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# Graham's law of diffusion

Graham's law of diffusion (also known as Graham's law of effusion) states that the rate of effusion a gas is inversely proportional to the square root of its molar mass. Often, it is used to compare the effusion rates of two gases. This is represented by the formula: $\frac{\text{rate of effusion A}}{\text{rate of effusion B}} = \sqrt{\frac{M_{B}}{M_{A}}}$ where $M$ refers to molar mass. Created by Rishi Desai.

## Want to join the conversation?

- Why are london forces found in all compounds?(8 votes)
- London forces are also called instantaneous dipole attraction, this is because of the polarity resulting from the transient unequal distribution of electrons in a bond or an electron cloud (in the case of single atoms or ions). Electrons are constantly moving and when they are concentrated at a given place of a bond/electron cloud a negative pole is created in that point in time (and a counterpart positive pole due to a local region of electron deficiency at the other end of the bond/atom). Since all chemical species possess electron clouds/bonding electron clouds they can all participate in dispersion/London forces.(9 votes)

- How did Graham, if he was the first to find out, find out how atoms had this weight that could slow or speed them up? I mean, you can't really see an atom unless you use modern technology, right?(10 votes)
- How would you solve for the molecular weight of an unknown gas instead of its rate? For example, if you knew that oxygen's rate of effusion was twice that of an unknown gas, how would you solve Graham's law to obtain the molecular weight (molar mass) of the unknown gas?(4 votes)
- This is how I would do it:

2x / x = sqrt( y / 32 ),

where x = rate of effusion of the unknown gas and y = molecular weight of the unkonwn gas.

2 = sqrt( y / 32)

4 = y / 32

y = 4 * 32

y = 128(6 votes)

- Doesn't that mean whatever you choose for rate 1 will always be faster? Also doesn't that mean what every has a lower Molar Mass is always faster? If so, there is no need for Graham's Law.(4 votes)
- Shouldn't this be Graham's Law of Effusion, as opposed to diffusion?(2 votes)
- No because Effusion is a process that occurs when a gas is permitted to "escape" its container through one or many small holes to another container whitout molecular collisions.

In this case, its diffusion becase there is appening af mixing of molecules by random motion, where molecular collisions occur.(2 votes)

- so, why did the Kinetic energy of the Oxygen molecules and the Kinetic energy of the Carbon dioxide molecules the same to make the equation? (obviously the larger the molecule, the larger the kinetic energy with the same speed; but we don't know the speed of the molecules;)(1 vote)
- Kinetic Energy of any gas is equal to 3/2RT (and not 3/2nRT), so in this case the temperature of atmosphere is same. Therefore, they have equal kinetic energy.(3 votes)

- At4:05, Rishi says that the energy formulas are the same for O2 and CO2 because they both absorb the same amount of energy. However, you don't have to heat up the pot for the molecules to start diffusing once the top is removed, and in fact the energy absorbed by different molecules is different for the same change of temperature. Can someone please confirm that the two sides of the equation are equal not because they are both absorbing the same amount of energy from the fire, but instead because they are both the same temperature and temperature is basically a measure of molecular kinetic energy?(2 votes)
- You're right on the money.

The internal energy of an O2 molecule is (5/2)kT, while the internal energy of a CO2 molecule is (6/2)kT. The internal energies are different because CO2 can store energy in vibrations in more ways than O2. This difference in internal energies means that CO2 will need to absorb more energy to be at the same temperature as O2. As you noted, the temperature is only related to the translational energy of the molecules, not the vibrational and rotational.(1 vote)

- Theoretically, how would these molecules have enough energy to move at 1000mph?(1 vote)
- The nitrogen molecules have enough energy to move at 1000mph simply because of their temperature. If you were to make the nitrogen gas hotter, then the nitrogen molecules would go even faster than 1000mph!(3 votes)

