Question

Four particles are in the insulated box of Fig. 20-17. What are (a) the least multiplicity, (b) the greatest multiplicity, (c) the least entropy, and (d) the greatest entropy of the four-particle system?

Short answer

1. The least multiplicity is 1
2. The greatest multiplicity is 6
3. The least entropy is 0 J/K
4. The greatest entropy of the four-particle system is$2.47\times {10}^{-23}\mathrm{J}/\mathrm{K}$

Step-by-step solution

The given data

Four particles are in an insulated box.

Understanding the concept of Boltzmann’s entropy equation

We can use the concept of Boltzmann’s entropy equation; we can get the direct relation of the multiplicity and entropy of the microstates.

Formula:

The entropy of a microstate using Boltzmann-entropy equation, $\mathrm{S}=\mathrm{klnW}$ (1)

(a) Calculations of the least multiplicity

By mathematically expanding the given terms for four particles in the insulated box, we get that

${\left(1+\mathrm{x}\right)}^4=1+4\mathrm{x}+6{\mathrm{x}}^2+4{\mathrm{x}}^3+{\mathrm{x}}^4.......................\left(2\right)$

The coefficients correspond to the multiplicities.

From above equation, it can be seen that the least multiplicity is given by 1.

Thus, the least multiplicity is 1

(b) Calculation of the greatest multiplicity

From equation (2), the greatest multiplicity of the equation can be seen as 6.

Hence, the greatest multiplicity is 6

(c) Calculation of the least entropy

Using equation (1) and the value of the least multiplicity W = 1 , we can get the least entropy as given (k=Boltzmann’s constant):

S = kln(1)

=0

The least entropy is s =0 J/K

(d) Calculation of the greatest entropy for the four particles system

Using equation (1) and the value of the greatest multiplicity W = 6 , we can get the greatest entropy as given (k=Boltzmann’s constant):

$$\begin{gathered} \mathrm{S}=(1.38\times {10}^{-23}\mathrm{J}/\mathrm{K})\ln \left(6\right) \\ =2.47\times {10}^{-23}\mathrm{J}/\mathrm{K} \end{gathered}$$

Hence, the value of the greatest entropy is $2.47\times {10}^{-23}\mathrm{J}/\mathrm{K}$