Chapter 13, Problem 43A

Interpretation Introduction

Interpretation: -Volume of a gas and partial pressure of each gas in a mixture needs to be calculated using variable values given.

Concept Introduction: - For volume calculation first of all one have to calculate number of moles (n ) of each gas.

  $\mathrm{n}\text{ = }\frac{\text{mass}}{\text{molar mass}}$

Then, volume is calculated by Ideal gas equation

  $\begin{array}{c}\mathrm{P}\mathrm{V}=\mathrm{n}\mathrm{R}\mathrm{T}\\ \mathrm{V}=\frac{\mathrm{n}}{\mathrm{P}}\mathrm{R}\mathrm{T}\end{array}$

For calculating partial pressure, Dalton’s law of partial pressure is used

  ${\mathrm{P}}_{\mathrm{T}}={\mathrm{P}}_1+{\mathrm{P}}_2+{\mathrm{P}}_3+{\mathrm{P}}_4$

Answer to Problem 43A

Volume of the mixture is 15.01mL

Partial pressure of gases:

  $\begin{array}{c}{\mathrm{P}}_{{\mathrm{O}}_2}=0.22\mathrm{a}\mathrm{t}\mathrm{m}\\ {\mathrm{P}}_{{\mathrm{N}}_2}=0.25\mathrm{a}\mathrm{t}\mathrm{m}\\ {\mathrm{P}}_{\mathrm{C}{\mathrm{O}}_2}=0.16\mathrm{a}\mathrm{t}\mathrm{m}\\ {\mathrm{P}}_{\mathrm{N}\mathrm{e}}=0.35\mathrm{a}\mathrm{t}\mathrm{m}\end{array}$

Explanation of Solution

There is a relation between volume of gas and molecules of gas present in mixture which is expressed by Avogadro’s law

  $\mathrm{V}\propto \mathrm{N}$

Now,

  $\mathrm{V}=\frac{\mathrm{n}}{{\mathrm{N}}_{\mathrm{A}}}$ { ${\mathrm{N}}_{\mathrm{A}}$ = Avogadro Number}

  $\mathrm{n}\propto \mathrm{N}$

After calculating number of moles, volume is calculated.

  $\text{No}\text{. of moles (}\mathrm{n}\text{) = }\frac{\text{mass}}{\text{molar mass}}$

Putting the values,

  $\begin{array}{c}\mathrm{N}\mathrm{o}.\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{s}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}{\mathrm{O}}_2\mathrm{}=\mathrm{}\frac{5}{32}\\ =0.15\mathrm{m}\mathrm{o}\mathrm{l}\\ {\mathrm{n}}_1=0.15\mathrm{m}\mathrm{o}\mathrm{l}.\end{array}$

Also,

  $\begin{array}{c}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{s}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}{\mathrm{N}}_2\mathrm{}=\mathrm{}\frac{5}{28.014}\\ =0.17\mathrm{m}\mathrm{o}\mathrm{l}\\ {\mathrm{n}}_2=0.17\mathrm{m}\mathrm{o}\mathrm{l}.\end{array}$

Or,

  $\begin{array}{c}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{s}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{C}{\mathrm{O}}_2\mathrm{}=\mathrm{}\frac{5}{44.01}\\ =0.11\mathrm{m}\mathrm{o}\mathrm{l}\\ {\mathrm{n}}_3=0.11\mathrm{m}\mathrm{o}\mathrm{l}.\end{array}$

Now,

  $\begin{array}{c}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{s}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{N}\mathrm{e}\mathrm{}=\mathrm{}\frac{5}{20.17}\\ =0.24\mathrm{m}\mathrm{o}\mathrm{l}\\ {\mathrm{n}}_4=0.24\mathrm{m}\mathrm{o}\mathrm{l}.\end{array}$

Therefore,

  $\begin{array}{l}{\mathrm{n}}_{\mathrm{T}}={\mathrm{n}}_1+{\mathrm{n}}_2+{\mathrm{n}}_3+{\mathrm{n}}_4\\ {\mathrm{n}}_{\mathrm{T}}=0.15+0.17+0.11+0.24\\ {\mathrm{n}}_{\mathrm{T}}=0.67\mathrm{m}\mathrm{o}\mathrm{l}.\end{array}$

Volume of mixture of gases will be:

  $\begin{array}{c}\mathrm{P}\mathrm{V}={\mathrm{n}}_{\mathrm{T}}\mathrm{R}\mathrm{T}\\ \mathrm{V}=\frac{{\mathrm{n}}_{\mathrm{T}}\mathrm{R}\mathrm{T}}{\mathrm{P}}\end{array}$

Putting the values,

  $\begin{array}{l}\mathrm{V}=\frac{0.67\mathrm{m}\mathrm{o}\mathrm{l}}{1\mathrm{a}\mathrm{t}\mathrm{m}}\times \frac{0.0821\mathrm{a}\mathrm{t}\mathrm{m}\mathrm{}\mathrm{L}}{\mathrm{K}\mathrm{m}\mathrm{o}\mathrm{l}}\times 273\mathrm{K}\\ \mathrm{V}=0.67\times 0.0821\times 273\\ \mathrm{V}=15.01\mathrm{L}\end{array}$

Volume of the mixture of gas is 15.01 L

For calculating Partial pressure of gases present in mixture, Dalton law is applied

  ${\mathrm{P}}_{\mathrm{T}}={\mathrm{P}}_1+{\mathrm{P}}_2+{\mathrm{P}}_3+{\mathrm{P}}_4$

For calculating partial pressure of individual gases.

