Chapter 1, Problem 1B.1P

Interpretation Introduction

Interpretation: The distribution of molecular velocities at $40\text{K}$ and $100\text{K}$ has to be calculated.  The validation of the theoretical prediction for a one-dimensional system for the given low pressure, collision free system has to be stated.

Concept introduction: The fraction of molecules that have the speeds over the range $v+dv$ is directly proportional to the width of the range, which is written as $f\left(v\right)dv$ , where $f\left(v\right)$ is called the distribution of speeds.

Answer to Problem 1B.1P

The values of $k$ and $I$ at $40\text{K}$ and different frequency are shown below.

    $\begin{array}{l}v\left(\text{Hz}\right)I\left(40\text{K}\right)kI\\ \begin{array}{rr}\hfill 20& \hfill 0.846\\ \hfill 40& \hfill 0.513\\ \hfill 80& \hfill 0.069\end{array}\begin{array}{r}\hfill 6.302361\\ \hfill 6.231897\\ \hfill 5.92696\end{array}\begin{array}{r}\hfill 0.846\\ \hfill 0.513\\ \hfill 0.069\end{array}\end{array}$

    $\begin{array}{rr}\hfill 100& \hfill 0.015\\ \hfill 120& \hfill 0.002\end{array}\begin{array}{rr}\hfill 5.587071& \hfill 0.015\\ \hfill 4.475205& \hfill 0.002\end{array}$

The values of $k$ and $I$ at $100\text{K}$ and different frequency are shown below.

    $\begin{array}{l}v\left(\text{Hz}\right)I\left(40\text{K}\right)kI\\ \begin{array}{rr}\hfill 20& \hfill 0.846\\ \hfill 40& \hfill 0.513\\ \hfill 80& \hfill 0.069\end{array}\begin{array}{r}\hfill 6.302361\\ \hfill 6.231897\\ \hfill 5.92696\end{array}\begin{array}{r}\hfill 0.846\\ \hfill 0.513\\ \hfill 0.069\end{array}\end{array}$

    $\begin{array}{rr}\hfill 100& \hfill 0.119\\ \hfill 120& \hfill 0.057\end{array}\begin{array}{rr}\hfill 6.073611& \hfill 0.119\\ \hfill 5.960063& \hfill 0.057\end{array}$

The values of calculated intensity correspond to the given values of intensity.

Explanation of Solution

The expression for the distribution of molecular velocity in the x-direction is,

    $f\left(v\right)={\left(\frac{M}{2\pi RT}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{e}^{\raisebox{1ex}{$-M{v}_{\text{x}}^2$}\!\left/ \!\raisebox{-1ex}{$2RT$}\right.}$        (1)

Where,

$M$ is the molar mass.

$R$ is the gas constant.

$T$ is the temperature.

$v_{\text{x}}$ is the speed.

The expression for the time required to displace the slots by $2°$ between neighbours is,

    $t=\left(\frac{2°}{360°}\right)\times \frac{1}{v}$

Where,

$v$ is the rotation rate.

$t$ is the time.

The expression to calculate the velocity in the x-direction with disk separation of $1\text{cm}$ is,

    $v_{\text{x}}=\frac{\text{Disk}\text{Separation}}{\text{Time}}$

Substitute the values of time and disk separation in the above equation.

    $\begin{array}{c}{v}_{\text{x}}=\frac{1\text{cm}}{\left(\frac{2°}{360°}\right)\times \frac{1}{v}}\\ =\left(180\times v\right)\text{cm}{\text{s}}^{-1}\end{array}$

The relative intensity is directly proportional to velocity distribution of the molecules.  Substitute the value of $v_{\text{x}}^2,R,$$M=83.80\text{g}$ to determine the exponential term of equation (1).

