Boltzmann Distribution LawB.T. Luke & Associates, Inc |
Main IndexGeneralized Boltzmann Models M. Lo Schiavo, N. Bellomo An Introduction to the Theory of the Boltzmann Equation (Paperback) Stewart Harris Lectures on Gas Theory (Paperback) Ludwig Boltzmann |
The diagram on the right shows two energy levels that are populated by
different numbers of molecules. The lower energy state contains
N_{0} molecules at an energy of
,
and the upper state contains N_{1} 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 j^{th} state
(N_{j}/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, 0^{th}, state has an energy of -18.0 kcal/mol, while the 1^{st} and 2^{nd} 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 N_{i} is the fraction of molecules (moles) that are in the i^{th} state.
The following table shows the calculation of N_{i} at 300K.
This table shows that at 300K, 83.20% of the molecules are in the 0^{th} state, 15.55% are in the 1^{st} state, and 1.25% are in the 2^{nd}.
If the temperature is increased to 1000K, the following results are
obtained.
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 2^{nd} 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 0^{th} state, D_{0}, is 1.872 debye, and D_{1}
and D_{2} are 2.031 and 2.218 for the 1^{st} and
2^{nd} 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 |