91Ó°ÊÓ

The system A has three particles in an insulated box.

The system B has five particles in an insulated box.

The gas molecules inside a box can be distributed in many different ways. The number of microstates in a configuration is called its multiplicity. For a system of N molecules that can be distributed between two regions, we can find multiplicity. Then by using Boltzmann’s entropy equation, we can calculate the greatest entropy in both systems.

Formula:

The multiplicity of the central configuration, $W_i=\frac{N!}{n_1!n_2!}$ (1)

where, $n_1$is the number of molecules in one region and $n_2$in another region.

The entropy of the central configuration, S = klnW (2)

where k is the Boltzmann’s constant having value $k=1.38×{10}^{-23}J/K$

The least multiplicity means that all the particles are in one region. For the system A , the total particles are N = 3

Thus, the multiplicity of configuration A using equation (1) is given as follows:

$$\begin{gathered} W_A=\frac{3!}{3!0!} \\ =1 \end{gathered}$$

Hence, the least multiplicity of A is 1

Similarly, for the system, the total number of particles is N = 5

Thus, the multiplicity of the configuration B using equation (1) is given as follows:

$$\begin{gathered} W_B=\frac{5!}{5!0!} \\ =1 \end{gathered}$$

Hence, the least multiplicity of configuration B is 1

The microstates are equally probable, so they are the most likely configuration for a system A is two particles in one region and one particle in another region. Therefore, the greatest multiplicity for the system A using equation (1) is given as:

$$\begin{gathered} W_A=\frac{3!}{2!1!} \\ =3 \end{gathered}$$

Hence, the greatest multiplicity of A is 3

Similarly, for the system B , the most likely configuration for 5 particles is three particles in one region and two particles in another region. So the greatest multiplicity is given using equation (1) as:

$$\begin{gathered} W_B=\frac{5!}{3!2!} \\ =\frac{5×4×3×2×1}{3×2×1×2×1} \\ =10 \end{gathered}$$

Hence, the greatest multiplicity of B is 10

The multiplicityof a configuration of a system and the entropy of the system in that configuration are related by Boltzmann’s entropy equation and thus, we can calculate the greatest entropy for the systemusing the greatest value of multiplicity,$W_A=3$in equation (2) as:

$$\begin{gathered} S_A=\left(1.38×{10}^{-23}J/K\right)\ln 3 \\ =1.38×{10}^{-23}J/K×1.098 \\ =1.5×{10}^{-23}J/K \end{gathered}$$

Hence, the value of the greatest entropy of A is $1.5×{10}^{-23}J/K$

The multiplicityof a configuration of a system and the entropy of the system in that configuration are related by Boltzmann’s entropy equation and thus, we can calculate the greatest entropy for the systemusing the greatest value of multiplicity,$W_B=10$in equation (2) as:

$$\begin{gathered} S_B=\left(1.38×{10}^{-23}J/K\right)×\ln 10 \\ =1.38×{10}^{-23}J/K×2.30 \\ =3.2×{10}^{-23}J/K \end{gathered}$$

Hence, the value of the greatest entropy of B is $3.2×{10}^{-23}J/K$