Transcript chem3542-2015

Statistical Thermodynamics
Driving force of chemical reactions
Boltzmann’s distribution
Link between quantum mechanics & thermodynamics
Contents:
Distribution of Molecular States
Partition Function
Perfect Gas
Fundamental Relations
Diatomic Molecular Gas
Non-ideal Gas
Maxwell’s Demon
Quantum Statistics
Transition State Theory
Text Book:
Atkins, “Physical Chemistry”
SchrÖdinger Equation
Hy=Ey
Wavefunction
Hamiltonian
H = (-h2/2m)2 - (h2/2me)ii2
+  ZZe2/r - i  Ze2/ri
+ i j e2/rij
Energy
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
Democracy among microscopic states !!!
a priori: as far as one knows
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
a demon
THE DISTRIBUTION OF
MOLECULAR STATES
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.
Configurations and Weights
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, ......}
{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! ...)
2. Calculate the number of ways of distributing
20 different molecules among six different
states with the configuration {1, 0, 3, 5, 10, 1}.
Answer:
20! / 1! 0! 3! 5! 10! 1! = 931170240
note: 0! = 1
Stirling’s Approximation:
When x is large, ln x!  x ln x - x
x
ln x!
x ln x - x
ln A
1
0.000
-1.000
0.081
2
0.693
-0.614
0.652
4
3.178
1.545
3.157
8 10.605
8.636
10.595
16 30.672
28.361
30.666
20 42.336
39.915
42.332
30 74.658
72.036
74.656
note, A = (2)1/2 (x+1/2)x e-x
ln W  ( N ln N - N ) -  ( ni ln ni - ni )
= N ln N -  ni ln ni
The Dominating Configuration
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
N = 8
W
60
40
20
0
0
1
2
3
4
k
5
6
7
8
12000
N =16
10000
W
8000
6000
4000
2000
0
0 1 2 3 4 5 6 7 8 9 10111213141516
k
6.00E+008
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
k
20
24
28
32
When N is even, the weight is maximum at k = N/2,
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)!]
The ratio of the two weights R(N) Wk=N/2 / Wk=N/4
is equal to
(N/4)! (3N/4)! / [(N/2)!]2
| N |
|R(N) |
4
8
16 32
256
1.5 2.5
7.1 57.1 3.5 x 1014
6.0 x 1023
2.6 x 103e+22
Therefore, for a macroscopic molecular system
( N ~ 1023 ), there are dominating configurations
so that the system is almost always found in the
dominating configurations, i.e. Equilibrium
Dominating Configuration: Equilibrium Configuration
To find the most important configuration, we
vary { ni } to seek the maximum value of W.
But how?
One-Dimensional Function: F(x) = x2
dF/dx = 0
X
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,... ???
1. The total energy is a constant, i.e.
 ni i = E = constant
2. The total number of molecules is conserved,
 ni = N = constant
How to maximize W or lnW ? How to maximize
a function ?
Let’s investigate the water melon’s surface:
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
Generally, to minimize or maximize a function
F(x1, x2, …, xn) under constraints,
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
JUSTIFICATION
dL = dF - i i dCi
under the constraints, dCi = 0, thus
dF = 0
i.e., F is at its maximum or minimum.
This method is called the method of Lagrangian
Multiplier, or Method of undetermined multipliers
Procedure
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
Since
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)
The Boltzmann Distribution
Pi = exp ( - i )
Interpretation of Boltzmann Distribution
 > 0, more molecules occupying the low
energy states.
1 = i ni / N = i exp( -i)
exp( ) = 1 / i exp(-i)
 = - ln [i exp(-i)]
Physical Meaning of 
Imagine a perfect gas of N molecules. Its total
energy
E = i ni i = i N exp(-i) i / i exp(-i)
= -Nq-1dq /d
= -N dlnq / d
where q = i exp(-i), ni and i are the
population and energy of a state i, respectively.
It will be shown later that
q = V / 3 and  = h (/2m)1/2
[ is called the thermal wavelength]
dlnq/d = -3dln/d = -3/2
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,
Physical Meaning of 
1 /  = kT
where k is the Boltzmann constant.
: the reciprocal temperature
Example 1:
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.
Answer
The energies of second, third and higher excited
states are much higher than those of ground and
first excited states, for instance, at T = 5,000 K,
the probability of finding the first excited state is,
p2 = exp(-2)
with 2 = -9.5 eV
And, the probability of finding the second excited
state is,
p3 = exp(-3)
with 3 = -1.0 eV
the ratio of probabilities between second & first
excited states is,
exp[-(-1.0+9.5) x 11600/5000] = exp(-20)
[ 1 eV = 1.60 x 10-19 J; k = 1.38 x 10-23 J K-1;
1 eV  11600 k K-1] i.e., compared to the first
excited state, the chance that the molecule is in
second excited state is exceedingly slim. Thus,
we consider only the ground and first excited
states, a two-state problem.
The probability of finding the molecule in the
first excited state is
p = exp(-2) / [exp(-1) + exp(-2)]
where, 1 = -10.0 eV, and 2 = -9.5 eV
(1) T = 300 K
 - 0.5  11600 
exp 

