Download Intermediate Physical Chemistry (CHEM2503)

G. H. CHEN
Department of Chemistry
University of Hong Kong
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Intermediate Physical Chemistry
Contents:
Distribution of Molecular States
Perfect Gas
Fundamental Relations
Diatomic Molecular Gas
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Intermediate Physical Chemistry
Configurations and Weights
Boltzmann Distribution, and Physical
Meanings of 
Molecular Partition Function and its
Interpretation
The Internal Energy and the Entropy
Independent Molecular and their
Partition Function
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Distribution of Molecular States
Partition Function
Energy
Heat Capacity
Pressure and the gas law
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Perfect Gas
Factorization of Partition Function
Rotational Partition Function
Vibrational Partition Function
Electronic Partition Function
Mean Energy and Heat Capacity
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Diatomic Molecular Gas
Helmholtz Energy
Pressure
Enthalpy
Gibbs Energy
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Fundamental Relations
Statistical Mechanics provides the link between the
microscopic properties of matter and its bulk
properties.
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Statistical Mechanics:
Consider a system composed of N molecules,
and its total energy E is a constant. These
molecules are independent, i.e. no interactions
exist among the molecules. Countless collisions
occur. It is hopeless to keep track positions,
momenta, and internal energies of all molecules.
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THE DISTRIBUTION OF MOLECULAR STATES
THE DISTRIBUTION OF MOLECULAR STATES
All possibilities for the distribution of energy
are equally probable provided the number of
molecules and the total energy are kept the
same.
That is, we assume that vibrational states of a
certain energy, for instance, are as likely to be
populated as rotational states of the same energy.
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Principle of equal a priori probabilities:
For instance, four molecules in a three-level
system: the following two conformations have
the same probability.
---------l-l-------- 2
---------l---------- 
---------l---------- 0
---------l--------- 2
---------1-1-1---- 
------------------- 0
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THE DISTRIBUTION OF MOLECULAR STATES
THE DISTRIBUTION OF MOLECULAR STATES
Imagine that there are total N molecules among
which n0 molecules with energy 0, n1 with
energy 1, n2 with energy 2, and so on, where 0
< 1 < 2 < .... are the energies of different states.
The specific distribution of molecules is called
configuration of the system, denoted as { n0, n1,
n2, ......}
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Configurations and Weights
THE DISTRIBUTION OF MOLECULAR STATES
STATE
10
1
14
2
3
15
1
6
7
2
13
12
11
4
3
5
16
9
8
17
The above configuration is thus, { 4, 6, 4, 3 }
4
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For instance, a system with 17 molecules,
and each molecule has four states.
{N, 0, 0, ......} corresponds that every molecule
is in the ground state, there is only one way to
achieve this configuration; {N-2, 2, 0, ......}
corresponds that two molecule is in the first
excited state, and the rest in the ground state,
and can be achieved in N(N-1)/2 ways.
A configuration { n0, n1, n2, ......} can be achieved in W
different ways, where W is called the weight of the
configuration. And W can be evaluated as follows,
W = N! / (n0! n1! n2! ...)
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THE DISTRIBUTION OF MOLECULAR STATES
Justification
1. N! different ways to arrange N molecules;
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THE DISTRIBUTION OF MOLECULAR STATES
2. ni! arrangements of ni molecules with energy i
correspond to the same configuration;
THE DISTRIBUTION OF MOLECULAR STATES
1. Calculate the number of ways of distributing
3 objects a, b and c into two boxes with the
arrangement {1, 2}.
Answer:
| a | b c |, | b | c a |, | c | a b |.
Therefore, there are three ways 3! / 1! 2!
To eliminate overcounting of these configurations
| a | c b |, | b | a c |, | c | b a |
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Example:
THE DISTRIBUTION OF MOLECULAR STATES
2. Calculate the number of ways of distributing
20 objects into six boxes with the
arrangement {1, 0, 3, 5, 10, 1}.
Answer:
20! / 1! 0! 3! 5! 10! 1! = 931170240
note: 0! = 1
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Example:
THE DISTRIBUTION OF MOLECULAR STATES
When x is large, ln x!  x ln x - x
x
1
2
4
6
8
10
ln x!
