0:04 the topic of this video is effusion and
0:06 defusion of gases the learning
0:08 objectives are on the screen and so you
0:10 can pause the video now to write those
0:14 down to begin this discussion of
0:17 diffusion and and effusion um I'll start
0:24 diffusion the defin definition of
0:36 molecules
0:50 to
0:53 differences in
0:55 concentration um as with other aspects
0:58 of this chapter on gases um a lot of
1:00 this is is sort of intuitive if you can
1:02 think about it in everyday
1:07 circumstances if um you spray perfume or
1:09 cologne um and one end of the room where
1:12 someone uses a very fragrant air
1:15 freshener candle something like that um
1:17 you know that at a certain point in time
1:18 you'll be able to smell that throughout
1:23 the room and that process of of a gas um
1:26 dispersing in space from an area of high
1:28 concentration at at whatever the source
1:33 uh of the gas is to the rest of the the
1:36 the room is the process of diffusion so
1:40 what does this uh look like um in you
1:42 know we can think about if we if we have
1:45 two gases separated by a valve um or in
1:47 this case a stopcock which is sort of
1:51 like a ground glass uh valve um if we
1:53 open up that valve over some amount of
1:56 time um those gases will become of equal
1:59 concentration in both vessels whereas
2:00 before they
2:04 um held to their own so gases want to um
2:05 disperse from areas of high
2:07 concentration to low concentration so
2:09 that the concentration is sort of the same
2:09 same
2:13 throughout um uh the we can also make a
2:15 comment here about the rate of diffusion
2:17 you know how fast can we expect
2:20 something um or how do we actually
2:26 Define uh this rate so rate of
2:30 diffusion is equal to
2:33 the amount of sort of gas particles or
2:35 we'll just say you know amount of gas it
2:36 could be moles
2:38 moles passing
2:48 through so um we have to also Define
2:50 whatever that area is so passing through
2:55 some area over time okay so clearly it's
2:57 going to be uh it depends this
2:59 definition of of the rate of of
3:02 diffusion depends on the area uh that
3:04 that gases can pass through now if you
3:06 restrict that area where gases can pass
3:09 through to a very small space this very
3:11 small opening then we have a different
3:14 process um so diffusion is sort of you
3:16 can think about the relatively
3:20 unrestricted dispersal of gases um in some
3:21 some
3:24 container okay so that's the case uh on
3:27 the left side over here um but e Fusion
3:29 is now going to be have the special
3:31 scenario where we have some sort of
3:33 barrier and there's a small opening in
3:36 that barrier that lets some of the gases
3:38 through but not uh but there is some
3:40 restriction um what What's Happening
3:42 Here is that we we envision all the gas
3:45 particles um as sort of bouncing off of
3:47 one another and just colliding around
3:49 and when we have this barrier oftentimes
3:51 the gas particles are hitting the
3:53 barrier itself and and staying on their
3:56 own side of the the the container but
3:58 every once in a while depending on the
4:01 um the the relative speed of those gases
4:03 some will make it
4:06 through um so we also have an expression
4:21 effusion and um it looks like this the
4:24 rate of effusion so again through some
4:28 tiny uh passageway in a container um the
4:30 rate of infusion is direct is
4:33 proportional to 1 over the square root
4:37 of capital M here where capital M is the
4:41 um the mass of
4:43 gas particle so pretty much the molar
4:47 mass of of the gas uh so you can see
4:51 that the rate of effusion then is going
4:52 to be
4:56 um if we have a very high molecular
4:58 weight gas or or relatively High
5:00 molecular weight gas then we would
5:03 expect a lower rate of diffusion and if
5:07 we have a a very light gas particle
5:09 something with a low molecular weight a
5:12 low uh mass per particle then we would
5:15 expect a higher rate of
5:19 effusion um so we can also uh uh take a
5:21 look then before I go over to that image
5:25 we can ask ourselves um how could we use
5:27 this to sort of predict what would have
5:29 a faster rate of infusion between two
5:31 gas sample so we can actually look at
5:33 Helium versus
5:36 argon and if we want to set this
5:38 expression up we can write that the rate
5:40 of infusion of helium over we can
5:43 actually you know take Graham's law and
5:45 and sort of sandwich two of them
5:47 together so we're dividing one Gams law
5:49 expression for helium by the grams law
5:51 expression for Argon so what that equals
5:54 is one over the square root of the mass
5:58 of helium over 1 over the square root of
6:01 the mass of argon of course that is a
6:03 terrible looking expression we can
6:05 simplify that by simply writing the
6:07 square root of the mass of argon over
6:10 the square root of the mass of helium so
6:12 that's going to be equal to square root
6:15 we're not going to use um uh all the
6:18 sigfigs in the periodic table um I'm
6:21 just going to sort of put in the
6:24 ballpark values here and um we get a
6:27 value of 3.2 according to this to to
6:29 using Graham's law so the the
6:31 interpretation here since the rate of
6:35 helium was on the numerator is that helium
6:37 helium
6:42 euses approximately three times faster
6:45 than argon okay so that would be the
6:47 conclusion from the application of
6:49 Graham's law here if we look at this
6:52 image um on the on the at the starting
6:55 point over here this is time equals
7:00 zero hours we have helium
7:02 filled in the in the orange balloon and
7:04 we have argon um that's that's been
7:07 filled Into the Blue balloon and um over
7:10 some amount of time I think that this is
7:13 supposed to be T equals 12 hours is when
7:16 this next uh photo was taken um you can
7:19 see here that clearly where they were
7:20 approximately the same size before if we
7:22 just sort of draw those lines we can see
7:25 that um the helium balloon has now
7:27 shrunk substantially and what's
7:30 happening there is that um the
7:32 material that balloons are made of are
7:35 uh are slightly porous they will allow
7:37 gases to escape through the rate of
7:41 effusion then depends on the molecular
7:43 weight of the gas and helium is a
7:45 lighter gas so it's eusing faster and
7:48 the balloon deflates
7:51 faster um just so you know we can also
7:54 uh compare um or we can use Graham's law
7:57 to also determine something like uh the
8:00 the the the M mass of an unknown gas so
8:02 in the interest of time I'm just going
8:04 to show you this practice problem so we
8:06 have an unknown gas that diffuses
8:09 1.66 times more rapidly than carbon
8:12 dioxide so we have that value that ratio
8:14 of grams law is
8:17 1.66 um and this you know instead of 3.2
8:18 like we just calculated for helium and
8:20 argon we also are given um carbon
8:22 dioxide so we can go and calculate the
8:24 mar mass of carbon dioxide so really all
8:26 you have to do is set up that same ratio
8:30 of of grams law where instead of of
8:32 solving for this um ratio it's provided as
8:33 as
8:35 1.66 um so then all you have to do is is
8:38 plug in the the M mass of carbon dioxide
8:42 um and uh the the marar mass of the
8:47 unknown and um making sure that uh we
8:49 you you keep track of which that there's
8:51 sort of this flip this inversion here
8:52 that that the rate of effusion of carbon
8:56 dioxide on the um denominator over here
8:57 but it's on the it's it's a square root
9:00 of the mass of carbon dioxide is on the
9:02 um numerator on the other side of the
9:05 equation because it's one over one over
9:08 something and uh if we just do a little
9:10 bit of algebra to isolate the M mass of
9:12 the unknown you calculate a m mass of
9:16 16.0 gr per mole and um there is no
9:18 noble gas for example that has that
9:20 specific molar mass but something like
9:23 methane one carbon four hydrogens would
