Kinetic theory of gases 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
 A: 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.