    $\begin{array}{c}\frac{M{v}_{\text{x}}^2}{2RT}=\frac{83.80\times {10}^{-3}\text{kg}\times {\left(180\times v\times {10}^{-2}\text{m}{\text{s}}^{-1}\right)}^2}{2\times 8.314{\text{JK}}^{-1}{\text{mol}}^{-1}\times T/\text{K}}\\ =\frac{2715120\times {10}^{-1}\times {\left(v/{\text{s}}^{-1}\right)}^2}{16.628\times T/\text{K}}\\ =\frac{0.0163\times {\left(v/{\text{s}}^{-1}\right)}^2}{T/\text{K}}\end{array}$

Substitute the exponential term in equation (1) so as to obtain the formula for relative intensity.

    $I\propto {\left(T/\text{K}\right)}^{-1/2}{e}^{-\frac{0.0163\times {\left(v/{\text{s}}^{-1}\right)}^2}{T/\text{K}}}$

    $I=k{\left(T/\text{K}\right)}^{-1/2}{e}^{-\frac{0.0163\times {\left(v/{\text{s}}^{-1}\right)}^2}{T/\text{K}}}$        (2)

    $k=\frac{I}{{\left(T/\text{K}\right)}^{-1/2}{e}^{-\frac{0.0163\times {\left(v/{\text{s}}^{-1}\right)}^2}{T/\text{K}}}}$        (3)

Substitute the values of $T$, $I$ and $v$ in equation (3) to calculate $k$.  When the temperature is $40\text{K}$ and $v$ is $20\text{Hz}$.

    $\begin{array}{c}k=\frac{0.846}{{\left(\text{40}\text{K}\right)}^{-1/2}{e}^{-\frac{0.0163\times {\left(20{\text{s}}^{-1}\right)}^2}{40\text{K}}}}\\ k=6.302361\end{array}$

Hence, the value of $k$ at $40\text{K}$ and $20\text{Hz}$ is $6.302361$.

Substitute the value of $k$ in equation (2)

    $\begin{array}{c}I=6.302361{\left(\text{40}\text{K}\right)}^{-1/2}{e}^{-\frac{0.0163\times {\left(20{\text{s}}^{-1}\right)}^2}{40\text{K}}}\\ =0.846\end{array}$

Hence, the value of $I$ at $40\text{K}$ and $20\text{Hz}$ is $0.846$.

Similarly, the values of $k$ and $I$ at $40\text{K}$ and different frequency are shown below.

    $\begin{array}{l}v\left(\text{Hz}\right)I\left(40\text{K}\right)kI\\ \begin{array}{rr}\hfill 20& \hfill 0.846\\ \hfill 40& \hfill 0.513\\ \hfill 80& \hfill 0.069\end{array}\begin{array}{r}\hfill 6.302361\\ \hfill 6.231897\\ \hfill 5.92696\end{array}\begin{array}{r}\hfill 0.846\\ \hfill 0.513\\ \hfill 0.069\end{array}\end{array}$

    $\begin{array}{rr}\hfill 100& \hfill 0.015\\ \hfill 120& \hfill 0.002\end{array}\begin{array}{rr}\hfill 5.587071& \hfill 0.015\\ \hfill 4.475205& \hfill 0.002\end{array}$

The values of $k$ and $I$ at $100\text{K}$ and different frequency are shown below.

    $\begin{array}{l}v\left(\text{Hz}\right)I\left(40\text{K}\right)kI\\ \begin{array}{rr}\hfill 20& \hfill 0.846\\ \hfill 40& \hfill 0.513\\ \hfill 80& \hfill 0.069\end{array}\begin{array}{r}\hfill 6.302361\\ \hfill 6.231897\\ \hfill 5.92696\end{array}\begin{array}{r}\hfill 0.846\\ \hfill 0.513\\ \hfill 0.069\end{array}\end{array}$

    $\begin{array}{rr}\hfill 100& \hfill 0.119\\ \hfill 120& \hfill 0.057\end{array}\begin{array}{rr}\hfill 6.073611& \hfill 0.119\\ \hfill 5.960063& \hfill 0.057\end{array}$

Thus, the values of calculated intensity correspond to the given values of intensity.