Download Statistical Thermodynamics

The average occupation
numbers
• In the above case, it means the average
number of students in room 1 or room 2.
• Let j = 1 or 2 , where N1 is the number of
students in room 1 and N2 is the number of
students in room 2; Let Njk be the number of
students in room j for the kth case

Nj
N W

W
jk
k
k
k
k
The value of thermodynamic probability WN1 will
become extremely large as the values of N and
N1 are increased.
To facilitate the calculation, Stirling’s
approximation becomes useful:
ln (n!) = n ln(n) – n
When n is more than 50, the error in using
Striling’s approximation becomes very small.
Extend the above example into the situation where
there are multiple rooms (n rooms)
Now, the macrostate will be defined by the #
of students in each room, say: N1, N2 …Nn
Note that
N1 + N2 + N3 + … + Nn-1 + Nn = N
The number of microstates for the above
macrostate can be calculated from
W=
x
=
N!
N1! N 2 ! N 3! N n !
x ... x
12.4 Thermodynamic Probability
and Entropy
Boltzman made the connection between the classical
concept of entropy and the thermodynamic probability
S = f (w)
f (w) is a single-valued, monotonically increasing function
(because S increases monotonically)
For a system which consists of two subsystems A and B
Or…
Stotal = SA + SB
(S is extensive)
f (Wtotal) = f (WA) + f (WB)
The configuration of the total system can be
calculated as Wtotal = WA x WB
thus: f (WA x WB) = f (WA) + f (WB)
The only function for which the above
relationship is true is the logarithm.
Therefore:
S = k · lnW
where k is the Boltzman constant with the
units of the entropy.
12.3 Assembly of distinguishable
particles
• An isolated system consists of N
distinguishable particles.
• The macrostate of the system is defined
by (N, V, U).
• Particles interact sufficiently, despite very
weakly, so that the system is in thermal
equilibrium.
• Two restrictive conditions apply here
n
(conservation of particles)
N N

j 1
n
j
(conservation of energy)
N
E

U
 j j
j 1
where Nj is the number of particles on the
energy level j with the energy Ej.
Example: Three distinguishable particles labeled
A, B, and C, are distributed among four energy
levels, 0, E, 2E, and 3E. The total energy is 3E.
Calculate the possible microstates and
macrostates.
Solution: The number of particles and their total
energy must satisfy
3
N
j 0
j
3
3
N E
j 0
j
j
(here the index j starts from 0)
 3E
# particles
on Level 0
# Particles
on Level 1 E
# particles
on Level 2E
# particles
on Level 3E
Case 1
2
0
0
1
Case 2
1
1
1
0
Case 3
0
3
0
0
So far, there are only THREE macrostates
satisfying the conditions provided.
Configurations for case 1
Level 0
Level 1E
Level 2E
Level 3E
A, B
C
A, C
B
B, C
A
Thermodynamic probability for case 1 is 3
Configurations for case 2
Level 0
Level 1E
Level 2E
A
B
C
A
C
B
B
A
C
B
C
A
C
A
B
C
B
A
Level 3E
Configuration for case 3
Level 0
Level 1E
Level 2E
Level 3E
A, B and C
Therefore, W1 = 3, W2 = 6, and W3 = 1
.
• The most “disordered” macrostate is the state
with the highest probability.
• The macrostate with the highest thermodynamic
probability will be the observed equilibrium state
of the system.
• The statistical model suggests that systems tend
to change spontaneously from states with low
thermodynamic probability to states with high
thermodynamic probability.
• The second law of thermodynamics is a
consequence of the theory of probability: the
world changes the way it does because it seeks
a state of probability.
12.5 Quantum States and Energy
Levels
To each energy level, there is one or more
quantum states described by a wave function
Ф.
When there are several quantum states that
have the same energy, the states are said to
be degenerate.
The quantum state associated with the lowest
energy level is called the ground state of the
system. Those that correspond to higher
energies are called excited states.
The energy levels can be thought of as a set
of shelves of different heights, while the
quantum states correspond to a set of
boxes on each shelf
•
•
•
•
E3
g3 = 5
E2
g2 = 3
E1
g1 = 1
For each energy level the number of
quantum states is given by the
degeneracy gj
a particle in a one-dimensional
box with infinitely high walls
• A particle of mass m
• The time-independent part of the wave function
Ф (x) is a measure of the probability of finding
the particle at a position x in the 1-D box.
• Ф (x) = A*sin(kx)
0≤x≤1
with k = n (п/L), n = 1, 2, 3
n=1
n=2
The momentum of the particle is
(de Broglie relationship)
where k is the wave number and h is the Planck
constant.
P=
·k
The kinetic energy is:
2
2
2
1 2 1P
1 hk
E  mv 

2
2
2 m 2m 4
• Plug-in k  n 
L
h 2 n 2 2
E
8m 2 L2
h2n2

8mL2
Therefore, the energy E is proportional to the
square of the quantum number, n.
In a three-dimensional box
E = h2 (
+
+
)
8m
where any particular quantum state is
designated by three quantum numbers nx, ny
and nz .
If Lx= Ly = Lz = L
then, we say nj2 = nx2 + ny2 + nz2
Where nj is the total quantum number for
states whose energy level is EJ.
An important result is that energy levels
depends only on the values of nj2 and not on
the individual values of integers (nx, ny, nz). nj
is the total quantum number for energy level
Ej.
Since
h2 1 2
2
2
Ej 
(
n

n

n
x
y
z)
2
8m L
L3 = V
L2 = V⅔
therefore,
h2 1
Ej 
8m V
2
2
2
(
n

n

n
x
y
z)
2/3
For the ground state, nx = 1, ny = 1, nz = 1
There is only one set of quantum number leads to this energy,
therefore the ground state is non-degenerate.
For the excited state, such as nx2 + ny2 + nz2 = 6
it could be
nx = 1
ny = 1
nz = 2
or
nx = 1
ny = 2
nz = 1
or
nx = 2
ny = 1
nz = 1
They all lead to
h2 1
Ej 
6
2/3
8m V
Note that increasing the volume would reduce the energy
difference between adjacent energy levels!
For a one-liter volume of helium gas
nJ ≈ 2 x 109
12.6. Density of Quantum
States
When the quantum numbers are large and
the energy levels are very close together,
we can treat n’s and E’s as forming a
continuous function.
2
h
1
For
E 
(n 2  n 2  n 2 )
j
8m V
2/3
x
y
z
nx2 + ny2 + nz2 =
· V⅔ ≡ R2
The possible values of the nx, ny, nz
correspond to points in a cubic lattice in
(n , n , n ) space
Let g(E) · dE = n (E + dE) – n(E) ≈
dn( E )
dE
dE
Within the octant of the sphere of Radius R;
Therefore,
g ( E )dE 
dn( E )
4 2V 3 / 2 1/ 2
dE 
m E dE
3
dE
h
Note that the above discussion takes into
account translational motion only.
4 2V 3 / 2 1/ 2
g ( E )dE   s
m E dE
3
h
Where γs is the spin factor, equals 1 for
zero spin bosons and equals 2 for spin
one-half fermions
• Example (12.8) Two distinguishable particles are
to be distributed among nondegenerate energy
levels, 0, ε, 2ε , such that the total energy U = 2
ε.
(a)What is the entropy of the assembly?
(b)If a distinguishable particle with ZERO energy is
added to the system, show that the entropy of
the assembly is increased by a factor of 1.63.