[ Prev | Up | Next ]

Maxwell-Boltzmann Distribution Function:

The Maxwell-Boltzmann probability distribution function is commonly used in statistical mechanics in order to determine the speeds of molecules. The reason statistics is used in order to do this is a direct result of the infinitesemal size of atoms and molecules. If a computer were to keep track of a sample of an ideal gas the size of a grain of salt at room temperature and pressure, it would need to dynamically account for the position and velocity vectors for a number of molecules on the order of 10¹⁵. This is too many operations for most modern computers to handle adequately. Other problems occur of course which stem from quantum mechanics and our increasing inability to precisely know the exact positions and velocities of particles the smaller the scale we choose to examine (Heisenberg uncertainty principle). We are forced therefore to concern ourselves with large populations of molecules. We are interested in determining a statistical representation which will acccurately predict the velocities of particles in a given sample of matter. In this case, we restrict ourselves to an ideal or hard-sphere gas in which there are no intermolecular interactions. To begin, we examine the number of collisions which a sample of gas exerts on a planar boundary surface, in this case orientated arbitrarily perpendicular to the x-axis.

eq.1

N is the number of collisions, r _N is the number density of particles, A is the area of the surface, D t is the time duration in which the particles are allowed to collide. The sum at the end occurs over all velocities which allow the particles to reach the surface in time D t from whatever distance they are located. Eq.1 may seem unintuitive, however, it is greatly simplified by noticing this is just density times volume, the v_xt is distance along the x-axis, A(v_xt) is volume. The change in momentum experienced by the molecules as they hit the wall will be two times their initial momentum prior to hitting it:

eq.2

Pressure is related to momentum, being equal to the change in momentum divided by the area:

eq.3

Since the gas is assumed to be stationary, we recognize that the sum of all velocities above zero is equal to half the sum of all the velocities from negative to positive infinity.

eq.4

At this point, we recognize an opportunity to relate the velocity of the particles to some macroscopic quantities which we already know. Rewriting the pressure using the ideal gas law, (note the assumption that the gas is ideal):

eq.5

The summation part on the right side can be expressed using some statisitical notation (which was used in the quantum mechanics section):

eq.6

The bracketed quantity represents the mean-square speed of the molecules. To keep going, the number density is written in terms of number and volume:

eq.7

Canceling volume:

eq.8

The number of molecules can be written in terms of the number of moles and Avogadro's number:

eq.9

Canceling moles and combining mass and Avogadro's number to get molecular mass:

eq.10

Equation 10 produces a relation between the speed of ideal gas molecules and temperature and molecular weight. In this case, it is the mean square speed of the molecules along the x-axis. Since chemists deal in three dimensions, we can expand this to three-dimensions without any loss of generality. We assume the gas is stationary so therefore:

eq.11

So consequently:

eq.12

V_rms stands for the root-mean-square speed of the molecules, given temperature and the molecular weight of the gas. Since we are restricted to dealing with molecules in a statistical manner, one will find that this is not the only method for representing the velocities of molecules, keeping in mind the true velocity of every particle is not obtainable. Using eq.12, we will formulate a relation to express the average kinetic energy of a gas, utilizing a new constant.

eq.13

K_b is the boltzmann constant defined exactly as it is in eq.13.

Its approximate value is 1.38x10^-23 J/K. Inserting this expressions into eq.12:

eq.14

Keeping in mind the typical expression for kinetic energy derived from elementary physics:

eq.15

The resulting expression, eq.15, states the mean kinetic energy of an ideal gas is dependent only on its temperature. It would be convenient now to derive a function which could relate several state variables of a gas to produce a distribution of speeds for the molecules of which the gas consists. This can be accomplished easily for an ideal gas. One begins by creating a distribution function g, which is equal to the product of three functions, each pertaining to distributions along only one axis:

eq.16

To derive some information from this, we take the derivative of g with respect to a coordinate variable, v_x in this case, and then using the chain rule:

eq.17

Taking the derivative on the right side results in:

eq.18

Dividing through by v_x:

eq.19

This is good. Now one can plug in the expression for g from eq.16:

eq.20

Taking the derivative with respect to v_x, keeping in mind two of the factors f are constants with respect to v_x:

eq.21

To continue, we have to look back at eq.17 for a moment to realize that we could have taken the derivative with respect to any variable; v_x, v_y, or v_z. Therefore, we can also state by analogy:

eq.22

And:

eq.23

From eqs.21, 22, and 23, we know that in order for all these to be true, they must be true for all values of v_x, v_y, and v_z. One can equate the left hand side of each equation with some constant, -k in this case will be used for convenience.

eq.24

Dividing eq.21, 22, and 23 simultaneously by g (imagine equating the left hand side of each equation to the other two on one line):

eq.25

This results in a seperable, first order differential equation which is easy to solve. Rewriting:

eq.26

And integrating:

eq.27

N is the integrating constant in this case. To determine its value, we realize that for any probability distribution, the integral of the distribution over the entire region of space it encompasses must be unity. Writing this explicitly:

eq.28

The integral in this equation cannot be solved in closed form. Its value can be determined however through numerical tables or from integral tables. Using such, one obtains:

eq.29

Rearranging:

eq.30

The function f is therefore:

eq.31

Now, the value of k needs to be determined. Going back to the definiton of the total distribution g:

eq.32

One can use eq.31 to write it out explicitly:

eq.33

This leaves one still with the problem of determining the constant k. To solve this problem, one goes back to eq.15 where the mean kinetic energy of an ideal gas was determined. Using the distribution g:

eq.34

Simplifing:

eq.35

The integral in eq.35 is another one which cannot be solved in closed form. Looking this one up too gives:

eq.36

Canceling stuff:

eq.37

Knowing that the value of the mean kinetic energy has been solved already in eq.15:

eq.38

Finally, having determined the value of that constant, the distribution f can be written:

eq.39