 # Calculating Pressure Of An Ideal Gas

Imagine an ideal gas contained in a container which is cubical in shape. Let one corner of the cube be the origin O, and let the x, y, z-axes along the edges. Let A1and A2 be the parallel faces perpendicular to the x-axis. Consider a molecule moving with velocity v in the container. The components of velocity along the axes are vx, vy, and vz. Now, when the molecule is colliding with face A1, the x component of velocity is reversed while the y and z component of velocity remains unchanged (as per our assumption that the collisions are elastic). The change in momentum of the molecule is:
$ΔP$ = $(-m v_x) – (m v_x)$ = $-2m v_x$ ……….. (1)

Since the momentum remains conserved, the change in momentum of the wall is 2mvx

After the collision, the molecule travels towards the face $A_2$ with x component of the velocity equal to $- v_x$.

Distance traveled from $A_1$ to $A_2$ = $L$
Therefore time = $\frac{L}{v_x}$

After collision with $A_2$, the molecule again travels to $A_1$. Thus the time between two collisions is $\frac{2L}{v_x}$

Therefore the number of collisions of the molecule per unit time:

$n$ = $\frac{v_x}{2L}$ ………………… (2)

Using equations 1 and 2,

The momentum imparted per unit time to the wall by the molecule:

$ΔF$ = $nΔp$

= $\frac{m}{L} v_x^2$

Therefore, total force on the wall A1 due to all the molecules is

$F$ = $Ʃ \frac{m}{L} vx^2$

= $\frac{m}{L} Ʃ{v_x}^2$

Now, $Ʃ{v_x}^2$ = $Ʃ{v_y}^2$ = $Ʃ{v_z}^2$ (symmetry)

= $\frac{1}{3} Ʃv^2$

Therefore, $F$ = $\frac{1}{3} \frac{m}{L} Ʃv^2$

Pressure is force per unit area so that

$P$ = $\frac{F}{L^2}$

= $\frac{1}{3} \frac{M}{L^{3}} \frac{\sum v^{2}}{N}$

= $\frac{1}{3} \rho \frac{\sum v^{2}}{N}$

Where

$M$= total mass of gas

$\rho$ = density of the gas

Now, $\frac{\sum v^{2}}{N}$ is written as $v^2$ and is called the mean square speed.
$P$= $\frac{1}{3}\rho v^{2}$