TITLE: Why is number of molecules going in any direction same according to kinetic theory
QUESTION [0 upvotes]: In my book it is stated that according to kinetic theory if that if in a particular $1 cm^3$ if 400 molecules have velocities in $x$ direction then there are also other 400 molecules which go in $y$ direction.
Further it is stated that all directions are equivalent and
$$\Sigma v_x=\Sigma v_y=\Sigma v_z$$
Please help me in understanding. This here $\Sigma v_x$ means summation x components of velocities of all molecules.
I am not getting it. Thank you in advance.

Edit:I am not able to get the underlined part.

REPLY [3 votes]: Simply, no direction is distinguished from any other. More specifically, it can be shown that the probability $p(v)$ of finding a molecule whose random velocity (the actual velocity relative to the average velocity) is close to $v$ is proportional to $$4\pi v^2\left(\frac{m}{2\pi kT}\right)^{3/2}\exp{\frac{-mv^2}{2kT}}$$
where $k$ is Boltzmanns constant, $m$ is the molecular mass, and $T$ is the absolute temperature. This is independent of direction. See Wikipedia on Maxwell distribution.
The derivation is based on the fact that if the speeds are equally likely to be in any direction before a collision, then they are equally likely to be in any direction afterwards. To put this argument on a sound footing, it has to be the case that collisions between molecules must be sufficiently frequent that they can be treated statistically. The random processes can then be described by bulk properties such as pressure, temperature, etc., and we speak of the continuum regime.
If the molecules are very far apart this is no longer true, and individual molecules must be followed. There is an intermediate situation where the statistical treatment is OK, but the probability distribution is no longer isotropic, but may be, as you suggest, biassed in some direction, such as the flow direction or the orientation of a nearby surface or shockwave.  
The is a dimensionless parameter called the Knudsen number, defined as the ratio of the mean free path $\sigma$ (average distance between collisions) and some typical size $L$ (length, radius of curvature)   $$Kn=\frac{\sigma}{L}$$
Roughly if $Kn<0.1$ then this continuum description is satisfactory, and if $Kn>10$ it is definitely unsatisfactory. Between these limits the situation is very complicated. This happens in the outer regions of the atmosphere.
THIS EDIT RESPONDS TO YOUR COMMENT
First let us be sure that you are reading $\Sigma v_x^2$ as $\Sigma (v_x)^2$ and not as $\left(\Sigma(v_x)\right)^2$, which would be zero. Several responders are puzzled how you can not "get" something that they find obvious. It may be that the notation is not clear to you. I will try to do the proof up to that point as nonmathematically as possible.
We have a cubical box with sides of length L containing a single molecule, anywhere in the box, with initial velocities $(v_x,v_y,v_z)$ When it collides with a face having $x$ constant it supplies an impulse to it that is twice its momentum in the $x$-direction ($2mv_x)$, where $m$ is the mass of the molecule). It does this twice in a time $2L/v_x$, so in unit time it applies to each wall $x=$ constant a total impulse of $(2mv_x)/(2L/v_x)=(mv_x^2)/L$. The area of each face is $L^2$ so the total force per unit area (pressure)is (m/L)$\Sigma v_x^2/(L^2) =(m/L^3)\Sigma v_x^2$, and you are probably still with me. 
Rewrite this as
$$(Nm/L^3)(1/N\Sigma v_x^2)=\rho\overline{v_x^2}$$
where the first term is the gas density and the second is the average over all molecules in the box of the square of the $x-$ component of the initial velocity. Since the labels $x,y,z$ are arbitrary, we expect that the average pressure on faces with $y=$ constant is $\rho\overline{v_y^2}$ with $\rho\overline{v_z^2}$ on $z=$ constant. But we also expect that the pressure will be same on all faces, so that $$p=\rho\overline{v_x^2}=\rho\overline{v_y^2}=\rho\overline{v_z^2}$$
and we have three equal results for the pressure, so we can also take the average of them $$p=\frac{1}{3}\rho\overline{v^2}$$
where $v^2=v_x^2+v_y^2+v_z^2$. This says that the pressure is 2/3 of the kinetic energy per unit volume.
The answer to your question why the velocities in each direction can be taken to behave similarly has two answers. "Why shouldnt they?" and "If they didnt the pressure would not be the same in all directions"

REPLY [1 votes]: What the book states in a colloquial way only holds when considering averages. Thus, if on average 400 molecules move in $x$ direction, then there are also on average 400 molecules moving in $y$ direction.
If the statement were not true, then one direction would be preferred over another. But if the problem is isotropic, that means symmetric, movement in every possible direction should be identical. Only the presence of e.g. an external field can break the symmetry and establish a preferred direction.
Furthermore, as motion in $x$ and $-x$ direction is identical
$$\sum v_x = \sum v_y = \sum v_z = 0$$
should hold in the limit of a large number of molecules.
{"set_name": "stack_exchange", "score": 0, "question_id": 345305}