- In the rewritten formula, why is it the molecular mass 2/ molecular mass 1 and not vice versa?(1 vote)
- Around seven minutes, I've watched this ten ten second loop for ten minutes straight. I'm still confused as to why he puts the square root of the molecular weight over one and says that it is inversely proportional to the diffusion rate.(1 vote)
- Imagine in the pink equation that MW2 and Rate2 are both constants. Then one could rearrange the equation into Rate1 = c * 1/sqrt(MW1) . Here c = sqrt(MW2)*Rate2 . As long as c is treated as a constant, Rate1 ~ 1/sqrt(MW1).(1 vote)

## Video transcript

Welcome to a new planet. This is actually a planet I'm
going to call Planet Graham. And you can already see the
blue alien on this planet. And the reason I want to call it
Planet Graham, I'll be honest, is because we're going to
talk about Graham's law. And I thought it would
be kind of a fun way to remember it and
think about it. So, before I came
to this planet, I actually brought along
with me a giant pot. And this pot is
like a cooking pot. But inside of it-- and I
put a lid on it of course-- but inside I've got some
molecules of oxygen, actually. From our home planet Earth,
I brought some molecules of oxygen. And I'm drawing them
as O22 molecules, or two atoms, rather, of
oxygen in one molecule of O2. And I've also got some
carbon dioxide here. And of course, carbon
dioxide-- in the name you can already hear
it-- it's got dioxide. So, it's got two
oxygens as well. It almost looks like
a little bow tie. So, I've got a few
of these molecules. And I was very careful actually,
made sure that in my pot I had 50% carbon
dioxides and 50% oxygens. So, I've got equal
proportions of both. So, I have a friend over
here, an alien friend, and I asked my friend to stand
some distance away from my pot. And you can see I'm
actually cooking my pot. I've got some fire with
firewood underneath it. And I say, please stand 10
feet away from my pot, sir. And this little alien friend
is a good friend of mine, so he says, no problem. And the reason I'm asking
him to help me with it is because he has a very
special nose, a very, very special nose. He's never in his a life smelled
oxygen or carbon dioxide. He's lived on Planet
Graham his whole life. And Planet Graham has these
little green molecules. But he has such a special
nose that he can actually detect whether he's smelling
carbon dioxide or oxygen. So, I'm going to actually
take this lid off. And I'm going to say,
hey, if you detect with your special nose either
one of these-- let's say that these molecules,
one of them goes over and kind of goes into
his nose-- if you can detect it, please let me know
which one you're smelling. And that's my test. And I want to know which
of these molecules, oxygen or carbon dioxide, is going
to reach his nose, which is 10 feet away, first. So it's basically a race. And you can make a
prediction right now as to which molecule
you think is going to get to his nose first, the
oxygen or the carbon dioxide. Now you might think,
oh, it's very easy. There's a direct path. But actually remember,
these molecules, these green molecules in
the planet atmosphere, are whizzing around. They're going in all sorts
of different directions. And as a result, they're
going to smack into our carbon dioxide or oxygen
molecules as they try to make their
way over there. And kind of a random fact,
but an interesting one to think about, is
that in our atmosphere we have a lot of nitrogen,
a lot of nitrogen gas. Now, if you took one nitrogen
gas molecule, which is N2, and let it go, and measured its
speed and kind of clocked it, it would be going at
about 1,000 miles an hour. But the only reason
it doesn't actually go that speed in reality
is because the molecules of nitrogen, they actually
will smack into each other and bounce off of each other
millions and millions of times every second. And so because they're smacking
and colliding constantly, they never really reach
those real potential speeds. They go much slower. So really what
we're talking about is when molecules are bouncing
and clanging into each other and slowly making progress
towards our little alien's nose, that is the
idea of diffusion. They're going to kind of
rattle around and slowly make their way over to his nose. And maybe if I came
back, let's say, 10 minutes later, maybe
this little oxygen would be right here. Maybe you might have a little
carbon dioxide right here. Slowly making progress
towards the nose. That's what we're
trying to figure out-- which one will get
over there first. So, you've had time
to think about it. And I'm actually
going to tell you how I think we should
