Question

A box contains gas molecules. Consider the box to be divided into three equal parts. (a) By extension of Eq. 20-20, write a formula for the multiplicity of any given configuration. (b) Consider two configurations: configuration Awith equal numbers of molecules in all three thirds of the box, and configuration B with equal numbers of molecules in each half of the box divided into two equal parts rather than three. What is the ratio ${\mathit{W}}_{\mathit{A}}\mathit{/}{\mathit{W}}_{\mathit{B}}$ of the multiplicity of configuration A to that of configuration B ? (c) Evaluate${\mathit{W}}_{\mathit{A}}\mathit{/}{\mathit{W}}_{\mathit{B}}$for N = 100. (Because is not evenly divisible by 3, put 34 molecules into one of the three box parts of configuration Aand 33 in each of the other two parts.)

Short answer

1. The formula for the multiplicity of any given configuration is $\frac{\mathrm{N}}{{\mathrm{n}}_{\mathrm{L}}!{\mathrm{n}}_{\mathrm{C}}!{\mathrm{n}}_{\mathrm{R}}!}$
2. The ratio of the multiplicity of any given configuration A to that of configuration B is

$\frac{\mathrm{N}}{\left({\displaystyle \frac{\mathrm{N}}{2}}\right)!\left({\displaystyle \frac{\mathrm{N}}{2}}\right)!}$

1. For N = 100 , then the value of $\frac{{\mathrm{W}}_{\mathrm{A}}}{{\mathrm{W}}_{\mathrm{B}}}$is $4.2\times {10}^{16}$.

Step-by-step solution

The given data

The gas molecules in a box are N and N = 100

Configuration A has equal number of molecules in all three thirds of the box.

Configuration B has equal number of molecules in each half of the box, box divided into two equal parts.

Understanding the concept of multiplicity of central configuration

We can use the concept of the multiplicity of configuration. The number of microstates in the configuration is the multiplicity of the configuration.

Formula:

The formula for multiplicity of a configuration, $\mathrm{W}=\frac{\mathrm{N}!}{{\mathrm{n}}_1!{\mathrm{n}}_2!{\mathrm{n}}_3!}$ (1)

(a) Calculation to find the formula of multiplicity of the configuration

A box is divided into three equal parts. Consider ${\mathrm{n}}_{\mathrm{L}}$molecules are in the left third. There are ${\mathrm{n}}_{\mathrm{C}}$ molecules are in the center third and ${\mathrm{n}}_{\mathrm{R}}$ molecules in the right third of the box.

Let N! be the rearrangements of N molecules

${\mathrm{n}}_{\mathrm{L}}!$is the rearrangements of${\mathrm{n}}_{\mathrm{L}}$ molecules in left third,

${\mathrm{n}}_{\mathrm{C}}!$is the rearrangements of role="math" localid="1661577603792" ${\mathrm{n}}_{\mathrm{C}}$molecules in left third,

${\mathrm{n}}_{\mathrm{R}}!$ is the rearrangements of ${\mathrm{n}}_{\mathrm{R}}$molecules in left third.

Substituting the above values to the formula of the multiplicity configuration that is equation (1), we get the formula of multiplicity as:

$\mathrm{W}=\frac{\mathrm{N}!}{{\mathrm{n}}_{\mathrm{L}}!{\mathrm{n}}_{\mathrm{C}}!{\mathrm{n}}_{\mathrm{R}}!}$

Hence, the multiplicity of the given configuration is$\mathrm{W}=\frac{\mathrm{N}!}{{\mathrm{n}}_{\mathrm{L}}!{\mathrm{n}}_{\mathrm{C}}!{\mathrm{n}}_{\mathrm{R}}!}$

(b) Calculation of the ratio WA/WB  of the multiplicity of any given configuration A to that of configuration B

Consider two configurations: For A box half of the molecules are in all three thirds of the box, then the multiplicity of the configuration A is given using equation (1) as:

${\mathrm{W}}_{\mathrm{A}}=\frac{\mathrm{N}!}{\left({\displaystyle \frac{\mathrm{N}}{3}}!\right)\left({\displaystyle \frac{\mathrm{N}}{3}}!\right)\left({\displaystyle \frac{\mathrm{N}}{3}}!\right)}.................................\left(2\right)$

For box, half molecules are on the right half and the other half is in the left half of the box. Then, the multiplicity of the configuration B using equation (1) is given as:

${\mathrm{W}}_{\mathrm{B}}=\frac{\mathrm{N}!}{\left({\displaystyle \frac{\mathrm{N}}{2}!}\right)\left({\displaystyle \frac{\mathrm{N}}{2}!}\right)}.................................\left(3\right)$

Thus, the value of the ratio is given by dividing equation (2) by equation (3) as given:

$$\begin{gathered} \frac{{\mathrm{W}}_{\mathrm{A}}}{{\mathrm{W}}_{\mathrm{B}}}=\frac{\mathrm{N}!}{\left({\displaystyle \frac{\mathrm{N}}{3}!}\right)\left({\displaystyle \frac{\mathrm{N}}{3}!}\right)\left({\displaystyle \frac{\mathrm{N}}{3}!}\right)} \\ \frac{\mathrm{N}!}{\left({\displaystyle \frac{\mathrm{N}}{2}}!\right)\left({\displaystyle \frac{\mathrm{N}}{2}}!\right)} \\ =\frac{\left({\displaystyle \frac{\mathrm{N}}{2}}!\right)\left({\displaystyle \frac{\mathrm{N}}{2}}!\right)}{\left({\displaystyle \frac{\mathrm{N}}{3}}!\right)\left({\displaystyle \frac{\mathrm{N}}{3}}!\right)\left({\displaystyle \frac{\mathrm{N}}{3}}!\right)} \end{gathered}$$

Hence, the value of the multiplicity of configuration A to B is $\frac{\left({\displaystyle \frac{\mathrm{N}}{2}!}\right)\left({\displaystyle \frac{\mathrm{N}}{2}!}\right)}{\left({\displaystyle \frac{\mathrm{N}}{3}!}\right)\left({\displaystyle \frac{\mathrm{N}}{3}!}\right)\left({\displaystyle \frac{\mathrm{N}}{3}!}\right)}$

(c) Calculation of WA/WB ratio for N = 100

The value of $\frac{{\mathrm{W}}_{\mathrm{A}}}{{\mathrm{W}}_{\mathrm{B}}}$for N = 100 can be given by using the above equation as:

$$\begin{gathered} \frac{{\mathrm{W}}_{\mathrm{A}}}{{\mathrm{W}}_{\mathrm{B}}}=\frac{\left({\displaystyle \frac{100}{2}}!\right)\left({\displaystyle \frac{100}{2}}!\right)}{\left({\displaystyle \frac{100}{3}}!\right)\left({\displaystyle \frac{100}{3}}!\right)\left({\displaystyle \frac{100}{3}}!\right)} \\ =\frac{\left(50!\right)\left(50!\right)}{\left(33!\right)\left(33!\right)\left(34!\right)} \\ =4.2\times {10}^{16} \end{gathered}$$

Hence, the value of the required ratio for N = 100 is$4.2\times {10}^{16}$