A box contains Nmolecules. Consider two configurations: Configuration Awith an equal division of the molecules between the two halves of the box, and configuration Bwith 60.0%of the molecules in the left half of the box and 40.0%in the right half. For N =50, what are (a) the multiplicity W_Aof configuration A, (b) the multiplicity${\mathit{W}}_{\mathbf{b}}$ of configuration B, and (c) the ratio${\mathit{f}}_{\mathbf{B}\mathbf{/}\mathbf{A}}$of the time the system spends in configuration Bto the time it spends in configuration A? For N =100, what are (d)${\mathit{W}}_{\mathbf{A}}$, (e)${\mathit{W}}_{\mathbf{B}}$, and (f)${\mathit{f}}_{\mathbf{A}\mathbf{/}\mathbf{B}}$? ForN =200, what are (g)${\mathit{W}}_{\mathbf{A}}$, (h)${\mathit{W}}_{\mathbf{B}}$, and (i)${\mathit{f}}_{\mathbf{A}\mathbf{/}\mathbf{B}}$? ( j) With increasingN, doesincrease, decrease, or remain the same?

A box contains N molecules where N=50,100 and 200

Configuration A with an equal division of the molecule between the two halves of the box.

Configuration B with 60% of the molecules in the left half of the box and 40% in the right half of the box.

The equation 20-18 gives the multiplicities for all the configurations. From the assumption of statistical mechanics, all the microstates are equally probable. We find the multiplicities of configuration A and B for different N values.

Formula:

The formula of the multiplicity of the central configuration of a body,

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

The ratio ofthe time the system spends in configurationB to the time it spends in configuration is given as

${\mathrm{f}}_{\mathrm{B}/\mathrm{A}}=\frac{{\mathrm{W}}_{\mathrm{B}}}{{\mathrm{W}}_{\mathrm{A}}}$ (2)

Configuration A contains an equal division of the molecules between the two halves of the box. Using the formula of equation (1), we can calculate the multiplicity for configuration A as given: (For configuration A,${\mathrm{n}}_1=\mathrm{N}/2$and ${\mathrm{n}}_2=\mathrm{N}/2$)

$$\begin{gathered} W_A=\frac{N!}{\left({\displaystyle \frac{N}{2}}\right)!\left({\displaystyle \frac{N}{2}}\right)!} \\ =\frac{50!}{25!25!} \\ =\frac{\left(3.04\times {10}^{64}\right)}{1.55\times {10}^{25}\times 1.55\times {10}^{25}} \\ =1.26\times {10}^{14} \end{gathered}$$

Hence, the value of the multiplicity is$1.26\times {10}^{14}$

The configurationcontains 60% of the molecules in the left half of the box and 40% in the right half of the box. That is given as:

$$\begin{gathered} {\mathrm{n}}_1=60\%\mathrm{of}\mathrm{N} \\ =0.60\mathrm{N} \end{gathered}$$

and

$$\begin{gathered} n_2=40\%\mathrm{of}\mathrm{N} \\ =0.40\mathrm{N} \end{gathered}$$

The multiplicity of configuration B using equation (1) is given as follows:

$$\begin{gathered} W_B=\frac{N!}{\left(0.60N\right)!\left(0.40\times 50\right)!} \\ =\frac{50!}{\left(0.60\times 50\right)!\left(0.40\times 50\right)!} \\ =\frac{50!}{\left(30\right)!\left(20\right)!} \\ =\frac{\left(3.04\times {10}^{64}\right)}{\left(2.65\times {10}^{32}\right)\left(2.43\times {10}^{18}\right)} \\ =4.71\times {10}^{13} \end{gathered}$$

Hence, the value of the multiplicity is$4.71\times {10}^{13}$

The ratio of the time spent in the configuration B and configurationA with all microstates being equally probable is given using equation (2), such that,

$$\begin{gathered} {\mathrm{f}}_{\mathrm{B}/\mathrm{A}}=\frac{4.71\times {10}^{13}}{1.26\times {10}^{14}} \\ =0.37 \end{gathered}$$ (3)

Hence, the value of the ratio is$0.37$

The configuration A contains an equal division of the molecule between the two halves of the box. Using the formula of equation (1), we can calculate the multiplicity for configurationas given: (For configuration A,${\mathrm{n}}_1=\frac{\mathrm{N}}{2}=50$and${\mathrm{n}}_2=\frac{\mathrm{N}}{2}=50$)

$$\begin{gathered} W_A=\frac{N!}{\left({\displaystyle \frac{N}{2}}\right)!\left({\displaystyle \frac{N}{2}}\right)!} \\ =\frac{100!}{50!50!} \\ =\frac{9.33\times {10}^{157}}{\left(3.04\times {10}^{64}\right)\times \left(3.04\times {10}^{64}\right)} \\ =1.01\times {10}^{29} \end{gathered}$$