0.000
0.693
3.178
6.579
10.605
15.104
x ln x - x
ln A
-1.000
0.081
-0.614
0.652
1.545
3.157
4.751
6.566
8.636 10.595
13.026
15.096
Note: A = (2)1/2 (x+1/2)x e-x
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Stirlings Approximation:
THE DISTRIBUTION OF MOLECULAR STATES
x
12
16
20
30
ln x!
19.987
30.672
42.336
74.658
x ln x - x
17.819
28.361
39.915
72.036
ln A
19.980
30.666
42.332
74.656
Therefore,
ln W  ( N ln N - N ) -  ( ni ln ni - ni )
= N ln N -  ni ln ni
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Stirlings approximation(cont’d):
Imagine that N molecules distribute among
two states. {N, 0}, {N-1, 1}, ..., {N-k, k}, ... ,
{1, N-1}, {0, N} are possible configurations,
and their weights are 1, N, ... , N! / (N-k)! k!,
... , N, 1, respectively. For instance, N=8, the
weight distribution is then
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The Dominating Configuration
N=8
W
60
40
20
0
0
1
2
3
4
k
5
6
7
8
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The Dominating Configuration
12000
N =16
10000
W
8000
6000
4000
2000
0
0 1 2 3 4 5 6 7 8 9 10111213141516
k
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The Dominating Configuration
The Dominating Configuration
N = 32
5.00E+008
4.00E+008
3.00E+008
2.00E+008
1.00E+008
0.00E+000
0
4
8
12
16 20 24 28 32
k
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6.00E+008
The Dominating Configuration
Wk=N/2 = N! / [N/2)!]2.
When N is odd, the maximum is at k = N/2  1
As N increases, the maximum becomes sharper!
The weight for k = N/4 is
Wk=N/4 = N! / [(N/4)! (3N/4)!]
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When N is even, the weight is maximum at k = N/2,
i.e.
The Dominating Configuration
R(N) Wk=N/2 / Wk=N/4
=(N/4)! (3N/4)! / [(N/2)!]2
| N | 4 | 8 | 16 | 32 | 256
| 6.0 x 1023
|R(N) | 1.5 | 2.5 | 7.1 | 57.1 | 3.5 x 1014 | 2.6 x 103e+22
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The ratio of the two weights is equal to
Therefore, for a macroscopic molecular system
( N ~ 1023 ), there are dominating configurations
so that the system is almost always found in or
near the dominating configurations, i.e. Equilibrium
If the system has more states, would we reach the same
conclusion as above? Why?
Two constraints for the system
 ni i = E = constant
2. The total number of molecules is conserved,
i.e.
 ni = N = constant
How to maximize W or lnW under
these constraints?
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1. The total energy is a constant, i.e.
The Boltzmann Distribution
ni / N = Pi = exp ( - i )
1 /  = kT
Meaning of : ensure total probability is ONE
1 = i ni / N = i exp( -i)
exp( ) = 1 / i exp(-i)
 = - ln [i exp(-i)]
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Interpretation of Boltzmann Distribution
The Boltzmann Distribution
Starting from the principle of equal a priori
probabilities, we evaluate the probabilities of
different configurations by simple counting,
and find that the dominating configuration
whose population obeys the Boltzmann
distribution which relates the macroscopic
observables to the microscopic molecular
properties, and is capable of explaining the
equilibrium properties of all materials.
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Summary
The Dominating Configuration
E.g. One-Dimensional Function: F(x) = x2
dF/dx = 0
Enough?
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To find the most important configuration, we vary { ni } to
seek the maximum value of W. But how?
Two-Dimensional Case: for instance, finding the
minimum point of the surface of a half water
melon F(x,y).
F/x = 0,
F/y = 0.
Multi-Dimensional Function: F(x1, x2, …, xn)
F/xi = 0, i = 1,2,…,n
To find the maximum value of W or lnW,
 lnW / ni = 0,
i=1,2,3,...
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The Dominating Configuration
The method of Lagrange Multipliers
cutting the watermelon
how to find the minimum or maximum
of F(x, y) under a constraint x = a ?
L = F(x, y) -  x
 L/x= 0
 L/y= 0
x=a
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Let’s investigate the water melon’s surface:
The
The
method
Dominating
of Lagrange
Configuration
Multipliers
C1(x1, x2, …, xn) = Constant1
C2(x1, x2, …, xn) = Constant2
.
