Question

A box contains Nidentical gas molecules equally divided between its two halves. For N = 50 , (a) What is the multiplicity of the central configuration, (b) What are the total number of microstates, and (c) What is the percentage of the time the system spends in the central configuration? For N = 100 , (d) What is of the central configuration, (e) What are the total number of microstates, and (f) What is the percentage of the time the system spends in the central configuration? For N = 200, (g) What is of the central configuration, (h) What are the total number of microstates, and (i) What is the percentage of the time the system spends in the central configuration? (j) Does the time spent in the central configuration increase or decrease with an increase in N?

Short answer

For N = 50

1. Multiplicity W of central configuration is$1.26\times {10}^{14}$
2. The total number of microstates is $1.13\times {10}^{15}$
3. The percentage of time spent in central configuration is 11.3 %

For N = 100

1. Multiplicity W of central configuration is$1.01\times {10}^{29}$
2. The total number of microstates is $1.27\times {10}^{30}$
3. The percentage of time spent in central configuration is 7.97 %

For N = 200

1. Multiplicity W of central configuration is $9.25\times {10}^{58}$
2. The total number of microstates is$1.61\times {10}^{60}$
3. The percentage of time spent in central configuration is$5.6\%$
4. Time spent in central configuration decreases as N increases.

Step-by-step solution

The given data

For the given three conditions, the numbers of molecules (N) are 50, 100, and 200 respectively.

The gas molecules are equally divided into two halves.

Understanding the concept of the central configuration of molecules

With the help of the multiplicity of the central configuration formula and the given number of molecules present in the gas, we can find the total number of microstates as well as the percentage of time spent in a microstate.

Formulae:

The multiplicity of the molecules:

$\mathrm{W}=\left(\frac{\mathrm{N}}{2},\mathrm{N}\right)=\frac{\mathrm{N}!}{{\mathrm{n}}_1!{\mathrm{n}}_2!},$

where N is the number of molecules,${\mathrm{n}}_1$is the number of molecules in one half,is the number of molecules in the other half

Available states of N-particle system, role="math" localid="1661575987114" ${\mathrm{n}}_{\mathrm{total}}=2^{\mathrm{N}}$=2

Percentage of time spent in the central configuration:

$\mathrm{p}\left(\frac{\mathrm{N}}{2},\mathrm{N}\right)\times 100=\frac{\mathrm{W}}{{\mathrm{N}}_{\mathrm{total}}}\times 100$

(a) Calculation for the multiplicity of  N =50

From the equation (1) of multiplicity, we get that

$$\begin{gathered} \mathrm{W}\left(25,50\right)=\frac{50!}{25!(50-25)!} \\ =1.26\times {10}^{14} \end{gathered}$$

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

(b) Calculation of the total number of microstates of N = 50

There are possible choices for each molecule – it can be either on side 1 or side 2 of the box. There are numbers of molecules, so the total number of states using equation (2) is given by:

$$\begin{gathered} {\mathrm{N}}_{\mathrm{total}}=2^{50} \\ =1.13\times {10}^{15} \end{gathered}$$

Hence, the value of microstates is$1.13\times {10}^{15}$

(c) Calculation for the percentage of time spent in a molecule of  N = 50

The percentage of time in question is equal to the probability for the system to be in the central configuration using equation (3) is given by:

$$\begin{gathered} \mathrm{p}\left(\frac{\mathrm{N}}{2},\mathrm{N}\right)=\frac{1.26\times {10}^{14}}{1.13\times {10}^{15}} \\ \mathrm{p}\%=\left(0.113\right)\times 100\% \\ =11.3\% \end{gathered}$$

Hence, the value of the percentage of time is 11.3 %

(d) Calculation for the multiplicity of N = 100

From the equation (1) of multiplicity, we get that

$$\begin{gathered} \mathrm{W}(50,100)=\frac{100!}{50!(100-50)!} \\ =\frac{9.332\times {10}^{157}}{9.247\times {10}^{128}} \\ =1.01\times {10}^{29} \end{gathered}$$

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

(e) Calculation of the total number of microstates of  N = 100

There are 2 possible choices for each molecule – it can be either on side 1 or side 2 of the box. There are numbers of molecules, so the total number of states using equation (2) is given by:

$$\begin{gathered} {\mathrm{N}}_{\mathrm{toatal}}=2^{100} \\ =1.27\times {10}^{30} \end{gathered}$$

Hence, the value of microstates is $1.27\times {10}^{30}$

(f) Calculation for a percentage of time spent in a molecule of  N = 100

The percentage of time in question is equal to the probability for the system to be in the central configuration using equation (3) is given by:

$$\begin{gathered} \mathrm{p}\left(\frac{\mathrm{N}}{2},\mathrm{N}\right)=\frac{1.01\times {10}^{29}}{1.27\times {10}^{30}} \\ =0.797 \end{gathered}$$

$$\begin{gathered} \mathrm{p}\%=\left(0.797\right)\times 100 \\ =7.97\% \end{gathered}$$

Hence, the value of the percentage of time spent is 7.97 %

(g) Calculation for a multiplicity of N = 200

From the equation (1) of multiplicity, we get that

$$\begin{gathered} \mathrm{W}(100,200)=\frac{200!}{100!(200-100)!} \\ =\frac{7.88\times {10}^{374}}{87.04\times {10}^{314}} \\ =9.25\times {10}^{58} \end{gathered}$$

Hence, the value of multiplicity is $9.25\times {10}^{58}$

(h) Calculation of the total number of microstates of  N = 200

There are 2 possible choices for each molecule – it can be either on side 1 or side 2 of the box. There are 200 numbers of molecules, so the total number of states using equation (2) is given by:

$$\begin{gathered} {\mathrm{N}}_{\mathrm{total}}=2^{200} \\ =1.61\times {10}^{60} \end{gathered}$$

Hence, the value of microstates is$1.61\times {10}^{60}$

(i) Calculation for the percentage of time spent in a molecule of  N = 200

The percentage of time in question is equal to the probability for the system to be in the central configuration using equation (3) is given by:

$$\begin{gathered} \mathrm{p}\left(\frac{\mathrm{N}}{2},\mathrm{N}\right)=\frac{9.1\times {10}^{58}}{1.606\times {10}^{60}} \\ =5.6\times {10}^{-2} \\ \mathrm{p}\%=\left(5.6\times {10}^{-2}\right)\times 100 \\ =5.6\% \end{gathered}$$

Hence, the value of the percentage of time spent is 5.6 %

(j) Checking the behavior of time spent in comparison to N

From part b, we can note that the time spent in the central configuration is nothing but the probability of the system being in the central configuration. As N increases, the probability decreases. Therefore, the time spent in the central configuration decreases.