know sometimes. They'll often say, you have 2
moles at standard temperature and pressure. How many liters is it
going to take up? Well, 1 mole will take up this
many, and so 2 moles at standard temperature and
pressure will take up twice as much, because you're just
taking PV equals nRT and just doubling. Everything else is being
held constant. The pressure, everything else
is being held constant, so if you double the number of
moles, you're going to double the volume it takes up. Or if you half the number of
moles, you're going to half the volume it takes up. So it's a useful thing to know
that in liters at standard temperature and pressure, where
standard temperature and pressure is defined as 1
atmosphere and 273 degrees Kelvin, an ideal gas will take
up 22.4 liters of volume.