.
.
Cm(x1, x2, …, xn) = Constantm
L = F(x1, x2, …, xn) - i iCi(x1, x2, …, xn)
L/xi = 0,
i=1,2, ..., n
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Generally, to minimize or maximize a function
F(x1, x2, …, xn) under constraints,
JUSTIFICATION
dL = dF - i i dCi
under the constraints, dCi = 0, thus
dF = 0
i.e., F is at its maximum or minimum.
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The
The
method
Dominating
of Lagrange
Configuration
Multipliers
The method of Lagrangian Multiplier
Construct a new function L,
L = lnW +  i ni -  i ni i
Finding the maximum of L by varying { ni }, 
and  is equivalent to finding the maximum of
W under the two constraints, i.e.,
L/ni = lnW/ni +  -i = 0
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Procedure
The method of Lagrangian Multiplier
ln W  ( N ln N - N ) - i ( ni ln ni - ni )
= N ln N - i ni ln ni
lnW/ni = (N ln N)/ni - (ni ln ni)/ ni
= - ln (ni/N)
Therefore,
ln (ni / N) +  -i = 0
ni / N = exp( -i)
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Since
Therefore, E = N <  > = 3N/2, where <  > is
the average kinetic energy of a molecule.
Therefore,
<  > = <mv2/2> = 3/2.
On the other hand, according to the Maxwell
distribution of speed, the average kinetic energy
of a molecule at an equilibrium,
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The Boltzmann Distribution
<mv2>/2 = 3kT/2
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The Boltzmann Distribution
(This is actually the definition of the temperature)
where k is the Boltzmann constant. Thus,
1 /  = kT
This is the physical meaning of , the reciprocal
temperature.
The Boltzmann Distribution
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Example 1:
Two friends A and B are drinking beer in a pub one
night. Out of boredom, the two start to play
“fifteen-twenty”. A mutual friend C steps in, and the
three play together. Of course, this time one hand is
used by each. A is quite smart, and studies chemistry
in HKU. He figures out a winning strategy. Before
long, B and C are quite drunk while A is still pretty
much sober. Now, what is A’s winning strategy?
When A plays with B and try the same strategy, he
finds that the strategy is not as successful. Why?
The Boltzmann Distribution
Consider a molecular whose ground state energy is
-10.0 eV, the first excited state energy -9.5 eV, the
second excited state energy -1.0 eV, and etc.
Calculate the probability of finding the molecule in
its first excited state T = 300, 1000, and 5000 K.
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Example 2:
The Molecular Partition Function
pi = exp(-i) / q
where pi is the probability of a molecule being found
in a state i with energy i. q is called the molecular
partition function,
q = i exp(-i)
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The Boltzmann distribution can be written as
The summation is over all possible states
(not the energy levels).
If the energy level is gi-fold degenerate, then the
molecular partition function can be rewritten as
q = i gi exp(-i)
As T  0,
q  g0,
i.e. at T = 0, the partition function is equal to the
degeneracy of the ground state.
As T  ,
q  the total number of states.
Therefore, the molecular partition function gives an
indication of the average number of states that are
thermally accessible to a molecule at the temperature of
the system. The larger the value of the partition function
is, the more the number of thermally accessible states is.
The relationship between q and :
exp() = q-1
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Interpretation of the partition function
Example 3
Consider a proton in a magnetic field B. The proton’s
spin (S=1/2) has two states: spin parallel to B and spin
anti-parallel to B. The energy difference between the
two states is  = pB where p is proton’s magneton.
Calculate the partition function q of the proton.
Example 4
Calculate the partition function for a uniform ladder
of energy levels
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The Boltzmann Distribution
Example 5
Calculate the proportion of I2 molecules in their ground,
first excited, and second excited vibrational states at
25oC. The vibrational wavenumber is 214.6 cm-1.
Statistical Thermodynamics
Partition function contains all the thermodynamic
information!
The relation between U and q
If we set the ground state energy 0 to zero, E
should be interpreted as the relative energy to
the internal energy of the system at T = 0,
E = i ni i = i Nexp(-i) i / q = - Ndlnq/d.
