Derive the kinetic gas equation I need a proof of high school level
I have confusion in taking probabilities in all directions not yet concluded this, and I need a proof using momentum i.e I need a high school level proof
 A: This treatment adapted from one I wrote for class and may eventually compile with others in the form of a book.

Introduction
The ``kinetic theory of ideal gasses'' provides an accessible
connection between the macroscopic variables of a diffuse,
single-component, atomic gas (pressure, volume, and temperature)
and a simple model of the average behavior of the atoms that make
up that gas. The ideal gas law can be recovered from the
kinematics of the gas molecules with few assumption beyond
Boltzmann's identification of the temperature with the average
kinetic energy of the atoms.
By way of an aside, I would like to note that this is one of the
most beautiful results in basic physics and almost alone among
other results of similar elegance in being decidedly accessible
to all comers.
Here we will see the marriage of the microscopic world---which we
shall assume follows the same rules for discrete particles we
state for classroom demos---with the macroscopic behavior of
smooth fluids through a simple statistical assumption.
Ideal Gas
For our purposes, an ideal gas is one made of identical,
spherical atoms so small that their volume and rotational
behavior can be neglected, and whose interactions are always
elastic and at once negligible over short time scales and
frequent enough to share the internal energy of the gas without
preference between all the atoms. Physical gasses that
approximate this definition are observed to obey a macroscopic
relation between their pressure $P$, volume $V$, and temperature
$T$ called the "ideal gas law" and written
\begin{align}
  PV  
  &= N k T \tag{1}\\
  &= n R T \,,
\end{align}
where $k \approx 1.381 \times 10^{-23} \,\mathrm{J/K}$ is
"Boltzmann's constant", $N$ is the number of atoms present,
$n = N/N_A$ is the number of moles present,
$R = N_A k \approx 8.314\,\mathrm{J/(mol \cdot K)}$ is the
"universal gas constant", and
$N_A \equiv 6.02214 \times 10^{23}\,\mathrm{objects/mol}$ is
Avogadro's Constant.
We'll write the mass of an individual atom of the gas $m$, which
makes the molar mass $m N_A$.
Kinetic Theory
In this section we will begin by considering the interaction of a
single atom in a cubical box of side $L$ and volume $V = L^3$
having mass much larger than the gas contained within it and from
the wall of which the atoms bounces elastically.1 We will then treat the
behavior of many such atoms to find a relationship between the
pressure of the gas, the volume of the box and the average
kinetic energies of the atoms.
From there we'll show that we can recover the ideal gas
law by giving a definition of temperature in terms of average atomic
kinetic energy.
One Atom in a Box
In a gas we expect to find atoms moving in all possible
directions with equal probability, and further to have a
distribution of speeds with many going at some typical speed,
some going faster or slower than that, and just a few going much
slower or much faster.
For the moment we'll concern ourselves, with one particular atom
with speed $v$. Moreover, we'll only ask about it's motion in the
$x$-direction, so we really only care about the $x$-component of
it's velocity $v_x$. The atom will typically have
  non-zero components of velocity in both $y$ and $z$
  directions as well, but the geometry of reflection insures that
  while all three component change sign from time to time (when
  they hit the wall) they keep their magnitude.2
At one particular moment we find the atom approaching the
right-hand wall, where it reflects elastically, picking up a change in
momentum of 
\begin{align*}
  \Delta p_x 
  &= J_x \\
  &= (-mv_x) - (mv_x) \\
  &= -2 m v_x \,.
\end{align*}
Of course the box feels the opposite force and opposite impulse
due to Newton's Third Law.  Before it touches the right hand wall
again, this atom must first coast to the left-hand wall, reflect
and coast back. The total distance is $2L$ which means this takes
time
\begin{align*}
  \Delta t 
  &= \frac{2L}{v_x} \,.
\end{align*}
Impulse is average force times average time, so we can compute
the ``average force'' on the right-hand wall due to this one
particle which is 
\begin{align*}
    \bar{F}_{OP} 
    &= \frac{J_x}{\Delta t} \\
    &= \frac{2 m v_x}{1} \frac{v_x}{2L} \\
    &= \frac{m\,v_x^2}{L} \,. 
\end{align*}
The "OP" here means "one particle", and we should keep in
mind that it represents an average, because most of the time the
particle is far from the wall and it only collides briefly on
each back and forth trip.
