Chapter 12, Problem 12.1P

(a)

Interpretation Introduction

Interpretation:

With the given data, the ratio of the populations of ${\text{CO}}_{\text{2}}$ molecules with $\text{J}\text{=}4$ and $\text{J}\text{=}2$ has to be calculated.

Concept Introduction:

The ratio of the population can be found using the Boltzmann distribution formula:

$\begin{array}{l}\frac{{\text{N}}_{\text{2}}}{{\text{N}}_{\text{1}}}\text{=}\frac{\raisebox{1ex}{${\text{Ng}}_{\text{2}}{\text{e}}^{\text{-}\raisebox{1ex}{${\in }_{\text{2}}$}\!\left/ \!\raisebox{-1ex}{$\text{kT}$}\right.}$}\!\left/ \!\raisebox{-1ex}{$\text{q}$}\right.}{\raisebox{1ex}{${\text{Ng}}_{\text{1}}{\text{e}}^{\text{-}\raisebox{1ex}{${\in }_{\text{1}}$}\!\left/ \!\raisebox{-1ex}{$\text{kT}$}\right.}$}\!\left/ \!\raisebox{-1ex}{$\text{q}$}\right.}\\ \text{Cancelling}\text{N}\text{and}\text{q}\\ \frac{{\text{N}}_{\text{2}}}{{\text{N}}_{\text{1}}}\text{=}\frac{{\text{g}}_{\text{2}}}{{\text{g}}_{\text{1}}}{\text{e}}^{\raisebox{1ex}{$-\left({\in }_{\text{2}}\text{-}{\in }_1\right)$}\!\left/ \!\raisebox{-1ex}{$\text{kT}$}\right.}\end{array}$

$\begin{array}{l}{\text{N}}_{\text{2}}\text{is}\text{the}\text{final}\text{population}\\ {\text{N}}_{\text{1}}\text{is}\text{the}\text{initial}\text{population}\\ {\text{g}}_{\text{2}}\text{degeneracy}\text{for}\text{the}\text{quantum}\text{level}\text{J}\text{=}\text{2}\\ {\text{g}}_{\text{1}}\text{degeneracy}\text{for}\text{the}\text{quantum}\text{level}\text{J}\text{=}\text{1}\end{array}$

$\begin{array}{l}{\in }_{\text{1}}=\text{Energy}\text{of}\text{the}\text{quantum}\text{level}\text{J}\text{=1}\\ {\in }_2=\text{Energy}\text{of}\text{the}\text{quantum}\text{level}\text{J}\text{=2}\\ \text{k=}\text{Boltzmann}\text{constant}\text{=}\text{1}\text{.381}{\text{×10}}^{\text{-23}}{\text{JK}}^{\text{-1}}\\ \text{T}\text{=}\text{Temperature}\end{array}$

Answer to Problem 12.1P

With the given data, the ratio of the populations of ${\text{CO}}_{\text{2}}$ molecules with $\text{J}\text{=}4$ and $\text{J}\text{=}2$ has been calculated as $1.7531$.

Explanation of Solution

Calculating the degeneracy and energy for the given quantum states $J=0$and$J=5$:

The given molecule is ${\text{CO}}_{\text{2}}$. It is a linear molecule. So, it’s degeneracy for any of its quantum state $\text{J}$ can be calculated as:

$\begin{array}{c}{\text{g}}_{\text{J}}\text{=}\text{2J}\text{+1}\\ \text{J}=\text{Quantum}\text{state}\\ {\text{g}}_{\text{J}}=\text{Degeneracy}\text{of}\text{the}\text{quantum}\text{state}\text{J}\end{array}$

For the quantum level or state $\text{J}\text{=}2$, the degeneracy is:

  $\begin{array}{c}{\text{g}}_2\text{=}\text{2}\left(2\right)\text{+1}\\ \text{=}\text{4}\text{+}\text{1}\\ \text{=}\text{ 5}\end{array}$

For the quantum level or state $\text{J}\text{=}4$, the degeneracy is:

  $\begin{array}{c}{\text{g}}_4\text{=}\text{2}\left(4\right)\text{+1}\\ \text{= 8}\text{+}\text{1}\\ \text{=}\text{ 9}\end{array}$

