Chapter 12.1, Problem 7PP

Interpretation Introduction

Interpretation:

The partial pressure of nitrogen, oxygen and argon gas in atmosphere needs to be determined, if the atmospheric pressure is 760 mmHg.

Concept introduction:

In a gas mixture, each gas component exerts some pressure. The sum of all gas pressure values is equal to the total pressure of the mixture. The pressure of each gas component in the gas mixture is called as partial pressure of that gas. The Dalton law of partial pressure of gas can be written as:

${\text{P}}_{\text{total}}{\text{= P}}_{\text{1}}{\text{+ P}}_{\text{2}}{\text{+ P}}_{\text{3}}\text{…}\text{.}$

Answer to Problem 7PP

$\begin{array}{l}{\text{Partial pressure N}}_{\text{2}}\text{ = 610 mmHg}\\ {\text{Partial pressure O}}_{\text{2}}\text{ = 144 mmHg}\\ \text{Partial pressure Ar = 5}\text{.49 mmHg}\end{array}$

Explanation of Solution

Given information :

Mass percent of Nitrogen = 78 %

Mass percent of Oxygen = 21%

Mass percent of Argon = 1 %

Total pressure = 760 mmHg

Mass percent of Nitrogen = 78 %

Mass percent of Oxygen = 21%

Mass percent of Argon = 1 %

Calculate moles of each gas:

$\begin{array}{l}{\text{N}}_{\text{2}}\text{ = }\frac{\text{78g}}{\text{28}\text{.0g/mol}}\text{= 2}\text{.78 moles}\\ {\text{O}}_{\text{2}}\text{ = }\frac{\text{21g}}{\text{32}\text{.0g/mol}}\text{= 0}\text{.656 moles}\\ \text{Ar = }\frac{\text{1 g}}{\text{39}\text{.99 g/mol}}\text{= 0}\text{.0250 moles}\end{array}$

Calculate partial pressures of each gas:

$\begin{array}{l}{\text{Partial pressure N}}_{\text{2}}{\text{ = χ × P}}_{\text{tota}}{}_{\text{l}}\text{ = }\frac{\text{ 2}\text{.78 moles}}{\text{ 2}\text{.78+0}\text{.656+0}\text{.0250 moles}}\text{× 760 mmHg = 610 mmHg}\\ {\text{Partial pressure O}}_{\text{2}}{\text{ = χ × P}}_{\text{tota}}{}_{\text{l}}\text{ = }\frac{\text{ 0}\text{.656 moles}}{\text{ 2}\text{.78+0}\text{.656+0}\text{.0250 moles}}\text{× 760 mmHg = 144 mmHg}\\ {\text{Partial pressure Ar = χ × P}}_{\text{tota}}{}_{\text{l}}\text{ = }\frac{\text{ 0}\text{.0250 moles}}{\text{ 2}\text{.78+0}\text{.656+0}\text{.0250 moles}}\text{× 760 mmHg = 5}\text{.49 mmHg}\end{array}$

Conclusion

Thus,

$\text{Partial pressure of gas = mole fraction of gas }\times {\text{ P}}_{\text{total}}$