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In the previous lecture, we developed

the postulates of the Kinetic Molecular Theory.

These postulates constituted description at the

particulate level, that is atoms and molecules.

About what we think the individual particules are doing in an ideal gas.

These postulates were based upon experimental observations,

most specifically, the ideal gas law and deviations

from the ideal gas law as well as things like Dalton's law of partial pressures.

They provide

a description at the particulate level that we would now like to be able

to use to make predictions about things that are true with the macroscopic level.

In other words, the most powerful thing we could say about the Kinetic Molecular

Theory is can it accurately predict the

experimental observations on which it is based?

And if so, then we would think we've got a very good model.

We're going to develop that this lecture. And we're going to base it upon the

experimen, the, Kinetic Molecular Theory Postulates

that we developed in the previous lecture.

Here is a review of those postulates.

Remember that we imagine that there are

gas particles flying around in constant random motion.

That the force which constitutes the pressure of that gas, is

a consequence of those particles running into the walls of the container.

Most significantly, we imagine, that these particles

are very far apart from one another.

And as

a consequence of their being very far apart, they don't exert

any attractive or repulsive forces on each other as they move about.

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Essentially, just based upon these postulates and a few

simple assumptions, we will be able to show that

these postulates lead to an accurate prediction of the

ideal gas law and that'll be very useful to us.

At the foundation then, we're going to start

with physics and the definition of pressure and the definition of force.

The definition of pressure is the force per unit area applied.

So if we could calculate the force exerted by particles colliding with the walls.

May be easy to divide by the area of the walls.

In turn the force itself is the mass

times acceleration of the particles since we know

the mass of the particles, then really what

this boils down to is if we could determine.

The acceleration of the particles we would be able to multiply that by the mass

to get the force, we divide that by the surface area to get the pressure.

Our primary assumption is going to be

that the particles are going to travel independently

back and forth between the walls of the container, colliding with the container.

Let's consider a really simple case then, where we're

talking about motion between two walls parallel to each other.

And I'm going to say that these are parallel to each other along the X axis.

And the distance between these two is going to be a distance L.

We're also going to imagine that the area on the end of the wall is A.

So that if we calculate the force of the particles impacting either of these walls,

we could divide by A, and that would tell us the pressure on those walls.

So let's imagine

then, that we have a particle here, flying along.

And it has vector in the x direction vx.

It has component in the x direction vx.

And we're also going to assume that it has a mass m

and that all of the particles have the same mass m.

As this particle travels, it will hit this wall, exerting a force on

the wall which can be detected from the fact that the wall in turn

exerts a force on the particle and those

forces are equal and opposite to each other.

We can tell that the particle has a

force exerted on it because it's going to accelerate.

In this particular case its acceleration will be to

reverse its direction and move towards the opposite wall.

So our goal here is going to be the calculate the force exerted by

the particle as it encounters the wall in the x ax, in the x direction.

And that'll be the mass of the particle

multiplied by its acceleration in the x direction.

In turn, the acceleration in the x direction,

is simply the change in the velocity in

the x direction, divided by the amount of

time over which that change in velocity occurs.

So this will be a simple way then to calculate the force.

Our task then is to calculate the change in the velocity and the

period of time over which that change takes place.

Changing, the, calculating the change in the

velocity in the x direction is relatively straightforward.

All we have to do is assume that when the particle encounters the wall, it does so

in what we call an elastic collision, meaning

that it will lose no energy to the wall.

That means that its speed before it hits the wall

is the same as the speed after it hits the wall.

And since it can only accelerate

in the x direction, then vx simply changes to minus vx.

Therefore, the change in the velocity in the x direction is equal to the

speed it had in the x direction before, minus the negative of that same

speed as it travels in the opposite direction, which is two times vx.

How about delta t?

With delta t, we can take advantage of the fact

that the particle is travelling with speed vx.

As it approaches this wall, speed vx as

it departs the wall and heads towards the opposite

end, it will then encounter that wall and

turn around and fly back to its original location.

