Chapter 11.7, Problem 13PPA

Interpretation Introduction

Interpretation:

The mole fraction and partial pressure of ${\text{CO}}_{\text{2}}\text{,}{\text{CH}}_{\text{4}}\text{and}\text{He}$ has to be determined.

Concept introduction:

• Dalton's law of partial pressure states to the sum of the partial pressure exerted each component of the mixture the total pressure exerted by a gas mixture.

${\text{P}}_{\text{Total}}\text{}{\text{=P}}_{\text{gas 1}}\text{}{\text{+P}}_{\text{gas 2}}\text{}{\text{+P}}_{\text{gas 3}}\text{}\mathrm{....}$

• A mole fraction is a unitless expression.  It is just a number of moles of a single component that is divided by total number of moles of considered solution.
• • ${\text{χ}}_{\text{i}}\text{=}\frac{{\text{P}}_{\text{i}}}{{\text{P}}_{\text{total}}}$

To determine: The mole fraction and partial pressure of ${\text{CO}}_{\text{2}}\text{,}{\text{CH}}_{\text{4}}\text{and}\text{He}$

Answer to Problem 13PPA

Solution:

The mole fraction of ${\text{CO}}_{\text{2}}$ is $\text{0}\text{.0495}$

The mole fraction of ${\text{CH}}_{\text{4}}$ is $\text{0}\text{.695}$

The mole fraction of $\text{He}$ is $\text{0}\text{.695}$

The partial pressure of ${\text{CO}}_{\text{2}}$ is $\text{0}\text{.286}\text{atm}$

The partial pressure of ${\text{CH}}_{\text{4}}$ is $\text{1}\text{.47}\text{atm}$

The partial pressure of $\text{He}$ is $\text{4}\text{.02}\text{atm}$

Explanation of Solution

Given,

The total pressure is $\text{5}\text{.78 atm}$

The mole fraction and partial pressure of ${\text{CO}}_{\text{2}}\text{,}{\text{CH}}_{\text{4}}\text{and}\text{He}$ is calculated below

To find the total moles of gas sample is sum of the moles of each component of the mixture

$\begin{array}{l}{\text{n}}_{\text{Total}}\text{=}{\text{n}}_{{\text{CO}}_{\text{2}}}{\text{+n}}_{{\text{CH}}_{\text{4}}}{\text{+n}}_{\text{He}}\\ \text{=}\text{(0}\text{.25}\text{mol}\text{+}\text{1}\text{.29}\text{mol}\text{+}\text{3}\text{.51}\text{mol)}\\ \text{=}\text{5}\text{.05}\text{mol}\end{array}$

To find the mole fraction of ${\text{CO}}_{\text{2}}$

$\begin{array}{l}{\text{χ}}_{{\text{CO}}_{\text{2}}}\text{=}\frac{{\text{n}}_{{\text{CO}}_{\text{2}}}}{{\text{n}}_{\text{total}}}\\ \text{=}\frac{\text{1}\text{.250}\text{mol}}{\text{5}\text{.05}\text{mol}}\\ \text{=}\text{0}\text{.0495}\end{array}$

To find the mole fraction of ${\text{CH}}_{\text{4}}$

$\begin{array}{l}{\text{χ}}_{{\text{CH}}_{\text{4}}}\text{=}\frac{{\text{n}}_{{\text{CH}}_{\text{4}}}}{{\text{n}}_{\text{total}}}\\ \text{=}\frac{\text{1}\text{.29}\text{mol}}{\text{5}\text{.05}\text{mol}}\\ \text{=}\text{0}\text{.255}\end{array}$

To find the mole fraction of $\text{He}$

$\begin{array}{l}{\text{χ}}_{\text{He}}\text{=}\frac{{\text{n}}_{\text{He}}}{{\text{n}}_{\text{total}}}\\ \text{=}\frac{\text{3}\text{.51}\text{mol}}{\text{5}\text{.05}\text{mol}}\\ \text{=}\text{0}\text{.695}\end{array}$

The partial pressure of each gas component in the mixture is the total pressure multiplied by its mole fraction

To find the partial pressure of ${\text{CO}}_{\text{2}}$

$\begin{array}{l}{\text{P}}_{{\text{CO}}_{\text{2}}}\text{=}{\text{χ}}_{{\text{CO}}_{\text{2}}}{\text{P}}_{\text{total}}\\ \text{=0}\text{.0495×5}\text{.78}\text{atm}\\ \text{=}\text{0}\text{.286}\text{atm}\end{array}$

To find the partial pressure of ${\text{CH}}_{\text{4}}$

$\begin{array}{l}{\text{P}}_{{\text{CH}}_{\text{4}}}\text{=}{\text{χ}}_{{\text{CH}}_{\text{4}}}{\text{P}}_{\text{total}}\\ \text{=0}\text{.255×5}\text{.78}\text{atm}\\ \text{=}\text{1}\text{.47}\text{atm}\end{array}$

To find the partial pressure of $\text{He}$

$\begin{array}{l}{\text{P}}_{\text{He}}\text{=}{\text{χ}}_{\text{He}}{\text{P}}_{\text{total}}\\ \text{=0}\text{.695×5}\text{.78}\text{atm}\\ \text{=}\text{4}\text{.02}\text{atm}\end{array}$

Conclusion

The mole fraction and partial pressure of ${\text{CO}}_{\text{2}}\text{,}{\text{CH}}_{\text{4}}\text{and}\text{He}$ was determined.