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The Internal Energy, the Heat Capacity & the Entropy
Statistical Thermodynamics
Therefore, the internal energy U may be expressed as
U = U(0) + E = U(0) - N (lnq/)V
Where, U(0) is the internal energy of the system at
T = 0. The above equation provides the energy as
a function of various properties of the molecular
system (for instance, temperature, volume), and
may be used to evaluate the internal energy.
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The Internal Energy, the Heat Capacity & the Entropy
Statistical Thermodynamics
Cv is the constant-volume heat capacity which measures
the ability of a system to store energy. It is defined as the
rate of internal energy change as the temperature T
varies while the volume is kept constant:
Cv  (U/T)V = N(1/kT2) (2lnq/2)V
Note that, d/dT = (d/dT) d/d = -(1/kT2) d/d
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The relation between CV and q
Statistical Thermodynamics
Calculate the constant-volume heat capacity of a
monatomic gas assuming that the gas is an ideal gas.
U = U(0) + 3N / 2 = U(0) + 3NkT / 2
where N is the number of atoms, and k is the Boltzmann constant.
Cv  (U/T)V = 3Nk / 2 =3nNAk/2= 3nR/2
where, n is the number of moles, R  NAk is the gas constant, and
NA = 6.02 x 1023 mol-1 is the Avogadro constant
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Example 6:
Statistical Thermodynamics
The Statistical Entropy
According to thermodynamics, entropy S is
some measurement of heat q. The change of
entropy S is proportional to the heat absorbed
by the system:
dS = dq / T
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The relation between S and the partition function q
The above expression is the definition of thermodynamic entropy.
Statistical Thermodynamics
S = k lnW
where, W is the weight of the most probable configuration of the system.
Boltzmann Formula
(1) indicates that the entropy is a measurement of the weight
(i.e. the number of ways to achieve the equilibrium
conformation), and thus a measurement of randomness,
(2) relates the macroscopic thermodynamic entropy of a system
to its distribution of molecules among its microscopic states,
(3) can be used to evaluate the entropy from the microscopic
properties of a system; and
(4) is the definition of the Statistical Entropy.
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Boltzmann Formula for the entropy
Statistical Thermodynamics
The energy of a molecular system U can be expressed as,
U = U(0) + i nii
where, U(0) is the internal energy of the system at T=0,
ni is the number of molecules which are in the state with
its energy equal to i
Now let’s imagine that the system is being heated while
the volume V is kept the same. Then the change of U
may be written as,
dU = dU(0) + i nidi + i idni = i idni
[dU(0) = 0 because U(0) is a constant; di = 0 because i
does not change as the temperature of the system arises.]
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JUSTIFICATION
According to the First Law of thermodynamics, the
change of internal energy U is equal to the heat
absorbed (q) and work received (w), i.e.,
dU = dq + dw
dq = TdS
(thermodynamic definition of entropy; or more the heat
absorbed, the more random the system)
dw = -PdV = -Force * distance
(as the system shrinks, it receives work from the environment)
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Statistical Thermodynamics
dU = TdS - PdV = TdS
(dV = 0)
dS = dU/T = k i idni
= k i (lnW/ni)dni + k i dni
= k i (lnW/ni)dni
= k dlnW
( lnW/ni = - ln (ni/N) = - +  i )
d(S - lnW) = 0
S = k lnW + constant
What is the constant?
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Statistical Thermodynamics
According to the Third Law of thermodynamics,
as T  0, S  0;
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Statistical Thermodynamics
as T  0, W  1 since usually there is only one ground state,
and therefore,
constant = 0.
Statistical Thermodynamics
S = k lnW = k ( N lnN -i ni lnni )
= k i ( ni lnN - ni lnni )
= - k i ni ln(ni /N)
= - Nk i (ni /N)ln(ni /N)
= - Nk i pi ln pi
since the probability pi = ni /N.
The above relation is often used to calculate the
entropy of a system from its distribution function.
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Relation between S and the Boltzmann distribution pi
Statistical Thermodynamics
According to the Boltzmann distribution,
ln pi = - i - ln q
Therefore,
S = - Nk i pi (- i - ln q)
= k i ni i + Nk ln qi pi
= E / T + Nk ln q
= [U-U(0)] / T + Nk ln q
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The relation between S and the partition function q
This relation may be used to calculate S from the known entropy q
Statistical Thermodynamics
Consider a system which is composed of N identical molecules.
