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 |vx| back and forth
between two infinitely-massive walls, which are a distance L apart. When
it hits the right wall its momentum changes from p1 = +m|vx|
to p2 = -m|vx|. The change in the molecule's
momentum is Δpmol = p2
- p1 = -2m|vx|. The time interval between successive
hits on the right wall is Δt = 2L/|vx|. So the average force
the wall exerts on this molecule is Fmol
= Δpmol/Δt = -2m|vx|/(2L/|vx|) = -mvx2/L.
By Newton's third law, the average force that the molecule exerts on the wall is
= mvx2/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 |vx|. 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<vx2>/L, where <vx2> is the average value of vx2.
The pressure is P = F/A = Nm<vx2>/V, since L*A = V. With ρparticle = N/V we have
P = F/A = ρparticlemvx2.
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.
<vx2> = <vy2> = <vz2>.
They are each equal to one-third of their sum, which is the square of the magnitude of the average velocity.
<v2> = <vx2> + <vy2>
<vx2> = (1/3)<v2>.
We may therefore write
P = (1/3)ρparticlem<v2> = (2/3)ρparticle(m<v2>/2).
This equation relates the pressure to the kinetic energy of the atoms or molecules, since m<v2>/2 is the kinetic energy of the center-of-mass or translational motion of an atom or molecule. Using ½m<v2> = (3/2)kBT and ρparticle = N/V from above we therefore find that
PV = (2/3)N(m<v2>/2).
PV = NkBT.
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 (vrms = <v2>1/2
= root mean square speed) are large, even at low temperatures. What is vrms
for helium atoms at 5.00 K, just one
degree above helium's liquefaction temperature?
½m<v2> = (3/2)kBT = (3/2)*1.381*10-23 J/K*(5 K) = 1.04*10-22 J.
<v2> = (2*1.04*10-22 J)/(4*1.66*10-27 kg) = 3.13*104 m2/s2.
vrms = 177 m/s.
(The mass of the 4He 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 vrms = <v2>1/2 = (3kBT/m)1/2.