300


p=

 - 0.5  11600 

1 + exp 
300



(2) T = 1000 K
 - 0.5 11600 
exp 

1000


p=

 - 0.5 11600 

1 + exp 
1000



(3) T = 5000 K
 - 0.5 11600 
exp 

5000


p=

 - 0.5 11600 

1 + exp 
5000



Molecular Partition Function
The Molecular Partition Function
The Boltzmann distribution can be written as
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)
The summation is over all possible states (not the energy
levels).
Interpretation of the partition function
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
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.
q = 1 + exp(-)
Example 4
Calculate the partition function for a uniform ladder of energy
levels
-
q=1+e
e
-
-2
+e
-
-2
q=e +e
=q-1
-3
+e
-3
+e
+ .......
+ ......
Therefore,
-
q=1/(1-e
)
Example 5:
Calculate the proportion of I2 molecules in their ground, first
o
excited, and second excited vibrational states at 25 C. The
-1
vibrational wavenumber is 214.6 cm .
n = (n+1/2) h
q = n exp(-n) = exp(-h/2) n exp(-n h)
= exp(-h/2)/[1- exp(-h)]
Portion of I2 is in n-th vibrational state is
pn = exp(-n)/q = exp(-nh) [1- exp(-h)]
o
o
Therefore, at T = 25 C = 298.15 K,
h = 1.036
p0 =0.645
p1 = 0.229
p2 = 0.081
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.
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.
The relation between S and the partition function q
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
The above expression is the definition of thermodynamic
entropy.
Boltzmann Formula for the entropy
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.
JUSTIFICATION
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.]
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)
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 constant?
According to the Third Law of thermodynamics,
as T  0, S  0;
as T  0, W  1 since usually there is only one ground state,
and therefore,
constant = 0.
Relation between S and the Boltzmann distribution pi
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.
The relation between S and the partition function q
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
This relation may be used to calculate S from the known
entropy q
Independent Molecules
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
The partition function Q
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. The above equation applies only to
molecules that are distinguishable, for instance, localized
molecules. However, if the molecules are identical and free to
move through space, we cannot distinguish them, and the
above equation is to be modified!
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.
Perfect Gas
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.,
T
=
i i
q = qT
i.e., there are only translational contribution to the energy and
the partition function.
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, …
Setting the lowest energy to zero, the relative energies can then
be expressed as,
n = (n2-1)
 = h2 / (8mX2)
with
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
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
q T = qx q y q z
= (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
Partition function of a perfect gas,
Q = (qT) N / N! = V N / [3N N!]
Energy
E = - (lnQ/)V = 3/2 nRT
where n is the number of moles, and R is the gas constant
Heat Capacity
Cv = (E/T)V = 3/2 nR
Relation between
the entropy S and the partition
function Q
Fundmental
