Why pressure stays the same when doubling both volume and temperature?

[lost captain]

From post #11:

You've been creating scenarios designed to imply chaotically varying force by looking at a tiny number of particles/collissions. This is far, far from the reality. The reality is that the number of collisions is huge, so the variation in force is tiny. If you're still doubting this, it might be instructive to calculate how many particles and how often there are collissions in real life.

Yeah i guess my initial point of view and how i imagined the particles and their collisions with the wall was wrong from the start

 

[A.T.]

Yeah i guess my initial point of view and how i imagined the particles and their collisions with the wall was wrong from the start

Key is that you don't have to know the details of how the particles collide with the wall, in order to answer the original question. All you need to know is how the average particle momentum, average rate of collisions and wall area change.

But the hand wavy explanation you mention in your OP is indeed not quite correct:

And the explanation for that is that the number of collisions per second stays the same. Now the atoms have double the kinetic energy but they also have the double space to move into.

As you see in post #14 the rate of collisions is not the same. And it is not that obvious that the various powers of 2 involved in the pressure ratio cancel. But they do.

 

[lost captain]

As you see in post #14 the rate of collisions is not the same.

Sorry, but where in post #14 ? Isn't the rate of collisions, the pressure? And pressure stays the same. Probably i didn't understand what you are referring to.
edit: wait you mean the time between collisions right?

 

[jbriggs444]

Sorry, but where in post #14 ? Isn't the rate of collisions, the pressure? And pressure stays the same. Probably i didn't understand what you are referring to.

The rate of collisions is not the pressure. It is just a rate of collisions. It is not even a force.

The [average] rate of collisions multiplied by the [average] momentum transferred per collision is not the pressure. It is a rate of momentum transfer. A force. [A force is a momentum transfer without an associated mass transfer].

The [average] rate of collisions multiplied by the [average] momentum transferred per collision and divided by the surface area over which the collisions are distributed is a force divided by an area. That is to say, a pressure.

edit: wait you mean the time between collisions right?

It does not really matter whether we describe the rate of collisions as how many collisions per unit time or how many units of time between collisions. Either way will work. One number is the reciprocal of the other.

One may need to be careful about how one computes an "average" rate in order to get the averages to match. But that is subject matter for a different thread.

 

[lost captain]

The [average] rate of collisions

Just a clarification: Average rate of collisions and average frequency u/L in post#3 is the same thing?

 

[A.T.]

Just a clarification: Average rate of collisions and average frequency u/L in post#3 is the same thing?

Yes, it's also the reciprocal of the $t$ in post #14, which is different in the primed case, and so is its reciprocal.