Obtain an expression for the momentum of a gas transferred to the walls of a container in time $\Delta t$.

(N/A) Consider an ideal gas in a cubical container of side length $l$ with perfectly elastic walls.
The area of each face is $A = l^2$.
Let a gas molecule have velocity $\vec{v} = (v_x, v_y, v_z)$.
When the molecule collides with a wall perpendicular to the $X$-axis,the $x$-component of velocity changes from $v_x$ to $-v_x$,while $v_y$ and $v_z$ remain unchanged.
The change in momentum of the molecule is $\Delta p_{molecule} = m(-v_x) - m(v_x) = -2mv_x$.
By the law of conservation of momentum,the momentum transferred to the wall is $\Delta p_{wall} = 2mv_x$.
In time $\Delta t$,only molecules within a distance $v_x \Delta t$ from the wall can collide with it.
The number of molecules hitting the wall in time $\Delta t$ is given by $\frac{1}{2} n A v_x \Delta t$,where $n$ is the number density of molecules.
The total momentum transferred to the wall is $P = (2mv_x) \times (\frac{1}{2} n A v_x \Delta t) = n m A v_x^2 \Delta t$.