CHE-20028: PHYSICAL & INORGANIC CHEMISTRY
STATISTICAL THERMODYNAMICS: LECTURE 2
Dr Rob Jackson
Office: LJ 1.16
r.a.jackson@keele.ac.uk
http://www.facebook.com/robjteaching
Statistical Thermodynamics:
topics for lecture 2
• Summary from lecture 1
• The molecular partition function
• Calculation of thermodynamic properties from
Statistical Thermodynamics
–
–
–
–
Internal energy
Heat capacity
Residual entropy
Entropy
che-20028: Statistical
Thermodynamics Lecture 2
2
The molecular partition function- 1
• What contributes
molecule?
to
the
energy
of
a
– Vibrational (V), translational (T) and rotational (R)
modes of motion
– Electron distribution (E)
– Electronic and nuclear spin (S)
Ei  E  E  E  E  E
V
i
T
i
R
i
E
i
che-20028: Statistical
Thermodynamics Lecture 2
S
i
3
The molecular partition function- 2
• Remembering from lecture 1 that the partition
function q is given by:
q   exp( Ei / kT )
i
• We can substitute for Ei in this expression:
q   exp( EiV / kT  EiT / kT  EiR / kT  EiE / kT  EiS / kT )
i
che-20028: Statistical
Thermodynamics Lecture 2
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The molecular partition function- 3
• We then use some mathematical trickery to
simplify the expression (ea+b = eaeb):
q
exp( EiV / kT )exp( EiT / kT )exp( EiR / kT )exp( EiE / kT )exp( EiS / kT )
i
i
i
i
i
• The total energy is the sum of contributions
from the different terms, and the total
partition function is the product of these
contributions:
q  qV qT q R q E q S
che-20028: Statistical
Thermodynamics Lecture 2
5
Note on electron and spin partition
functions
• While it is not possible to obtain expressions
for these, as we have done for the other
terms, we note the following:
– For closed shell molecules, excited states are so
high in energy that only the ground state is
occupied, and qE= 1
– Electron spin makes an important contribution
when there are unpaired electrons, since the
electron can occupy either spin state, and then
qS=2
che-20028: Statistical
Thermodynamics Lecture 2
6
Calculation of internal energy from
the partition function - 1
• It can be shown that the internal energy is
obtained from the derivative of the partition
function with respect to temperature:
NkT 2 dq
E
q dT
• Note that E is the energy with respect to the
lowest energy state of the molecule
che-20028: Statistical
Thermodynamics Lecture 2
7
Calculation of internal energy from
the partition function - 2
• So, in general terms we should write the
internal energy U as:
U = E + U(0), where U(0) is the zero point
energy
e.g. for a harmonic oscillator U(0) = ½ h
• We can use these expressions to calculate
the internal energy for some example
systems:
che-20028: Statistical
Thermodynamics Lecture 2
8
The internal energy of a
monatomic gas - 1
• Only need to consider qT (and assume qE=1)
• Remember that qT = aT3/2
(where a= (2mk)3/2 V/h3) (lecture 1 slide 23)
• So
dq T
3
 aT 1 / 2
dT
2
– (from slide 6)
• And so
ET 
NkT 2
aT 3 / 2
3
3
x aT 1 / 2  NkT
2
2
che-20028: Statistical
Thermodynamics Lecture 2
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The internal energy of a
monatomic gas - 2
• So if N=NA, ET = (3/2)RT

U = U(0) + (3/2)RT
• Example: calculate the internal energy of
Argon gas at 300 K
 U = U(0) + (3/2) x 8.314 x 300 J mol-1
 = 3741.3 J mol-1
• What would be the result for another
monatomic gas?
che-20028: Statistical
Thermodynamics Lecture 2
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The internal energy of a diatomic
gas - 1
• Neglecting vibrational motion, we just need to
calculate the energy contribution of rotational
motion, ER:
• From lecture 1 slide 20, qR = bT
• (where b=k/(hB))
 dqR/dT = b
NkT 2
R
E

