Boltzmann Distribution Law
B.T. Luke & Associates, Inc
Generalized Boltzmann Models
M. Lo Schiavo,
An Introduction to the Theory of the Boltzmann Equation
Lectures on Gas Theory
The diagram on the right shows two energy levels that are populated by
different numbers of molecules. The lower energy state contains
N0 molecules at an energy of
and the upper state contains N1 molecules at an energy of
. The Boltzmann distribution law
is then given by the expression
If there are a total of N molecules in the system that can reside in many
different energy states, the above expression can be rearranged and used to
determine the fraction of the molecules in the jth state
(Nj/N). This fraction, which is the more common form of the
Boltzmann distribution law, is given by the equation
By looking at the NIST Physical Reference Data, one can find that the current value for the Boltzmann constant, k, is 1.38065×10-23 J K-1. By using Avagadros number and realizing that 1 calorie is exactly 4.184J, the Boltzmann constant is also 1.9872 cal mol-1 K-1, which is just the gas constant, R.
Assume that there is a molecule with three low-lying conformations. The lowest, 0th, state has an energy of -18.0 kcal/mol, while the 1st and 2nd have energies of -17.0 and -15.5 kcal/mol. respectively. We can now determine the fraction of molecules in each state at a given temperature.
Before we run off and do this, we should first realize that we are only interested in energy differences, not the absolute value of the actual energies. Therefore, we can first shift the energies so that these three states have relative energies of 0.0, 1.0 and 2.5 kcal/mol, respectively. In addition, we can assume that we have a total of 1 mol of these molecules so that Ni is the fraction of molecules (moles) that are in the ith state.
The following table shows the calculation of Ni at 300K.
This table shows that at 300K, 83.20% of the molecules are in the 0th state, 15.55% are in the 1st state, and 1.25% are in the 2nd.
If the temperature is increased to 1000K, the following results are
As expected, as the temperature increases, the fraction of molecules in the higher-energy states increases. Increasing the temperature from 300K to 1000K causes the fraction in the 2nd state to increase from 1.25% to 15.05%.
Calculating the fraction of molecules in each state at a given temperature
can also be used to calculate the Boltzmann average of physical properties
at a given temperature. This is done by using the following formula
To continue with our example, assume that we want to calculate the average
of the dipole moment, <D>. If the dipole moment of the molecule in
the 0th state, D0, is 1.872 debye, and D1
and D2 are 2.031 and 2.218 for the 1st and
2nd states, respectively, the terms in the above equation are
given in the following table for the system at 1000K.
Therefore, the average dipole moment at 1000K is
<D> = 3.730/1.88879 = 1.975 debye