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# Avogadro'sÂ law

## Video transcript

Voiceover: So continuing the history of the ideal gas equation here. We get to the 19th century
with an Italian chemist name Amedeo Avogadro. Actually his name was Lorenzo Romano Amedeo Carlo Avogadro di
Quaregna e di Cerreto, but we're going to call him Amedeo. Amedeo spent a bit of
his time experimenting with tiny particles. In honor of his experiments, the number of particles
in one mole of something was named Avogadro's Number which is approximately
6.02 times 10 to the 23. One thing that Avogadro postulated was that equal volumes of gas at the same temperature and pressure contain equal number of particles whether those particles
are atoms or molecules. For instance, if you
filled up four balloons to exactly one liter at 25 degrees Celsius with different gases, so let's have a green balloon, and we can say that this
green balloon is argon, and we'll have a pink balloon, and I'll try to make
it the exact same size as the previous one, and
we'll say it's nitrogen. Then we've got a blue balloon that we'll say is hydrogen. Again, I'm off a little bit in the sizes, but we're saying that all of these have exactly one liter volume. Then our fourth balloon will be yellow, and we'll say it's filled with methane. What Avogadro was saying was that at one liter each one
of these four balloons would have the same number of particles. I'm using five particles, but the point is that
all of these balloons would have the same number of particles. At one liter, each of these balloons would contain .041 moles. If we use Avogadro's Number here, that's the same thing as saying that each of these balloons would be filled with 2.5 times
10 to the 22 particles. If you think about it, this is the exact same
thought behind our value for the molar volume at STP. We said that for any gas, one mole at STP would take up a volume
of about 22.4 liters. The fact that we know
this is true for any gas not just kind of a particular gas, is credited to the work
of Amedeo Avogadro. Avogadro used this idea and some intuition to develop his law which
is V is equal to A, a constant A, times the number of moles which means that the moles of gas, the number of particles
of gas in the system varies directly with the volume. Or the quotient of V and
N is equal to a constant. I'm sure that you've
used this same intuition when you've blown up a balloon. As you put in more air particles, as you blow more air into the balloon, the balloon gets larger. For example, if we put in
one mole of air particles, the balloon would be about 22.4 liters. As we blew more air into it, say another third of a mole, we would increase the
volume to 29.88 liters. Another third of a mole of air gets us up to 29.88 liters. Our volume is getting bigger. If we were to blow even more air into this system, we would increase the volume even more. So another third of a mole of air would bring our volume up to 37.35 liters. Our volume has expanded proportionally to the amount of air that we've increased in this system. Avogadro's law validates the V,N part of the ideal gas equation
because the volume and the moles are directly proportional to each other. Let's put this principle
to work in an example. When .15 moles of helium gas are added to a piston containing
.82 miles of another gas, by what percent does the
total volume increase assuming isothermal and
isobaric conditions? Isothermal is same temperature, and isobaric is same pressure. We have same temperature, same pressure, and we're looking for the percent of volume increase related
to the molar increase. This is a perfect opportunity
to use Avogadro's Law. Let's start with V1 divided by N1 is equal to V2 divided by N2. We know this is true
because Avogadro's Law says that the quotient of the volume and the number of moles is
constant for an ideal gas. So the initial conditions would equal the final conditions. We have a ratio set up here, and we can rearrange this ratio to say that V1 over V2
is equal to N1 over N2. Here we've got kind of an interesting part that we can actually start using to solve this problem because what we see is that the increase in volume is directly proportional
to the increase in moles. For that reason, the
percent change in volume looks like the final volume
minus the initial volume divided by the initial volume. This is just the formula
kind of for percent change which is pretty intuitive. The final volume minus the initial volume, the change, divided by the initial volume is going to give us the percent change. We know that this is equal to the percent change in the number of moles so we can say N2 minus N1 divided by N1. This really gets us to
a point where we can use the values in the formula
because it gives us an initial number of
particles, an initial moles and it gives us the change in moles. We had the final moles as well. To find the final moles
because it just says that .15 moles are added to .82. We'll need to add those together so .82 plus .15 is equal to .97. So .97 is our final number of moles. That would be our N2. Let's substitute these values in. Our final moles is .97
minus our initial moles which is .82 divided by our initial moles. That's going to give
us .15 divided by .82. Now we're at a point
where we can solve this, and we can just kind of
evaluate this expression and that would give us the total, or the percent of the
total volume increase, but I want to take a second and I want to think about how we could think about this kind of
with some common sense and use rounding to give
us an approximate answer. So .15 divided by .82 is
between .15 divided by 1, and .15 divided by .75. And .15 divided by 1 is 15%, and .15 divided by .75 it goes into .75 five times so that would give us 20% because one-fifth is 20%. If we had to think about this expression in a hurry, say on a
test, and we didn't have a calculator, we could use rounding to give us a pretty approximate answer. We know that the answer
is between 15 and 20%. It turns out that that actual answer if you do use a calculator is really close to 18.3%. So when .15 moles of helium gas are added to a piston containing .82 moles of another gas, by what percent does the
total volume increase? It's just about exactly 18.3%. We were able to solve this problem with a little bit of critical thinking based on the principles of Avogadro's Law.