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.