We may generalize the molecular partition function q to the
partition function of the system Q
Q = i exp(-Ei)
where Ei is the energy of a state i of the system, and summation
is over all the states. Ei can be expressed as assuming there is no
interaction among molecules,
Ei = i(1) + i(2) +i(3) + … + i(N)
where i(j) is the energy of molecule j in a molecular state i
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Independent Molecules
Statistical Thermodynamics
Q = i exp[-i(1) - i(2) - i(3) - … -i(N)]
= {i exp[-i(1)]}{i exp[-i(2)]} … {i exp[-i(N)]}
= {i exp(-i)}N
= qN
where q  i exp(-i) is the molecular partition function. The
second equality is satisfied because the molecules are independent
of each other.
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The partition function Q
The relation between U and the partition function Q
U = U(0) - (lnQ/)V
The relation between S and the partition function Q
S = [U-U(0)] / T + k ln Q
The above two equations are general because they not only apply to
independent molecules but also general interacting systems.
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Statistical Thermodynamics
Statistical Thermodynamics
Perfect gas is an idealized gas where an individual molecule is
treated as a point mass and no interaction exists among molecules.
Real gases may be approximated as perfect gases when the
temperature is very high or the pressure is very low.
The energy of a molecule i in a perfect gas includes only
its kinetic energy, i.e.,
 i =  iT
q = qT
i.e., there are only translational contribution to the energy
and the partition function.
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Perfect Gas
Translational Partition Function of a molecule qT
Although usually a molecule moves in a three-dimensional space,
we consider first one-dimensional case. Imagine a molecule of mass
m. It is free to move along the x direction between x = 0 and x = X,
but confined in the y- and z-direction. We are to calculate its
partition function qx.
The energy levels are given by the following expression,
En = n2h2 / (8mX2)
n = 1, 2, …
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Statistical Thermodynamics
Setting the lowest energy to zero, the relative energies
can then be expressed as,
n = (n2-1) with  = h2 / (8mX2)
qx = n exp [ -(n2-1) ]
 is very small, then
qx = 1 dn exp [ -(n2-1) ] = 1 dn exp [ -(n2-1) ]
= 0 dn exp [ -n2 ] = (2m/h22)1/2 X
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Statistical Thermodynamics
Now consider a molecule of mass m free to move in a
container of volume V=XYZ. Its partition function qT
may be expressed as
qT = qx qy qz
= (2m/h22)1/2 X (2m/h22)1/2 Y (2m/h22)1/2 Z
= (2m/h22)3/2 XYZ = (2m/h22)3/2 V
= V/3
where,  = h(/2m)1/2, the thermal wavelength. The thermal
wavelength is small compared with the linear dimension of the
container. Noted that qT  as T . qT  2 x 1028 for an O2 in
a vessel of volume 100 cm3,  = 71 x 10-12 m @ T=300 K
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Statistical Thermodynamics
Statistical Thermodynamics
Q = (qT) N = V N / 3N
Energy
E = - (lnQ/)V = 3/2 nRT
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Partition function of a perfect gas,
where n is the number of moles, and R is the gas constant
Heat Capacity
Cv = (E/T)V = 3/2 nR
Statistical Thermodynamics
Consider an equilibrium system which is consistent of N
interacting molecules. These molecules may or may not
be the same.
Relation between energy and partition function
U = U(0) + E = U(0) - (lnQ/)V
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Fundamental Thermodynamic Relationships
Relation between the entropy S and the partition function Q
S = [U-U(0)] / T + k lnQ
Helmholtz energy
The Helmholtz free energy A  U - TS. At constant temperature
and volume, a chemical system changes spontaneously to the
states of lower Helmholtz free energy, i.e., dA  0, if possible.
