Chemistry 103A; Sections 5, 6, 7, 8; Lecture 35, 17 Nov 00

Recall:

We were talking about:

Kinetic Molecular Theory of Gases

The kinetic molecular theory of gases is an example of model building. That is, we build a (theoretical) model of how we think nature operates and then calculate the properties of our model to see how they compare with nature.

If the properties of our model agree with what we see in nature we keep the model and try to improve it. If the properties of our model do not agree with nature we throw it away (or make drastic revisions).

The model:

Our model of a gas contains

N molecules
of mass, m,
in a volume V,
at temperature, T.

The molecules have a diameter, d.

The average distance between molecules is much greater than d.

The molecules are in constant random motion.

The molecules collide (elastically) with the walls of the container to produce a pressure, p. The molecules also collide with each other.

Molecules do not lose or gain energy in their collisions with the walls or with each other.

Since we are talking about individual molecules instead of moles, we write the gas constant in terms of molecules instead of in moles. That is,

. kB is called Boltzmann’s constant in honor of Ludwig Boltzmann, who helped invent the kinetic theory of gases.

The value of Boltzmann's constant is given by


It is easy to show that,

,
so that the ideal gas law, in terms of molecules becomes,
We are now going to derive an equation that gives the pressure exerted by the collisions of the molecules with the walls. (This derivation makes some simplifying approximations, two of which are in error, but they cancel each other. However, we get the correct answer and we gain some physical insight as to how pressure is produced.) First, let's remind ourselves that a pressure is a force per unit area. That is, pressure is the force the molecules exert on the wall by their collisions divided by the area of the wall.

Second, let's remind ourselves of Newton's second law of mechanics. You have probably seen Newton's second law as,

F = ma.
That is "force equals mass times acceleration."

This form of Newton's second law is not how Newton wrote his law. To see how F = ma relates Newton's law we must realize that

acceleration = rate of change of velocity.
With this in mind let's rewrite Newton's 2nd law as
F = m ´ (rate of change of velocity).
Now let's rewrite this equation by bringing the mass, m inside the parentheses as
F = (rate of change of m ´ velocity)
or
F = rate of change of (m ´ velocity).
(Since m is a constant this is not a problem.

But m ´ velocity equals momentum, which leads us to Newton's form of his 2nd law,

F = rate of change of momentum.
Now we can use this to talk about how molecules produce a force against a wall. A molecule moving toward the wall with momentum, mv, will hit the wall and bounce off with the same velocity in the reverse direction. So, after hitting the wall the molecule will have momentum, -mv and the change in momentum for the molecule will be
D momentum = final momentum - initial momentum

= -mv - (+mv) = - 2mv.

Newton's third law says (indirectly) that momentum is conserved. So, if the change of momentum of the particle which collides with the wall is - 2mv then an amount of momentum, +2mv must be transferred to the wall in each collision (so that the sum of momentum changes in the particle and in the wall is zero).

Consider a cubical box with the area of one side equal to A . Further, the box contains a gas with a density of N/V molecules per cubic meter. Let's suppose that all the molecules are moving with some average value of speed, v. Further, lets suppose that 1/6 of the molecules are moving in each of the possible plus or minus x, y, and z directions. In one second any molecule that is within a distance v´(1 second), will hit the wall. This distance, times the area, A, of the wall forms a small volume, v´1´A.

This small volume contains v´1´A´ N/V molecules and one sixth of these molecules will hit the wall in 1 second. When these molecules hit the wall and bounce off they will transfer an amount of momentum

,
to the wall in 1 second. This momentum transfer, or rate of change of momentum, amounts to a force against the wall and, if we divide it by the area of the wall, A, we can get the pressure against the wall,
.
But we know from the ideal gas law that
So, equating the two expressions for p we get


We can use the next to last expression to calculate the average kinetic energy for a molecule,


The kinetic energy for N molecules is N times the kinetic energy for one molecule,

These equations provide a lot of information. For example we can calculate the average velocity of a molecule with
Notice that the average velocity is proportional to the square root of the temperature and to the inverse of the square root of the mass of the molecule. We will use this latter fact when we talk about diffusion.

In calculating the average speed it is easier to use the formula weight of the molecules than mass of one molecule. Remembering that

,
we can rewrite the expression for average velocity as
.
The only thing we have to be careful about here is the make sure the formula weight is expressed in kg/mol instead of g/mol. R, of course, must be in SI units, namely 8.31451 J/K mol.

Example, calculate the average velocity of a N2 molecule at room temperature, 298 K.
 

Diffusion

If we release a small amount of ammonia gas in a quiet room (no air currents or turbulence or mixing) the ammonia will eventually spread throughout the room, due to the motion of the molecules.

Looking at our equation for average molecular velocities we see that the average speed is inversely proportional to the square root of the formula weight. In equation form,

.
The rate at which a molecule diffuses through another gas is proportional to its average velocity. Thus the diffusion rate is inversely proportional to the square root of the formula weight.

We will not need to know how to calculate absolute diffusion rates in this course, but we will need to be able to calculate relative diffusion rates. The ratio of the diffusion rate of gas A (RateA) to the diffusion rate of gas B (RateB) is, from the above discussion,

Example:

Calculate how much faster He will diffuse than oxygen gas at 298 K.