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## Class 11 Physics (India)

### Course: Class 11 Physics (India) > Unit 17

Lesson 1: Ideal gas equation- The ideal gas law (PV = nRT)
- Worked example: Using the ideal gas law to calculate number of moles
- Worked example: Using the ideal gas law to calculate a change in volume
- Calculations using the ideal gas equation
- Derivation of gas constants using molar volume and STP
- Boyle's law
- Charles's law
- Avogadro's law
- Gas mixtures and partial pressures
- Worked example: Calculating partial pressures
- The Maxwell–Boltzmann distribution
- Dalton's law of partial pressure
- Gas phase questions

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# Avogadro's law

Created by Ryan Scott Patton.

## Want to join the conversation?

- Where does .041mol come from @1:53(12 votes)
- A great question! It took me some time to find the answer and like the most of us watching the video I just took it to notice and didn't think about it. but dividing 1 through 22,4 gives a slightly different number of moles,

But let me explain it what i found out:

He is saying starting @0:52*"... if you filled up four balloons to exactly one liter at 25 degrees celsius..."*

and the key point is the**one liter at 25°C**and it took me some browsing on google looking for how much volume takes one mole of ideal gas at 25°C.

- We know that one mole of ideal gas at STP (Standard Temperature and Pressure- which means 0°C and 1 atmosphere of pressure) takes 22,4 liters

But how much of volume increase would it be if we heat up our balloons from 0°C to room temperature or more exactly 25°C (presuming that we are filling the four balloons somewhere at the garage or some simple room and not inside a pressure chamber where you can change the atmospheric pressure, so our pressure would stay 1 atmosphere.

And finally I found this at wikipedia over the ideal gases:

"The molar volume of an ideal gas at one atmosphere of pressure is

22,414 dm3/mol at 0°C

24,465 dm3/mol at 25 °C"

dm3 equals to liters so by rearranging we can figure out that if we would have 1 mol of gas at one atmosphere and 25°C it would make our balloon 24,465 liters big, but here we have a 1 liter big balloon so it contains 1 / 24,465 moles which is 0,0408 rounded up to....

yes exactly here it is where it comes from**the 0,041 moles***References:*

http://en.wikipedia.org/wiki/Molar_volume(32 votes)

- this may be a dumb question but im asking anyways..

Different gas molecules have different sizes, then why do the different gases at equal volume have same number of molecules ....??

To make it more clear take an example of hydrogen and methane ...

How can same number of molecules of methane fit in 1 litre of container as by hydrogen when methane molecule is obviously bigger than hydrogen molecule ?

plz answer fast cause my exams in a week(12 votes)- You're right - size would seem to be a potentially limiting factor at glance. However, these molecules are still very small when compared to the available space or volume around them. So small, that an assumption we make with ideal gas laws, is that these molecules occupy no space or volume.

Contrast this with real gas laws, where we DO assume that these molecules do occupy space.

Check out the next video in this playlist: https://www.khanacademy.org/test-prep/mcat/physical-processes/gas-phase/v/real-gases-and-the-van-der-waals-equation(4 votes)

- How Did He Get That .97 in the problem as n2(3 votes)
- He added 0.15 and 0.82 because the question says ".15 moles of helium gas are ADDED to a piston containing .82 moles of another gas" therefor you started with .82 moles (n1) and added .15 moles leading to the equation .82 + .15 = .97 so your final number of moles is .97 (n2)(6 votes)

- but what about ,force=mass *acc., so, molecules having more mass should have more pressure as compared to oters having less mass or less volume?(4 votes)
- Yes you are right for real gases, but in this case we are talking about ideal gases, and according to the definition the ideal gases have randomly moving point particles that do not interact except when they collide elastically (
*http://en.wikipedia.org/wiki/Ideal_gas*) but the fact is that many real gases at normal conditions act like ideal gases- when the molecules are too big they don't, but methane is still tiny, and all the rest as well, even if they are not point shaped. The key point is, that one mol of each of them contains 6.022* 10^23 number of particles and takes in about 22,4 liters volume at 0°C and 24,5 liters at 25°C (http://en.wikipedia.org/wiki/Molar_volume the section about ideal gases)(3 votes)

- boi someone explain more simple I don't get it(3 votes)
- Based on your comment, I think all you need to know from this video is that the amount of one "mole" of something is the same as 6.022 X 10^23 atoms of that substance. And when you blow up a balloon, it gets bigger.(4 votes)

- Okay so, I'm using this video to revise for my Chemistry end-of-year, and I wanted to ask a bit of a stupid question: Where is s.t.p. applicable? Usually, questions I've encountered about molar gas volume involve compounds at r.t.p., so where is s.t.p. used? Is s.t.p. a better measure than r.t.p.?(3 votes)
- S.T.P is used when the gas is present at these conditions. In which when gas is at 1 atm and 273K then the gas will also be 22.4 L per mole. Usually, within problems, it is clearly stated when gas is at S.T.P. This is usually used within combined gas laws, ideal gas law, as well as problems relating to specific laws (i.e. Boyle's law, Avogadro's law, etc.). Between S.T.P and R.T.P, usually, S.T.P is mainly used within gas law problems especially.(2 votes)

- Thank You for explaining this better than my teacher(3 votes)
- 22.5 liters of something in the gas phase at STP = 1 mol does not make any sense because:

1) Molar mass is different between 1 gas and another and the size of the molecules is also different(Like CH4 vs O2 for example with CH4 being larger than O2)

2) What if you lower the pressure of something so that it becomes a gas and then raise it to STP? The molar volume while still in the gas phase would be different because of pressure initially being lowered.

3) every gas has a different melting point and boiling point so that N2 vs O2 will have one of them(The O2 I think) moving around really fast and the other a little slower and that would effect the molar volume.(2 votes)- Consider this:

22.4 L of something in the gas phase at STP = 1 mol is correct because

1) We don't take into consideration molar mass of the gas, we look at the NUMBER OF MOLES. Now for any gas, 1 mole means the exact same thing. If its O2, then in mole O2 there are

6.022* 10^23 O2 molecules. In 1 mole of CH4, there are 6.022* 10^23 CH4 molecules and so on. That doesn't change.

Also, one of the assumptions of the ideal gas equation is that we consider each and any molecule or atom (if the gas is momoatomic) a point mass. We consider its atomic/molecular volume negligible. I think this one in one of the videos... :)

2) Since Volume is inversely proportional to Pressure (Boyle's Law) once you take away the pressure, the volume (which decreased when pressure increased) should return to 22.4 L again (if n = 1 mole). The two work together.

3) You are right about boiling point. But if you remember, we standardised the Temperature. Temperature is basically the average kinetic energy of a gas (they have a video on this too). So if temperature of both gases is same, their kinetic energy is same, so their molecules will have equal amount of energy - so that will not affect molar volume.

I hope that answered all your questions :D

Cheers.(3 votes)

- can you please explain Gay-Lussac law of combining volume(2 votes)
- I'm a little confused where the 0.041 mol came from. Was it give or something calculated?(2 votes)

## 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.