Chapter 1, Problem 1B.4P

(i)

Interpretation Introduction

Interpretation: The proportions of gas, according to the Maxwell-Boltzmann distribution have speeds more than the root mean square speed has to be calculated.

Concept introduction: Root mean square speed is the average of squares of speed in the square root.  This speed is represented as,

  $v_{\text{rms}}=\sqrt{\frac{3RT}{M}}$

Where,

$R$ is the gas constant.

$T$ is the temperature.

$M$ is the molar mass of the gas.

Answer to Problem 1B.4P

The proportion of molecules that have speed more than the mean square speed, $c$ is $\underset{\_}{0.39}$.

Explanation of Solution

The expression for the Maxwell-Boltzmann distribution is,

    $f\left(v\right)=4\pi {\left(\frac{M}{2\pi RT}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{v}^2{e}^{\raisebox{1ex}{$-M{v}^2$}\!\left/ \!\raisebox{-1ex}{$2RT$}\right.}$

The proportion of molecules that have speed less than the mean square speed, $c$ is,

    $P={\displaystyle \underset{0}{\overset{c}{\int }}}f\left(v\right)dv=4\pi {\left(\frac{m}{2\pi kT}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\displaystyle \underset{0}{\overset{c}{\int }}}{v}^2{e}^{\raisebox{1ex}{$-m{v}^2$}\!\left/ \!\raisebox{-1ex}{$2kT$}\right.}dv$

Let

  $a=\frac{m}{2kT}$

Define $x^2=a{v}^2,dv=a^{-1/2}dx$

  $P=-4\pi {\left(\frac{a}{\pi }\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{d}{da}\left\{\frac{1}{a^{1/2}}{\displaystyle \underset{0}{\overset{c{a}^{1/2}}{\int }}{e}^{-x^2}dx}\right\}$

    $P=-4\pi {\left(\frac{a}{\pi }\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left\{\frac{-1}{2}{\left(\frac{1}{a}\right)}^{3/2}{\displaystyle \underset{0}{\overset{c{a}^{1/2}}{\int }}{e}^{-x^2}dx}+{\left(\frac{1}{a}\right)}^{1/2}\frac{d}{da}{\displaystyle \underset{0}{\overset{c{a}^{1/2}}{\int }}{e}^{-x^2}dx}\right\}$        (1)

The identity used in the above expression is,

  ${\displaystyle \underset{0}{\overset{c{a}^{1/2}}{\int }}{e}^{-x^2}}dx=\left(\raisebox{1ex}{${\pi }^{1/2}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)erf\left(c{a}^{1/2}\right)$        (2)

  $\frac{d}{da}{\displaystyle \underset{0}{\overset{c{a}^{1/2}}{\int }}{e}^{-x^2}}dx=\left(\frac{dc{a}^{1/2}}{da}\right)\times e^{-c^{2}a}=\frac{1}{2}\left(\frac{c}{a^{1/2}}\right)e^{-c^{2}a}$        (3)

Substitute the equations (2) and (3) in equation (1).

    $P=-4\pi {\left(\frac{a}{\pi }\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left[\frac{-1}{2}{\left(\frac{1}{a}\right)}^{3/2}\left(\raisebox{1ex}{${\pi }^{1/2}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)erf\left(c{a}^{1/2}\right)+{\left(\frac{1}{a}\right)}^{1/2}\frac{1}{2}\left(\frac{c}{a^{1/2}}\right)e^{-c^{2}a}\right]$

    $P=\text{erf}\left(c{a}^{1/2}\right)-\left(\raisebox{1ex}{$2c{a}^{1/2}$}\!\left/ \!\raisebox{-1ex}{${\pi }^{1/2}$}\right.\right)e^{-c^{2}a}$        (4)

The value of $c$ is ${\left(\raisebox{1ex}{$3kT$}\!\left/ \!\raisebox{-1ex}{$m$}\right.\right)}^{1/2}$, thus,

  $c{a}^{1/2}={\left(\raisebox{1ex}{$3kT$}\!\left/ \!\raisebox{-1ex}{$m$}\right.\right)}^{1/2}\times {\left(\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$2kT$}\right.\right)}^{1/2}={\left(\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}^{1/2}$

Substitute the above values in equation (4).

