Introduction statistical thermodynamics.pptx

To introduce the topic of statistical thermodynamics specially to M. Sc students and make them understand the concepts

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 = (n x 2 + n y 2 + n z 2 ) h 2 /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 1 T , hence the number of configuration is 2.

8. Probability T he 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 N , N 1 , N 2 , … etc., group of molecules be palced in energy levels  ,  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. N­ = 1, N­ 1 = 1, N­ 2 = 1, that is each energy level is occupied by one molecule. Solution: The probability W is given by Here N = 3; N = N 1 = N 2 = 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, N 1 = 1, N 2 = 1 Configurations I II III IV V VI  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  , 1 molecule in the  1 energy level, 1 molecule in the  2 energy level, and zero in the  3 energy level i.e. N = 2, N 1 = 1, N 2 = 1, N3 = 0, Solution : The probability equation gives  Energy state N = 2, N 1 = 1, N 2 = 1, N 3 = 0 Configuration I II III IV V VI VII VIII IX X XI XII  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

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 = l N = 2 2 = 4 N = 2; l = 3; W = l N = 3 2 = 3 x 3 = 9