Chapter 5, Problem 5.69QP

Dry air near sea level has the following composition by volume: N₂, 78.08 percent; O₂, 20.94 percent; Ar, 0.93 percent; CO₂, 0.05 percent. The atmospheric pressure is 1.00 atm. Calculate (a) the partial pressure of each gas in atm and (b) the concentration of each gas in moles per liter at 0°C. (Hint: Because volume is proportional to the number of moles present, mole fractions of gases can be expressed as ratios of volumes at the same temperature and pressure.)

Interpretation Introduction

Interpretation:

The partial pressures of each gas in atmosphere 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 5.69QP

Partial pressure of the gases in atmosphere is given below:

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

Explanation of Solution

The mole fraction of the gases in atmosphere is given as follows:

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

Given that volume is proportional to the number of moles of gas present, we can directly convert the volume percent to mole fractions.  Thus, the partial pressure of the gases are:

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

Interpretation Introduction

Interpretation:

The concentration of each gas have to be calculated.

Concept Introduction:

Refer part “(a)”.

Answer to Problem 5.69QP

Concentration of each gas is:

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

Explanation of Solution

The concentration of given gas is related to pressure as follows:

  $\begin{array}{c}\text{c}\text{=}\frac{\text{n}}{\text{V}}\\ \text{= }\frac{\text{P}}{\text{RT}}\end{array}$

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

Therefore, the concentration of the gases is calculated as follows:

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

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

The concentration of given gas was:

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

Conclusion

The concentration of each gases was calculated.