Chapter 1, Problem 1A.13P

(a)

Interpretation Introduction

Interpretation: The molar concentration of ${\text{CCl}}_{\text{3}}\text{F}$ and ${\text{CCl}}_2{\text{F}}_2$ at the mid-latitude troposphere has to be calculated.

Concept introduction: The perfect gas equation is given for the gases which follow a specific rule under certain conditions.  This is represented by the formula given below as,

  $pV=nRT$

Where,

• Ø  $p$ is the total pressure.
• Ø  $V$ is the volume.
• Ø  $n$ is the number of moles.
• Ø  $R$ is the gas constant.
• Ø  $T$ is the temperature.

The partial pressure of any gas in a mixture of gases is calculated using the total pressure of the mixture and the mole fraction of the gas in the mixture.  This is represented by the formula given below as,

  $p_{\text{total}}=\frac{p_{\text{a}}}{x_{\text{a}}}$

Answer to Problem 1A.13P

The molar concentration of ${\text{CCl}}_{\text{3}}\text{F}$ and ${\text{CCl}}_2{\text{F}}_2$ at the mid-latitude troposphere has been calculated as $\underset{\_}{1.124\times 10^{-11}mold{m}^{-3}}$ and $\underset{\_}{2.19\times 10^{-11}mold{m}^{-3}}$ respectively.

Explanation of Solution

The perfect gas equation is given by the formula as,

    $pV=nRT$

Where,

• Ø  $p$ is the total pressure.
• Ø  $V$ is the volume.
• Ø  $n$ is the number of moles.
• Ø  $R$ is the gas constant.
• Ø  $T$ is the temperature.

Rearrange the above equation by substituting the relation, $p=x_{\text{a}}{p}_{\text{a}}$, for number of moles $\left(n\right)$ by volume as follows.

    $\begin{array}{c}pV=nRT\\ \frac{n}{V}=\frac{x_{\text{a}}{p}_{\text{a}}}{RT}\end{array}$

At mid-latitude troposphere, the temperature, $T=\left(10+273.15\right)\text{K}$  and pressure, $p=1.0\text{atm}$.  The mole fraction of ${\text{CCl}}_{\text{3}}\text{F}$ and ${\text{CCl}}_2{\text{F}}_2$ are given as $261\text{ppt}$ and $509\text{ppt}$ by volume.

Substitute the values in the above equation for ${\text{CCl}}_{\text{3}}\text{F}$ as follows.

    $\begin{array}{c}\frac{n}{V}=\frac{x_{{\text{CCl}}_{\text{3}}\text{F}}{p}_{{\text{CCl}}_{\text{3}}\text{F}}}{RT}\\ =\frac{\left(261\times {10}^{-12}\right)\left(1.0\text{atm}\right)}{\left(0.0821\text{atm}{\text{dm}}^{\text{3}}{\text{K}}^{-1}{\text{mol}}^{-1}\right)\left(10+273.15\right)\text{K}}\\ =\underset{\_}{1.124\times 10^{-11}mold{m}^{-3}}\end{array}$

Substitute the values in the above equation for ${\text{CCl}}_2{\text{F}}_2$ as follows.

    $\begin{array}{c}\frac{n}{V}=\frac{x_{{\text{CCl}}_{\text{3}}\text{F}}{p}_{{\text{CCl}}_{\text{3}}\text{F}}}{RT}\\ =\frac{\left(509\times {10}^{-12}\right)\left(1.0\text{atm}\right)}{\left(0.0821\text{atm}{\text{dm}}^{\text{3}}{\text{K}}^{-1}{\text{mol}}^{-1}\right)\left(10+273.15\right)\text{K}}\\ =\underset{\_}{2.19\times 10^{-11}mold{m}^{-3}}\end{array}$

Thus, the molar concentration of ${\text{CCl}}_{\text{3}}\text{F}$ and ${\text{CCl}}_2{\text{F}}_2$ at the mid-latitude troposphere is calculated as $\underset{\_}{1.124\times 10^{-11}mold{m}^{-3}}$ and $\underset{\_}{2.19\times 10^{-11}mold{m}^{-3}}$ respectively.

