Q.On the basis of kinetic theory of gases obtain an expression for pressure exerted by gas molecules enclosed in a container on its walls.

Kinetic-theory derivation of gas pressure from molecular collisions with the walls.

Consider $N$ identical gas molecules, each of mass $m$, in a cubical container of side $L$ (volume $V=L^3$). Take one molecule with velocity components $(v_x,v_y,v_z)$.

Collision with a wall (⊥ to x-axis): on elastic collision, the x-component of velocity reverses ($v_x \to -v_x$), so the change in momentum of the molecule is $-2mv_x$, and the momentum imparted to the wall is $2mv_x$.

Rate of collisions: the molecule travels a distance $2L$ (to the opposite wall and back) between successive collisions with this wall, taking time $2L/v_x$. So the number of collisions per second with this wall is $v_x/2L$, and the average force exerted by this one molecule on the wall is

$f = 2mv_x \times \frac{v_x}{2L} = \frac{mv_x^2}{L}$

Summing over all N molecules: total force on the wall $F = \dfrac{m}{L}\sum v_x^2 = \dfrac{mN\overline{v_x^2}}{L}$, using the mean-square x-velocity. Pressure is force per unit area ($A=L^2$): …