That means that it accelerates by an amount two vx, everytime

it completes one circuit, between the left wall and the right wall.

How long does it take to get from the left wall to the right wall and back again?

Well, it must travel a distance two L and it's travelling with speed vx,

so the amount of time that it takes to move between the walls is two L over vx.

That means we can calculate the acceleration in the x direction.

It is simply two vx divided by two L, divided by vx.

The twos cancel out. And the acceleration is

vx squared, divided by L. That was relatively straightforward.

Now what we want to do is calculate the force exerted

in the x direction, by this one particle hitting the wall.

And the answer will be the mass of that particle, times vx

squared, divided by the length of the box in the x direction.

Now an interesting challenge here is really two fold.

One, that's only a single particle.

There are of course N particles inside this box.

Each one of which is on occasion encountering the

the wall on the left in contributing to the force.

But secondly, all of those particles don't have the same speed vx.

They don't have that same component.

Even if they were all travelling at the same speed, they may be

travelling in various directions with various

amounts of component in the vx direction.

So when we shift this over to calculate the total force in the x direction,

by multiplying by the number the of particles, we will replace the speed vx

with the average of the square of the x, rather than the x itself.

Because each particle will be travelling with a different component x.

But on average, they'll be hitting the wall with the

component which is v x squared on average.

Now we need to figure out how large is this vx squared?

How big do we think the component in the x direction will be?

To calculate that, we'll remember that the overall magnitude of a vector v is given

by, when we square it, is given by

the squares of the three components added together.

That's just the magnitude

of a vector v.

So if we take the average of v squared, that's the average of vx

squared plus the average of v y squared plus the average of vz squared.

Now we'll use the postulate that tells us that the particles

are travelling at random directions. As such the average

in the three different directions is going to

be exactly the same direction so if

we sum these together we get three times the average of v x squared.

That now allows us to replace vx squared

in the previous relationship with the average speed squared.

Let's do that as we move forward. Get a new piece of paper here.

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If this was one half mv squared, we would be

right where we would want to be with the kinetic energy.

So let's actually put that one half in there which requires

us then to put a two in front of the N.

So we have two N over three v multiplied by one half times m times the

average of v squared.

This is actually essentially our final result.

Notice what it does is it relates the pressure of

the gas to the kinetic energy of the gas particles.

That's a very interesting result, because it will now allow us to relate the

properties to the individual mole particles to the pressure of the gas.

In fact we'll move on to the next

slide here where we capture this result.

This is actually the same result that I have written on the page here except,

I multiplied v on both sides to get our familiar p times v being a constant.

Notice that according to the calculation we've done, the pressure times the volume

is constant provided only that the kinetic energy is constant for all gas molecules.

Furthermore, the pressure times the volume

is proportional to N just as it is in the ideal gas law.

Think a bit more though, in the idea of gas law we have the temperature.

The question is what happened to the temperature in this derivation.

Why have we not actually shown the

real, full, ideal gas law involving the temperature.

The answer is pretty straightforward. Temperature is in fact a bolt quantity.

Its an average quantity of the property of many materials.

And individual particle in general doesn't

have a temperature.

Most notably of course, the temperature doesn't show

up in any of Newton's equations of motion.

Consequently, you'll recall that in the ide, in deriving the ideal

gas law, we use the definition of the temperature, which was

based on a relatively arbitrary scale, in which it turned out

that the pressure was proportional to the temperature for constant volume.

Or the volume was proportional to the temperature for constant pressure.

Therefore, we can actually go and figure out

what the temperature means, in a molecular sense.

Let's do that.

This term here, is the average kinetic energy of an individual molecule.

If I multiply by, that by the number of molecules,

together these terms are the total kinetic energy of the molecules.

Let's multiply both sides of the equation then by three

halves, and what we wind up with is, let's see,

p times v multiplied by three halves is equal to the total kinetic energy

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We can now say, since the total

kinetic energy of the molecules are proportional to

T, then T is a measure of the kinetic energy of a mole of gas molecules.

