Transcript 13.7 The Connection between Classical and Statistical

Summary
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Boltzman statistics:
Fermi-Dirac statistics:
Bose-Einstein statistics:
Maxwell-Boltzmann statistics:
Problem 13-4: Show that for a system of N particles
obeying Maxwell-Boltzmann statistics, the occupation
number for the jth energy level is given by
  ln Z 

N j   NkT
  
j T

13.7 The Connection between Classical
and Statistical thermodynamics
U = Σ NJ EJ
where EJ = EJ (x)
x is an extensive property
such as volume (see eqn 12.21!)
Differentiate the above equation
Since
define
For two states with the same X
For classical thermodynamics
dU  TdS  PdV
or
 TdS  YdX
Since dQ = T·dS
and
dW = PdV = YdX
The first equation illustrates that the heat transfer
is energy resulting in a net redistribution of
particles among the available energy levels
involving NO WORK!
The equation
could be interpreted as: An increase
in the system’s internal energy could be brought by a decrease
in volume with an associated increase in the EJ. the energy
levels are shifted to higher values with no redistribution of the
particles among the levels.
For an open system, the change in the internal energy is
where μ is the chemical potential defined on a per particle.
The Helmholtz function (F = U –TS) can be
written as
For MB statistics

S  k ln( WMB )  k  ln( g j )
Nj
 ln( N j !)

F  U  TS   NkT (ln Z  ln N  1)
Now the MB distribution can be rewritten as
N  E j kT
 e

gj Z
Nj
1
( E j u )
e
kT
(similar to those derived in FD
and BE statistics)
13.8
Nj
gj
Comparison of the three
Distributions
1

E j u
e
kT
a = 1 for FD statistics
a
a = -1 for BE statistics
a = 0 for MB statistics
• BE curve: The distribution is undefined for x < 0.
particles tend to condense in regions where Ej is small,
that is, in the lower energy state.
• FD curve: At the lower levels with Ej – u negative the
quantum states are nearly uniformly populated with
one particle per state.
• MB curve: lies between BE and FD curves and is only
valid for the dilute gas region: many states are
unoccupied.
• Statistical equilibrium is a balance between the
randomizing forces of thermal agitation, tending to
produce a uniform population of the energy levels,
and the tendency of mechanical systems to sink to the
states of lowest energy.
Alternative Statistical Models
• Microcanonical ensemble: treats a single material sample of
volume V consisting of an assembly of N particles with fixed
total energy U. The independent variables are V, N, and U.
• The canonical ensemble: considers a collection of Na identical
assemblies, each of volume V. A single assembly is assumed to
be in contact through a diathermal wall with a heat reservoir of
the remaining Na -1 assemblies. The independent variables are
V, N, and T, where T is the temperature of the reservoir.
• Grand canonical ensemble: consists of open assemblies that
can exchange both energies and particles with a reservoir. This
is the most general and most abstract model. The independent
variables are V, T and u, where u is the chemical potential.
Further comments on degeneracy
• A microstate is said to be non-degenerate if no two
particles have the same energy.
• Alternative way of defining a non-degenerate system
is to say that the number of quantum states with E
<kT, >> N, where k is Boltzmann constant, T is the
temperature and N is the number of particles.
• Equation 12.24 can be employed to calculate the
number of states contained within the octant.
• Example: Calculate the total number of
accessible microstates in a system where 100
units of energy have been distributed among
three distinguishable particles of zero spin.
• Solution:
• 13-10
(a) using results from chapter 9, show that
 S 
u  T

 N U ,V
(b) It follows from the statistical definition of the entropy that
u
 ln W  
N
kT
Consider a system with a chemical potential u = - 0.3eV. By
what factor is the number of possible microstates of the
system increased when a single particle is added to it at room
temperature?
(k = 8.617 x 10-5 eVK-1)
• Solution: (a) using equation 13.53 directly
Chapter 14:
The Classical Statistical
Treatment of an Ideal Gas
14.1 Thermodynamic properties from
the Partition Function
• All the thermodynamic properties can be
expressed in terms of the logarithm of the
partition function and its derivatives.
• Thus, one only needs to evaluate the partition
function to obtain its thermodynamic
properties!
• From chapter 13, we know that for dilute gas
system, M-B statistics is applicable, where
S= U/T + Nk (ln Z – ln N +1)
F= -NkT (ln Z – ln N +1)
μ=
= – kT (lnZ – lnN)
• Now one can derive expressions for other
thermodynamic properties based on the above
relationship.
1. Internal Energy:
n
U   N j j
and
j 1
N  j kT
 e
gj Z
Nj
 N  j
 N
kT

U 
e g j j  

 Z
j  Z

We have …
Z   g je
j
 j
kT
 j

j  g j j e kT





differentiating the above Z equation with respect to T, we
have …
 j
 Z 

   g j e kT
 T V
j
  j
d 
 kT
dT

j
1
= kT 2  g j j e kT
j
(keeping V constant means ε(V) is constant)
Therefore,
or
N
2  Z 
U  (kT ) 
Z
 T V
  ln Z 
U  NkT 

 T V
2


 j
  g e kT ( (  1 ))
j j
j
kT 2
2. Gibbs Function
since
G=μN
G = - NkT (ln Z – ln N)
3. Enthalpy
G = H – TS
→
H = G + TS
H =-NkT(lnZ - ln N) + T(U/T + NklnZ – NklnN + Nk)
= -NkTlnZ + NkTlnN + U + NkTlnZ – NkTlnN + NkT
= U + NkT = NkT2
= NkT (1 + T ·
+ NkT
)
4. Pressure
 F 
P

 V T
 ( NkT (ln Z  ln N 1)) 
P 

V

T
 (ln Z  ln N 1) 
P  NkT

V

T
 (ln Z )

P  NkT
 0  0
 V
T
 (ln Z ) 
P  NkT

 V T