The energy for any of the quantum level $\text{J}$ can be calculated as:

${\in }_{\text{J}}=\text{hBJ}\left(\text{J+1}\right)$

  $\begin{array}{c}\text{B}\text{=}\text{Rotational}\text{constant}\\ \text{h}\text{=}{\text{Plank}}^{\text{'}}\text{s}\text{constant=}\text{6}\text{.626}{\text{×10}}^{\text{-34}}\text{J}\text{s}\\ \text{J}=\text{Quantum}\text{state}\\ {\text{g}}_{\text{J}}=\text{Degeneracy}\text{of}\text{the}\text{quantum}\text{state}\text{J}\end{array}$

  ${\in }_{\text{J}}=\text{Energy}\text{of}\text{the}\text{quantum}\text{state}\text{J}$

The energy for the quantum level $\text{J}\text{=}\text{2}$ can be calculated as:

  $\begin{array}{c}{\in }_2=\text{hB}\left(\text{2}\right)\left(\text{2+1}\right)\\ =2\text{hB}\times 3\\ =6\text{hB}\end{array}$

The energy for the quantum level $\text{J}\text{=}4$ can be calculated as:

$\begin{array}{c}{\in }_4=\text{hB}\left(4\right)\left(\text{4+1}\right)\\ =4\text{hB}\times 5\\ =20\text{hB}\end{array}$

The ratio of the population of the levels with $J=2$ and $J=4$ at the given temperature of $25{}^{o}C$ can be determined as shown here:

Converting temperature from ${}^{\text{o}}\text{C}$ to $\text{kelvin}$ scale:

$\begin{array}{l}\text{25}{}^{\text{o}}\text{C}\text{+}\text{273}\\ \text{=}\text{298}\text{K}\\ \therefore \text{T}\text{=}\text{298}\text{K}\end{array}$

Given: The rotational constant is $\text{11}\text{.70}\text{GHz}$.

$\begin{array}{c}\text{B}\text{=}\text{11}\text{.70}\text{GHz}\\ \text{=}\text{11}\text{.70}\text{×}{\text{10}}^{\text{9}}{\text{s}}^{\text{-1}}\because \text{GHz}={\text{10}}^{\text{9}}{\text{s}}^{\text{-1}}\end{array}$

$\begin{array}{c}\frac{{\text{N}}_4}{{\text{N}}_2}\text{=}\frac{{\text{g}}_4}{{\text{g}}_2}{\text{e}}^{\raisebox{1ex}{$-\left({\in }_4\text{-}{\in }_2\right)$}\!\left/ \!\raisebox{-1ex}{$\text{kT}$}\right.}\\ =\frac{9}{5}{\text{e}}^{\raisebox{1ex}{$-\left(\text{20hB-6hB}\right)$}\!\left/ \!\raisebox{-1ex}{$\text{1}\text{.381}{\text{×10}}^{\text{-23}}{\text{JK}}^{\text{-1}}\times \text{298}\text{K}$}\right.}\\ =\frac{9}{5}{\text{e}}^{\raisebox{1ex}{$-\left(\text{14hB}\right)$}\!\left/ \!\raisebox{-1ex}{$\text{1}\text{.381}{\text{×10}}^{\text{-23}}{\text{JK}}^{\text{-1}}\times \text{298}\text{K}$}\right.}\\ =\frac{9}{5}{\text{e}}^{\raisebox{1ex}{$-\left(14\times \text{6}\text{.626}{\text{×10}}^{\text{-34}}\text{J}\text{s}\times \text{11}\text{.70}\text{×}{\text{10}}^{\text{9}}{\text{s}}^{\text{-1}}\right)$}\!\left/ \!\raisebox{-1ex}{$\text{1}\text{.381}{\text{×10}}^{\text{-23}}{\text{JK}}^{\text{-1}}\times \text{298K}$}\right.}\end{array}$

  $\begin{array}{c}=\frac{9}{5}{\text{×e}}^{\text{-(0}\text{.0264)}}\\ \text{=}\frac{9}{5}\text{×0}\text{.9739}\\ =1.7531\end{array}$

Therefore, the ratio of the populations of ${\text{CO}}_{\text{2}}$ molecules with $\text{J}\text{=}4$ and $\text{J}\text{=}2$ has been calculated as $1.7531$.

