The document discusses statistical thermodynamics, emphasizing the connection between thermodynamics and quantum mechanics through statistical mechanics. It explains key concepts such as probability, partition functions, microstates, macrostates, and the distribution of particles among energy levels. Various examples illustrate how to calculate thermodynamic probabilities and configurations of particles in different scenarios.

1. STATISTICAL THERMODYNAMICS
Presented by
Dr. Sharayu M. Thorat
Shri Shivaji College of Arts, Commerce &
Science, Akola (Maharashtra)

2. Introduction
The limitation of the classical treatment is that we are not certain about the
rate and mechanism of the process under investigation. The relevant
information can be had provided we know the link between the
macroscopic properties and the microscopic properties of the system.
Such types of correlation are studied by statistical mechanics.
The nature and details of the microscopic properties (like speed,
momentum, energy of each particle) are provided by Quantum
mechanics. The macroscopic properties of the system are determined
by thermodynamic principles. Statistical mechanics acts as a bridge
between thermodynamics and quantum mechanics.

3. Thus the study of link between thermodynamics and quantum mechanics
is called statistical thermodynamics.
Summarizing ,we can state that quantum mechanics provides information
about the energy of the molecular system, statistical mechanics tells us
about the possible arrangement of the energy among various molecules
of the system, and introduces the concept of probability & partition functions.
Statistical thermodynamics deals with the relationships between the
probability, partition functions and the thermodynamic properties.

4. The Role of Statistical Mechanics
The energy expressions suggest that the energy of the molecules present
in a system will depend on quantum numbers. For a lower
quantum number, the energy of the state is lower than that for a higher
quantum number.
Now the question is that how the molecules are distributed amongst the
possible energy levels?
In other words, we want to know that how many molecules are present
in the lowest energy level, highest energy level, and how many are
present in various intermediate energy levels.
That is to say, we wish to have information about the best possible
arrangements of the molecules in various quantum or energy levels.
Such a description is given by the term probability denoted by W.

5. Common Terms
Assembly A number N of identical entities is called an assembly. If the entities
are single particles then the assembly is called simply as a system.
If the entities are assemblies of particles then we call the number N as assembly
of assemblies or an ensemble.
An ensemble consists of a large number of replicas of the system under
consideration. Each member of the ensemble has same number of molecules (N),
same volume V, and same energy E. It is called a micro canonical ensemble.
Canonical ensemble The ensemble consisting of a large number of
assemblies (systems) each having the same value of N, V, and T is called a
canonical ensemble. At equilibrium each assembly (member) of the canonical
ensemble has the same temperature, but not necessarily the same energy E.
The value of E will fluctuate about the ensemble average value.
Occupation number It is the number of system in that particular state. The
set of occupation number is called a distribution.

6. Suppose we have to distribute 100 distinguishable balls into five boxes in such
a way that each box contains 20 balls. This over all distribution is called a
macrostate. The detailed description of the distribution that balls numbered 1
to 20 can appear in one box, 21 to 40 in the second, 41 to 60 in the third etc, is
called the microstate. The number of microstates which correspond to a
macrostate is referred to as thermodynamic probability.

7. Statistical weight factor g It is the degree of degeneracy of a particular energy
level, and is equal to the energy states of any energy level. For example, the energy
level of a particle in the three dimensional box is given by 1 = (nx
2 + ny
2 + nz
2 )
h2/8mV. The energy of the quantum states 211, 121, and 112 is the same, but the
three states are distinct. Hence the degree of degeneracy is 3, and the statistical
weight factor g is also 3.
Configuration Various equivalent ways of achieving a state is called a
configuration of the system. For example, consider two coins a and b. The state of
showing 1 head (H) and 1 tail (T) can be achieved in two ways:
•Coin a shows a head and coin b tail.
•Coin a shows a tail and coin b head.
There are two ways to arrive at the same state of 1 H and 1T, hence the
number of configuration is 2.

8. Probability The probability of a state of a system is defined as the
number of configurations leading to that particular state divided by the
total number of configurations possibly available to the system. For
instance consider the tossing of a coin. It can either show head or tail.
Thus, total number of possible configurations of the state of the coin is
two (1 head + 1 tail), and the probability of showing head is one out of
two configurations i.e. ½, similarly the probability of showing a tail is
½.

9. Thermodynamic Probability
The thermodynamic probability of system is equal to the number of
ways of realizing the distribution. The symbol for probability is W.
Expression of Thermodynamic Probability
Consider an assembly of N identical particles of a gas at a temperature T,
volume V and total energy E. Let N0, N1, N2, … etc., group of molecules be palced in
energy levels 0, 1, 2, … etc. in such a way that the total number of molecules and
total energy are constant.
The number of ways in which the molecules can be distributed into
different energy level is calculated by the principle of permutation and
combination. Thus, the probability W is expressed as
… (6.5).
Here N! is N factorial and is written as
N! = N x (N- 1) (N- 2) x … x 4 x 3 x 2 x 1
Similarly, the terms in the denominators are different.

10. Permutation and combination. Suppose there are 25 particles which
have to be put in groups of 12, 6, 4, 2, 1.
The number of ways of arranging there are
This can be calculated as follows. If there is no restriction on
group placement, the number of ways would be 25 x 24 x 23 ….. x 1
which is denoted by 25! Since one has to place the particles in groups, the
number has to be divided b y 12 ! 6 ! 4 ! 2 ! 1 !.
The methods of evaluating the configurations for systems of different
nature are summarized as follows.
i) Number of ways of selecting n distinguishable objects from the N
distinguishable objects (n < N) is
ii)The number of ways of arranging N indistinguishable objects into l
distinguishable locations with only one object in a location is (l ≥ N) is

11. iii) The number of configurations for N indistinguishable objects into l
distinguishable locations without any restriction on occupation is
iv) The number of ways of putting N distinguishable objects into l locations
with no restriction is
v) In writing the factorials we keep in mind that

12. Example 6.1. Calculate the number of ways of distributing distinguishable
molecules a, b, c, between three energy levels so as to obtain the following set
of occupation number.
N0 = 1, N1 = 1, N2 = 1, that is each energy level is occupied by one molecule.
Solution: The probability W is given by
Here N = 3; N0 = N1 = N2 = 1,
There are six ways of distributing the three molecules as required in the
problem. The same result may be obtained by considering the following
treatment:

13. Energy state
N = 1, N1 = 1,
N2 = 1
Configurations
I II III IV V VI
0 a a b b c c
1 b c c a a b
2 c b a c b a
There are six possible configuration

14. Example 6.2. Calculate the number of ways of distributing four
molecules in four energy levels so as there are 2 molecules in the level
0, 1 molecule in the 1 energy level, 1 molecule in the 2 energy level,
and zero in the 3 energy level i.e.
N0 = 2, N1 = 1, N2 = 1, N3 = 0,
Solution : The probability equation gives

Energy state
N0 = 2, N1 = 1,
N2 = 1, N3 = 0
Configuration
I II III IV V VI VII VIII IX X XI XII
0 ab ab ac ac ad ad bc bc dc dc bd bd
1 c d b d c b a d a b a c
2 d c d b b c d a b a c a
3 0 0 0 0 0 0 0 0 0 0 0 0

15. Example 6.3. Calculate the number of ways to distribute
•Two distinguishable objects in two boxes
•Two distinguishable objects in three boxes
•Two indistinguishable objects in two boxes
•Two indistinguishable objects in three boxes
Solution :
•Number of objects = N = 2
Number of boxes (location) = l = 2
Number of configurations = W = lN = 22 = 4
• N = 2; l = 3;
W = lN = 32 = 3 x 3 = 9