# Derivation of gas constants using molar volume and STP | Physical Processes | MCAT | Khan Academy

Voiceover: All right. So,
we just put together an equation based off some
basic observations of gas, and we called it the ideal gas
equation, or PV equals NRT. For the most part, the
equation is pretty intuitive, but I remember when I was
learning it for the first time, the part that confused me the most was this constant value R. That’s what I want to try
to clear up for you now. For starters, R is a constant. What that means is that as
the rest of the variables change with whatever
situation we’re looking at, the R value is going to stay the same. Let me show you what I’m
talking about. Let’s isolate R. When we divide both sides by
the moles and the temperature, we’re going to get R
is equal to PV over NT. Which means that for any ideal gas, if you multiply the
pressure and the volume, and you divide that product
by the number of moles, and by the temperature in the system, you’re going to get the same number. Again for any ideal gas,
you’ll always get the value R. I’m going to show you how this works. First, I need to make sure
we’re on the same page about a couple of different things. First, I need you to
know that when I say STP, I mean standard temperature and pressure. The standard temperature is 273 Kelvin, which is the same thing
as zero degrees Celsius. The standard pressure is one atmosphere. So, standard temperature
and pressure is just kind of a nice theoretical condition
that we can perform kind of situational experiments with. The second thing I need to clarify is that we find experimentally,
that for any ideal gas, one mole of gas takes up a volume
of approximately 22.4 liters. We’re going to use
these conditions to find the ideal gas constant. Let’s start with R equals PV over NT. In four condition’s, we’re
going to talk about one mole at standard temperature and pressure. For our pressure, we have one atmosphere, and then we have one mole. We know that one mole is 22.4 liters, so, that’s our volume. Again, we know that at standard
temperature and pressure we’re talking about 273 Kelvin. This should equal R. If we solve this out, our
ones are going to cancel, and we really just need
to divide 22.4 by 273. That’s going to give us .0821, and our units are going to
be atmospheres times liters. So, atmospheres times liters
divided by moles times Kelvin. The constant R is equal to .0821 atmospheres times liters
divided by moles Kelvin. This is the ideal gas constant. It’s going to be the
same for all ideal gases, as long as we’re dealing
with pressure in atmospheres, and volume and liters. It’s probably the one most
often used in general chemistry. I mentioned this earlier,
sometimes we deal with pressure in a unit called a pascal. We deal with volume in cubic meters, and we use both of these. Their both based off of SI Units, and we might have another value for R, which looks like this. We’re still going to start
with R equals PV over NT. We’re still going to use standard
temperature and pressure. For our standard pressure,
instead of using atmospheres, we’re going to use pascals. One atmosphere is equal
to 101,325 pascals. 101,325 pascals is going
to be our pressure. We’re still going to use one mole of gas. Instead of measuring the
volume of one mole in liters, instead of saying that
it’s equal 22.4 liters, I’ll want to do it in cubic meters, because that’s another SI Unit. One cubic meter is equal to 1,000 liters. This means that if we take 22.4 liters, and we do a dimensional analysis, we’re going to get .0224 meters cubed. That’s the value that we’re
going to use in our formula. We’re still talking
about one mole, so .0224. Last but not least,
we’re still talking about standard temperature and pressure, which in Kelvin the
temperature would be 273. We have our pressure, we have our volume, we have our moles, and
we have our temperature. If we run this through
a dimensional analysis to see if we can cancel any units, we would start with 101,325 pascals. Keep in mind that a
pascal is the same thing as saying a newton per meter squared. That’s a SI Unit of pressure. We have 101,325 newtons per meter squared. We would take our volume, and
that’s .0224 meters cubed, and so, we would put that in there. Most of our meters would cancel. Two of the meters on top, and two of the meters on the bottom, leaving us with just a meter on top. Then, we would add in
a division by one mole, because we’re dividing by one mole, and we’ll put that on the bottom. Then we’ll finish it off by
putting 273 Kelvin on the bottom. If we simplify this down, we’d say that 101,325 times 0.0224 divided by 273, and that would give us a value of R equals 8.314 newton meters per mole Kelvin. We could simplify the
units just to take more, because newton meters are
the same thing as joules. So, we could just insert a joule, and we’d have joules per mole kelvin. To be sure, the constants
look different here, but keep in mind that the only
thing changing are the units. The values here
[unintelligible] are always the same for any ideal gas. For any ideal gases, and
the product of the pressure, and the volume divided
by the number of moles and the temperature is equal to R, which might look like 8.21 times 10 to the negative 2 atmosphere
liters per mole Kelvin, or it might look like 8.314
joules per mole Kelvin. Again, this is the same value. Think about what this means. If R is constant, then
R is equal to PV over NT for the initial state of the gas, so for the initial state of the gas. R is also equal to PV over NT
for the final state of a gas. So, for a final state of gas. This is true, because
PV over NT always equals R for any ideal gas. What we’re really saying here is that the initial PV over NT is equal to the final PV over NT which opens up
all sorts of neat, kind of, predictive possibilities say if we were to change one of the variables. We’re going to explore these possibilities in the next few videos.

• akash nagi

Nice video….helped me a lot….:)

• Diana Anna

Helped me a lot! Thanks.

• Ano Varghese

wat abt 8.314 Kj/kmolk

• Jan Roy Malto

how to convert mols to kg?