Kinetic theory

What is the relationship between temperature and pressure?

imageAssume 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

imageThe 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 Fwall = 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> + <vz2>.

<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.

Problem:

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?

Solution:
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.