2. Entropy revisited

### 2.1 Thermodynamic laws

Thermodynamic laws can be summarized as follows:


• The second law: An element of irreversible heat transferred, δQ, is a product of the temperature T and the increment of its conjugate variable S (i.e., δQ ¼ TdS). where the coefficient of the polynomial expansion can be written as follows:

<sup>y</sup><sup>1</sup> � <sup>y</sup><sup>2</sup> � <sup>y</sup><sup>3</sup> � <sup>⋯</sup> � ym <sup>¼</sup> <sup>Y</sup><sup>m</sup>

Example 2: Imagine that we have three containers and ten balls. Each container has enough room to

<sup>¼</sup> <sup>3628800</sup>

A thermodynamic system is assumed to have a number of small micro-systems.

Say that there are N micro-systems and mð Þ ≤ N thermodynamic states. This situation is similar to Nð Þ ¼ 10 balls in mð Þ ¼ 3 containers. The number of balls in container 1, 2, and 3 is N1, N2, and N3, respectively. Then the total number of different configurations of micro-systems in m micro-states is defined as

> <sup>Ω</sup><sup>N</sup> <sup>¼</sup> <sup>N</sup>! Q<sup>m</sup>

Boltzmann proposed a representation of entropy of the entire ensemble as

configurations are available to put ten balls into the three containers? If N<sup>1</sup> ¼ 2, N<sup>2</sup> ¼ 3, and N<sup>3</sup> ¼ 5, then

<sup>¼</sup> <sup>10</sup>! 2!3!5! <sup>¼</sup> <sup>N</sup>! Q<sup>3</sup>

k¼1

<sup>k</sup>¼<sup>1</sup> Nk! (8)

th container. How many different

ð Þ<sup>2</sup> ð Þ <sup>6</sup> ð Þ <sup>120</sup> <sup>¼</sup> <sup>2520</sup> (10)

<sup>k</sup>¼<sup>1</sup> Nk! (11)

SB ¼ kB ln Ω<sup>N</sup> (12)

m k¼0

ln Nk!

¼ ln N! � ∑

yk (9)

N! N1!N2!N3!

Fundamentals of Irreversible Thermodynamics for Coupled Transport

using the product notation of

DOI: http://dx.doi.org/10.5772/intechopen.86607

the equation is with the answer being 2520:

satisfying N ¼ N<sup>1</sup> þ N<sup>2</sup> þ N<sup>3</sup> ¼ 10:

2.2 Definitions

2.2.2 Gibbs entropy

81

2.2.1 Boltzmann's entropy

hold all ten balls. Let Ni (for i ¼ 1 � 3) be the number of balls in i

The Gibbs entropy can be written using Ω, as

<sup>¼</sup> ln <sup>Ω</sup><sup>N</sup> <sup>¼</sup> ln <sup>N</sup>!

Q<sup>m</sup> <sup>k</sup>¼<sup>0</sup> Nk!

ln N! ¼ N ln N=e

S kB

and using Stirling's formula as

for a large Nð Þ ≫ 1 , to derive

N! N1!N2!N3!

• The third law: As T ! 0, S ! constant, and S ¼ kB ln Ω, where Ω is the number of microstates.

The entropy S is defined in the second thermodynamic laws, and its fundamental property is described in the third law, linking the macroscopic element of irreversible heat transferred (i.e., δQÞ and the microstates of the system.

Suppose you have N objects (e.g., people) and need to position them in a straight line consisting of the same number of seats. The first and second objects have N and N � 1 choices, respectively; similarly, the third one has N � 2; the fourth one has N � 3 choices; and so on. The total number of ways of this experiment is as follows:

$$N \cdot (N-1) \cdot (N-2) \cdot (N-3) \cdot \dots \cdot 2 \cdot 1 = N! \tag{2}$$

Example 1: In a car, there are four seats including a driver's. Three guests will occupy the same number of seats. How many different configurations are available? There are three people, A, B, and C, and three seats, S1, S2, and S3. If A can chose a seat first, then A has three choices. Then, B and C have, in a sequence, two and one choices. Then, the total number of possible configurations are 3 � 2 � 1 ¼ 3! ¼ 6.

