kinetic theory

Physics II

Class Notes, 2/01/99

The kinetic theory of an ideal gas provides a successful model of the behavior of an ideal gas, summarized by the Ideal Gas Law PV = n R T, and by the specific heats of ideal monatomic and diatomic gases.

We can understand the kinetic theory through a series of models.

The first model consists of a single molecule bouncing back and forth with velocity v along the axis of a cylindrical container with cross-sectional area A and length L, with the molecule making elastic collisions with the ends of the container. We wish to determine the average pressure exerted by the molecule on one of the ends.

The average pressure exerted on an end will be equal to the average force exerted by the collisions of the molecule with the end divided by the area of the end.
The average force exerted on an end can be determined by the Impulse-Momentum Theorem, where we divide the momentum change associated with a collision by the time between collisions.

The momentum change of a collision with the left -hand end of the cylinder, if we assume that the positive direction is to the right, will be the difference between the initial momentum -mv and the momentum +mv after collision; this difference is `dp = mv - (-mv) = 2 m v. Note that this momentum change is positive, corresponding to a force directed to the right.

The time between collisions with this end of the container is the time to make a round trip from the left-hand end to the other end and back.

The distance traveled is thus 2 L, and the speed is v, so the time between collisions is `dt = 2 L / v.

The average force exerted on an end is therefore Fave = `dp / `dt = 2 m v / (2L / v), as indicated in the figure below.

This expression simplifies to Fave = mv^2 / L
Since mv^2 = 2 KE, where KE is the kinetic energy of the molecule, we see that Fave = 2 KE / L.

We easily find the average pressure Pave by dividing Fave by the cross-sectional area A.

Since the denominator of the resulting expression is A * L, which is equal to the volume of the cylinder, we see that Pave = 2 KE / V.

We can take as our definition of temperature any quantity which is proportional to the average kinetic energy of the molecules in a substance.

We therefore say that KE = const. * T.

From this follows the assertion that Pave = const * T / V, or Pave * V = const * T (note that for constant n, the Ideal Gas Law tells us that P V = n R T = const * T).

Our second model replaces the single molecule with N molecules, all moving in the same direction (along the axis) at velocity v.

Our intuition tells us that N molecules will make N times the number of collisions in any time interval and will therefore exert N times the force of a single molecule.

We can see that all the molecules will collide with a given end in the time interval `dt = 2 L / v required for a round trip, so that in this time interval there will be N collisions as compared to the single collision of our first single-molecule model.
We could also say that the average `dt between collisions is 1 / N times is great, so to find ave. force we will be dividing momentum change by 1/N of the `dt we used before.
In any case we see that the average force is N times as great so that the average pressure will be N times as great as in the first model.

We therefore obtain Pave = N * const * T / V.

Our third model takes account of the 3-dimensional nature of the velocities found in a real gas.

The 2-dimensional billiard-ball simulation shows how when all velocities are initially directed in one direction, say the x direction, it takes only a few collisions of each molecule to completely mix up all the kinetic energy and divide it randomly between the x and y directions.

In three dimensions the kinetic energy of the molecules is divided equally, on the average, between the x, y and z directions, with 1/3 of the kinetic energy in each of the three independent directions.
If we choose the x direction to be along the axis of the cylinder, then only the x kinetic energy has an effect on the average pressure on an end (motion in the y or z direction will not be affected by a collision on an end; only the x velocity is reversed, so only the x kinetic energy will appear in the expression for the force on the end).
Since the kinetic energy is divided equally among three independent directions, the average x kinetic energy is KExAve = 1/3 KEave, where KEave is the average kinetic energy of a molecule.

It follows that the pressure on an end will be P = N * 2 KExAve / V = N * * 2(1/3 KEave) / V, as shown above, which is equal to N * 2/3 KEave / V, as shown below.

We thus see that PV = N * 2/3 KEave = N * const. * 2/3 T, began assuming the proportionality KEave = const. * T.

2/3 * const, for the constant of this proportionality, is usually denoted k and is called Boltzman's Constant. Its value is 1.38 * 10^-23 J / (particle Kelvin).

We thus obtain the relationship PV = N k T, where N is the number of particles.

We can easily compare this with the Ideal Gas Law PV = n R T, where n is the number of moles.

We let Navag stand for Avagadro's Number, which is the number of particles in a mole.
N / Navag will therefore be the number n of moles in the gas.
Our equation is therefore rearranged as shown at the bottom of the figure below to read P V = n (k Navag) * T.
We conclude that, since from experimental measurements PV = n R T, R = k * Navag.

We can solve the relationship N k T = N ( 2/3 KEaveTrans) for KEaveTrans to see that the average translational kinetic energy of a molecule of a gas at temperature T must be KEaveTrans = 3/2 kT.

We can solve the relationship PV = N ( 2/3 KEaveTrans) for total kinetic energy total translational kinetic energy of a volume V of a gas at pressure P must be N * KEaveTrans = 3/2 P V.

We can easily find the average translational kinetic energy of an oxygen molecule at 300 K; we can then use this kinetic energy to find the average velocity (more properly the root-mean-square or RMS velocity) of an oxygen molecule.

The translational kinetic energy is 3/2 k T, which at 300 K is on the order of 10^-21 J (you should find the accurate value).
The mass of a mole of the substances is equal to its molecular mass in grams.

Oxygen has atomic number 16 so diatomic oxygen has molecular mass number 32 and therefore molecular mass approximately 32 g / mole.

The mass of the single oxygen atom is therefore 32 grams divided by Navag, the number of particles in a mole.

We find for oxygen that this molecular mass is on the order of 10^-25 Kg (again you should find the accurate value).

Setting 1/2 m v^2 = 10^-21 J, we easily solve for v, which we find to be on the order of 100 meters/second (the actual value will be somewhat greater; you should compute it).

We can find a total translational kinetic energy in a gas given its pressure in volume. For example in 2 liters of gas atmospheric pressure we have the total kinetic energy shown in the figure below, 300 Joules.

Since the translational kinetic energy is equally split among the x, y and z directions we have 100 J of total kinetic energy associated with each of these directions.

Each of these independent directions is called a 'degree of freedom' in which a particle may move.

As discussed earlier, for a diatomic molecule there are two independent degrees of freedom associated with the rotation of the molecule.

These rotational degrees of freedom consists of rotation about the two axes which are perpendicular to the axis between the two atoms making up the molecule, and also perpendicular to each other.

Each rotational degree of freedom also gets its 100 Joules (kinetic energy is distributed equally among degrees of freedom).

Thus the total internal kinetic energy of the diatomic gas at the given pressure and volume is U = 500 Joules, as indicated below.