Chapter 1, Problem 1B.9P

Interpretation Introduction

Interpretation: The fraction of ${\text{O}}_{\text{2}}$ molecules with the speeds in the range $100\text{m}{\text{s}}^{-1}$ to $200\text{m}{\text{s}}^{-1}$ in a gas at $300\text{K}$ and $1000\text{K}$ has to be calculated.

Concept introduction: The speeds of individual molecules spread across a wide range in a gas and the gas collisions ensure that their speeds change constantly.  The distribution of speeds is identified by recognizing that the molecules possess kinetic energy and the distribution of energy over the molecules is determined with the help of Maxwell Boltzmann distribution.

Answer to Problem 1B.9P

The fraction of molecules with the speeds in the range $100\text{m}{\text{s}}^{-1}$ to $200\text{m}{\text{s}}^{-1}$ in a gas at $300\text{K}$ and $1000\text{K}$ is shown below.

Velocity, $v$	Temperature, $T$	Fraction
$100\text{m}{\text{s}}^{-1}$	$300\text{K}$	$0.000344$
$110\text{m}{\text{s}}^{-1}$	$300\text{K}$	$0.000411$
$120\text{m}{\text{s}}^{-1}$	$300\text{K}$	$0.000482$
$130\text{m}{\text{s}}^{-1}$	$300\text{K}$	$0.000556$
$140\text{m}{\text{s}}^{-1}$	$300\text{K}$	$0.000634$
$150\text{m}{\text{s}}^{-1}$	$300\text{K}$	$0.000714$
$160\text{m}{\text{s}}^{-1}$	$300\text{K}$	$0.000797$
$170\text{m}{\text{s}}^{-1}$	$300\text{K}$	$0.000881$
$180\text{m}{\text{s}}^{-1}$	$300\text{K}$	$0.000965$
$190\text{m}{\text{s}}^{-1}$	$300\text{K}$	$0.00105$
$200\text{m}{\text{s}}^{-1}$	$300\text{K}$	$0.001135$

Velocity, $v$	Temperature, $T$	Fraction
$100\text{m}{\text{s}}^{-1}$	$1000\text{K}$	$5.91156\times {10}^{-5}$
$110\text{m}{\text{s}}^{-1}$	$1000\text{K}$	$7.12414\times {10}^{-5}$
$120\text{m}{\text{s}}^{-1}$	$1000\text{K}$	$8.44087\times {10}^{-5}$
$130\text{m}{\text{s}}^{-1}$	$1000\text{K}$	$9.85875\times {10}^{-5}$
$140\text{m}{\text{s}}^{-1}$	$1000\text{K}$	$0.000113746$
$150\text{m}{\text{s}}^{-1}$	$1000\text{K}$	$0.000129849$
$160\text{m}{\text{s}}^{-1}$	$1000\text{K}$	$0.00014686$
$170\text{m}{\text{s}}^{-1}$	$1000\text{K}$	$0.000164742$
$180\text{m}{\text{s}}^{-1}$	$1000\text{K}$	$0.000183453$
$190\text{m}{\text{s}}^{-1}$	$1000\text{K}$	$0.000202953$
$200\text{m}{\text{s}}^{-1}$	$1000\text{K}$	$0.000223197$

Explanation of Solution

The molecular distribution of speeds is given by,

    $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.}$

Where,

• • $M$ is the molar mass.
• • $R$ is the gas constant.
• • $T$ is the temperature.
• • $v$ is the speed.

It is given that,

• • $M=32\text{g}{\text{mol}}^{-1}$.
• • $T=300\text{K}$.

Take $v=100\text{m}{\text{s}}^{-1}$.

Substitute the values of $M,T,v$ in the above equation to calculate the fraction of molecules with velocity, $100\text{m}{\text{s}}^{-1}$.