approach the problem, which is thinking back
to kinetic energy. We're heating this
thing up, so we're putting thermal, or heat
energy, into the molecules. Both types of molecules are
getting the same amount. I've got the oxygen getting
some kinetic energy. I'm going to put a
little o for oxygen. And remember, the formula is
1/2 mass times velocity squared. And it's going to
equal, or should equal, the amount of energy that my
carbon dioxide is getting. And I'm going to do that as a
little c for carbon dioxide. So these two
molecule types should be getting the same
amount of energy. Now remember, it's not like it's
one molecule we're thinking of. We're thinking of
many, many molecules. So first, I'm going
to have to change these units a little
bit. m, or mass, is going to change
to molecular weight. Because again, I'm thinking
about the individual molecule. So I've got to figure out
what these molecules weigh. And v is going to change over
to rate, or diffusion rate. And the reason I'm doing
that is because, again, I'm thinking about the overall
diffusion of the gas. It's not like I'm betting
on any one molecule. I'm betting on the entire
population of carbon dioxide molecules beating out the
population of oxygen molecules, or vice versa, the oxygen
molecules beating out the carbon dioxide molecules. But not an individual molecule. So I have to think
of the average rate that those molecules are moving. So, let me rewrite
this equation. It's going to now be 1/2
times molecular weight, I put it in parentheses,
times diffusion rate. I'm just going to call it rate. And we'll call it rate 1. And 1 will be for the oxygen. In fact, molecular weight 1
can be for the oxygen as well. And over here, I'm
going to say it equals 1/2 times the
molecular weight 2. And 2 refers to carbon
oxide, and rate 2 refers to carbon
dioxide as well. And I really don't need to keep
carrying on with these halves. I can just multiply both
sides of the equation by 2 and get rid of them. So that makes it a
little bit easier. And I almost forgot, I
have to square both sides. That would have been a mistake. I forgot to square them earlier. So now I've squared them. And let me actually
rearrange it to make it a little bit
neater in a new color. So let's do this. So let me write it
out nice and neat. And this is actually
going to be Graham's law. So all I'm doing is
rearranging the formula. I've got rate 1. This is the diffusion rate
of one molecule divided by the diffusion rate
of a second molecule, and then the molecular
weight on the other side of the second molecule divided
by the molecular weight of the first molecule. And you do a square
root of this side. So, that's just a
rearrangement of the formula. But what I've written out
for you is now Graham's law. It's basically taking
the kinetic energy rule and rearranging it to
make sense for molecules. And let me make a little
bit of space here. And so that an
extension of this would be if you're just thinking about
one molecule, then the rate, the diffusion rate-- when I say
rate I mean diffusion rate-- is going to be proportional
to the square root of the molecular weight. So, let's figure out
how to apply this to our little riddle. We wanted to know
whether oxygen or carbon monoxide is going
to diffuse faster. And I can now go back to
our the periodic table and look up oxygen. And I know that the molecular
weight of 16 here and carbon is 12. And that means that
O2 is just 16 times 2. So the molecular weight is 32. And carbon dioxide is going
to be 2 oxygens plus 12 more. So it's going to be 44. So these are the
molecular weights of carbon dioxide and oxygen. So basically what I do
is I just plug them in. And I say, OK, let's plug
them into the formula. Let's use this one right here. And I'm going to call
rate 1 my oxygen rate. So what's happening with rate 1? We'll say, well, rate
1 is rate of oxygen-- I'm going to write
a big o here-- equals the square root
of-- let's make sure I stay consistent--
I said 1 was oxygen, so it's going to be 32
down here and 44 up here. And then that's going
to be multiplied by rate of carbon dioxide. And I'll put a c
for carbon dioxide. So what does this
work out to be? That's 1.17. I just punched it
into the calculator. So really, the
diffusion rate of oxygen is 1.17 times faster--
this is our answer-- than the rate of carbon dioxide. So that's our answer. The oxygen is going
to be the winner. So it's going to move faster. This is going to move
a little bit faster. And it's going to get to our
alien friend's nose first. So this is the power
of Graham's law. It's basically
telling us that, hey, if you have a small
molecular weight, you're going to be able
to diffuse pretty fast.