If the momentum transferred by gas molecules to the walls of a container is $p_1 = nmAv_x^2 \Delta t$,then derive the equation for the pressure of the gas.

(N/A) The momentum transferred by gas molecules to the walls of the container is $\Delta P = m n V_x^2 A \Delta t$.
The force exerted is $F = \frac{\Delta P}{\Delta t} = \frac{m n V_x^2 A \Delta t}{\Delta t} = m n V_x^2 A$.
Now,the pressure $P$ is defined as $P = \frac{F}{A} = \frac{m n V_x^2 A}{A} = m n V_x^2$.
Since all molecules in the gas do not have the same velocity,we take the average value of $V_x^2$ to get the total pressure:
$P = m n \langle V_x^2 \rangle$.
Here,$m n = \rho$ (density of the gas) and $\langle V_x^2 \rangle$ is the average value of $V_x^2$.
Therefore,$P = \rho \langle V_x^2 \rangle$.
Since gas molecules move randomly,their average behavior in all directions is the same:
$\langle V_x^2 \rangle = \langle V_y^2 \rangle = \langle V_z^2 \rangle$ .... $(1)$
Moreover,the mean square speed is $\langle V^2 \rangle = \langle V_x^2 \rangle + \langle V_y^2 \rangle + \langle V_z^2 \rangle$ .... $(2)$
From $(1)$ and $(2)$,we get $\langle V^2 \rangle = 3 \langle V_x^2 \rangle$,which implies $\langle V_x^2 \rangle = \frac{1}{3} \langle V^2 \rangle$.
Substituting this into the pressure equation:
$P = \rho \left( \frac{1}{3} \langle V^2 \rangle \right) = \frac{1}{3} \rho \langle V^2 \rangle$,where $\langle V^2 \rangle$ is the mean square speed of the gas molecules.