# Kinetic theory of gases [duplicate]

While deriving expression for pressure of gas

We assume that gas molecule has travelled the between two opposite walls , and time taken to travel this distance is Δt= 2l/u .

Then we say that , Force is change in momentum per unit time .

The change change is momentum is 2mu and time in which momentum is changed is 2l/u

Hence force = 2mu/(2l/u)

And we further calculate pressure.

But my question is : the time ∆t= 2l/u is not the time in which momentum is changed , it is the time in which molecule had travelled between one wall to another wall. Then why we take ∆t as the time in which momentum is changed?

i.e , the time in which momentum is changed should be equal to the time for which molecule remain in contact with wall , but in this derivation we take ∆t as the time required to move between two wall.

Why is this so?

I am beginner in statistical mechanics, pls don't use high level mathematics

• @RogerVadim , the question suggested by you is related to with equipartition theorem. I am struggling with some another doubt
– user324098
Dec 28, 2022 at 15:31
• I think the question I linked and the answer given do answer your question - but they take time to read and digest. Dec 28, 2022 at 16:37

We are not trying to find the force exerted by a molecule as it collides with the wall, but the mean force exerted over a 'long' time by the molecule. That long time will include the very short intervals while the collisions with the wall that change the momentum by $$2mu$$ are taking place, but will be dominated by the long intervals while the molecule is travelling between contacts with the walls. Hence $$2l/u$$ is the appropriate time per collision. You might find the argument more convincing if we consider a time during which the molecule makes $$n$$ collisions with one of the walls... $$\text{Mean Force} = \frac {\text{total change in momentum}}{\text{total time}}=\frac{n\times 2mu}{n\times 2l/u}=\frac{mu^2}l$$ Note that this still applies if the molecule collides elastically with the 'side' walls on its journey to and fro, as the velocity component of magnitude $$u$$ will be conserved.
There are more general ways to derive $$pV=\frac 13 Nm\overline {c^2}$$ than the one you are struggling with, but the molecule banging back and forth between walls in a cuboidal container is an undemanding first method.