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deriving kinetic theory of gases

This is a cube with each side having a length of $l$ containing a certain amount of molecules in it.

Let's say a molecule of mass m is traveling in the $y$-direction with the velocity $v$ and hits the wall 1. Obviously, just before the molecule hits the wall 1, its momentum is $mv$. After hitting the wall 1, the direction of the molecule will be reversed, and its velocity and momentum will be $-v$ and $-mv$, respectively. So the change in momentum is $2mv$. So far, so good.

Now, let's say the time the molecule is in contact with the wall 1 is $t$. In other words, the wall 1 applies a certain force on the molecule for the time $t$, so to get the force the wall 1 exerts on the molecule, we simply divide the change in momentum $2mv$ by $t$.

However, this is not what is done in the proofs I have seen.

As the distance between wall 1 and wall 2 is $l$, and the velocity of the molecule is $v$, after hitting the wall 1, the time it takes for the molecule to travel back to the wall 2 and come back and hit the wall 1 again is, $t_1 = \frac{2l}{v}$.

And they calculate the force the wall 1 exerts on the molecule by dividing the change in momentum $2mv$ by $t_1$.

I do not understand it. Someone please help.

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It's from the impulse formula $force \times time = \delta mv$

Where $\delta mv$ stands for change in momentum. They consider it over 1 second, so the time used is 1 second, not the actual time of contact of the molecule with the wall. There are many collisions of the molecule with wall 1 and 2 in one second. The time between collisions is $\frac{l}v$, and the number of collisions per second is $\frac{v}l$, and the number of collisions with wall 1 per second is $\frac{v}{2l}$. So the formula above becomes

$force \times time = 2mv\times\frac{v}{2l}$ with $time =1$

so the force on wall 1 is $force = \frac{mv^2}{l}$

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