  $\frac{\mathrm{P}}{{\mathrm{P}}_{\mathrm{T}}}=\frac{\mathrm{n}}{{\mathrm{n}}_{\mathrm{T}}}$

Where P = Partial Pressure of gas

  ${\mathrm{P}}_{\mathrm{T}}$ = Total pressure

  $\mathrm{n}$ = no. of moles of gas

  ${\mathrm{n}}_{\mathrm{T}}$ = Total moles.

1) Partial Pressure of Oxygen

  $\begin{array}{c}\frac{{\mathrm{P}}_{{\mathrm{O}}_2}}{{\mathrm{P}}_{\mathrm{T}}}=\frac{{\mathrm{n}}_{{\mathrm{O}}_2}}{{\mathrm{n}}_{\mathrm{T}}}\\ {\mathrm{P}}_{{\mathrm{O}}_2}=\frac{\mathrm{n}{\mathrm{o}}_2}{{\mathrm{n}}_{\mathrm{T}}}\times {\mathrm{P}}_{\mathrm{T}}\\ {\mathrm{P}}_{{\mathrm{O}}_2}=\frac{0.15}{0.67}\times 1\mathrm{a}\mathrm{t}\mathrm{m}\\ =0.22\mathrm{a}\mathrm{t}\mathrm{m}\end{array}$

2) Partial Pressure of ${\mathrm{H}}_2$

  $\begin{array}{c}\frac{{\mathrm{P}}_{{\text{H}}_{\text{2}}}}{{\mathrm{P}}_{\mathrm{T}}}=\frac{{\mathrm{n}}_{{\text{H}}_{\text{2}}}}{{\mathrm{n}}_{\mathrm{T}}}\\ {\mathrm{P}}_{{\text{H}}_{\text{2}}}=\frac{{\mathrm{n}}_{{\text{H}}_{\text{2}}}}{{\mathrm{n}}_{\mathrm{T}}}\times {\mathrm{P}}_{\mathrm{T}}\\ {\mathrm{P}}_{{\text{H}}_{\text{2}}}=\frac{0.17}{0.67}\times 1\mathrm{a}\mathrm{t}\mathrm{m}\\ =0.25\mathrm{a}\mathrm{t}\mathrm{m}\end{array}$

1) Partial Pressure of ${\text{CO}}_{\text{2}}$

  $\begin{array}{c}\frac{{\mathrm{P}}_{\mathrm{C}{\mathrm{O}}_2}}{{\mathrm{P}}_{\mathrm{T}}}=\frac{{\mathrm{n}}_{\mathrm{C}{\mathrm{O}}_2}}{{\mathrm{n}}_{\mathrm{T}}}\\ {\mathrm{P}}_{\mathrm{C}{\mathrm{O}}_2}=\frac{{\mathrm{n}}_{\mathrm{C}{\mathrm{O}}_2}}{{\mathrm{n}}_{\mathrm{T}}}\times {\mathrm{P}}_{\mathrm{T}}\\ {\mathrm{P}}_{\mathrm{C}{\mathrm{O}}_2}=\frac{0.11}{0.67}\times 1\mathrm{a}\mathrm{t}\mathrm{m}\\ =0.16\mathrm{a}\mathrm{t}\mathrm{m}\end{array}$

4) Partial Pressure of Ne

  $\begin{array}{c}\frac{{\mathrm{P}}_{\mathrm{N}\mathrm{e}}}{{\mathrm{P}}_{\mathrm{T}}}=\frac{{\mathrm{n}}_{\mathrm{N}\mathrm{e}}}{{\mathrm{n}}_{\mathrm{T}}}\\ {\mathrm{P}}_{\mathrm{N}\mathrm{e}}=\frac{{\mathrm{n}}_{\mathrm{N}\mathrm{e}}}{{\mathrm{n}}_{\mathrm{T}}}\times {\mathrm{P}}_{\mathrm{T}}\\ {\mathrm{P}}_{{\mathrm{O}}_2}=\frac{0.24}{0.67}\times 1\mathrm{a}\mathrm{t}\mathrm{m}\\ =0.35\mathrm{a}\mathrm{t}\mathrm{m}\end{array}$

Conclusion

Partial pressure of gases:

  $\begin{array}{c}\mathrm{a})\mathrm{}{\mathrm{P}}_{{\mathrm{O}}_2}=0.22\mathrm{a}\mathrm{t}\mathrm{m}\\ \mathrm{b})\mathrm{}{\mathrm{P}}_{{\mathrm{N}}_2}=0.25\mathrm{a}\mathrm{t}\mathrm{m}\\ \mathrm{c})\mathrm{}{\mathrm{P}}_{\mathrm{C}{\mathrm{O}}_2}=0.16\mathrm{a}\mathrm{t}\mathrm{m}\\ \mathrm{d})\mathrm{}{\mathrm{P}}_{\mathrm{N}\mathrm{e}}=0.35\mathrm{a}\mathrm{t}\mathrm{m}\end{array}$

Volume of mixture = 15.01L