Thermodynamic
Relationships
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
Pressure
dA = dU - d(TS) = dU - TdS - SdT
dU = dq + dw
dq = TdS
dw = -pdV
dA = - pdV - SdT
Therefore, pressure may be evaluated by the following expression,
p = -(A/V)T
p = kT( lnQ/V)T
This expression may be used to derive the equation of state for a
chemical system.
the entropy may also be expressed as
S = -(A/T)V
= klnQ + kT(lnQ/T)V
= klnQ -(lnQ/)V / T
= klnQ + [U-U(0)] / T
Consider a perfect gas with N molecules. Its partition function Q is
evaluate as
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.
The enthalpy
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
The Gibbs energy
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
Example 6:
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,
23
-1
and NA = 6.02 x 10 mol is the Avogadro constant
Example 7
Calculate the translational partition function of an H2 molecule
confined to a 100-cm3 container at 25oC
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
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?
Example 11
Calculate the ratio of the translational partition functions of D2 and
H2 at the same temperature and volume.
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 13
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.
Diatomic Gas
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.
Factorization of Molecular Partition Function
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.
q = i exp( - i ) = i exp[ -  ( iT +  iR +  iV +  iE )]
= [i exp( - iT )][ i exp( - iR )][ i exp( - iV )][ i exp( - iE )]
= qT q R qV q E
The translational partition function of a molecule
qT  i exp( - iT )
ì sums over all the translational states of a molecule.
The rotational partition function of a molecule
q R  i exp( - iR )
 ì sums over all the rotational states of a molecule.
The vibrational partition function of a molecule
qV  i exp( - iV )
 ì sums over all the vibrational states of a molecule.
The electronic partition function of a molecule
q E  i exp( - iE )
 ì sums over all the electronic states of a molecule.
q = qT qV q R
qT = V / 3
w / qE = 1
where
 = h(  / 2M )1/ 2
 = 1 / kT
Vibrational Partition Function
Two atoms vibrate along an axis connecting the two atoms. The
vibrational energy levels:
 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
 nV = nhv
5--------------5hv
4--------------4hv
3--------------3hv
2--------------2hv
1--------------hv
0--------------0
kT
 = hv
Then the molecular partition function can be evaluated
q v = n exp( - n ) = n exp( -nhv) = 1 /[1 - exp( - hv)]
q v = 1 + e -  + e -2  + e -3  ...
e -  q v = e -  + e -2  + e -3  + .... = q v - 1
Therefore,
1
q = 1 /(1 - e
)=
1 - e - hv
Consider the high temperature situation where kT >>hv, i.e.,
v
- 
hv  1, q v  1 / hv = kT / hv
Vibrational temperature v
k v  hv
High temperature means that T>>v
 e - hv  1 - h
v/K
v/cm-1
I2
309
215
F2
HCl H2
1280 4300 6330
892 2990 4400
m
v
=
where
k
Rotational Partition Function
If we may treat a heteronulcear diatomic molecule as a rigid rod, besides
its vibration the two atoms rotates. The rotational energy
 JR = hcBJ ( J + 1)
where B is the rotational constant. J =0, 1, 2, 3,…
q R = all rotationalstates exp[ - JR ]
= all rotationalenergy levels g J exp[ -  JR ]
= J (2 J + 1) exp[ - hcBJ ( J + 1)]
B=h/8cI2
where gJ is the degeneracy of rotational energy level εJ
hcB<<1
Usually hcB is much less than kT,

