91影视

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.

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:

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

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

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

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

There are possible choices for each molecule 鈥� 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:

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

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

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:

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

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

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

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

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

There are 2 possible choices for each molecule 鈥� 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:

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

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

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:

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

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

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

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

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

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

There are 2 possible choices for each molecule 鈥� 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:

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

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

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:

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

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

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.