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