Chapter 13.2, Problem 5RQ

Interpretation Introduction

Interpretation:

The partial pressures of each gas have to be calculated.

Concept Introduction:

Dalton’s law of partial pressures for a mixture containing three gases can be expressed as-

  ${\mathrm{P}}_{\mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}}={\mathrm{P}}_1+{\mathrm{P}}_2+{\mathrm{P}}_3$

Where the pressures P₁, P₂, and P₃ represents the partial pressures of each gas. Each gas is responsible for the total pressure.

  $\begin{array}{l}{\mathrm{P}}_1=\frac{{\mathrm{n}}_1\mathrm{R}\mathrm{T}}{\mathrm{V}},{\mathrm{P}}_2=\frac{{\mathrm{n}}_2\mathrm{R}\mathrm{T}}{\mathrm{V}}\mathrm{a}\mathrm{n}\mathrm{d}{\mathrm{P}}_3=\frac{{\mathrm{n}}_3\mathrm{R}\mathrm{T}}{\mathrm{V}}\end{array}$

       $\begin{array}{c}{\mathrm{P}}_{\mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}}={\mathrm{P}}_1+{\mathrm{P}}_2+{\mathrm{P}}_3\end{array}$

       $\begin{array}{c}{\mathrm{P}}_{\mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}}=\frac{{\mathrm{n}}_1\mathrm{R}\mathrm{T}}{\mathrm{V}}+\frac{{\mathrm{n}}_2\mathrm{R}\mathrm{T}}{\mathrm{V}}+\frac{{\mathrm{n}}_3\mathrm{R}\mathrm{T}}{\mathrm{V}}\end{array}$

        $\begin{array}{c}=\frac{\mathrm{R}\mathrm{T}}{\mathrm{V}}({\mathrm{n}}_1+{\mathrm{n}}_2+{\mathrm{n}}_3)\end{array}$

       $\begin{array}{c}={\mathrm{n}}_{\mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}}\left(\frac{\mathrm{R}\mathrm{T}}{\mathrm{V}}\right)\end{array}$

Where n_total is the sum of the numbers of moles of the gases in the mixture.

Answer to Problem 5RQ

The partial pressure of CO₂, O₂ and N₂ in the mixture is 1266.7 torr, 2533.33 torr and 3800 torr respectively.

Explanation of Solution

Given Information: n₁ = 1.0 mole, n₂ = 2.0 mole and n₃ = 3.0 mole

Total pressure = 10.0 atm

n₁ and P₁ for CO₂

n₂ and P₂ for O₂

n₃ and P₃ for N₂

Total number of moles ‘n_total’ = (1.0 + 2.0 + 3.0) mol = 6.0 mole

  $\begin{array}{c}{\mathrm{P}}_{\mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}}={\mathrm{n}}_{\mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}}\left(\frac{\mathrm{R}\mathrm{T}}{\mathrm{V}}\right)\end{array}$

  $\begin{array}{c}\left(\frac{\mathrm{R}\mathrm{T}}{\mathrm{V}}\right)=\frac{10}{6}\mathrm{a}\mathrm{t}\mathrm{m}/\mathrm{m}\mathrm{o}\mathrm{l}\end{array}$

The partial pressure of CO₂ in the mixture:

  $\begin{array}{c}{\mathrm{P}}_1={\mathrm{n}}_1\left(\frac{\mathrm{R}\mathrm{T}}{\mathrm{V}}\right)\end{array}$

     $\begin{array}{c}=1.0\mathrm{m}\mathrm{o}\mathrm{l}\times \frac{10}{6}\mathrm{a}\mathrm{t}\mathrm{m}/\mathrm{m}\mathrm{o}\mathrm{l}\end{array}$

    $\begin{array}{c}=1.67\mathrm{a}\mathrm{t}\mathrm{m}\end{array}$

   $\begin{array}{c}=1.67\times 760\mathrm{t}\mathrm{o}\mathrm{r}\mathrm{r}\end{array}$

    $\begin{array}{c}=1266.7\mathrm{t}\mathrm{o}\mathrm{r}\mathrm{r}\end{array}$

The partial pressure of O₂ in the mixture:

  $\begin{array}{c}{\mathrm{P}}_2={\mathrm{n}}_2\left(\frac{\mathrm{R}\mathrm{T}}{\mathrm{V}}\right)\end{array}$

     $\begin{array}{c}=2.0\mathrm{m}\mathrm{o}\mathrm{l}\times \frac{10}{6}\mathrm{a}\mathrm{t}\mathrm{m}/\mathrm{m}\mathrm{o}\mathrm{l}\end{array}$

     $\begin{array}{c}=3.33\mathrm{a}\mathrm{t}\mathrm{m}\end{array}$

    $\begin{array}{c}=3.33\times 760\mathrm{t}\mathrm{o}\mathrm{r}\mathrm{r}\end{array}$

     $\begin{array}{c}=2533.3\mathrm{t}\mathrm{o}\mathrm{r}\mathrm{r}\end{array}$

The partial pressure of N₂ in the mixture:

  $\begin{array}{c}{\mathrm{P}}_3={\mathrm{n}}_3\left(\frac{\mathrm{R}\mathrm{T}}{\mathrm{V}}\right)\end{array}$

    $\begin{array}{c}\text{=}\text{3}\text{.0}\text{mol}\text{×}\frac{\text{10}}{\text{6}}\text{atm/mol}\end{array}$

    $\begin{array}{c}\text{=}\text{5}\text{.0}\text{atm}\end{array}$

    $\begin{array}{c}\text{=}\text{5}\text{.0}\text{×}\text{760}\text{torr}\end{array}$

    $\begin{array}{c}\text{=}\text{3800}\text{torr}\end{array}$