Next, when the N objects are divided into two groups. Group 1 and group 2 can contain N<sup>1</sup> and N<sup>2</sup> objects, respectively. Then, the total number of the possible ways to place N objects into two groups is

$$\frac{N!}{N\_1! N\_2!}$$

which is equal to the number of combinations of N objects taking N1objects at a time

$$\mathbf{C}\_{N\_1}^{N} = \frac{N!}{N\_1!(N-N\_1)!} \tag{3}$$

For example, consider the following equation of a binomial expansion

$$(\mathbf{x} + \mathbf{y})^3 = \mathbf{1} \cdot \mathbf{x}^3 + \mathbf{3} \mathbf{x}^2 \mathbf{y} + \mathbf{3} \mathbf{x} \mathbf{y}^2 + \mathbf{1} \cdot \mathbf{y}^3 = \sum\_{n=0}^{3} a\_n \mathbf{x}^n \mathbf{y}^{3-n} \tag{4}$$

where a<sup>0</sup> ¼ a<sup>3</sup> ¼ 1 and a<sup>1</sup> ¼ a<sup>2</sup> ¼ 3. For the power of N, the equation exists as

$$\left(\left(\mathbf{x}\_1 + \mathbf{x}\_2\right)^N = \sum\_{N\_1=0} \sum\_{N\_2=0} \frac{N!}{N\_1! N\_2!} \mathbf{x}\_1^{N\_1} \mathbf{x}\_2^{N\_2} = \sum\_{k=0}^N \mathbf{C}\_N^k \mathbf{x}\_1^k \mathbf{x}\_2^{N-k} \tag{5}$$

where N<sup>1</sup> þ N<sup>2</sup> ¼ N and

$$\mathbf{C}\_{N}^{k} = \frac{N!}{k!(N-k)!} = \mathbf{C}\_{N}^{N-k} \tag{6}$$

If we add <sup>x</sup><sup>3</sup> with a constraint condition of <sup>N</sup> <sup>¼</sup> <sup>∑</sup><sup>3</sup> <sup>k</sup>¼<sup>1</sup>Nk, then

$$(\varkappa\_1 + \varkappa\_2 + \varkappa\_3)^N = \sum\_{N\_1=0} \sum\_{N\_2=0} \sum\_{N\_3=0} \frac{N!}{N\_1! N\_2! N\_3!} \varkappa\_1^{N\_1} \varkappa\_2^{N\_2} \varkappa\_3^{N\_3} \tag{7}$$

Fundamentals of Irreversible Thermodynamics for Coupled Transport DOI: http://dx.doi.org/10.5772/intechopen.86607

where the coefficient of the polynomial expansion can be written as follows:

$$\frac{N!}{N\_1! N\_2! N\_3!} = \frac{N!}{\prod\_{k=1}^3 N\_k!} \tag{8}$$

using the product notation of

$$\left\langle \boldsymbol{\chi}\_1 \cdot \boldsymbol{\chi}\_2 \cdot \boldsymbol{\chi}\_3 \cdot \cdots \cdot \boldsymbol{\chi}\_m = \prod\_{k=1}^m \boldsymbol{\chi}\_k \right\rangle \tag{9}$$

Example 2: Imagine that we have three containers and ten balls. Each container has enough room to hold all ten balls. Let Ni (for i ¼ 1 � 3) be the number of balls in i th container. How many different configurations are available to put ten balls into the three containers? If N<sup>1</sup> ¼ 2, N<sup>2</sup> ¼ 3, and N<sup>3</sup> ¼ 5, then the equation is with the answer being 2520:

$$\frac{N!}{N\_1! N\_2! N\_3!} = \frac{10!}{2! 3! 5!} = \frac{3 \pounds 2 \pounds 000}{(2)(6)(120)} = 2 \pounds 20\tag{10}$$

satisfying N ¼ N<sup>1</sup> þ N<sup>2</sup> þ N<sup>3</sup> ¼ 10:
