Why does the air pressure at the surface of the earth exactly equal the weight of the entire air column above it Why does the air pressure at the surface of the earth (resulting from collisions of molecules on the surface of the earth which has to do with the velocity of the particles) exactly equal the weight of the entire air column above it (which just has to do with the number and mass of the molecules in the air column)? 
Here are the beliefs that give rise to my perplexity:


*

*The weight of the atmosphere is the mass of the molecules (in a column with cross section 1 square inch, let's say) times the force of gravity.

*Thus if we cooled the air column to the point where it was a solid and set the solid on a scale, we would get the weight of that column of air which, I am told, should read 14.7 lbs.

*The pressure from the air on a patch of ground 1 square inch in area is due to the collisions of the air molecules with that patch of ground.  This is a function of the number of particles hitting the ground per unit time (which is itself a function of the density of the air at the ground) and the average kinetic energy of those molecules (their temperature).  The air pressure I am told is 14.7 lbs/square inch.  

*If I enclose some the air at the ground in a rigid container (say I screw the cap on to a glass jar on the ground), thus isolating it from the effects of the column of air above, the pressure of the gas inside the jar is 14.7 psi and would remain that (assuming the container doesn't change shape/volume and I keep it at the same temperature) even if I took it too the top of a mountain or in to space.

*The molecules of air are sufficiently far apart so that inter-molecular forces are negligible.  
So why does the air pressure at the surface of the earth (resulting from collisions of molecules on the surface of the earth which has to do with the density and velocity of the particles) exactly equal the weight of the entire air column above it (which just has to do with the number and mass of the molecules in the air column)? 
 A: 
Why does the air pressure at the surface of the earth (resulting from collisions of molecules on the surface of the earth which has to do with the velocity of the particles) exactly equal the weight of the entire air column above it (which just has to do with the number and mass of the molecules in the air column)?

That's not exactly true. Deviations from hydrostatic equilibrium do exist. The generic names for these deviations are "weather" (short term and local deviations) and "climate" (longer term and global deviations).
It is however approximately true. The reasons are twofold, hydrostatic equilibrium and the second law of thermodynamics.
Consider a small cuboid shaped parcel of air with two horizontal faces (top and bottom) of area $A$ and height $\Delta z$. The vertical forces on this packet are the downward force of gravitation and the upward or downward force of buoyancy.  Buoyancy is the pressure difference between the top and bottom faces times the area of the top and bottom faces. Defining $\Delta p \equiv p_\text{top} - p_\text{bottom}$, the downward force of buoyancy is $F_b = A \Delta p$. Adding the gravitational force $mg$ yields the net downward force on the parcel, $F = A \Delta p + mg$.  Dividing by mass gives the downward acceleration: $a = \frac {A \Delta p} m + g$. Since mass is volume times density, this becomes $a = \frac 1 {\rho} \frac {\Delta p}{\Delta z} + g$. In the limit $\Delta z \to 0$ this becomes $a = \frac 1 \rho \left(\frac {dp}{dz} + \rho g\right)$. The parenthesized term must be zero for the parcel to remain stationary:
$$\frac {dp}{dz} + \rho g = 0$$
This is the hydrostatic equilibrium condition. A parcel of air that is not in hydrostatic equilibrium with the air around it will move up or down until it is in hydrostatic equilibrium.
The other reason is the second law of thermodynamics. A consequence of this law is that a system tends toward a state that minimizes the total potential energy if there is a path that allows this to happen. Hydrostatic equilibrium minimizes the potential energy of the atmosphere. The upward and downward motions that result from deviations from hydrostatic equilibrium provide the self-correcting that moves the atmosphere toward hydrostatic equilibrium.
As noted in my opening paragraph, deviations do exist. That the Earth is heated by the Sun during the daytime and cools at nighttime, and that the heating per unit area varies with latitude forces these deviations to occur. These deviations range from small and local (afternoon convective rain showers) to many hundreds of kilometers (hurricanes and typhoons) to global (the Hadley, Ferrel, and polar cells). The forcing from the Sun means the Earth's atmosphere can never be exactly in hydrostatic equilibrium. It is extremely close to being in hydrostatic equilibrium after averaging out those deviations over time and over the face of the Earth.
A: Here's a simple, non-mathematical, answer. Although the pressure at the surface does depend on the velocity of air molecules that's not the whole picture. It is more precise to say that it depends on the rate of collisions. The collision rate depends on the velocity of the molecules, i.e. the temperature. But it also depends on density of molecules. Higher density means more collision. In turn, it should be obvious that the density at the surface depends on the mass of the air column. Therefore the pressure depends on the mass of the air column and the local temperature (velocity) of the air. 
A: Suppose the pressure at the Earth's surface is $P$.
Consider an air column of cross-sectional area $A$.
The upward force on the column is $F_{\text{up}}=PA$.
Denote the weight of the column as $W$.
By definition of "weight", the downward force on the column is $F_{\text{down}}=W$.
Suppose the pressure is too low, such that $F_{\text{up}}<F_{\text{down}}.$
The column of air will then fall downward.
As it does, it more air molecules are arriving at the surface of the Earth, increasing the density of air and therefore also increasing the pressure.
Since the pressure increases, so does $F_{\text{up}}$.
This will continue until $F_{\text{up}} = F_{\text{down}}$, at which time the system is in equilibrium and stays the same.
In other words, the pressure is such as to balance the weight of the column because that's the only situation which won't immediately change.
