# Deriving pressure from particle velocity: missing 1/3 factor

Let us assume we have a perfect gas, isotropic, homogeneous, inside a cube of side a.

We want to calculate the pressure as a function of the velocities of the particles inside.

We must find the total force this particles impart as a function of their speed. We know $$F*dt = m*dv \rightarrow F = m*dv/dt$$.

$$p = \frac{F}{S}$$

Let me discuss a single particle only in the x axis because the system is isotropic:

From a hit of a wall and the hit of the opposite wall it passes

$$v = ds/dt = a / dt \rightarrow dt=a/v \rightarrow 1/dt = v/a$$

So in an arbitrary time $$dT$$ we have $$dT/dt$$ collisions.

A single particle considering only the x axis transfers this much quantity of motion in a time $$dT \rightarrow Q_x = 2mv_x * dT/dt$$

$$Q_x = 2mv_x$$ for a single collision because v changes sign but not magnitude (elastic collision)

So in the given time a particle inflicts on the wall a force of:

$$F_x = Q/dT = 2mv_x * dT/dt / dT = 2mv_x / dt = 2mv_x * v_x / a = 2m*v_x^2 /a$$

summing all the particles (there are N of them, m is the same for all of them, avg is short for average, $$\rho$$ is density, S=2a the two opposite vertical walls):

$$F_{x_{tot}} = \sum 2m*v_x^2/a = 2m \sum v_x^2 /a= 2mN * avg(v_x^2)/a$$

$$p_x = \frac{F{x_tot}}{S} = 2mN * avg(v_x^2)/2a*a^2 = \rho * avg(v_x^2)$$

Given that pressure is a scalar to get total pressure I sum all the pressures, (x,y,z) and they are all the same because the system is isotropic.

$$p = \rho * avg(v_x^2) + \rho * avg(v_y^2) + \rho * avg(v_z^2) = \rho * avg(v^2)$$

But while the physical values are correct I am missing a factor of 1/3 in the final equation, what is the correct way to sum the components of the pressure? Did I make other errors? Also given that pressures is a scalar how am I allowed to use $$p_x$$? What does it mean?

What you call $$p_x$$ is the force per unit area acting on the $$y-z$$ wall of your container and likewise for $$p_y$$ and $$p_z$$. Pressure is the same in all directions and therefore, this quantity is all you want (you don't need to add the contributions of the other walls). That is, the statement made by Mike i.e. $$p_x = p_y = p_z = p_{avg}$$
$$v^2 = v_x^2 + v_y^2 + v_z^2$$
Since all directions are equivalent, $$v_x^2 = \frac{1}{3}v^2$$. But there is nothing special about using the $$x$$ direction of course - so this is true for any direction.
Everything is good until the last step. Then $$p_x=p_y=p_z=p_{\rm average},$$ so $$p_{\rm average}= \frac 13 (p_x+p_y+p_z).$$