  $\begin{array}{c}f\left(v\right)=4\times 3.14{\left(\frac{32\times {10}^{-3}\text{kg}{\text{mol}}^{-1}}{2\times 3.14\times 8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\times 300\text{K}}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\left(100\text{m}{\text{s}}^{-1}\right)}^2\\ e^{\raisebox{1ex}{$-\left(32\times {10}^{-3}\text{kg}{\text{mol}}^{-1}\right){\left(100\text{m}\text{s}\right)}^2$}\!\left/ \!\raisebox{-1ex}{$2\left(8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\right)\left(300\text{K}\right)$}\right.}\\ =0.000344\end{array}$

Similarly the velocity values are taken upto $200\text{m}{\text{s}}^{-1}$ in intervals of $10\text{m}{\text{s}}^{-1}$ and calculate the fraction of molecules.  The calculated values are shown below.

Velocity, $v$	Temperature, $T$	Fraction
$100\text{m}{\text{s}}^{-1}$	$300\text{K}$	$0.000344$
$110\text{m}{\text{s}}^{-1}$	$300\text{K}$	$0.000411$
$120\text{m}{\text{s}}^{-1}$	$300\text{K}$	$0.000482$
$130\text{m}{\text{s}}^{-1}$	$300\text{K}$	$0.000556$
$140\text{m}{\text{s}}^{-1}$	$300\text{K}$	$0.000634$
$150\text{m}{\text{s}}^{-1}$	$300\text{K}$	$0.000714$
$160\text{m}{\text{s}}^{-1}$	$300\text{K}$	$0.000797$
$170\text{m}{\text{s}}^{-1}$	$300\text{K}$	$0.000881$
$180\text{m}{\text{s}}^{-1}$	$300\text{K}$	$0.000965$
$190\text{m}{\text{s}}^{-1}$	$300\text{K}$	$0.00105$
$200\text{m}{\text{s}}^{-1}$	$300\text{K}$	$0.001135$

It is given that,

• • $M=32\text{g}{\text{mol}}^{-1}$.
• • $T=1000\text{K}$.

Take $v=100\text{m}{\text{s}}^{-1}$.

Substitute the values of $M,T,v$ in the above equation to calculate the fraction of molecules with velocity, $100\text{m}{\text{s}}^{-1}$.

  $\begin{array}{c}f\left(v\right)=4\times 3.14{\left(\frac{32\times {10}^{-3}\text{kg}{\text{mol}}^{-1}}{2\times 3.14\times 8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\times 1000\text{K}}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\left(100\text{m}{\text{s}}^{-1}\right)}^2\\ e^{\raisebox{1ex}{$-\left(32\times {10}^{-3}\text{kg}{\text{mol}}^{-1}\right){\left(100\text{m}\text{s}\right)}^2$}\!\left/ \!\raisebox{-1ex}{$2\left(8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\right)\left(1000\text{K}\right)$}\right.}\\ =5.91156\times {10}^{-5}\end{array}$

Similarly the velocity values are taken upto $200\text{m}{\text{s}}^{-1}$ in intervals of $10\text{m}{\text{s}}^{-1}$ and calculate the fraction of molecules.  The calculated values are shown below.

Velocity, $v$	Temperature, $T$	Fraction
$100\text{m}{\text{s}}^{-1}$	$1000\text{K}$	$5.91156\times {10}^{-5}$
$110\text{m}{\text{s}}^{-1}$	$1000\text{K}$	$7.12414\times {10}^{-5}$
$120\text{m}{\text{s}}^{-1}$	$1000\text{K}$	$8.44087\times {10}^{-5}$
$130\text{m}{\text{s}}^{-1}$	$1000\text{K}$	$9.85875\times {10}^{-5}$
$140\text{m}{\text{s}}^{-1}$	$1000\text{K}$	$0.000113746$
$150\text{m}{\text{s}}^{-1}$	$1000\text{K}$	$0.000129849$
$160\text{m}{\text{s}}^{-1}$	$1000\text{K}$	$0.00014686$
$170\text{m}{\text{s}}^{-1}$	$1000\text{K}$	$0.000164742$
$180\text{m}{\text{s}}^{-1}$	$1000\text{K}$	$0.000183453$
$190\text{m}{\text{s}}^{-1}$	$1000\text{K}$	$0.000202953$
$200\text{m}{\text{s}}^{-1}$	$1000\text{K}$	$0.000223197$