(b)

Interpretation Introduction

Interpretation: The molar concentration of ${\text{CCl}}_{\text{3}}\text{F}$ and ${\text{CCl}}_2{\text{F}}_2$ at the Antarctic stratosphere has to be calculated.

Concept introduction: The perfect gas equation is given for the gases which follow a specific rule under certain conditions.  This is represented by the formula given below as,

  $pV=nRT$

Where,

• Ø  $p$ is the total pressure.
• Ø  $V$ is the volume.
• Ø  $n$ is the number of moles.
• Ø  $R$ is the gas constant.
• Ø  $T$ is the temperature.

The partial pressure of any gas in a mixture of gases is calculated using the total pressure of the mixture and the mole fraction of the gas in the mixture.  This is represented by the formula given below as,

  $p_{\text{total}}=\frac{p_{\text{a}}}{x_{\text{a}}}$

Answer to Problem 1A.13P

The molar concentration of ${\text{CCl}}_{\text{3}}\text{F}$ and ${\text{CCl}}_2{\text{F}}_2$ at the Antarctic stratosphere has been calculated as $\underset{\_}{7.95\times 10^{-13}mold{m}^{-3}}$ and $\underset{\_}{1.55\times 10^{-12}mold{m}^{-3}}$ respectively.

Explanation of Solution

The perfect gas equation is given by the formula as,

    $pV=nRT$

Where,

• Ø  $p$ is the total pressure.
• Ø  $V$ is the volume.
• Ø  $n$ is the number of moles.
• Ø  $R$ is the gas constant.
• Ø  $T$ is the temperature.

Rearrange the above equation by substituting the relation, $p=x_{\text{a}}{p}_{\text{a}}$, for number of moles $\left(n\right)$ by volume as follows.

    $\begin{array}{c}pV=nRT\\ \frac{n}{V}=\frac{x_{\text{a}}{p}_{\text{a}}}{RT}\end{array}$

At Antarctic stratosphere, the temperature, $T=200\text{K}$  and pressure, $p=0.05\text{atm}$.  The mole fraction of ${\text{CCl}}_{\text{3}}\text{F}$ and ${\text{CCl}}_2{\text{F}}_2$ are given as $261\text{ppt}$ and $509\text{ppt}$ by volume.

Substitute the values in the above equation for ${\text{CCl}}_{\text{3}}\text{F}$ as follows.

    $\begin{array}{c}\frac{n}{V}=\frac{x_{{\text{CCl}}_{\text{3}}\text{F}}{p}_{{\text{CCl}}_{\text{3}}\text{F}}}{RT}\\ =\frac{\left(261\times {10}^{-12}\right)\left(0.05\text{atm}\right)}{\left(0.0821\text{atm}{\text{dm}}^{\text{3}}{\text{K}}^{-1}{\text{mol}}^{-1}\right)200\text{K}}\\ =\underset{\_}{7.95\times 10^{-13}mold{m}^{-3}}\end{array}$

Substitute the values in the above equation for ${\text{CCl}}_2{\text{F}}_2$ as follows.

    $\begin{array}{c}\frac{n}{V}=\frac{x_{{\text{CCl}}_{\text{3}}\text{F}}{p}_{{\text{CCl}}_{\text{3}}\text{F}}}{RT}\\ =\frac{\left(509\times {10}^{-12}\right)\left(0.05\text{atm}\right)}{\left(0.0821\text{atm}{\text{dm}}^{\text{3}}{\text{K}}^{-1}{\text{mol}}^{-1}\right)200\text{K}}\\ =\underset{\_}{1.55\times 10^{-12}mold{m}^{-3}}\end{array}$

Thus, the molar concentration of ${\text{CCl}}_{\text{3}}\text{F}$ and ${\text{CCl}}_2{\text{F}}_2$ at the mid-latitude troposphere is calculated as $\underset{\_}{7.95\times 10^{-13}mold{m}^{-3}}$ and $\underset{\_}{1.55\times 10^{-12}mold{m}^{-3}}$ respectively.