    $\begin{array}{c}P=\text{erf}\left(\sqrt{\frac{3}{2}}\right)-{\left(\frac{6}{\pi }\right)}^{1/2}{e}^{-3/2}\\ =0.92-0.31\\ =0.61\end{array}$

Hence, the proportion of molecules that have speed more than the mean square speed, $c$ is $1-0.61=\underset{\_}{0.39}$.

(ii)

Interpretation Introduction

Interpretation: The proportions of gas, according to the Maxwell-Boltzmann distribution have speeds less than the root mean square speed has to be calculated.

Concept introduction: The compression factor is defined as the ratio of molar volume of a gas to the molar volume of a perfect gas.  This is represented by the formula given below as,

  $Z=\frac{V_{\text{m}}}{V_{\text{m}}^{\text{°}}}$

Answer to Problem 1B.4P

The proportion of molecules that have speed less than the mean square speed, $c$ is $\underset{\_}{0.61}$.

Explanation of Solution

The expression for the Maxwell-Boltzmann distribution is,

    $f\left(v\right)=4\pi {\left(\frac{M}{2\pi RT}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{v}^2{e}^{\raisebox{1ex}{$-M{v}^2$}\!\left/ \!\raisebox{-1ex}{$2RT$}\right.}$

The proportion of molecules that have speed less than the mean square speed, $c$ is,

    $P={\displaystyle \underset{0}{\overset{c}{\int }}}f\left(v\right)dv=4\pi {\left(\frac{m}{2\pi kT}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\displaystyle \underset{0}{\overset{c}{\int }}}{v}^2{e}^{\raisebox{1ex}{$-m{v}^2$}\!\left/ \!\raisebox{-1ex}{$2kT$}\right.}dv$

Let

  $a=\frac{m}{2kT}$

Define $x^2=a{v}^2,dv=a^{-1/2}dx$

  $P=-4\pi {\left(\frac{a}{\pi }\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{d}{da}\left\{\frac{1}{a^{1/2}}{\displaystyle \underset{0}{\overset{c{a}^{1/2}}{\int }}{e}^{-x^2}dx}\right\}$

    $P=-4\pi {\left(\frac{a}{\pi }\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left\{\frac{-1}{2}{\left(\frac{1}{a}\right)}^{3/2}{\displaystyle \underset{0}{\overset{c{a}^{1/2}}{\int }}{e}^{-x^2}dx}+{\left(\frac{1}{a}\right)}^{1/2}\frac{d}{da}{\displaystyle \underset{0}{\overset{c{a}^{1/2}}{\int }}{e}^{-x^2}dx}\right\}$        (1)

The identity used in the above expression is,

  ${\displaystyle \underset{0}{\overset{c{a}^{1/2}}{\int }}{e}^{-x^2}}dx=\left(\raisebox{1ex}{${\pi }^{1/2}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)erf\left(c{a}^{1/2}\right)$        (2)

  $\frac{d}{da}{\displaystyle \underset{0}{\overset{c{a}^{1/2}}{\int }}{e}^{-x^2}}dx=\left(\frac{dc{a}^{1/2}}{da}\right)\times e^{-c^{2}a}=\frac{1}{2}\left(\frac{c}{a^{1/2}}\right)e^{-c^{2}a}$        (3)

Substitute the equations (2) and (3) in equation (1).

    $P=-4\pi {\left(\frac{a}{\pi }\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left[\frac{-1}{2}{\left(\frac{1}{a}\right)}^{3/2}\left(\raisebox{1ex}{${\pi }^{1/2}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)erf\left(c{a}^{1/2}\right)+{\left(\frac{1}{a}\right)}^{1/2}\frac{1}{2}\left(\frac{c}{a^{1/2}}\right)e^{-c^{2}a}\right]$

    $P=\text{erf}\left(c{a}^{1/2}\right)-\left(\raisebox{1ex}{$2c{a}^{1/2}$}\!\left/ \!\raisebox{-1ex}{${\pi }^{1/2}$}\right.\right)e^{-c^{2}a}$        (4)

The value of $c$ is ${\left(\raisebox{1ex}{$3kT$}\!\left/ \!\raisebox{-1ex}{$m$}\right.\right)}^{1/2}$, thus,

  $c{a}^{1/2}={\left(\raisebox{1ex}{$3kT$}\!\left/ \!\raisebox{-1ex}{$m$}\right.\right)}^{1/2}\times {\left(\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$2kT$}\right.\right)}^{1/2}={\left(\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}^{1/2}$

Substitute the above values in equation (4).