If we know the average force we can compute the average pressure
on the right-hand wall, which comes to
\begin{align*}
    \bar{P}_{OP} 
    &= \frac{\bar{F}_{OP}}{A} \\
    &= \frac{\bar{F}_{OP}}{L^2} \\
    &= \frac{m\, v_x^2}{L^3} \\
    &= \frac{m\,v_x^2}{V} \,.
\end{align*}
So far we've only worried about the pressure on one side of the
box, but we can also ask about the pressure on the others. By
symmetry, the left hand side will be just like the right hand
side. The top and bottom will be the same as each other and
similar to the left and right, but using $v_y$. The front and
back will also be the same as each other and will use $v_z$.
Which leaves us with a little problem: it doesn't make sense for
some sides of the box to have different pressure than the
others. This problem doesn't actually bother us as long as it
is a single atom, because it's be silly to treat one atoms as a
fluid, but we should keep it in mind as we begin to look at a
system that can be considered a fluid.
Many Atoms in a Box
The total pressure on the walls of our box due to many atoms
should just be the sum of the pressures due to a single
atoms.3 But our atoms are not all going in the same
direction, nor are they all going at the same speed regardless of
their headings.
However, if we could find the right average speed, we should be
able to get the right answer anyway. In that case we could say
the total pressure on the wall was
\begin{align*}
  P = N P_{avg} \,,
\end{align*}
for $N$ particles where $P_{avg}$ is the correctly chosen
average4 all the
one-particle pressures, and the average will take care of not
only the different speeds, but also the different directions
(which means different ratios of $v_x$ to $v_y$ to $v_z$).
By assumption all the atoms have the same mass $m$, and they're
all in the same box so they all use the same volume $V$, so the
average is going to concern itself only with the speed
$v_{avg}$. Which depends on the speeds in all possible
directions; but notice we've assumed all direction are equally
likely, so we expect the average $v_x$ and the average $v_y$ and
the average $v_z$ to all be the same:5
\begin{align*}
  v_{avg}^2 
  &= v_{avg,x}^2 + v_{avg,y}^2 + v_{avg,z}^2\\
  &= 3 v_{avg,x}^2 \,.
\end{align*}
Now we proceed by writing
\begin{align}
    P 
    &= N P_{avg} \\
    &= N \frac{m v_{avg,x}^2}{V} \\
    &= N \frac{m v_{avg}^2}{3V} \\
    &= N \frac{2 \, K_{avg}}{3V} \tag{2} \,.
\end{align}
Suddenly the pressure depends on the average kinetic
  energy $K_{avg}$ of the atoms. I'm just going to tell you
(without justifying it) that we can use the arithmetic mean of
the kinetic energies $\bar{K}$, which means that we use the
root-mean-square6 the velocities if we
want to express it that way.
We should admit to a hidden assumption here: that the
distribution of the speeds and directions of the atoms stays the
same over time. In a thin enough gas this can be true simply
because the atoms don't collide with each other during the time
we examine the system, but for generality we would have to prove
that the averages we're taking are constant as the atoms collide
with one another over and over again. This is a difficult
computation that was first done by Boltzmann.
Temperature
If we rearrange EQN~(2), we get
\begin{align}
  PV = \frac{2}{3} N \overline{K} \,,
\end{align}
which is very similar to the ideal gas law
(EQN~(1)).7 
In fact, EQN~(2) would be exactly the ideal gas law if
we had $\frac{2}{3} \overline{K} = kT$, and that would agree
well with our hand-waving model of temperature being related to
the total amount of energy being used to jiggle atoms and
molecules about inside a
substance.
This is also very interesting because we can recognize $U = N
\overline{K}$ as the internal energy of our ideal gas, which
is a fairly special thing because most types of matter have a
very complicated internal energy and formulas for them are hard
to find or even not known with any precision.
Writing that in it's conventional form we have
\begin{align}
  \overline{K} = \frac{3}{2} k T \tag{3} \,,
\end{align}
as a definition of the temperature of an ideal gas in
terms of the kinetic energy of the constituent atoms. Similarly
\begin{align*}
  U = \frac{3}{2} N k T \,.
\end{align*}
Generalization
At this point our mathematical expressions have no explicit
reference to the original assumption that the box was
rectangular, and so we drop the assumption that the box even
exists and treat the results here as simply describing gasses
that meet the assumptions we made for an ideal gas.
Energy and the Speed of Air
What is the typical kinetic energy and speed of an air molecule?