Notice, that for one mole of gas, regardless of what that type of gas is,

the total energy is fixed when the temperature is fixed.

And in fact, the temperature and the

kinetic energy are proportional to one another.

That gives us an interpretation of the temperature.

The temperature measures, depending upon your perspective,

either the total kinetic energy of everything

in the gas, or the average kinetic energy of the individual particles in the gas.

Notice that the temperature is proportional to kinetic energy.

It's not proportional

to the speed of the molecules.

In particular, we know that the speed of the molecules is depends upon

the masses of the molecules because the kinetic energy is one half mv squared.

We can combine our expression here, the total kinetic energy with the fact that

the total kinetic energy is equal to, N times one half M average of v squared.

And we could solve

this equation for v squared to try to

determine what the average speed of the molecules are.

And when we do that, we discover of

course that the average speed depends upon the mass.

And it particular we'll wind up with the equation written here

solving the above equation for v squared taking the square root

that v squared is equal to three times the gas constant

R, times the temperature divided by the molar mass of the gas.

Notice that tells us that the speed

increases like the square root of the

temperature, not as the temperature.

And it decreases as the square root

of the molar mass of the gas, not

as a function of the mass. What that tells us is,

the speed of the gas molecules does depend on the molecular masses of the gases.

So, even though, for example, at 298K,

a mole of methane and a mole of hydrogen and

a mole of nitrogen all have the same kinetic energy.

The individual gas molecules are moving at different speeds

because they have different masses of the individual particles.

We're actually going to use these equations in the next

lecture to do some predictions of some experimental observations.

What we would like to do in the remainder

of this particular lecture is use this derivation to

help us interpret the ideal gas law.

In particular, there's some interesting questions here.

Why did it happen that the pressure depends not upon v, but upon v squared.

Why is it that the pressure, according to the equation we've written here, that the

pressure is proportional to the square of

the speed rather than just the speed itself.

The answer is found from the derivation that we went through.

Remember why

the speed showed up twice.

I'll pull the sheet of paper back up, to show

that the acceleration depended upon the square of the speed.

That's because the change in the speed depended on the speed.

The faster the particles are going, the harder they hit the wall.

And the frequency with which they hit the wall increases as we increase the speed.

That is the length of time required for

the particles to encounter the wall, is smaller

when the particles are moving more rapidly.

As a consequence, the force exerted in total by the particles

on the wall, depends not upon v but on v squared.

And therefore the pressure depends upon v squared.

It's worth keeping that in mind, that the pressure is a result, both

of the frequency of the collisions, as well as how hard those collisions are.

In turn, that's why the temperature depends upon mv squared, and not

upon v itself.

Because the temperature's proportional to pressure.

And the pressure is dependent is proportional to the force.

We see that the temperature then depends upon both the frequency with which

the particles hit the wall as well as how hard they hit the wall.

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Or, if the dimensions of the box are larger such that the

particles take longer to transit from one end of the box to the

other, such that they hit the wall less frequently than with a larger

volume, the frequency of collisions of the particles with the wall is smaller.

Hence the consequence the pressure they exert will be smaller.

That explains then why the pressure is proportional to n over v

and therefore inversely proportional to the

volume as we observed with Boyle's law.

And then, lastly, why did it matter in our

postulates that the gas particles didn't interact with each other?

The answer is, that as we did our, through our derivation, we calculated the

trajectories that the particles were travelling with

as though they never encountered another particle.

They were simply flee, freely travelling with constant speed,

back and forth, between the ends of the box.

As a consequence, we assumed that they would not have their

pads changed by forces exerted by other particles in the box.

Well, what would happen if they did interact with each other?

Well, clearly that would disrupt the calculation that we've done.

One of the tests that we then have ahead of us,

is to try to understand if the particles do interact

with each other, what impact does that have on the pressure?

We're going to pick that up in the next lecture.