1078 25 Equilibrium Statistical Mechanics. I. The Probability Distribution for Molecular Statesd.What approximately is the effect on the vibrational
partition function of a typical small molecule if the
Kelvin temperature is doubled from 150 K to 300 K?
e.What approximately is the effect on the electronic
partition function of a typical small molecule if the
Kelvin temperature is doubled from 150 K to 300 K?25.43Choose a diatomic gas and compute its translational,
rotational, vibrational, and electronic partition
functions at 298.15 K and 1.000 bar, looking up
parameters as needed in Table A.22 in Appendix A
or in some more complete table.^10 Unless you choose
NO or a similar molecule with an unpaired electron,
assume that only the ground electronic state needs to be
included.
25.44The ground electronic level of NO is a^2 Π 1 / 2 term. The
first excited level is a^2 Π 3 / 2 term with an energy 2.380×
10 −^21 J above the ground-level. Both these states have
degeneracy equal to 2. All other electronic states are more
than 7× 10 −^19 J above the ground-level and can be
neglected. For the ground-levelν ̃e 1904 .20 cm−^1 , and
B ̃e 1 .67195 cm−^1. Assume that the ground-level values
ofν ̃eandB ̃ecan be used for the excited
state.
a. Find the electronic factor in the molecular partition
function of NO at 298.15 K and 1.000 atm.
b. Find the fraction of NO molecules in the first excited
electronic state at 298.15 K.
25.45Identify the following statements as either true or false. If
a statement is true only under special circumstances, label
it as false.
a.Dilute gases are the only systems that can be treated
with statistical mechanics.
b.Dilute gases are the only systems for which molecular
partition functions apply.
c.It is impossible for a state of higher energy to have a
larger population than a state of lower energy.
d.The value of the molecular partition function can be
interpreted as the effective number of states available
to the molecule at the temperature and pressure of the
system.
e.Sums over states and sums over levels can be used
interchangeably.
(^10) K. P. Huber and G. Herzberg,Molecular Spectra and Molecu-
lar Structure. Vol. IV. Constants of Diatomic Molecules, Van Nostrand
Reinhold, New York, 1979.
f.For a system of macroscopic size, the average
distribution of molecular states and the most probable
distribution are essentially the same.
g.Fermion and boson probability distributions become
more and more like the Boltzmann distribution as the
occupation becomes more dilute.
h.Fermion and boson probability distributions become
more and more like the Boltzmann distribution as the
energy increases.
i.Molecular partition functions are exactly factored into
translation, rotational, vibrational, and electronic
factors.
j.Negative temperatures are not an equilibrium concept.
25.46 a.Represent the electron in a hydrogen atom as an
electron in a cubic box 1. 00 × 10 −^10 m on a side. Find
the electronic partition function at 298.15 K, using
Eq. (25.3-21) for the translational partition function.
There are two spin states, so that each state has a
degeneracy of 2.
b. There is a problem with part a in that the integral
approximate used to derive Eq. (25.3-21) is inaccurate,
because adjacent terms in the sum are not nearly equal
to each other. Sum the electronic partition function by
adding terms explicitly. Discontinue the sum when an
additional term changes the value of the sum by less
than one part in 1000. Set the energy of the
ground-state equal to zero so that energy of an excited
level is
Enx,ny,nz
h^2
8 ma^2
(
nx 2 +ny 2 +nz 2 − 3
)
c.There is a serious conceptual problem with the
electronic partition function of the hydrogen atom,
since there are infinitely many bound states with finite
energy eigenvalues. Explain what this appears to do to
the electronic partition function. Some have proposed
as a solution to this problem that those states should be
omitted that correspond to an expectation value of the
atomic radius equal or greater than the size of the
known universe.
d.Sum the electronic partition function explicitly for a
temperature of 298.15 K, using the energy levels and
degeneracies from Chapter 17, including states up to
and including then3 level, but setting the energy of
then1 level equal to zero so that the energy of the
nth level is
En(13.60 eV)
(
1 −
1
n^2
)