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?

Mike has answered your question perfectly already but it may help to add one detail since you asked.

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. $$p_x = p_y = p_z = p_{avg}$$

When you express it in terms of the average speed of the particles (and not just their x-component), use Pythagoras theorem in 3D to show that the speed of the particles squared is given by

$$v^2 = v_x^2 + v_y^2 + v_z^2$$

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.

• "What you call px is the force per unit area acting on the y−z wall of your container and likewise for py and pz. 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)." My problem is that I calculated px only considering vx, how Is it possible that It is equal to the complete result if I ignored the other two components of velocity? – Caridorc Jan 23 at 10:05
• Pressure is defined in terms of the force acting perpendicular to a plane. You are free to pick any plane you like and the y-z plane therefore is a perfectly good choice. For this plane, only vx matters since vy and vz are parallel to the wall and cannot exert the force on the wall. – user1936752 Jan 23 at 11:03

Everything is good until the last step. Then $$p_x=p_y=p_z=p_{\rm average},$$ so $$p_{\rm average}= \frac 13 (p_x+p_y+p_z).$$