Therefore, the Helmholtz free energy can be employed to assess
whether a chemical reaction may occur spontaneously. A system
at constant temperature and volume reaches its equilibrium when
A is minimum, i.e., dA=0. The relation between the Helmholtz
energy and the partition function may be expressed as,
A - A(0) = -kT ln Q
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Statistical Thermodynamics
Statistical Thermodynamics
dA = dU - d(TS) = dU - TdS - SdT
dU = dq + dw
dq = TdS
dw = -pdV
dA = - pdV - SdT
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Pressure
Therefore, pressure may be evaluated by the following expression,
p = -(A/V)T
= kT( lnQ/V)T
This expression may be used to derive the equation of state
for a chemical system.
Statistical Thermodynamics
Q = (1/N!) (V / 3)N
the pressure p is then
p = kT( lnQ/V)T
= kT N ( lnV/V)T
= NkT / V
pV = NkT = nNAkT = nRT
which is the equation of the state for the perfect gas.
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Consider a perfect gas with N molecules. Its partition
function Q is evaluate as
Statistical Thermodynamics
During a chemical reaction, the change in internal energy is
not only equal to the heat absorbed or released. Usually, there
is a volume change when the reaction occurs, which leads
work performed on or by the surroundings. To quantify the
heat involved in the reaction, a thermodynamic function, the
enthalpy H, is introduced as follows,
H  U + pV
Therefore,
H - H(0) = -( lnQ/)V + kTV( lnQ/V)T
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The enthalpy
Statistical Thermodynamics
Usually chemical reactions occur under constant temperature. A
new thermodynamic function, the Gibbs energy, is introduced.
G  A + pV
At constant temperature and pressure, a chemical system changes
spontaneously to the states of lower Gibbs energy, i.e., dG  0, if
possible. Therefore, the Gibbs free energy can be employed to
access whether a chemical reaction may occur spontaneously. A
system at constant temperature and pressure reaches its
equilibrium when G is minimum, i.e., dG = 0. The relation
between the Helmholtz energy and the partition function may be
expressed as,
G - G(0) = - kT ln Q + kTV( lnQ/V)T
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The Gibbs energy
Statistical Thermodynamics
Example 8
Calculate the entropy of a collection of N independent harmonic
oscillators, and evaluate the molar vibraitional partition function
of I2 at 25oC. The vibrational wavenumber of I2 is 214.6 cm-1
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Example 7
Calculate the translational partition function of an H2 molecule
confined to a 100-cm3 container at 25oC
Example 9
What are the relative populations of the states of a two-level system
when the temperature is infinite?
Example 10
Evaluate the entropy of N two-level systems. What is the entropy
when the two states are equally thermally accessible?
Statistical Thermodynamics
Example 12
A sample consisting of five molecules has a total energy 5. Each
molecule is able to occupy states of energy j with j = 0, 1, 2, ….
(a) Calculate the weight of the configuration in which the molecules
share the energy equally. (b) Draw up a table with columns headed
by the energy of the states and write beneath then all configurations
that are consistent with the total energy. Calculate the weight of
each configuration and identify the most probable configuration.
Example 14
Given that a typical value of the vibrational partition function of one
normal mode is about 1.1, estimate the overall vibrational partition
function of a NH3.
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Example 11
Calculate the ratio of the translational partition functions of D2 and
H2 at the same temperature and volume.
Example 15
Consider Stirling’s approximation for lnN! In the derivation of
the Boltzmann distribution. What difference would it make if
the following improved approximation is used ?
x! = (2)1/2 (x+1/2)x e-x
Example 16
Consider N molecules in a cube of size a.
(a) Assuming molecules are moving with the same speed v
along one of three axises (x-, y-, or z-axis). The motion may be
along the positive or negative direction of an axis. On the
average, how many molecules move parallel to the positive x
direction?
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Statistical Thermodynamics
(b)When a molecule collides with one of six sides of the
cube, it is reflected. The reflected molecule has the same
speed v but moves in the opposite direction. How many
molecules are reflected from the side within a time interval
t ? What is the corresponding momentum change of these
molecules?
(c) Force equals the rate of momentum change. Calculate the
force that one side of the cube experiences.
Pressure is simply the force acting on a unit area. What is
the pressure of the gas? Assuming the average kinetic energy
of a molecule is 3kT/2, derive the equation of state for the
system.
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Statistical Thermodynamics
Statistical Thermodynamics
Consider a diatomic gas with N identical molecules. A molecule is
made of two atoms A and B. A and B may be the same or different.