    $\begin{array}{c}P=\text{erf}\left(\sqrt{\frac{3}{2}}\right)-{\left(\frac{6}{\pi }\right)}^{1/2}{e}^{-3/2}\\ =0.92-0.31\\ =\underset{\_}{0.61}\end{array}$

Hence, the proportion of molecules that have speed less than the mean square speed, $c$ is $\underset{\_}{0.61}$.

(iii)

Interpretation Introduction

Interpretation: The proportions of gas, according to the Maxwell-Boltzmann distribution have greater and less speeds than the mean speed has to be calculated.

Concept introduction: The expression for the mean speed is derived from Maxwell Boltzmann distribution.  The mean speed for the gas particles is represented as,

  $v_{\text{mean}}={\left(\frac{8RT}{\pi M}\right)}^{1/2}$

Answer to Problem 1B.4P

The proportion of molecules that have speed less than the mean speed, $\overline{c}$ is $\underset{\_}{0.533}$.  The proportion of molecules that have speed more than the mean speed, $\overline{c}$ is $\underset{\_}{0.467}$.

Explanation of Solution

The expression for the Maxwell-Boltzmann distribution is,

    $f\left(v\right)=4\pi {\left(\frac{M}{2\pi RT}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{v}^2{e}^{\raisebox{1ex}{$-M{v}^2$}\!\left/ \!\raisebox{-1ex}{$2RT$}\right.}$

The proportion of molecules that have speed less than the mean square speed, $c$ is,

    $P={\displaystyle \underset{0}{\overset{c}{\int }}}f\left(v\right)dv=4\pi {\left(\frac{m}{2\pi kT}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\displaystyle \underset{0}{\overset{c}{\int }}}{v}^2{e}^{\raisebox{1ex}{$-m{v}^2$}\!\left/ \!\raisebox{-1ex}{$2kT$}\right.}dv$

Let

  $a=\frac{m}{2kT}$

Define $x^2=a{v}^2,dv=a^{-1/2}dx$

  $P=-4\pi {\left(\frac{a}{\pi }\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{d}{da}\left\{\frac{1}{a^{1/2}}{\displaystyle \underset{0}{\overset{c{a}^{1/2}}{\int }}{e}^{-x^2}dx}\right\}$

    $P=-4\pi {\left(\frac{a}{\pi }\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left\{\frac{-1}{2}{\left(\frac{1}{a}\right)}^{3/2}{\displaystyle \underset{0}{\overset{c{a}^{1/2}}{\int }}{e}^{-x^2}dx}+{\left(\frac{1}{a}\right)}^{1/2}\frac{d}{da}{\displaystyle \underset{0}{\overset{c{a}^{1/2}}{\int }}{e}^{-x^2}dx}\right\}$        (1)

The identity used in the above expression is,

  ${\displaystyle \underset{0}{\overset{c{a}^{1/2}}{\int }}{e}^{-x^2}}dx=\left(\raisebox{1ex}{${\pi }^{1/2}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)erf\left(c{a}^{1/2}\right)$        (2)

  $\frac{d}{da}{\displaystyle \underset{0}{\overset{c{a}^{1/2}}{\int }}{e}^{-x^2}}dx=\left(\frac{dc{a}^{1/2}}{da}\right)\times e^{-c^{2}a}=\frac{1}{2}\left(\frac{c}{a^{1/2}}\right)e^{-c^{2}a}$        (3)

To calculate the proportion of molecules that have speed less than the mean speed, $\overline{c}$, replace the mean square speed by mean speed, $c$ as shown below.

    ${\left(\frac{3RT}{M}\right)}^{1/2}={\left(\frac{8RT}{\pi M}\right)}^{1/2}$

Therefore,

    $c{a}^{1/2}=2/{\pi }^{1/2}$

Hence,

  $\begin{array}{c}P=\text{erf}\left(\overline{c}{a}^{1/2}\right)-\left(2\overline{c}{a}^{1/2}/{\pi }^{1/2}\right)\times \left(e^{-{\overline{c}}^{2}a}\right)\\ =\text{erf}\left(2/{\pi }^{1/2}\right)-\left(4/\pi \right)e^{-4/\pi }\\ =0.889-0.356\end{array}$

    $P=\underset{\_}{0.533}$

Hence, the proportion of molecules that have speed less than the mean speed, $\overline{c}$ is $\underset{\_}{0.533}$.  The proportion of molecules that have speed more than the mean speed, $\overline{c}$ is $1-0.533=\underset{\_}{0.467}$