Air, being mostly nitrogen (and most of the rest oxygen) we'll
take the molar mass to by $28\,\mathrm{g}$ which makes the
molecular mass $m_{\mathrm{N}_2} = 4.6 \times
10^{-26}\,\mathrm{kg}$. Using EQN~(3) and taking $T =
290\,\mathrm{K}$ we find
\begin{align*}
  \overline{K}_{\mathrm{N}_2} = 6.0 \times 10^{-21}\,\mathrm{J}\;,
\end{align*}
and
\begin{align*}
  \overline{v}_{\text{RMS},\mathrm{N}_2} = 510\,\mathrm{m/s} \;.
\end{align*}
This is slightly more than the speed of sound,8 and gives us confidence that neglecting gravitational interaction
on the scale of humans is a safe move.
In addition to asking "Is it OK to neglect gravity in the lab?"
we could ask "At what height is it clear that we definitely need
to be taking gravity into account?" We could take
that to be the height over which the average kinetic energy drops
by 10% (or equivalently when the average speed drops by 5%).
Conserving energy between a molecule moving upward from the floor
with a speed of $v_i = 510\,\mathrm{m/s}$ and seeking the
height at which it has speed $v_f = 485\,\mathrm{m/s}$ we get
\begin{align*}
  E_f &= E_i \\
  \frac{1}{2}m v_f^2 + mgh &= \frac{1}{2} m v_i^2 \\
  v_f^2 + 2gh &= v_i^2 \\
  h &= \frac{v_i^2 - v_f^2}{2g} \\
  &\approx 1270\,\mathrm{m} \;.
\end{align*}
For the sake of being conservative we might say that
$1000\,\mathrm{m}$ elevation changed demand that we worry about
gravity. For higher precision work we might want to insure that
velocity changes don't exceed 1%, and begin to worry around
$200\,\mathrm{m}$.
Conclusion
A model that treats atoms almost as if they were tiny billiard
balls (using the same rules we use in class for macroscopic
objects) was employed to arrive at the ideal gas law for
sufficiently thin gasses. Along the way we relied on a few
additional assumptions: a choice of an averaging procedure (take
the arithmetic mean of kinetic energies), an assertion that the situation described represents and equilibrium, and an identification of
that mean kinetic energy of the constituent atoms with the
temperature of the gas.
This is one of several results from around the turn of the
twentieth century that linked atomic theory to experimentally
confirmed macroscopic behaviors,9 leading to the acceptance
of atoms as real physical objects in the first couple of decades
of the twentieth century.

1 We do
  have one limitation on the value of $L$: it must be small
  enough that we can ignore the static pressure of the gas in the
  box so that we can treat the top and bottom of the box as the
  same. We'll show later that the kinetic energy of the atoms
  must be much larger than the potential energy change between
  the top and bottom $v^2 \gg Lg$.
2 We chose the coordinate system so that
  the sides of the box are aligned with the coordinate axes for
  simplicity.
3 Recall that we assumed the atoms rarely
  interact so they mostly fly back and forth unperturbed. This is
  not actually a great assumption for air at STP, and ought to be
  replaced by an argument about the statistical distribution of
  directions and speeds remaining the same over time. Proving
  this is quite a difficult task and well beyond the scope of
  this document.
4 There are many ways to take averages. The one
  you are used to ("add up all the numbers then divide by how
  many there are"; written
$$ \bar{x} = \frac{1}{N} \left( \sum_{i=1}^N x_i \right)$$ in
mathematics) is called the arithmetic mean, but sometimes
different averages are useful. For the moment we'll leave the
question of which average to use undecided.
5 This equality
  also forces the pressure to be the same on all sides of the
  box, putting to rest a concern that cropped up earlier.
6 "Root-mean-square" is a way
  of averaging. You square all the numbers, add up the squares,
  divide by how many there are and then take the square root.
  $$v_\text{rms} = \sqrt{\frac{1}{N}\left(
      \sum_{i=1}^N v_i^2\right) }\,.$$
7 Here is where we see the importance of
  the condition $v^2 \gg Lg$: in order for the pressure to
  uniform the gas molecules must have the same average kinetic
  energy at the bottom of the box as at the top (to the level of
  approximation that we require).
8 The
  speed of individual atoms should be faster than the speed of
  sound because sound waves are propagated by the motion of
  individual molecules and can't go faster than their carriers.
9 Though it
  seems strange from the modern perspective, the assorted Laws of
  Proportions from Chemistry were not historically taken as
  sufficient to prove the existence of atoms. Even the theory
  presented here left some prominent critics unconvinced. Albert
  Einstein's quantitative theory of Brownian motion published in
  1905 seems to have sealed the deal.
