Chapter 3
3.1 If we assume that the equipartition principle is valid for these degrees of freedom, then each O2
molecule has Nep = 3 for translation, Nep = 2 for rotation (because O2 is linear), and Nep = 1 × 2
for vibration (1 vibrational mode with kinetic and potential energy terms). For each mole of O2 , the
equipartition principle predicts that the contribution to the energy will be Nep RT /2, so we multiply
these values by 3.50 mol to obtain the energy contribution to our system:
E = Etrans + Erot + Evib
(3.50 mol)(8.3145 J K−1 mol−1 )(355 K)
(3 + 2 + 2) =
(3 + 2 + 2)
These contributions come to: trans: 15.5 kJ; rot: 10.3 kJ; vib: 10.3 kJ.
3.2 We need to solve for P(v = 1), where v here is the vibrational quantum number, based on the
vibrational constant (which with v will give us the energy) and the temperature (which with ωe will
give us the partition function). We combine the vibrational partition function (Eq. 3.26)
qvib (T ) =
e−ωe /(kB T )
e−(1)(891 cm−1 )/[(0.6950 cm−1 / K)(428 K)]
= 1.05
with the vibrational energy expression Evib = vωe in the canonical probability distribution given by
Eq. 2.32:
P(v) =
g(v)e−Evib /(kB T )
qvib (T )
(1)e−(1)(891 cm
)/[(0.6950 cm−1 / K)(428 K)]
= 0.0475.
Note that the vibration of a diatomic is a nondegenerate mode, so we can always set g = 1 for the
vibration of a diatomic.
A couple of quick checks are available here. First, we notice that ωe is more than twice the thermal
energy kB T (as a very rough guide, the thermal energy in cm−1 is about 1.5 times the temperature in
K). That means that we expect most of the molecules to be in the ground state, because few will have
enough energy to get across the gap between v = 0 and v = 1. Sure enough, qvib = 1.05 is very close
to one, meaning that only one quantum state (the ground state) is highly populated. Secondly, the
partition function is only about 5% bigger than 1.0, which suggests that about 5% of the population is
in excited states. Since the closest excited state is v = 1, it makes sense that the probability of being
in v = 1 turns out to be 0.0475, which is just about 5%.
3.3 Asking for the fraction of molecules, the population in a given quantum state, the number of
molecules or moles (out of some total in the system) at a particular energy—all of these are ways of
asking us to find the probability of an individual state or an energy level using the canonical distribution
Eq. 2.32. To do this, we will always need three things: the degeneracy of the energy level (unless we
are looking for a particular state among several that share the same energy), the energy expression, and
the partition function. For rotations of any linear molecule (which includes all diatomic molecules), the
expressions we need are these:
grot = 2J + 1
Erot = Be J(J + 1)
qrot =
kB T
We can quickly verify that the integral approximation for the partition function is valid, because Be kB T = (0.6950 cm−1/ K)(428 K) = 297 cm−1. Then we put all this into Eq. 2.32 to get the probability:
P(J = 4) =
g(J)e−Erot /(kB T )
qrot (T )
(2J + 1)e−Be J(J+1)/(kB T )
kB T /Be
(9)e−20(20.956 cm )/(297 cm )
= 0.155.
(297 cm−1 )/(20.956 cm−1)
In this problem, we expect that the molecules are spread out over a large number of quantum states,
because the rotational constant Be is small compared to the thermal energy kB T . A fraction of 15.5%
for the J = 4 energy level is as high as it is only because Erot = 20Be = 419 cm−1 is fairly close to the
thermal energy of 297 cm−1, meaning that there is a high probability of molecules colliding with enough
energy to get to this energy level. The fact that the degeneracy increases with J also helps, because it
means that a collision that lands in any of the g = 2J + 1 = 9 quantum states that correspond to the
J = 4 energy level contribute to this probability.
3.4 The average of the momentum vector p = mv should be zero for physical reasons, because every
particle has an equal probability of traveling in either direction along any Cartesian axis (unless we
add forces of some type that push or pull the molecules along a particular direction). To show that
this average is zero mathematically, we would use the classical integrated average, which is obtained
by integrating over all space the property times its probability distribution, which in this case is the
velocity vector distribution Pv3 (v ) given by Eq. 1.15. For each vector component of the momentum,
we would need to solve an integral of the form (shown here just for the X component)
∞ 1/2
pX = m
e−a(vX ) vX dvX .
−∞ π
But this integral is always zero because the Gaussian function e−a(vX ) is symmetric about zero whereas
vX is antisymmetric. For every value of vX from −∞ to +∞, the integrand is equal and opposite to
the value of the integrand at the point −vX . The integral sums all these values together and gets zero.