bT
b  NkT
• If N=NA, ER = RT
che-20028: Statistical
Thermodynamics Lecture 2
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The internal energy of a diatomic
gas - 2
• So to summarise:
U = U(0) + ET + ER = U(0) + (3/2)RT + RT
 U = U(0) + 5/2RT
• Note: both expressions, for the monatomic
gas and the diatomic gas, assume the gas to
be perfect.
• The vibrational contribution has been
neglected for the diatomic gas.
che-20028: Statistical
Thermodynamics Lecture 2
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How can we make comparisons
with experimental values?
• Heat capacities can be calculated, and
compared with experimental values.
• Remember that CV= dU/dT at constant
volume.
– For a monatomic gas:
U = U(0) + (3/2)RT, dU/dT= (3/2)R= 12.47 J mol-1K-1
– For a diatomic gas:
U = U(0) + (5/2)RT, dU/dT= (5/2)R= 20.79 J mol-1K-1
che-20028: Statistical
Thermodynamics Lecture 2
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Heat capacities
• Good agreement for inert
gases and for nitrogen;
agreement gets worse for
halogens with increasing
departure from perfect
gas behaviour.
• Why does this happen?
Experimental heat capacities
che-20028: Statistical
Thermodynamics Lecture 2
Gas
Cv/J K-1 mol-1
He
12.5
Ar
12.5
Xe
12.5
N2
20.8
F2
23.2
I2
28.6
14
Entropy and Statistical
Thermodynamics
• If we can calculate entropy, this is an
important step to being able to calculate
further thermodynamic properties
• We are already familiar with the connection
between entropy and the disorder of a system
• Boltzmann suggested that the entropy, S, of a
system should be given by the expression
S = k ln W
(W is the number of different ways the molecules in a
system can be arranged to give the same energy)
che-20028: Statistical
Thermodynamics Lecture 2
15
Boltzmann's grave in the
Zentralfriedhof, Vienna,
with bust and entropy
formula.
che-20028: Statistical
Thermodynamics Lecture 2
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Calculating entropy - 1
• The formula makes sense because S = 0 if
there is only one way of achieving a given
energy (if W = 1, S = ln (1) = 0).
• Similarly the formula predicts a high entropy if
there are many arrangements with the same
energy (i.e. if W is large).
• In most cases W = 1 when T= 0 because
there is only one way to achieve zero energy:
put all the molecules into the lowest energy
level.
che-20028: Statistical
Thermodynamics Lecture 2
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Calculating entropy - 2
• In this case S = 0 when T = 0, which agrees
with the third law of thermodynamics (the
entropy of all perfectly crystalline materials is
zero at 0 K).
• There are, however, cases where this is not
observed.
• This occurs when there is more than one
energetically equivalent arrangement of
molecules when T = 0.
che-20028: Statistical
Thermodynamics Lecture 2
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Residual entropy - 1
• There are several examples of molecules that
can have more than one energetically
equivalent arrangement at 0 K
– Examples include linear molecules where they
may be no energetic difference between the
arrangement
AB AB AB AB and the arrangement AB BA AB AB
– e.g. solid carbon monoxide, where there is no
energetic difference between the arrangements
CO CO CO CO and CO CO OC CO
che-20028: Statistical
Thermodynamics Lecture 2
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Residual entropy - 2
• In this case we say that the substance has a
residual entropy.
• For CO, there are 2 orientations of equal
energy, CO or OC, and if there are N
molecules, the number of ways of getting the
same overall energy is 2N

So S = k ln W = k ln 2N = N k ln 2 = R ln 2
• Where we assume that we have one mole of
molecules.
che-20028: Statistical
Thermodynamics Lecture 2
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Residual entropy - 3
• In this case, S = 8.314 x ln 2= 5.76 J K-1 mol-1
• The experimental value is 5.0 JK-1 mol-1
• Other examples of molecules having a
residual entropy include N2O and H2O.
• The summarised expression is:
S = R ln (no. of orientations of equal energy)
• Example questions are included in the
problem sheet.
che-20028: Statistical
Thermodynamics Lecture 2
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The entropy of a monatomic gas
• To calculate the entropy of a monatomic gas
we use an equation called the SackurTetrode* equation:
3/2
5/2


e
k
T
2

mk
T



B
B
Sm  R ln 

 

2
 p

h




• All the symbols have their usual meanings,
and p = 105 Pa, (noting that kB = k).
* Derived in 1912, anniversary in 2012.
che-20028: Statistical
Thermodynamics Lecture 2
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Sackur-Tetrode equation example
• Calculate the molar entropy of argon gas
(M = 39.948 g mol-1) at 298 K and 105 Pa
Divide the calculation into sections and rewrite the
equation as: Sm = R ln (A B3/2)
A = e5/2 kT/p = 12.182 x 1.381x10-23 x 293 /105 = 5.014 x 10-25
B = 2mkT/h2 = 2 x (39.948 x 1.661 x 10-27) x 1.381 x 10-23 x
298)/(6.626 x 10-34)2
= 3.908 x 1021, so B3/2 = 2.443 x 1032
S = R ln (5.014 x 10-25 x 2.443 x 1032) = 154.84 J mol-1 K-1
(Experimental value is 154.6 J mol-1 K-1)
che-20028: Statistical
Thermodynamics Lecture 2
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Advice on calculations
• Don’t try to do the whole calculation in one
go!
• This is because most calculators have a
maximum index value of 10+/- 99.
• Don’t try to calculate h3 (or h4) in one go!
• If your calculation comes out as zero, it’s
probably because the calculator can’t handle
the numbers. Break down the calculation.
che-20028: Statistical
Thermodynamics Lecture 2
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Summary
• The molecular partition function has been
defined.
• Internal energy and heat capacity have been
obtained from the partition function.
• Residual entropy has been introduced and
calculated.
• The Sackur-Tetrode equation, for calculating
entropy, has been introduced and used.
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Thermodynamics Lecture 2
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