Chapter 11, Problem 11.9QP

Interpretation Introduction

Interpretation:

The root-mean-square speeds of given molecules in the given region has to be calculated.

Concept Introduction:

Root mean-square speed is the parameter to measure the speed of particles in a gas. M is the molecular mass of the molecule at temperature T, root mean-square speed (${\text{u}}_{\text{rms}}$) has to be calculate using the formula

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

Answer to Problem 11.9QP

$\begin{array}{l}{\text{N}}_{\text{2}}{\text{: u}}_{\text{rms}}\text{=}\text{471}\text{m/s, }\\ \\ {\text{O}}_{\text{2}}\text{:}{\text{u}}_{\text{rms}}\text{=}\text{441}\text{m/s}\\ \\ {\text{O}}_{\text{3}}\text{:}{\text{u}}_{\text{rms}}\text{=}\text{360m/s}\end{array}$

Explanation of Solution

Record the given data

Mass of ${\text{N}}_{\text{2}}$ is $14.01\text{g/mole}$

Mass of ${\text{O}}_2$ is $16.00\text{g/mole}$

The Mass of ${\text{O}}_{\text{2}}$, and Mass of ${\text{N}}_{\text{2}}$ are recorded as shown above.

First, let’s calculate the molar masses of ${\text{N}}_{\text{2}}{\text{, O}}_{\text{2}}{\text{, and O}}_{\text{3}}$.

$\begin{array}{l}{\text{M}}_{{\text{N}}_{\text{2}}}\text{=2(14}\text{.01}\text{g/mole)}\text{=28}\text{.02}\text{g/mole}\\ \text{28}\text{.02}\frac{\text{g}}{\text{mol}}\text{×}\frac{\text{1}\text{kg}}{\text{1000}\text{g}}\text{=}\text{0}\text{.02802}\text{kg/mole}\end{array}$

$\begin{array}{l}{\text{M}}_{{\text{O}}_{\text{2}}}\text{=2(16}\text{.00}\text{g/mole)}\text{=32}\text{.00}\text{g/mole}\\ \text{32}\text{.00}\frac{\text{g}}{\text{mol}}\text{×}\frac{\text{1}\text{kg}}{\text{1000}\text{g}}\text{=}\text{0}\text{.03200}\text{kg/mole}\end{array}$

$\begin{array}{l}{\text{M}}_{{\text{O}}_{\text{3}}}\text{=3(16}\text{.00}\text{g/mole)}\text{=48}\text{.00}\text{g/mole}\\ \text{48}\text{.00}\frac{\text{g}}{\text{mol}}\text{×}\frac{\text{1}\text{kg}}{\text{1000}\text{g}}\text{=}\text{0}\text{.04800}\text{kg/mole}\end{array}$

The molar masses of ${\text{N}}_{\text{2}}{\text{, O}}_{\text{2}}{\text{, and O}}_{\text{3}}$ by plugging in the values of Mass of ${\text{O}}_{\text{2}}$, and Mass of ${\text{N}}_{\text{2}}$.  The masses found to be $\text{0}\text{.02802}\text{kg/mole,}\text{0}\text{.0320}\text{kg/mole}\text{and}\text{0}\text{.0480}\text{kg/mole}$ respectively.

To calculate the root-mean-square speed of  ${\text{N}}_{\text{2}}$

${\text{N}}_{\text{2}}{\text{: u}}_{\text{rms}}\text{=}\sqrt{\frac{\text{3(8}\text{.314}\frac{\text{J}}{\text{K}\text{.mol}}\text{)(-23+273)K}}{\text{0}\text{.02802}\frac{\text{kg}}{\text{mol}}}}\text{=}\text{471}\text{m/s}$

The root-mean-square speed of ${\text{N}}_{\text{2}}$ is compared by plugging in the values of Molar mass of ${\text{N}}_{\text{2}}$ and temperature of ${\text{N}}_{\text{2}}$.  The root-mean-square speed of ${\text{N}}_{\text{2}}$ are found to be $471\text{m/s}$

To calculate the root-mean-square speed of  ${\text{O}}_{\text{2}}$

${\text{O}}_{\text{2}}{\text{: u}}_{\text{rms}}\text{=}\sqrt{\frac{\text{3(8}\text{.314}\frac{\text{J}}{\text{K}\text{.mol}}\text{)(-23+273)K}}{\text{0}\text{.03200}\frac{\text{kg}}{\text{mol}}}}\text{=}\text{441m/s}$

The root-mean-square speed of ${\text{O}}_{\text{2}}$ is compared by plugging in the values of Molar mass of ${\text{O}}_{\text{2}}$ and Temperature of ${\text{O}}_{\text{2}}$.  The root-mean-square speed of ${\text{O}}_{\text{2}}$ are found to be $441\text{m/s}$

To calculate the root-mean-square speed of  ${\text{O}}_3$

${\text{O}}_{\text{3}}{\text{: u}}_{\text{rms}}\text{=}\sqrt{\frac{\text{3(8}\text{.314}\frac{\text{J}}{\text{K}\text{.mol}}\text{)(-23+273)K}}{\text{0}\text{.04800}\frac{\text{kg}}{\text{mol}}}}\text{=}360\text{m/s}$

The root-mean-square speed of ${\text{O}}_3$ is compared by plugging in the values of Molar mass of ${\text{O}}_3$ and Temperature of ${\text{O}}_3$.  The root-mean-square speed of ${\text{O}}_3$ are found to be $360\text{m/s}$

Conclusion

The root-mean-square speeds of given molecules in this region was calculated.