0:03 the topic of this video is the kinetic
0:06 molecular theory the learning objectives
0:08 are on the screen so go ahead and pause
0:09 the video and you can write those down
0:12 in your notes to kick off this
0:14 discussion of kinetic molecular theory
0:16 I'll go ahead and um actually type out
0:19 the the five postulates and you can
0:21 write these down as you take your notes
0:25 the first is that gases are composed of
0:28 molecules that are in
0:31 continuous motion
0:35 traveling in straight lines and changing
0:39 direction only when they collide with
0:45 other molecules or with the walls of a
0:48 container okay so essentially just think
0:51 about them as little tiny spheres that
0:52 um are zipping around and they can zip
0:54 in a straight line until they smash into
0:56 something else and then they are
0:59 deflected um and uh that includes with
1:02 other uh gas particles and also the
1:04 walls of the container the second
1:07 postulate is that the um
1:13 molecules composing the gas are
1:18 negligibly small compared to the
1:21 distances between them so essentially
1:23 that the gas yeah molecules are just
1:24 really really tiny and there's huge
1:28 amounts of space in between them um the
1:35 pressure exerted I a gas in a container
1:37 results from
1:40 collisions between the gas
1:47 molecules and the container walls okay
1:49 so this is an important one so the
1:51 pressure exerted by gas results from the
1:53 the collisions between the gas molecules
1:55 and the container walls so the more
1:57 collisions per second the higher the
2:00 pressure we would predict there okay so
2:05 so uh again the number of of collisions
2:08 then is directly proportional to the
2:10 pressure exerted by the gas in a in a
2:14 container four gas
2:16 molecules exert
2:19 no attractive or
2:23 repulsive forces on each other or the container
2:24 container
2:44 elastic that is they do not involve a
2:45 loss of
2:48 energy okay so this is this is
2:50 definitely an oversimplification of
2:52 what's actually happening but but but
2:54 what we're pretending is that um the
2:56 collisions are perfectly the the energy
2:59 um is not lost upon um an interaction
3:00 between gas
3:02 particles um or with the wall of the
3:04 container the fifth postulate is that
3:09 the average kinetic energy of the
3:15 gas of the gas molecules is proportional
3:18 to the Kelvin should be capitalized
3:22 right the Kelvin temperature of the gas
3:25 okay so um uh this this is this makes
3:27 sense so the higher the the Calin
3:29 temperature the higher the average
3:30 kinetic energy these are the five
3:32 postulates of the kinetic molecular
3:36 theory um we can actually uh go back and
3:38 start to think about some of the gas
3:40 laws that we discussed in a prior video
3:42 within the the lens now of the kinetic molecular
3:44 molecular
3:46 theory we can look at the um three
3:49 example laws here amonton's law or the
3:53 GAC law O's law and avagadro's
3:57 law so for amonton's law um this is
3:59 where the temperature has increased the
4:01 volume is Con but the volume is held
4:04 constant okay so that piston up here is
4:07 not allowed to move up or down so the
4:09 volume is held constant when you heat it
4:11 up the average kinetic molecular uh the
4:13 average kinetic energy is going to
4:14 increase the speed of those particles
4:17 increases there's more collisions uh per
4:20 given time collisions per second for
4:22 example and each Collision contributes
4:24 to the pressure in that container so
4:26 then we would expect that the um that
4:30 that there's an increased pressure um
4:33 uh in boils law if we um uh decrease the
4:37 volume okay uh but the amount of the
4:40 material is the same
4:43 um uh what what we end up seeing is that
4:46 um there is much less space for the um
4:49 particles to travel before they collide
4:51 with another wall so essentially that
4:55 too similar to in amonton's law um the
4:57 heat is causing more collisions per
4:59 second in boils law when we reduce the
5:02 volume there's less available free space
5:05 for the particles to uh ZIP through they
5:07 collide with the wall of the container
5:08 more frequently which causes the
5:12 pressure to increase okay um and then we
5:15 can also look at uh the situation where
5:18 the container pressure is constant okay
5:21 but we have um we we put more gas
5:23 molecules into the Container okay so to
5:24 if the pressure is constant that means
5:27 when we put more gas particles in um we
5:30 would have to allow the volume to go up
5:33 so that the pressure stays constant
5:36 um but why that's the case is because if
5:39 you were to cram twice as many particles
5:41 into the into the container on the left
5:43 you would have twice as many collisions
5:45 per second and that would increase the
5:47 pressure but in order to alleviate that
5:49 you have to increase the space available
5:51 to the gas so that the number of
5:53 collisions per second um is about the
5:57 same in both cases so that's um what
5:58 happens that's another way to think
6:00 about avagadro's law so really all of
6:02 the behavior of Ideal gases is
6:04 explainable within kinetic molecular
6:08 theory so um within now the postulates
6:09 of kinetic molecular theory and the
6:12 other gas laws um we've been talking
6:14 about energy talking about speed so now
6:16 there's um a couple of Expressions that
6:19 I want to write out the first is that
6:21 the the average kinetic energy so
6:25 Capital ke kinetic energy subscript AVG
6:30 for average um is equal to 12 capital n
6:33 u r Ms
6:35 Ms
6:38 squared okay so this is now an equation
6:43 that um relates the kinetic energy uh to
6:45 the average kinetic energy is equal to
6:48 1/2 of the mass of the particles okay so
6:50 that's capital M the molar mass of a
6:55 given gas multiplied by the the um this
6:58 variable which is lowercase U which is
7:01 root means Square velocity so the U is
7:03 the velocity the RMS here this sorry
