Chapter 11, Problem 11.8QP

Interpretation Introduction

Interpretation: The root means square of the given set of compounds has to be compared

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 (${\text{u}}_{\text{rms}}$) can be calculate using the formula

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

Answer to Problem 11.8QP

(a)

$\begin{array}{l}{\text{O}}_{\text{2}}{\text{: u}}_{\text{rms}}\text{=}\text{528}\text{m/s}\\ {\text{O}}_{\text{3}}\text{:}{\text{u}}_{\text{rms}}\text{=}\text{431}\text{m/s}\end{array}$

(b)

$\begin{array}{l}{\text{F}}_{\text{2}}{\text{: u}}_{\text{rms}}\text{=}\text{485}\text{m/s}\\ {\text{Cl}}_{\text{2}}\text{:}{\text{u}}_{\text{rms}}\text{=}\text{355}\text{m/s}\end{array}$

(c)

$\begin{array}{l}{\text{CH}}_{\text{4}}{\text{: u}}_{\text{rms}}\text{=}\text{746}\text{m/s}\\ {\text{CCl}}_{\text{4}}\text{:}{\text{u}}_{\text{rms}}\text{=}\text{241}\text{m/s}\end{array}$

Explanation of Solution

(a)

Recorded the given data

Temperature is ${\text{85}}^{\text{ο}}\text{C}$

Molecular mass of ${\text{O}}_{\text{2}}$ is $\text{32}{\text{×10}}^{\text{-3}}\text{kg/mole}$

Molecular mass of ${\text{O}}_{\text{3}}$ is $48{\text{×10}}^{\text{-3}}\text{kg/mole}$

The Molecular mass of ${\text{O}}_{\text{2}}$, Molecular mass of ${\text{O}}_{\text{3}}$ and Temperature of ${\text{O}}_{\text{2}}{\text{ and O}}_{\text{3}}$ are recorded as shown above.

To calculate and compared the root-mean-square speed of ${\text{O}}_{\text{2}}{\text{ and O}}_{\text{3}}\text{.}$

$\begin{array}{l}{\text{O}}_{\text{2}}{\text{: u}}_{\text{rms}}\text{=}\sqrt{\frac{\text{3(8}\text{.314J/K}\text{.mol)(85+273)K}}{{\text{32×10}}^{\text{-3}}\text{kg/mol}}}\text{=}\text{528}\text{m/s}\\ {\text{O}}_{\text{3}}\text{:}{\text{u}}_{\text{rms}}\text{=}\sqrt{\frac{\text{3(8}\text{.314J/K}\text{.mol)(85+273)K}}{{\text{48×10}}^{\text{-3}}\text{kg/mol}}}\text{=}\text{431}\text{m/s}\end{array}$

The root-mean-square speed of ${\text{O}}_{\text{2}}{\text{ and O}}_{\text{3}}$ is compared by plugging in the values of Molecular mass of ${\text{O}}_{\text{2}}$, Molecular mass of ${\text{O}}_{\text{3}}$ and Temperature of ${\text{O}}_{\text{2}}{\text{ and O}}_{\text{3}}$

The root-mean-square speed of ${\text{O}}_{\text{2}}{\text{ and O}}_{\text{3}}$ are found to be $\text{528}\text{m/s}$ and $\text{431}\text{m/s}$ respectively.

(b)

Recorded the given data

Temperature is ${\text{85}}^{\text{ο}}\text{C}$

Molecular mass of ${\text{F}}_{\text{2}}$ is $\text{38}{\text{×10}}^{\text{-3}}\text{kg/mole}$

Molecular mass of ${\text{Cl}}_{\text{2}}$ is $70.90{\text{×10}}^{\text{-3}}\text{kg/mole}$

The Molecular mass of ${\text{F}}_{\text{2}}$, Molecular mass of ${\text{Cl}}_{\text{2}}$ and Temperature of ${\text{F}}_{\text{2}}{\text{ and Cl}}_2$ are recorded as shown above.

To calculate and compared the root-mean-square speed of ${\text{F}}_{\text{2}}{\text{ and Cl}}_2$

$\begin{array}{l}{\text{F}}_{\text{2}}{\text{: u}}_{\text{rms}}\text{=}\sqrt{\frac{\text{3(8}\text{.314J/K}\text{.mol)(85+273)K}}{{\text{38×10}}^{\text{-3}}\text{kg/mol}}}\text{=}\text{485}\text{m/s}\\ {\text{Cl}}_{\text{2}}\text{:}{\text{u}}_{\text{rms}}\text{=}\sqrt{\frac{\text{3(8}\text{.314J/K}\text{.mol)(85+273)K}}{\text{70}{\text{.90×10}}^{\text{-3}}\text{kg/mol}}}\text{=}\text{355}\text{m/s}\end{array}$

Explanation:

The root-mean-square speed of ${\text{F}}_{\text{2}}{\text{ and Cl}}_2$ is compared by plugging in the values of Molecular mass of ${\text{F}}_{\text{2}}$, Molecular mass of ${\text{Cl}}_{\text{2}}$ and Temperature of ${\text{F}}_{\text{2}}{\text{ and Cl}}_2$

The root-mean-square speed of ${\text{F}}_{\text{2}}{\text{ and Cl}}_2$ are found to be $485\text{m/s}$ and $355\text{m/s}$ respectively.

(c)

Recorded the given data

Temperature is ${\text{85}}^{\text{ο}}\text{C}$

Molecular mass of ${\text{CH}}_{\text{4}}$ ${\text{O}}_{\text{2}}$ is $16.04{\text{×10}}^{\text{-3}}\text{kg/mole}$

Molecular mass of ${\text{CCl}}_{\text{4}}$ is $153.8{\text{×10}}^{\text{-3}}\text{kg/mole}$

The Molecular mass of ${\text{CH}}_{\text{4}}$, Molecular mass of ${\text{CCl}}_{\text{4}}$ and Temperature ${\text{CH}}_{\text{4}}{\text{ and CCl}}_{\text{4}}$ are recorded as shown above.

To calculate and compared the root-mean-square speed of ${\text{CH}}_{\text{4}}{\text{ and CCl}}_{\text{4}}$

$\begin{array}{l}{\text{CH}}_{\text{4}}{\text{: u}}_{\text{rms}}\text{=}\sqrt{\frac{\text{3(8}\text{.314J/K}\text{.mol)(85+273)K}}{\text{16}{\text{.04×10}}^{\text{-3}}\text{kg/mol}}}\text{=}\text{746}\text{m/s}\\ {\text{CCl}}_{\text{4}}\text{:}{\text{u}}_{\text{rms}}\text{=}\sqrt{\frac{\text{3(8}\text{.314J/K}\text{.mol)(85+273)K}}{\text{153}{\text{.8×10}}^{\text{-3}}\text{kg/mol}}}\text{=}\text{241}\text{m/s}\end{array}$

The root mean square speed of ${\text{CH}}_{\text{4}}{\text{ and CCl}}_{\text{4}}$ is compared by plugging in the values of Molecular mass of ${\text{CH}}_{\text{4}}$, Molecular mass of ${\text{CCl}}_{\text{4}}$ and Temperature ${\text{CH}}_{\text{4}}{\text{ and CCl}}_{\text{4}}$.

The root-mean-square speed of ${\text{CH}}_{\text{4}}{\text{ and CCl}}_{\text{4}}$ are found to be $746\text{m/s}$ and $\text{241}\text{m/s}$ respectively.

Conclusion

The root means square of the given set of compounds was compared