Question

The standard entropy values $\left(S°\right)$for ${\mathit{H}}_{\mathbf{2}}\mathit{O}\left(l\right)$and ${\mathit{H}}_{\mathbf{2}}\mathit{O}\left(g\right)$are $\mathbf{70}\mathit{J}{\mathit{K}}^{\mathbf{-}\mathbf{1}}\mathit{m}\mathit{o}{\mathit{l}}^{\mathbf{-}\mathbf{1}}$ and $\mathbf{189}\mathit{J}{\mathit{K}}^{\mathbf{-}\mathbf{1}}\mathit{m}\mathit{o}{\mathit{l}}^{\mathbf{-}\mathbf{1}}$respectively. Calculate the ratio of ${\mathit{\Omega }}_{\mathbf{g}}$to ${\mathit{\Omega }}_{\mathbf{l}}$for water using Boltzmann's equation. (See Exercise 22)

Short answer

The ratio of ${\mathit{\Omega }}_{\mathbf{g}}$to ${\mathit{\Omega }}_{\mathbf{f}}$for water using Boltzmann's equation is $\mathbf{1}\mathbf{.}\mathbf{6}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}$.

Step-by-step solution

Introduction to the Concept

The Boltzmann constant $\left({\text{k}}_{\text{B}}\right)$ describes the relationship between temperature and energy. It's an essential tool in thermodynamics, the study of heat and how it interacts with other forms of energy.

Step 2: Solution Explanation

The equation below employs molecular entropy values instead of molar entropy values.

$\mathit{S}\mathbf{=}\mathit{k}\mathit{l}\mathit{n}\mathit{\Omega }$

S stands for molar entropy.

k is the Boltzmann constant

$\Omega$is the number of different ways a state can exist.

To get molar entropy, multiply the right side by Avogadro's number$\left(N_A\right)$.

$\mathit{S}\mathbf{=}{\mathit{N}}_{\mathbf{A}}\mathbf{\times }\mathit{k}\mathbf{\times }\mathit{l}\mathit{n}\mathit{\Omega }$

As shown below, the entropy values are subtractedfrom one another:

$S_g-S_l=\left(N_A\times k\times ln{\Omega }_g\right)-\left(N_A\times k\times ln{\Omega }_l\right)$

$S_g-S_l=N_A\times k\times ln\left(\frac{{\Omega }_g}{{\Omega }_l}\right)$

By solving the ratio,

$\frac{S_g-S_l}{N_A\times k}=ln\left(\frac{{\Omega }_g}{{\Omega }_l}\right)$

$\left(\frac{{\Omega }_g}{{\Omega }_l}\right)=e^{\frac{S_g-S_l}{N_A\times k}}$

Let’s substitute the given values,

$\left(\frac{{\Omega }_g}{{\Omega }_l}\right)=e^{\frac{\left(189-70\right)J{K}^{-1}mo{l}^{-1}}{\left(6.023\times {10}^{23}{\text{mol}}^{-1}\right)\times \left(1.38\times {10}^{-23}J{K}^{-1}\right)}}$

$\left(\frac{{\Omega }_g}{{\Omega }_l}\right)=e^{14.3}$

$\left(\frac{{\Omega }_g}{{\Omega }_l}\right)=1.6\times {10}^6$

Therefore the ratio is$\mathbf{1}\mathbf{.}\mathbf{6}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}$