7:07 this is RMS is root mean Square velocity
7:08 there's a mathematical expression for
7:10 how to calculate root mean squares
7:12 that's not something I'm going to get
7:14 into in this video the other expression
7:16 so this allows us to think about energy
7:18 with respect to speed and and and mass
7:20 well we can also think about um average
7:22 kinetic energy as equal to the um
7:26 expression three Hales multiplied by
7:28 RT okay so that's an ideal gas constant
7:30 times the temperature so if we set both
7:32 of these equations equal to one another
7:35 we get the expression that looks like uh
7:38 this or we can um get an expression that
7:41 can then be rearranged to this the root
7:43 mean Square
7:46 velocity is equal to the square root of
7:52 3 RT over um m is a molar mass here the
7:54 um gas constant that we're going to be
7:56 using is
8:00 8.314 jewles per mole Kelvin
8:03 um and the uh you might be wondering how
8:05 in the world do we use these units
8:08 there's a really um interesting uh unit
8:10 equivalency that a jewel is actually
8:12 equal to a
8:16 kilogram met squared per second squared
8:17 um and so we'll I'll show you a practice
8:20 problem with how to handle that in one
8:22 second um what does this look like in
8:24 terms of root mean Square velocity
8:26 versus most probable velocity if we look
8:28 at a plot of the number of molecules at
8:31 a given speed as a function of the speed
8:33 itself in meters per second so number of
8:36 molecules is a function of speed we see
8:39 that um for a sample of gas there's a
8:41 distribution so in any given sample
8:42 there's going to be distribution some
8:44 gas particles will have a low speed some
8:45 gas particles will have a really high
8:48 speed but the most uh the vast majority
8:51 will sort of be you know within a
8:53 certain um galaxian
8:56 distribution of of sort of the the most
8:59 probable or the average um so so the
9:01 most probable is of course the velocity
9:04 the most probable velocity U subp is the
9:06 the highest position that's the velocity
9:08 with the most number of gas particles at
9:11 that speed the root mean Square as you
9:13 can see by the way that it's calculated
9:16 is actually um in uh slightly higher
9:19 than that it doesn't represent the most
9:22 probable um uh but it reflects the the
9:30 distributions okay another thing that I
9:32 want to point out here is what happens
9:33 if you take a gas and heat it up then
9:36 considering kinetic molecular theory um
9:38 we see here at 100 Kelvin we have this
9:41 sort of narrow distribution but as you
9:43 increase the temperature more and more
9:46 to now 200 Kelvin you see that the um
9:48 the root mean Square velocity is is
9:51 increasing we also have that this tail
9:52 that's now dipping up to higher and
9:54 higher speeds um if we increase to 500
9:56 Kelvin you see like the the the height
9:59 of the of the plot or
10:01 is is is now coming down but it's it's
10:03 spreading out quite a bit we have a
10:05 greater distribution of speeds so on and
10:08 so forth up until 1,00
10:11 Kelvin and you can also look not as a
10:12 function of different temperatures but
10:14 as a function of different molecular
10:15 weights the lowest molecular weight
10:18 noble gas helium um is going to uh have
10:20 some of the particles that are traveling
10:22 some of the fastest compared to um
10:25 heavier molecular weight uh noble gases
10:28 who will have relatively um slower uh
10:31 root means Square speed and we'll also
10:33 have much fewer particles moving at high
10:36 rates of speed so let's go ahead and do
10:39 one quick practice problem um we're
10:40 going to
10:43 be uh we're going to calculate the root
10:48 mean Square velocity of nitrogen gas at
10:52 30° C so I'm going to use this
10:55 expression U sub RMS is equal to sare <
11:00 TK 3rt over capital m m Mass
11:03 um again the R value here is
11:06 8.314 Jew per mole
11:10 Kelvin and um I'll use the jewel unit
11:12 equivalency in a second um but you
11:14 notice before I even get to that the
11:16 temperature although it's provided as
11:18 30° C we do need to convert that to
11:20 Kelvin um in pretty much every case to
11:23 use it for the ideal gas uh constants
11:27 and this gives us a value of 303 Kelvin
11:31 the molar mass um is 28 G per mole for
11:33 dinitrogen um but we need it in
11:35 kilograms and and you'll see why when I
11:36 get there in a second but you can do a
11:41 quick conversion 28 uh grams um and and
11:43 there are 1,00 GR in 1 kilogram so we
11:45 dividing by a th000 so this is the same
11:47 as writing
11:52 0.028 um kilograms per mole so let's go
11:55 ahead and plug this in it's uh urms is
12:01 equal to 3 multiplied 8.314 4 Jew per
12:06 mole Kelvin multiplied by 303 Kelvin uh
12:08 Clos the square root and then the
12:10 denominator is 0.028
12:12 0.028
12:14 kilogram per
12:17 mole some units that do cancel out right
12:20 away um we have Kelvin that cancel out
12:23 with Kelvin over here we have one over
12:25 Mo which is over one over mole so it's
12:28 mole over mole so those cancel out and
12:30 we're left with um Jew per kilogram um
12:33 so I'm going to go ahead and and uh
12:35 combine these these values here uh
12:36 simplify this expression a bit so it
12:41 becomes 2.70 * 10 5 um jewles per
12:44 kilogram and that's all square root now
12:47 now remember that um a jewel is equal to
12:51 a kilogram met squar per second squared
12:54 so if we have a jewel over a kilogram
12:56 that's like multiplying this by 1 over
12:59 kilog so kilog cancel out and we're left
13:02 with with me s/s squar but keep in mind
13:05 that this is um so what I can do is
13:07 actually rewrite these units Jews over
13:10 kilograms is the equivalent of met squar
13:12 per second squared but it's square root
13:16 so the units will actually be um uh uh
13:18 become meters per second which is a unit of
13:19 of
13:21 velocity uh and what we get here is a
13:25 value of 519 m/s so that is a velocity
