Chapter 21, Problem 53QP

On a smoggy day in a certain city. the ozone concentration was 0.42 ppm by volume. Calculate the partial pressure of ozone (in atm) and the number of ozone molecules per liter of air if the temperature and pressure were $\text{20}\text{.0}°\text{C}$ and 748 mmHg, respectively.

Interpretation Introduction

Interpretation:

The partial pressure of ozone and the number of ozone molecules per liter of air at temperature 20°C and at pressure 748 mmHg are to becalculated.

Concept introduction:

The Ideal Gas Law has the volume (V) occupied by n moles of any gas also has a pressure (P) at temperature (T) in Kelvin and this ideal gas equation is represented as follows:

$PV=n\text{R}T$

Here, $P$ is the pressure, $n$ is the number of moles, $R$ is the universal gas constant, $V$ is the volume, and $T$ is the absolute temperature.

Molal fraction is defined as the amount of a constituentdivided by the total amount of all constituents in a mixture.

Ozone is a molecule made up of three oxygen atoms. Ozone is called a secondary pollutant, which is formed from the primary pollutants like nitrogen oxide.

Answer to Problem 53QP

Solution: The partial pressure of ${\text{O}}_3$ is $4.1\times {10}^{-7}\text{ atm}$ and the number of ozone molecules per liter of airis $1\times {10}^{16}\text{ molecule/L}$.

Explanation of Solution

The number of moles of gas is directly proportional to the volume of the gas. Thus, $0.42\text{ ppm}$ by volume can be represented as mole fraction.

$X_{{\text{O}}_3}=\frac{n_{{}_{{\text{O}}_3}}}{n_{\text{total}}}$

Substitute the values in the above equation:

$\begin{array}{l}{X}_{{\text{O}}_3}=\frac{0.42}{1\times {10}^6}\\ X_{{\text{O}}_3}=4.2\times {10}^{-7}\end{array}$

Therefore, the mole fraction of ${\text{O}}_3$ is $4.2\times {10}^{-7}$.

The partial pressure of ${\text{O}}_3$ can be calculated with the help of total pressure and mole fraction as follows:

$P_{{\text{O}}_3}=X_{{\text{O}}_3}\times P_{\text{T}}$

Here, $X_{{\text{O}}_3}$ is the mole fraction of ${\text{O}}_3$ and $P_{\text{T}}$ is the total pressure.

Substitute the values in the above equation:

$\begin{array}{l}{P}_{{\text{O}}_3}=\left(4.2\times {10}^{-7}\right)\times \left(748\text{ mm Hg}\right)\\ P_{{\text{O}}_3}=\left(3.14\times {10}^{-4}\cancel{\text{mm Hg}}\right)\times \left(\frac{1\text{ atm}}{760\cancel{\text{mm Hg}}}\right)\\ P_{{\text{O}}_3}=4.1\times {10}^{-7}\text{ atm}\end{array}$

Therefore, the partial pressure of ${\text{O}}_3$ is $4.1\times {10}^{-7}\text{ atm}$.

The moles of ozone can be calculated by ideal gas equation as follows:

$n_{{\text{O}}_3}=\frac{P_{{}_{{\text{O}}_3}}V}{\text{R}T}$

Here, $P$ is the pressure, $n$ is the number of moles, $R$ is the universal gas constant, $V$ is the volume, and $T$ is the temperature.

Substitute the values of $V,P_{{}_{{\text{O}}_3}},\text{R, and }T$ in the above ideal gas equation as follows:

$\begin{array}{l}{n}_{{\text{O}}_3}=\frac{\left(4.1\times {10}^{-7}\cancel{\text{atm}}\right)\left(1\cancel{\text{L}}\right)}{\left(0.0821\cancel{\text{L}}\text{.}\cancel{\text{atm}}\text{/mol}\text{.}\cancel{\text{K}}\right)\left(293\cancel{\text{K}}\right)}\\ n_{{\text{O}}_3}=1.7\times {10}^{-8}\text{ mol}\end{array}$

Therefore, the moles of ozone are $1.7\times {10}^{-8}\text{ mol}$.

The number of ozone molecules per liter is calculated as follows:

$\begin{array}{l}{n}_{{\text{o}}_3}=\left(1.7\times {10}^{-8}\cancel{{\text{mol O}}_3}\right)\times \left(\frac{6.022\times {10}^{23}\text{ molecules}}{1\cancel{{\text{mol O}}_3}}\right)\\ n_{{\text{o}}_3}=1\times {10}^{16}\text{ molecule/L}\end{array}$

Conclusion

The partial pressure of ozone is $4.1\times {10}^{-7}\text{ atm}$ and the number of ozone molecules per liter of air at temperature 20°C and pressure 748mmHg is $1\times {10}^{16}\text{ molecule/L}$.