Question

Entropy can be calculated by a relationship proposed by Ludwig Boltzmann:

$\mathit{S}\mathbf{=}{\mathit{k}}_{\mathbf{B}}\mathit{I}\mathit{n}\mathit{\Omega }$

where ${\mathit{k}}_{\mathbf{B}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{38}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{23}}\mathit{J}\mathbf{/}\mathit{K}$ and $\mathit{\Omega }$ is the number of ways a particular state can be obtained. (This equation is engraved on Boltzmann's tombstone.) Calculate role="math" localid="1658744937449" $\mathit{S}$ for the three arrangements of particles in Table 10.2.

Short answer

The Entropy for Arrangement I is 0.

The Entropy for Arrangement II is $\mathbf{1}\mathbf{.}\mathbf{91}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{23}}\mathit{J}{\mathit{K}}^{\mathbf{-}\mathbf{1}}$

The Entropy for Arrangement III is$\mathbf{2}\mathbf{.}\mathbf{47}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{23}}\mathit{J}{\mathit{K}}^{\mathbf{-}\mathbf{1}}$

Step-by-step solution

Step 1: Introduction to the Concept

Theratio of thermal energy to the temperature at which work cannot be done is known as entropy. It's also known as the measure of a system's molecular disarray. It has a lot of properties and state functions.

The number of microstates for a system is connected to entropy. Microstate is defined as the number of possible arrangements for the system.

The following is the relationship between entropy and probability:

$\mathit{S}\mathbf{=}{\mathit{k}}_{\mathbf{B}}\mathit{I}\mathit{n}\mathit{\Omega }$

Where ${\mathit{k}}_{\mathbf{B}}$is the Boltzmann constant ( $1.38\times {10}^{-23}J{K}^{-1}$)

$\mathit{\Omega }$ isthe total number of microstates in a given configuration.

Step 2: Determination of first arrangement

Let’s calculate the entropy for three arrangements as follows,

1^st Arrangement:

$S=k_{B}In\Omega$

Let’s substitute $\Omega =1$, and $k_B=1.38\times {10}^{-23}J{K}^{-1}$

$S=1.38\times {10}^{-23}J{K}^{-1}In\left(1\right)$

$S=0$

Therefore the entropy for theArrangement Iis 0.

Step 3: Determination of second arrangement

2^nd Arrangement:

$S=k_{B}In\Omega$

Let’s substitute $\Omega =4$, and $k_B=1.38\times {10}^{-23}J{K}^{-1}$

$S=1.38\times {10}^{-23}J{K}^{-1}In\left(4\right)$

$S=1.91\times {10}^{-23}J{K}^{-1}$

Therefore the entropy for the Arrangement II is $\mathbf{1}\mathbf{.}\mathbf{91}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{23}}\mathit{J}{\mathit{K}}^{\mathbf{-}\mathbf{1}}$

Step 4: Determination of third arrangement

3^rd Arrangement:

$S=k_{B}In\Omega$

Let’s substitute $\Omega =6$, and $k_B=1.38\times {10}^{-23}J{K}^{-1}$

$S=1.38\times {10}^{-23}J{K}^{-1}In\left(6\right)$

$S=2.47\times {10}^{-23}J{K}^{-1}$

Therefore the entropy for theArrangement III is $\mathbf{2}\mathbf{.}\mathbf{47}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{23}}\mathit{J}{\mathit{K}}^{\mathbf{-}\mathbf{1}}$.