When A and B are he same, the molecule is a homonuclear diatomic
molecule; when A and B are different, the molecule is a heteronuclear
diatomic molecule. The mass of a diatomic molecule is M. These
molecules are indistinguishable. Thus, the partition function of the gas
Q may be expressed in terms of the molecular partition function q,
Q  q N / N!
The molecular partition q
q  i exp(  i )
where, i is the energy of a molecular state i, β=1/kT, and ì is the
summation over all the molecular states.
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Diatomic Gas
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Statistical Thermodynamics
Statistical Thermodynamics
The energy of a molecule j is the sum of contributions from its
different modes of motion:
 ( j)   T ( j)   R ( j)   V ( j)   E ( j)
where T denotes translation, R rotation, V vibration, and E the
electronic contribution. Translation is decoupled from other
modes. The separation of the electronic and vibrational motions
is justified by different time scales of electronic and atomic
dynamics. The separation of the vibrational and rotational
modes is valid to the extent that the molecule can be treated as a
rigid rotor.
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Factorization of Molecular Partition Function
Statistical Thermodynamics
q  i exp(  i )  i exp[   ( iT   iR   iV   iE )]
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Factorization of Molecular Partition Function
 [i exp(  iT )][ i exp(  iR )][ i exp(  iV )][ i exp(  iE )]
 qT q R qV q E
Statistical Thermodynamics
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The translational partition function of a molecule
qT  i exp(  iT )
ì sums over all the translational states of a molecule. q  qT qV q R
The rotational partition function of a molecule
q R  i exp(  iR )
 ì sums over all the rotational states of a molecule.
qT  V / 3
The vibrational partition function of a molecule
qV  i exp(  iV )
 ì sums over all the vibrational states of a molecule.
  1 / kT
The electronic partition function of a molecule
q E  i exp(  iE )
 ì sums over all the electronic states of a molecule.
where
  h(  / 2M )1/ 2
w / qE  1
Vibrational Partition Function
 nV  (n  1 / 2)hv
n= 0, 1, 2, …….
If we set the ground state energy to zero or measure energy
from the ground state energy level, the relative energy levels
can be expressed as
V
5--------------5hv
4--------------4hv
3--------------3hv
2--------------2hv
1--------------hv
0--------------0
 n  nhv
kT
  hv
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Two atoms vibrate along an axis connecting the two atoms.
The vibrational energy levels:
Vibrational Partition Function
e   q v  e    e 2   e 3   ....  q v  1
q v  1  e    e 2   e 3  ...
q v  n exp(  n )  n exp( nhv)  1 /[1  exp(  hv)]
Therefore, q  1 /(1  e
v
 
1
)
1  e  hv
Consider the high temperature situation where kT >>hv, i.e.,
hv  1, q v  1 / hv  kT / hv  e  hv  1  h
Vibrational temperature v
k v  hv
High temperature means that T>>v
v/K
v/cm-1
I2
309
215
F2
HCl H2
1280 4300 6330
892 2990 4400
m
v

where
k
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Then the molecular partition function can be evaluated
Rotational Partition Function
 JR  hcBJ ( J  1)
where B is the rotational constant. J =0, 1, 2, 3,…
q  all rotationalstates exp[   ]
 all rotationalenergy levels g J exp[   JR ]
R
 J (2 J  1) exp[  hcBJ ( J  1)]
where gJ is the degeneracy of rotational energy level εJR
Usually hcB is much less than kT,
hcB<<1

q   (2 J  1) exp[  hcBJ ( J  1)]dJ
R
Bh/8cI2
R
J
0

 (1 / hcB) d{exp[  hcBJ ( J  1)]} / dJ dJ
0
 (1 / hcB){exp[  hcBJ ( J  1]}l0
=kT/hcB
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If we may treat a heteronulcear diatomic molecule as a rigid rod,
besides its vibration the two atoms rotates. The rotational energy
c: speed of light
I: moment of Inertia
2
i
r im   I
i
Note: kT>>hcB
Rotational Partition Function
q R  kT / hcB
Generally, the rotational contribution to the molecular partition
function,
q R  kT / 2hcB
Where  is the symmetry number.
