Chapter 11, Problem 11.70QP

Interpretation Introduction

Interpretation:  The partial pressures of the given gases have to be calculated.

Concept Introduction:

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.

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

The number of molecules of gases is known. Since the mole fraction is:

$\frac{{\text{n}}_{\text{i}}}{{\text{n}}_{\text{total}}}\text{=}\frac{\frac{\text{molecules}\text{of}\text{i}}{\text{6}\text{.022×}\text{10}{}^{\text{23}}\text{molecules/mol}}}{\frac{\text{molecules}\text{total}}{\text{6}\text{.022×}\text{10}{}^{\text{23}}\text{molecules/mol}}}\text{=}\frac{\text{molecules}\text{of}\text{i}}{\text{molecules}\text{total}}$

• 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.

Answer to Problem 11.70QP

The partial pressure of ${\text{CH}}_{\text{4}}$ gas was found to be $\text{0}\text{.54}\text{atm}$

The partial pressure of ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}$ gas was found to be $\text{0}\text{.44}\text{atm}$

The partial pressure of ${\text{C}}_{\text{3}}{\text{H}}_{\text{8}}$ gas was found to be $\text{0}\text{.51}\text{atm}$

Explanation of Solution

First, we calculate the mole fraction Methane from the mixture.  Then, we can calculate the partial pressure of each component.

The number of moles of the combined gases is:

${\text{n=n}}_{{\text{CH}}_{\text{4}}}\text{+}{\text{n}}_{{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}}\text{+}{\text{n}}_{{\text{C}}_{\text{3}}{\text{H}}_{\text{8}}}\text{=}\text{0}\text{.31}\text{mol}\text{+}\text{0}\text{.25}\text{mol}\text{+}\text{0}\text{.29}\text{mol}\text{=}\text{0}\text{.85}\text{mol}$

$\begin{array}{l}{\text{χ}}_{{\text{CH}}_{\text{4}}}\text{=}\frac{\text{0}\text{.31}\text{mol}}{\text{0}\text{.85}\text{mol}}\text{=}\text{0}\text{.36}\\ \text{The partial pressure of Methane}\\ {\text{P}}_{{\text{CH}}_{\text{4}}}\text{=}{\text{χ}}_{{\text{CH}}_{\text{4}}}\text{×}{\text{P}}_{\text{total}}\text{=}\text{0}\text{.36}\text{×}\text{1}\text{.50}\text{atm}\text{=}\text{0}\text{.54}\text{atm}\end{array}$

The partial pressure of ${\text{CH}}_{\text{4}}$ gas is calculated by plugging in the values of the given pressure of gas and number of moles of the combined gas.  The partial pressure of ${\text{CH}}_{\text{4}}$ gas was found to be $\text{0}\text{.54}\text{atm}$

Second, we calculate the mole fraction Ethane from the mixture.  Then, we can calculate the partial pressure of each component.

The number of moles of the combined gases is:

${\text{n=n}}_{{\text{CH}}_{\text{4}}}\text{+}{\text{n}}_{{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}}\text{+}{\text{n}}_{{\text{C}}_{\text{3}}{\text{H}}_{\text{8}}}\text{=}\text{0}\text{.31}\text{mol}\text{+}\text{0}\text{.25}\text{mol}\text{+}\text{0}\text{.29}\text{mol}\text{=}\text{0}\text{.85}\text{mol}$

$\begin{array}{l}{\text{χ}}_{{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}}\text{=}\frac{\text{0}\text{.25}\text{mol}}{\text{0}\text{.85}\text{mol}}\text{=}\text{0}\text{.29}\\ \text{The partial pressure of Ethane}\\ {\text{P}}_{{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}}\text{=}{\text{χ}}_{{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}}\text{×}{\text{P}}_{\text{total}}\text{=}\text{0}\text{.29}\text{×}\text{1}\text{.50}\text{atm}\text{=}\text{0}\text{.44}\text{atm}\end{array}$

The partial pressure of ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}$ gas is calculated by plugging in the values of the given pressure of gas and number of moles of the combined gas.  The partial pressure of ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}$ gas was found to be $\text{0}\text{.44}\text{atm}$

Third, we calculate the mole fraction Propane from the mixture.  Then, we can calculate the partial pressure of each component.

The number of moles of the combined gases is:

${\text{n=n}}_{{\text{CH}}_{\text{4}}}\text{+}{\text{n}}_{{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}}\text{+}{\text{n}}_{{\text{C}}_{\text{3}}{\text{H}}_{\text{8}}}\text{=}\text{0}\text{.31}\text{mol}\text{+}\text{0}\text{.25}\text{mol}\text{+}\text{0}\text{.29}\text{mol}\text{=}\text{0}\text{.85}\text{mol}$

$\begin{array}{l}{\text{χ}}_{{\text{C}}_{\text{3}}{\text{H}}_{\text{8}}}\text{=}\frac{\text{0}\text{.29}\text{mol}}{\text{0}\text{.85}\text{mol}}\text{=}\text{0}\text{.34}\\ \text{The partial pressure of Propane}\\ {\text{P}}_{{\text{C}}_{\text{3}}{\text{H}}_{\text{8}}}\text{=}{\text{χ}}_{{}_{{\text{C}}_{\text{3}}{\text{H}}_{\text{8}}}}\text{×}{\text{P}}_{\text{total}}\text{=}\text{0}\text{.34}\text{×}\text{1}\text{.50}\text{atm}\text{=}\text{0}\text{.51}\text{atm}\end{array}$

The partial pressure of ${\text{C}}_{\text{3}}{\text{H}}_{\text{8}}$ gas is calculated by plugging in the values of the given pressure of gas and number of moles of the combined gas.  The partial pressure of ${\text{C}}_{\text{3}}{\text{H}}_{\text{8}}$ gas was found to be $\text{0}\text{.51}\text{atm}$

Conclusion

The partial pressures of the given gases were calculated.