(b)

Interpretation Introduction

Interpretation:

With the given data, the ratio of the populations of ${\text{CH}}_4$ molecules with $\text{J}\text{=}4$ and $\text{J}\text{=}2$ has to be calculated.

Concept Introduction:

The ratio of the population can be found using the Boltzmann distribution formula:

  $\begin{array}{l}\frac{{\text{N}}_{\text{2}}}{{\text{N}}_{\text{1}}}\text{=}\frac{\raisebox{1ex}{${\text{Ng}}_{\text{2}}{\text{e}}^{\text{-}\raisebox{1ex}{${\in }_{\text{2}}$}\!\left/ \!\raisebox{-1ex}{$\text{kT}$}\right.}$}\!\left/ \!\raisebox{-1ex}{$\text{q}$}\right.}{\raisebox{1ex}{${\text{Ng}}_{\text{1}}{\text{e}}^{\text{-}\raisebox{1ex}{${\in }_{\text{1}}$}\!\left/ \!\raisebox{-1ex}{$\text{kT}$}\right.}$}\!\left/ \!\raisebox{-1ex}{$\text{q}$}\right.}\\ \text{Cancelling}\text{N}\text{and}\text{q}\\ \frac{{\text{N}}_{\text{2}}}{{\text{N}}_{\text{1}}}\text{=}\frac{{\text{g}}_{\text{2}}}{{\text{g}}_{\text{1}}}{\text{e}}^{\raisebox{1ex}{$-\left({\in }_{\text{2}}\text{-}{\in }_1\right)$}\!\left/ \!\raisebox{-1ex}{$\text{kT}$}\right.}\end{array}$

$\begin{array}{l}{\text{N}}_{\text{2}}\text{is}\text{the}\text{final}\text{population}\\ {\text{N}}_{\text{1}}\text{is}\text{the}\text{initial}\text{population}\\ {\text{g}}_{\text{2}}\text{degeneracy}\text{for}\text{the}\text{quantum}\text{level}\text{J}\text{=}\text{2}\\ {\text{g}}_{\text{1}}\text{degeneracy}\text{for}\text{the}\text{quantum}\text{level}\text{J}\text{=}\text{1}\end{array}$

$\begin{array}{l}{\in }_{\text{1}}=\text{Energy}\text{of}\text{the}\text{quantum}\text{level}\text{J}\text{=1}\\ {\in }_2=\text{Energy}\text{of}\text{the}\text{quantum}\text{level}\text{J}\text{=2}\\ \text{k=}\text{Boltzmann}\text{constant}\text{=}\text{1}\text{.381}{\text{×10}}^{\text{-23}}{\text{JK}}^{\text{-1}}\\ \text{T}\text{=}\text{Temperature}\end{array}$

Answer to Problem 12.1P

With the given data, the ratio of the populations of ${\text{CH}}_4$ molecules with $\text{J}\text{=}4$ and $\text{J}\text{=}2$ has been calculated as $2.2743$

Explanation of Solution

The given molecule is ${\text{CH}}_4$. It is a spherical molecule. So, it’s degeneracy for of its any quantum state $\text{J}$ can be calculated as:

$\begin{array}{c}{\text{g}}_{\text{J}}\text{=}{\left(\text{2J}\text{+1}\right)}^2\\ \text{J}=\text{Quantum}\text{state}\\ {\text{g}}_{\text{J}}=\text{Degeneracy}\text{of}\text{the}\text{quantum}\text{state}\text{J}\end{array}$.

For the quantum level or state $\text{J}\text{=}2$, the degeneracy is:

$\begin{array}{c}{\text{g}}_2\text{=}{\left(\text{2}\left(2\right)\text{+1}\right)}^2\\ \text{=}{\left(\text{4}\text{+}\text{1}\right)}^2\\ \text{=}{\left(\text{5}\right)}^2\\ =25\end{array}$

For the quantum level or state $\text{J}\text{=}4$, the degeneracy is:

$\begin{array}{c}{\text{g}}_4\text{=}{\left(\text{2}\left(4\right)\text{+1}\right)}^2\\ \text{= }{\left(\text{8}\text{+}\text{1}\right)}^2\\ \text{=}{\left(\text{9}\right)}^2\\ =81\end{array}$