q R =  (2 J + 1) exp[ - hcBJ ( J + 1)]dJ
0
c: speed of light
R I: moment of Inertia
2
i
r im  = I
i

= -(1 / hcB) d{exp[ - hcBJ ( J + 1)]} / dJ dJ
0
= -(1 / hcB){exp[ - hcBJ ( J + 1]}l0
=kT/hcB
Note: kT>>hcB
For a homonuclear diatomic molecule
q R = kT / 2hcB
Generally, the rotational contribution to the molecular partition function,
q R = kT / hcB
Where  is the symmetry number.

H 2O NH3 CH 4
2
3 12
Rotational temperature R
k R  hcB
Electronic Partition Function
q E = all electronicstates exp[ -  Ej ] = all electronicenergies g j exp[ - Ej ]
 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
Mean Energy and Heat Capacity
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)
 (7 / 2) N kT (T>>1)
Contribution of a molecular to the total energy
Translational contribution
(1/2)kT x 3 = (3/2)kT
Rotational contribution
(1/2)kT x 2 = kT
Vibrational contribution
(1/2)kT + (1/2)kT = kT
kinetic
potential
the total contribution is (7/2)kT
qV = kT/hv
qR = kT/hcB
The rule: at high temperature, the
contribution of one degree of freedom
to the kinetic energy of a molecule
(1/2)kT
the constant-volume heat capacity
Cv = (U / T ) v
= (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
Thus, the total contribution of a molecule to the heat capacity is (7/2) k
Summary
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.
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
Partition Function
q = i exp(-i) = j gjexp(-j)
Q = i exp(-Ei)
Energy E= N i pi i = U - U(0) = - (lnQ/)V
Entropy S = k lnW = - Nk i pi ln pi = k lnQ + E / T
A= A(0) - kT lnQ
H = H(0) - (lnQ/)V + kTV (lnQ/V)T
Q = qN or (1/N!)qN
q=q q q q
T
R
V
E
Non-Ideal Gas
Now let’s derive the equation of state for real
gases.
Consider a real gas with N monatomic molecules
in a volume V. Assuming the temperature is T,
and the mass of each molecule is m. So the
canonical partition function Q can be expressed
as
Q = i exp(-Ei / kT)
where the sum is over all possible state i, and Ei
is the energy of state i.
In the classical limit, Q may be expressed as
Q =(1/N!h3N) … exp(-H / kT) dp1 … dpN dr1 … drN
where, H = (1/2m) i pi2 + i>j V(ri,rj)
Q = (1/N!) (2mkT / h2)3N/2 ZN
ZN = … exp(-i>j V(ri,rj) / kT) dr1 … drN
N
N
=
V
[ note: for ideal gas, ZNZ=
V
, and
N
Q = (1/N!) (2m kT / h2)3N/2 VN ]
The equation of state may be obtained via
p = kT( lnQ/V)T
We have thus,
p / kT = ( lnQ/V)T = ( (lnZ
 lnZ
V)
N / N/V)
T T
ZN = … { 1 + [ exp(-i>j V(ri,rj) / kT) - 1 ] } dr1 … drN
= VN + … [ exp(-i>j V(ri,rj) / kT) - 1 ] dr1 … drN
 VN + (1/2) VN-2 N(N-1)  [ exp(- V(r1,r2) / kT) - 1 ] dr1 dr2
 VN { 1 - (1/2V2) N2  [ 1 - exp(- V(r1,r2) / kT) ] dr1 dr2 }
= VN { 1 - B N2 / V }
where, BB==(1/2V)
(1/2V)[ [11- -exp(exp(-V(r
V(r1,r
kT)] ]dr
dr1 1dr
dr2 2
1,r
2)2)/ /kT)
Therefore, the equation of state for our gas:
p / kT = N / V + (N / V)2 B
= n + B n2
Comparison to the Virial Equation of State
The equation of state for a real gas
P / kT = n + B2(T) n2 + B3(T) n3 + …
This is the virial equation of state, and the
quantities B2(T), B3(T), … are called the
second, third, … virial coefficients.
Thus,
1  
 - V (r1 , r2 ) 
B2 (T ) = B =  V    1 - exp 
 dr1dr2
kT
2  


1 
 - V (r12 ) 
=    1 - exp 
 dr12
2 
 kT 
A. HARD-SPHERE POTENTIAL

r12 < 
0
r12 > 
U(r12) =
2 2dr
(1/2)004r
4r
dr
B2(T) ==(1/2)
3/3
==2
23/3
B. SQUARE-WELL POTENTIAL

U(r12) = -
0
r12 < 
 < r12 < 
r12 > 
(1/2) 00 4r22 dr
dr
B2(T) = (1/2)
3/3)[1
== (2
(23/3)
[1- -((3 3-1)
-1)( (ee- -11)])]
C. LENNARD-JONES POTENTIAL
U(r) = 4 [ (/r)12 - (/r)6 ]
1 
B2(T) = ( ) 0 { 1 - exp{(
2
With
- 4)[
kT
   ]}
  } 4r2 dr
  - 
r r
12
6
x = /r, T* = kT / 
B2 (T ) 
*) ( x12 - x6 )] } x2 dx
=

