Construct a table like Table 20 - 1for eight molecules

Eight molecules are given, n=8 to 1

With the help of the multiplicity of central configurations formula and the corresponding entropy formula, we can generate the required table.

Formulae:

The formula of multiplicity of microstates according to multiplicity of central configurations,

$\mathrm{W}({\mathrm{n}}_1,\mathrm{N})=\frac{\mathrm{N}!}{{\mathrm{n}}_1!{\mathrm{n}}_2!}$ (1)

The formula of entropy according to multiplicity of central configurations,

S = klnW (2)

Where, k = Boltzmann constant, that is $1.38\times {10}^{-23}\mathrm{J}/\mathrm{k}$particles.

For Label I, N = 8, n1 = 8

The multiplicity of microstates using equation (1):

$$\begin{gathered} \mathrm{W}=\frac{8!}{8!0!} \\ =1 \end{gathered}$$

Therefore, the entropy using equation (2):

$$\begin{gathered} \mathrm{S}=(1.38\times {10}^{-13}\mathrm{J})\ln \left(1\right) \\ =0 \end{gathered}$$

For Label II, N = 8, n1 = 7

The multiplicity of microstates using equation (1):

$$\begin{gathered} \mathrm{w}=\frac{8!}{7!1!} \\ =8 \end{gathered}$$

Therefore, the entropy using equation (2):

$$\begin{gathered} \mathrm{S}=(1.38\times {10}^{-23}\mathrm{J})\ln \left(8\right) \\ =2.9\times {10}^{-23}\mathrm{J} \end{gathered}$$

For Label III, N = 8, n1 = 6

The multiplicity of microstates using equation (1):

$$\begin{gathered} \mathrm{W}=\frac{8!}{6!2!} \\ =28 \end{gathered}$$

Therefore, the entropy using equation (2):

$$\begin{gathered} \mathrm{S}=(1.38\times {10}^{-23}\mathrm{J})\ln \left(28\right) \\ =4.6\times {10}^{-23}\mathrm{J} \end{gathered}$$

For Label IV, N = 8, n1 = 5

The Multiplicity of microstates using equation (1):

$$\begin{gathered} \mathrm{W}=\frac{8!}{5!3!} \\ =56 \end{gathered}$$

Therefore, the entropy using equation (2):

$$\begin{gathered} \mathrm{S}=(1.38\times {10}^{-23}\mathrm{J})\ln \left(56\right) \\ =5.6\times {10}^{-23}\mathrm{J} \end{gathered}$$

For Label V, N = 8, n1 = 4

Multiplicity of microstates using equation (1):

$$\begin{gathered} \mathrm{W}=\frac{8!}{4!4!} \\ =70 \end{gathered}$$

Therefore, the entropy using equation (2):

$$\begin{gathered} \mathrm{S}=(1.38\times {10}^{-23}\mathrm{J})\ln \left(70\right) \\ =5.9\times {10}^{-23}\mathrm{J} \end{gathered}$$

For Label VI, N = 8, n1 = 3

Multiplicity of microstates using equation (1):

$$\begin{gathered} \mathrm{W}=\frac{8!}{3!5!} \\ =56 \end{gathered}$$

Therefore, the entropy using equation (2):

$$\begin{gathered} \mathrm{S}=(1.38\times {10}^{-23}\mathrm{J})\ln \left(56\right) \\ =5.9\times {10}^{-23}\mathrm{J} \end{gathered}$$

For Label VII, N = 8, n1 = 2

Multiplicity of microstates using equation (1):

$$\begin{gathered} \mathrm{W}=\frac{8!}{2!6!} \\ =28 \end{gathered}$$

Therefore, the entropy using equation (2):

$$\begin{gathered} \mathrm{S}=(1.38\times {10}^{-23}\mathrm{J})\ln \left(28\right) \\ =4.6\times {10}^{-23}\mathrm{J} \end{gathered}$$

For Label VIII, N = 8, n1 = 1

Multiplicity of microstates using equation (1):

$$\begin{gathered} \mathrm{W}=\frac{8!}{1!7!} \\ =8 \end{gathered}$$

Therefore, the entropy using equation (2):

$$\begin{gathered} \mathrm{S}=(1.38\times {10}^{-23}\mathrm{J})\ln \left(8\right) \\ =2.9\times {10}^{-23}\mathrm{J} \end{gathered}$$

For Label IX, N = 8, n1 = 1

Multiplicity of microstates using equation (1):

$$\begin{gathered} \mathrm{W}=\frac{8!}{0!8!} \\ =1 \end{gathered}$$

Therefore, Entropy using equation (2):

$$\begin{gathered} \mathrm{S}=(1.38\times {10}^{-23}\mathrm{J})\ln \left(1\right) \\ =0\mathrm{J} \end{gathered}$$

Hence, the above table represents the central configuration of eight molecules.