The energy for any of the quantum level $\text{J}$ can be calculated as:

${\in }_{\text{J}}=\text{hBJ}\left(\text{J+1}\right)$

$\begin{array}{c}\text{B}\text{=}\text{Rotational}\text{constant}\\ \text{h}\text{=}{\text{Plank}}^{\text{'}}\text{s}\text{constant=}\text{6}\text{.626}{\text{×10}}^{\text{-34}}\text{J}\text{s}\\ \text{J}=\text{Quantum}\text{state}\\ {\text{g}}_{\text{J}}=\text{Degeneracy}\text{of}\text{the}\text{quantum}\text{state}\text{J}\end{array}$

${\in }_{\text{J}}=\text{Energy}\text{of}\text{the}\text{quantum}\text{state}\text{J}$

The energy for the quantum level $\text{J}\text{=}\text{2}$ can be calculated as:

$\begin{array}{c}{\in }_2=\text{hB}\left(\text{2}\right)\left(\text{2+1}\right)\\ =2\text{hB}\times 3\\ =6\text{hB}\end{array}$

The energy for the quantum level $\text{J}\text{=}4$ can be calculated as:

$\begin{array}{c}{\in }_4=\text{hB}\left(4\right)\left(\text{4+1}\right)\\ =4\text{hB}\times 5\\ =20\text{hB}\end{array}$

The ratio of the population of the levels with $J=2$ and $J=4$ at the given temperature of $25{}^{o}C$ can be determined as shown here:

Converting temperature from ${}^{\text{o}}\text{C}$ to $\text{kelvin}$ scale:

$\begin{array}{l}\text{25}{}^{\text{o}}\text{C}\text{+}\text{273}\\ \text{=}\text{298}\text{K}\\ \therefore \text{T}\text{=}\text{298}\text{K}\end{array}$

Given: The rotational constant is $\text{157}\text{GHz}$.

$\begin{array}{c}\text{B}\text{=}\text{157}\text{GHz}\\ \text{=}\text{157}\text{×}{\text{10}}^{\text{9}}{\text{s}}^{\text{-1}}\because \text{GHz}={\text{10}}^{\text{9}}{\text{s}}^{\text{-1}}\end{array}$

$\begin{array}{c}\frac{{\text{N}}_4}{{\text{N}}_2}\text{=}\frac{{\text{g}}_4}{{\text{g}}_2}{\text{e}}^{\raisebox{1ex}{$-\left({\in }_4\text{-}{\in }_2\right)$}\!\left/ \!\raisebox{-1ex}{$\text{kT}$}\right.}\\ =\frac{81}{25}{\text{e}}^{\raisebox{1ex}{$-\left(\text{20hB-6hB}\right)$}\!\left/ \!\raisebox{-1ex}{$\text{1}\text{.381}{\text{×10}}^{\text{-23}}{\text{JK}}^{\text{-1}}\times \text{298}\text{K}$}\right.}\\ =\frac{81}{25}{\text{e}}^{\raisebox{1ex}{$-\left(\text{14hB}\right)$}\!\left/ \!\raisebox{-1ex}{$\text{1}\text{.381}{\text{×10}}^{\text{-23}}{\text{JK}}^{\text{-1}}\times \text{298}\text{K}$}\right.}\\ =\frac{81}{25}{\text{e}}^{\raisebox{1ex}{$-\left(14\times \text{6}\text{.626}{\text{×10}}^{\text{-34}}\text{J}\text{s}\times \text{157}\text{×}{\text{10}}^{\text{9}}{\text{s}}^{\text{-1}}\right)$}\!\left/ \!\raisebox{-1ex}{$\text{1}\text{.381}{\text{×10}}^{\text{-23}}{\text{JK}}^{\text{-1}}\times \text{298K}$}\right.}\end{array}$

$\begin{array}{c}=\frac{81}{25}{\text{×e}}^{\text{-(0}\text{.3539)}}\\ \text{=}\frac{81}{25}\text{×0}\text{.7019}\\ =2.2743\end{array}$

Therefore, the ratio of the populations of ${\text{CH}}_4$ molecules with $\text{J}\text{=}4$ and $\text{J}\text{=}2$ has been calculated as $2.2743$.