Hence, the value of the multiplicity is$1.01\times {10}^{29}$

The configuration B contains 60% of the molecule in the left half of the box and 40% in the right half of the box. That is given as:

$$\begin{gathered} {\mathrm{n}}_1=60\%\mathrm{of}\mathrm{N} \\ =0.60\mathrm{N} \end{gathered}$$

and

$$\begin{gathered} {\mathrm{n}}_2=40\%\mathrm{of}\mathrm{N} \\ =0.40\mathrm{N} \end{gathered}$$

The multiplicity of configuration B using equation (1) is given as follows:

$$\begin{gathered} W_B=\frac{N!}{\left(0.60N\right)!\left(0.40N\right)!} \\ =\frac{100!}{\left(0.60\times 100\right)!\left(0.40\times 100\right)!} \\ =\frac{100!}{\left(60\right)!\left(40\right)!} \\ =\frac{9.33\times {10}^{157}}{\left(8.32\times {10}^{81}\right)\left(8.15\times {10}^{47}\right)} \\ =\frac{\left(9.33\times {10}^{29}\right)}{8.32\times {10}^{28}} \\ =1.37\times {10}^{28} \end{gathered}$$

Hence, the value of the multiplicity is $1.37\times {10}^{28}$

The ratio of the time spent in the configuration B and configuration A with all microstates being equally probable is given using equation (2) as:

$$\begin{gathered} {\mathrm{f}}_{\mathrm{B}/\mathrm{A}}=\frac{1.37\times {10}^{28}}{\left(1.01\times {10}^{29}\right)} \\ =0.14 \end{gathered}$$ (4)

Hence, the value of the ratio is 0.14

The configuration A contains an equal division of the molecule between the two halves of the box. Using the formula, we can calculate the multiplicity for configuration A using equation (1) as given: (For configuration A,${\mathrm{n}}_1=\frac{\mathrm{N}}{2}=100$and${\mathrm{n}}_2=\frac{\mathrm{N}}{2}=100$)

$$\begin{gathered} W_A=\frac{N!}{\left({\displaystyle \frac{N}{2}}\right)!\left({\displaystyle \frac{N}{2}}\right)!} \\ =\frac{200!}{100!100!} \\ =\frac{7.88\times {10}^{157}}{\left(9.33\times {10}^{157}\right)\times \left(9.33\times {10}^{157}\right)} \\ =\frac{7.88}{9.33\times 9.33}\times {10}^{60} \\ =9.05\times {10}^{58} \end{gathered}$$

Hence, the value of multiplicity is$9.05\times {10}^{58}$

The configuration B contains 60% of the molecule in the left half of the box and 40% in the right half of the box. That is given as:

$$\begin{gathered} {\mathrm{n}}_1=60\%\mathrm{of}\mathrm{N} \\ =0.60\mathrm{N} \end{gathered}$$

and

$$\begin{gathered} {\mathrm{n}}_2=40\%\mathrm{of}\mathrm{N} \\ =0.40\mathrm{N} \end{gathered}$$

The multiplicity of configuration B using equation (1) is given as follows:

$$\begin{gathered} W_B=\frac{N!}{\left(0.60N\right)!\left(0.40N\right)!} \\ =\frac{200!}{\left(0.60\times 200\right)!\left(0.40\times 00\right)!} \\ =\frac{200!}{\left(120\right)!\left(80\right)!} \\ =\frac{7.88\times {10}^{374}}{\left(6.68\times {10}^{198}\right)\left(7.14\times {10}^{118}\right)} \\ =\frac{\left(9.33\right)}{6.68\times 715}\times {10}^{58} \\ =1.64\times {10}^{57} \end{gathered}$$

Hence, the value of multiplicity is$1.64\times {10}^{57}$

The ratio of the time spent in the configuration B and configuration A with all microstates being equally probable is given using equation (2) as:

$$\begin{gathered} f_{B/A}=\frac{1.64\times {10}^{57}}{9.05\times {10}^{58}} \\ =0.018 \end{gathered}$$ (5)

Hence, the value of the ratio is 0.018

From equations (3), (4), and (5), we can conclude that as increases then the ratio of configuration to decreases, i.e., f decreases.