Question

System Aof three particles and system B of five particles are in insulated boxes like that in Fig. 20-17. What is the least multiplicity W of (a) system A and (b) system B ? What is the greatest multiplicity Wof (c) A and (d) B ? What is the greatest entropy of (e) A and (f) B?

Short answer

a) The least multiplicity of A is 1

b) The least multiplicity of B is 1

c) The greatest multiplicity of A is 3

d) The greatest multiplicity of B is 10

e) The greatest entropy of A is $1.5\times {10}^{-23}J/K$

f) The greatest entropy of B is $3.2\times {10}^{-23}J/K$

Step-by-step solution

The given data

The system A has three particles in an insulated box.

The system B has five particles in an insulated box.

Understanding the concept of central configurations

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 $k=1.38\times {10}^{-23}J/K$

a) Calculation of least multiplicity of configuration A

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

b) Calculation of least multiplicity of configuration B

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

c) Calculation of greatest multiplicity of configuration A

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

d) Calculation of greatest multiplicity of configuration B

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:

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

Hence, the greatest multiplicity of B is 10

e) Calculation of greatest entropy of configuration A

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:

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

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

f) Calculation of greatest entropy of configuration B

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:

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

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