This content explains the concepts of gas diffusion and effusion, defining them as the movement of gas molecules from areas of high to low concentration and the escape of gas through a small opening, respectively. It highlights that the rate of these processes is inversely proportional to the square root of the gas's molar mass, as described by Graham's Law.
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the topic of this video is effusion and
defusion of gases the learning
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diffusion and and effusion um I'll start
diffusion the defin definition of
molecules
to
differences in
concentration um as with other aspects
of this chapter on gases um a lot of
this is is sort of intuitive if you can
think about it in everyday
circumstances if um you spray perfume or
cologne um and one end of the room where
someone uses a very fragrant air
freshener candle something like that um
you know that at a certain point in time
you'll be able to smell that throughout
the room and that process of of a gas um
dispersing in space from an area of high
concentration at at whatever the source
uh of the gas is to the rest of the the
the room is the process of diffusion so
what does this uh look like um in you
know we can think about if we if we have
two gases separated by a valve um or in
this case a stopcock which is sort of
like a ground glass uh valve um if we
open up that valve over some amount of
time um those gases will become of equal
concentration in both vessels whereas
before they
um held to their own so gases want to um
disperse from areas of high
concentration to low concentration so
that the concentration is sort of the same
same
throughout um uh the we can also make a
comment here about the rate of diffusion
you know how fast can we expect
something um or how do we actually
Define uh this rate so rate of
diffusion is equal to
the amount of sort of gas particles or
we'll just say you know amount of gas it
could be moles
moles passing
through so um we have to also Define
whatever that area is so passing through
some area over time okay so clearly it's
going to be uh it depends this
definition of of the rate of of
diffusion depends on the area uh that
that gases can pass through now if you
restrict that area where gases can pass
through to a very small space this very
small opening then we have a different
process um so diffusion is sort of you
can think about the relatively
unrestricted dispersal of gases um in some
some
container okay so that's the case uh on
the left side over here um but e Fusion
is now going to be have the special
scenario where we have some sort of
barrier and there's a small opening in
that barrier that lets some of the gases
through but not uh but there is some
restriction um what What's Happening
Here is that we we envision all the gas
particles um as sort of bouncing off of
one another and just colliding around
and when we have this barrier oftentimes
the gas particles are hitting the
barrier itself and and staying on their
own side of the the the container but
every once in a while depending on the
um the the relative speed of those gases
some will make it
through um so we also have an expression
effusion and um it looks like this the
rate of effusion so again through some
tiny uh passageway in a container um the
rate of infusion is direct is
proportional to 1 over the square root
of capital M here where capital M is the
um the mass of
gas particle so pretty much the molar
mass of of the gas uh so you can see
that the rate of effusion then is going
to be
um if we have a very high molecular
weight gas or or relatively High
molecular weight gas then we would
expect a lower rate of diffusion and if
we have a a very light gas particle
something with a low molecular weight a
low uh mass per particle then we would
expect a higher rate of
effusion um so we can also uh uh take a
look then before I go over to that image
we can ask ourselves um how could we use
this to sort of predict what would have
a faster rate of infusion between two
gas sample so we can actually look at
Helium versus
argon and if we want to set this
expression up we can write that the rate
of infusion of helium over we can
actually you know take Graham's law and
and sort of sandwich two of them
together so we're dividing one Gams law
expression for helium by the grams law
expression for Argon so what that equals
is one over the square root of the mass
of helium over 1 over the square root of
the mass of argon of course that is a
terrible looking expression we can
simplify that by simply writing the
square root of the mass of argon over
the square root of the mass of helium so
that's going to be equal to square root
we're not going to use um uh all the
sigfigs in the periodic table um I'm
just going to sort of put in the
ballpark values here and um we get a
value of 3.2 according to this to to
using Graham's law so the the
interpretation here since the rate of
helium was on the numerator is that helium
helium
euses approximately three times faster
than argon okay so that would be the
conclusion from the application of
Graham's law here if we look at this
image um on the on the at the starting
point over here this is time equals
zero hours we have helium
filled in the in the orange balloon and
we have argon um that's that's been
filled Into the Blue balloon and um over
some amount of time I think that this is
supposed to be T equals 12 hours is when
this next uh photo was taken um you can
see here that clearly where they were
approximately the same size before if we
just sort of draw those lines we can see
that um the helium balloon has now
shrunk substantially and what's
happening there is that um the
material that balloons are made of are
uh are slightly porous they will allow
gases to escape through the rate of
effusion then depends on the molecular
weight of the gas and helium is a
lighter gas so it's eusing faster and
the balloon deflates
faster um just so you know we can also
uh compare um or we can use Graham's law
to also determine something like uh the
the the the M mass of an unknown gas so
in the interest of time I'm just going
to show you this practice problem so we
have an unknown gas that diffuses
1.66 times more rapidly than carbon
dioxide so we have that value that ratio
of grams law is
1.66 um and this you know instead of 3.2
like we just calculated for helium and
argon we also are given um carbon
dioxide so we can go and calculate the
mar mass of carbon dioxide so really all
you have to do is set up that same ratio
of of grams law where instead of of
solving for this um ratio it's provided as
as
1.66 um so then all you have to do is is
plug in the the M mass of carbon dioxide
um and uh the the marar mass of the
unknown and um making sure that uh we
you you keep track of which that there's
sort of this flip this inversion here
that that the rate of effusion of carbon
dioxide on the um denominator over here
but it's on the it's it's a square root
of the mass of carbon dioxide is on the
um numerator on the other side of the
equation because it's one over one over
something and uh if we just do a little
bit of algebra to isolate the M mass of
the unknown you calculate a m mass of
16.0 gr per mole and um there is no
noble gas for example that has that
specific molar mass but something like
methane one carbon four hydrogens would
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