Assume we have a collection of gas molecules in gravity-free space in a container with volume V at absolute temperature T.

Then each molecule moves along with constant velocity in a straight line, until it hits another molecule, or a container wall. A collision between two molecules is similar to a collision between two balls. The molecules exchange momentum, but the total momentum of the two molecules is conserved. When a molecule hits a wall, it bounces back. Its momentum changes. To change the molecule's momentum, the wall must exert a force on the molecule. Newton's third law tells us that the molecule exerts a force on the wall. The greater the number of molecules hitting a wall, the greater is the force on the wall. In a container with different size walls, the bigger walls will receive more hits than the smaller walls and therefore experience a greater force. The pressure in the container is the magnitude of the normal force F on a wall divided by the surface area A of the wall.

P = F/A

The faster the molecules move in the container, the greater is the change in
momentum when they bounce off a wall, and the more often do they hit the walls.
Assume a molecule moves horizontally with speed |v_{x}| back and forth
between two infinitely-massive walls, which are a distance L apart. When
it hits the right wall its momentum changes from p_{1 }= +m|v_{x}|
to p_{2 }= -m|v_{x}|. The change in the molecule's
momentum is Δp_{mol }= p_{2 }
- p_{1 }= -2m|v_{x}|. The time interval between successive
hits on the right wall is Δt = 2L/|v_{x}|. So the average force
the wall exerts on this molecule is F_{mol
}= Δp_{mol}/Δt = -2m|v_{x}|/(2L/|v_{x}|) = -mv_{x}^{2}/L.
By Newton's third law, the average force that the molecule exerts on the wall is
F_{wall
}= mv_{x}^{2}/L, it is proportional to the square of the
speed of the molecule or its kinetic energy.

Assume that there are N molecules in the volume V, moving horizontally with
speed |v_{x}|. Not all the molecules have the same
kinetic energy. The force
exerted by the molecules on the walls of a container is therefore F = Nm<v_{x}^{2}>/L,
where <v_{x}^{2}> is the average value of v_{x}^{2}.

The pressure is P = F/A = Nm<v_{x}^{2}>/V,
since L*A = V. With ρ_{particle} = N/V we have

P = F/A = ρ_{particle}mv_{x}^{2}.

There is nothing special about the x-direction. The atoms can move up and down, back and forth, in and out. The average velocity components in all directions are all going to be equal to each other.

<v_{x}^{2}> = <v_{y}^{2}> = <v_{z}^{2}>.

They are each equal to one-third of their sum, which is the square of the magnitude of the average velocity.

<v^{2}> = <v_{x}^{2}> + <v_{y}^{2}>
+ <v_{z}^{2}>.

<v_{x}^{2}> = (1/3)<v^{2}>.

We may therefore write

P = (1/3)ρ_{particle}m<v^{2}> = (2/3)ρ_{particle}(m<v^{2}>/2).

This equation relates the pressure to the kinetic energy of the
atoms or molecules, since m<v^{2}>/2 is the kinetic energy of the center-of-mass or
translational motion of an atom or molecule. Using ½m<v^{2}> =
(3/2)k_{B}T and ρ_{particle} = N/V from above we therefore find that

PV = (2/3)N(m<v^{2}>/2).

PV = Nk_{B}T.

The pressure in a container is proportional to the average kinetic energy of the molecules and therefore to the absolute temperature T of the gas.

If all the molecules in a container would be at rest, their kinetic energy would be zero and the pressure would be zero.

Average atomic and molecular speeds (v_{rms} = <v^{2}>^{1/2}
= root mean square speed) are large, even at low temperatures. What is v_{rms}
for helium atoms at 5.00 K, just one
degree above helium's liquefaction temperature?

Solution:

½m<v^{2}> = (3/2)k_{B}T = (3/2)*1.381*10^{-23
}J/K*(5 K) = 1.04*10^{-22} J.

<v^{2}> = (2*1.04*10^{-22} J)/(4*1.66*10^{-27}
kg) = 3.13*10^{4} m^{2}/s^{2}.^{}v_{rms} = 177 m/s.

(The mass of the
^{4}He atom is 4 atomic mass units = 4*1.66*10^{-27} kg.)

The root-mean-square speed of the atoms or molecule swith mass m is v_{rms} = <v^{2}>^{1/2}
= (3k_{B}T/m)^{1/2}.