Chapter 5, Problem 5.59QP

(a)

Interpretation Introduction

Interpretation:

The partial pressures of each gas in atmospheres and concentration of each gas has to be calculated.

Concept Introduction:

• In a mixture of gases, every gas has an incomplete pressure which is the theoretical stress of that gas if it alone engaged the entire volume of the original combination at the same temperature.
• The sum pressure of an ideal gas mixture is the amount of the partial pressures of the gases in the mixture.

$\begin{array}{l}\text{Concentration (mol/L) is }\\ \text{c}\text{=}\frac{\text{n}}{\text{V}}\text{ = }\frac{\text{P}}{\text{RT}}\\ \end{array}$

• A mole fraction is the unit less proportion of the number of moles of a mixture constituent and the total number of moles in the mixture.

${\text{χ}}_{\text{i}}\text{=}\frac{{\text{P}}_{\text{i}}}{{\text{P}}_{\text{total}}}$

• STP in chemistry is the short form for Standard Temperature and Pressure.  STP most usually is used when performing arts calculations on gases, such as gas density.  The normal temperature is $\text{273 K }\left(\text{0° Celsius or 32° Fahrenheit}\right)$ and the standard pressure is $\text{1 atm}$ pressure. This is the freezing point of pure water at sea level atmospheric pressure.

Answer to Problem 5.59QP

${\text{P}}_{{\text{N}}_{\text{2}}}\text{= 0}{\text{.781 atm, P}}_{{\text{O}}_{\text{2}}}\text{= 0}{\text{.209 atm, P}}_{\text{Ar }}\text{=9}\text{.3 ×}{\text{10}}^{\text{-3}}\text{atm,}{\text{P}}_{{\text{CO}}_{\text{2}}}\text{=}\text{5×}{\text{10}}^{\text{-4}}\text{atm}$

Explanation of Solution

$\begin{array}{l}{\text{χ}}_{{\text{N}}_{\text{2}}}\text{=}\text{0}\text{.7808}\\ {\text{χ}}_{{\text{O}}_{\text{2}}}\text{=}\text{0}\text{.2094}\\ {\text{χ}}_{\text{Ar}}\text{=}\text{0}\text{.0093}\\ {\text{χ}}_{{\text{CO}}_{\text{2}}}\text{=}\text{0}\text{.0005}\end{array}$

${\text{P}}_{{\text{N}}_{\text{2}}}\text{= 0}{\text{.781 atm, P}}_{{\text{O}}_{\text{2}}}\text{= 0}{\text{.209 atm, P}}_{\text{Ar }}\text{=9}\text{.3 ×}{\text{10}}^{\text{-3}}\text{atm,}{\text{P}}_{{\text{CO}}_{\text{2}}}\text{=}\text{5×}{\text{10}}^{\text{-4}}\text{atm}$

Given that volume is proportional to the number of moles of gas present, we can directly convert the volume percents to mole fractions.

(b)

Interpretation Introduction

Interpretation:

The partial pressures of each gas in atmospheres and concentration of each gas have to be calculated.

Concept Introduction:

• In a mixture of gases, every gas has an incomplete pressure which is the theoretical stress of that gas if it alone engaged the entire volume of the original combination at the same temperature.
• The sum pressure of an ideal gas mixture is the amount of the partial pressures of the gases in the mixture.

$\begin{array}{l}\text{Concentration (mol/L) is }\\ \text{c}\text{=}\frac{\text{n}}{\text{V}}\text{ = }\frac{\text{P}}{\text{RT}}\\ \end{array}$

• A mole fraction is the unit less proportion of the number of moles of a mixture constituent and the total number of moles in the mixture.

${\text{χ}}_{\text{i}}\text{=}\frac{{\text{P}}_{\text{i}}}{{\text{P}}_{\text{total}}}$

• STP in chemistry is the short form for Standard Temperature and Pressure.  STP most usually is used when performing arts calculations on gases, such as gas density.  The normal temperature is $\text{273 K }\left(\text{0° Celsius or 32° Fahrenheit}\right)$ and the standard pressure is $\text{1 atm}$ pressure. This is the freezing point of pure water at sea level atmospheric pressure.

Answer to Problem 5.59QP

${\text{c}}_{{\text{N}}_{\text{2}}}\text{= 3}\text{.48}\text{×}{\text{10}}^{\text{-2}}{\text{M, c}}_{{\text{O}}_{\text{2}}}\text{= 9}\text{.32}\text{×}{\text{10}}^{\text{-3}}{\text{M, c}}_{\text{Ar }}\text{=4}\text{.1}\text{×}{\text{10}}^{\text{-4}}\text{M,}{\text{c}}_{{\text{CO}}_{\text{2}}}\text{=}\text{2}\text{×}{\text{10}}^{\text{-5}}\text{M}$

Explanation of Solution

The concentration of given gas

$\begin{array}{l}\text{Concentration (mol/L) is }\\ \text{c}\text{=}\frac{\text{n}}{\text{V}}\text{ = }\frac{\text{P}}{\text{RT}}\\ \text{Therefore, we have:}\end{array}$

$\begin{array}{l}{\text{c}}_{{\text{N}}_{\text{2}}}\text{=}\frac{\text{0}\text{.781}\text{atm}}{\left(\text{0}\text{.0821}\frac{\text{L}\text{.}\text{atm}}{\text{K}\text{.}\text{mol}}\right)\text{(273K)}}\text{=}\text{3}\text{.48}\text{×}{\text{10}}^{\text{-2}}\text{M}\\ \\ {\text{c}}_{{\text{O}}_{\text{2}}}\text{=}\frac{\text{0}\text{.209}\text{atm}}{\left(\text{0}\text{.0821}\frac{\text{L}\text{.}\text{atm}}{\text{K}\text{.}\text{mol}}\right)\text{(273K)}}\text{=}\text{9}\text{.32}\text{×}{\text{10}}^{\text{-3}}\text{M}\\ \\ {\text{c}}_{\text{Ar}}\text{=}\frac{\text{0}\text{.0093}\text{atm}}{\left(\text{0}\text{.0821}\frac{\text{L}\text{.}\text{atm}}{\text{K}\text{.}\text{mol}}\right)\text{(273K)}}\text{=}\text{4}\text{.1}\text{×}{\text{10}}^{\text{-4}}\text{M}\\ \\ {\text{c}}_{{\text{CO}}_{\text{2}}}\text{=}\frac{\text{0}\text{.0005}\text{atm}}{\left(\text{0}\text{.0821}\frac{\text{L}\text{.}\text{atm}}{\text{K}\text{.}\text{mol}}\right)\text{(273K)}}\text{=}\text{2}\text{×}{\text{10}}^{\text{-5}}\text{M}\end{array}$

The concentration of given gas is calculated by plugging in the values of the given pressure of gas and temperature of gas.  The concentration of given gas was found to be

${\text{c}}_{{\text{N}}_{\text{2}}}\text{= 3}\text{.48}\text{×}{\text{10}}^{\text{-2}}{\text{M, c}}_{{\text{O}}_{\text{2}}}\text{= 9}\text{.32}\text{×}{\text{10}}^{\text{-3}}{\text{M, c}}_{\text{Ar }}\text{=4}\text{.1}\text{×}{\text{10}}^{\text{-4}}\text{M,}{\text{c}}_{{\text{CO}}_{\text{2}}}\text{=}\text{2}\text{×}{\text{10}}^{\text{-5}}\text{M}$