H 2O NH3 CH 4
 2
3 12
Rotational temperature R
k R  hcB
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For a homonuclear diatomic molecule
Electronic Partition Function
 g 0 exp[  0E ]
=g0
=gE
where, gE = g0 is the degeneracy of the electronic ground state,
and the ground state energy 0E is set to zero.
If there is only one electronic ground state qE = 1, the partition
function of a diatomic gas,
Q  (1 / N!)(V / 3 ) N (kT / hcB) N (1  e hv )  N
At room temperature, the molecule is always in its ground state
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q E  all electronicstates exp[   Ej ]  all electronicenergies g j exp[  Ej ]
Mean Energy and Heat Capacity
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The internal energy of a diatomic gas (with N molecules)
U  U (0)  3N (1n /  )v  N ( ln  /  )v  N[1n(1  e hv ) /  ]v
 (3 / 2) N1 /   N1 /   Nhv /( e hv  1)
 (5 / 2) NkT  Nhv /( e hv  1)
(T>>1)
 (7 / 2) N kT
The rule: at high temperature, the contribution of one
degree of freedom to the kinetic energy of a molecule
(1/2)kT
qV = kT/hv
qR = kT/hcB
Mean Energy and Heat Capacity
the constant-volume heat capacity
 (5 / 2) N k  N K ( hv) 2 e hv /( e hv  1) 2
 ( 7 / 2) N k
(T>>1)
Contribution of a molecular to the heat capacity
Translational contribution
(1/2) k x 3 = (3/2) k
Rotational contribution
(1/2) k x 3 = k
Vibrational contribution
(1/2) k + (1/2) k = k
kinetic potential
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Cv  (U / T ) v
Thus, the total contribution of a molecule to the heat capacity is (7/2) k
Mean Energy and Heat Capacity
n2h2 1
2
 

mVx
8mx2 2
T
n
Quantum Classical
h
Vx  n
2mx
where n = 1, 2, …
n is a measurement of the speed of the molecule
n
Vx
1
h
2mx
2 ....
h
mx
100 ....
50h
mx
h
 1.9 10 5 m / sec for a H in a one-dimonsional box x= 1cm
2mx
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Translational energy
Mean Energy and Heat Capacity
1
 n  (n  2 )hV
v
Quantum
1 2
 kA
2
Classical
A is the amplitude of the vibration
1 hv
hv
A  2(n  )  (2n  1)
2 k
k
Here the vibrational quantum number n is a measurement
of the vibrational amplitude.
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Vibrational energy
Mean Energy and Heat Capacity

R
J
 hcBJ ( J  1) 
Quantum
 is the rotational angulor velocity

h
2I
J ( J  1)
1 2
I
2
Classical
mj=-J,-J+1…….
J is a measurement of angular velocity
q
R

e

J , mj

 
(
2
J

1
)

e
R
R
J
J
J
mJ is a measurement of the projection of the angular velocity of the zaxis. i.e. a measurement of the rotation’s orientation.
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Rotational energy
Molecule with N atoms
Degree of freedom:
Translation:
3
Rotation:
3
nonlinear
2
linear
vibration:
3N – 6 nonlinear
3N – 5 linear
Diatomic Molecule A-B
Symmetry  =
0 if A  B
1 if A = B
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Supplementary note
Summary
A configuration { n0, n1, n2, ......} can be achieved in W
different ways or the weight of the configuration
W = N! / (n0! n1! n2! ...)
Dominating Configuration vs Equilibrium
The Boltzmann Distribution
Pi = exp (-i ) / q
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Principle of equal a priori probabilities:
All possibilities for the distribution of energy are equally
probable provided the number of molecules and the total
energy are kept the same.
Summary
q = i exp(-i) = j gjexp(-j)
Q = i exp(-Ei)
Energy
E= N i pi i = U - U(0) = - (lnQ/)V
Heat Capacity
CV = (E/T)V = k2 (2lnQ/2)V
Entropy
S = k lnW = - Nk i pi ln pi = k lnQ + E / T
Helmholtz energy
A= A(0) - kT lnQ
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Partition Function
Enthalpy
H = H(0) - (lnQ/)V + kTV (lnQ/V)T
Molecular partition function
Q = qN or (1/N!)qN
Factorization of Molecular Partition Function
q = qTqRqVqE
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Summary