{
1
exp[(-4/T
2 3 0
Maxwell’s Demon (1867)
Thermal Fluctuation (Smolochowski, 1912)
In his talk “Experimentally Verifiable Molecular Phenomena that
Contradict Ordinary Thermodynamics”,… Smoluchowski showed
That one could observe violations of almost all the usual statements
Of the second law by dealing with sufficiently small systems. …
the increase of entropy… The one statement that could be upheld…
was the impossibility of perpetual motion of the second kind. No
device could be ever made that would use the existing fluctuations
to convert heat completely into work on a macroscopic scale …
subject to the same chance fluctuations….
-----H.S. Leff & A.F. Rex, “Maxwell’s Demon”
Szilard’s one-molecule gas model (1929)
To save the second law, a measure of where-about of the
molecule produces at least entropy > k ln2
Measurement via light signals (L. Brillouin, 1951)
h  >> k T
A Temporary Resolution !!!???
Mechanical Detection of the Molecule
Counter-clockwise rotation always !!!
A Perpetual Machine of second kind ???
Bennett’s solution (1982)
To complete thermodynamic circle,
Demon has to erase its memory !!!
Memory eraser needs minimal
Entropy production of k ln2
(R. Landauer, 1961)
Demon’s memory
Feynman’s Ratchet and
Pawl System (1961)
T1=T2, no net rotation
Feynman’s Lecture Notes
A honeybee stinger
potential
coordinate
A Simplest Maxwell’s demon
door
Average over 200 trajectories
No temperature difference!!!
T
t
A cooler demon
T1 > T 2
door
TL > TR !!!
TL
TR
Number of
particles
in left side
Our simple demon
No. of particles: 60
The door’s moment of inertia: 0.2
Force constant of the string: 10
Maxwell’s demon
No. of particles: 60
Threshold energy: 20
Hatch of an egg
Quantum Statistics
Quantum Particle:
Fermion (S = 1/2, 3/2, 5/2, ...)
e.g. electron, proton, neutron, 3He nuclei
Boson
(S = 0, 1, 2, ...)
e.g. deuteron, photon, phonon, 4He nuclei
Pauli Exclusion Principle:
Two identical fermions can not occupy the same
state at the same time.
Question: what is the average number particles or
occupation of a quantum state?
Fermi-Dirac Statistics
System: a fermion’s state with an
energy  (  - / )
occupation
energy
probability
-----------------n=0
0
exp(0)
--------l--------n=1

exp[-(-)]
There are only two states because of the Pauli
exclusion principle.
Thus, the average occupation of the quantum
state ,

0  exp (0 ) + 1 exp -  ( -  )
n(ε ) =
exp (0) + exp -  ( -  )
exp -  ( -  )
=
exp (0) + exp -  ( -  )
exp -  ( -  )
=
1 + exp -  ( -  )
= 1 / {exp[(-)] + 1}
Therefore, the average occupation number n()
of a fermion state whose energy is ,
n() = 1 / {exp[(-)] + 1}
 is the chemical potential. When  = , n = 1/2
For instance, distribution of electrons
Bose-Einstein Statistics
System: a boson’s state with an energy 
Occupation of the system may be 0, 1, 2, 3, …,
and correspondingly, the energy may be 0, , 2,
3, …. Therefore, the average occupation of the
boson’s state,

0 exp (0) + 1exp -  ( -  ) + 2 exp - 2  ( -  ) + 3 exp - 3 ( -  ) + ...
n( ) =
exp (0) + exp -  ( -  ) + exp - 2 ( -  ) + exp - 3 ( -  ) + ...
exp -  ( -  )
=
1 - exp -  ( -  )
= 1 / {exp[(-)] - 1}
Therefore, the average occupation number n() of
a boson state whose energy is ,
n() = 1 / {exp[(-)] - 1}
the chemical potential  must less than or equal
to the ground state energy of a boson, i.e.   0,
where 0 is the ground state energy of a boson.
This is because that otherwise there is a negative
occupation which is not physical. When  = 0,
n()  , i.e., the occupation number is a
macroscopic number. This phenomena is called
Bose-Einstein Condensation!
superfluid: when T  Tc = 2.17K, 4He fluid
flows with no viscosity.
4He
Classical or Chemical Statistics
When the temperature T is high enough or the
density is very dilute, n() becomes very small,
i.e. n() << 1. In another word, exp[(-)] >> 1.
Neglecting +1 or -1 in the denominators, both
Fermi-Dirac and Bose-Einstein Statistics become
n() = exp[-(-)]
The Boltzmann distribution!