Chapter 5, Problem 49E

(a)

Interpretation Introduction

Interpretation:Mole fraction of ${\text{CH}}_{\text{4}}$ and ${\text{O}}_{\text{2}}$ gas is to be calculated.

Concept introduction: The assimilation of various gas laws namely Boyle’s law, Charles law and Avogadro’s results in formulation of ideal gas equation as given below:

  $PV=nRT$

Where,

• $R$ is gas constant.
• $V$ denotes the volume.
• $n$ denotes number of moles.
• $T$ denotes temperature.
• $P$ denotes pressure.

The formula to convert degree Celsius to kelvin is as follows:

  $T\left(\text{K}\right)=T\left(°\text{C}\right)+273\text{ K}$

Answer to Problem 49E

Mole fraction of ${\text{CH}}_{\text{4}}$ and ${\text{O}}_{\text{2}}$ is 0.4117 and 0.588 respectively.

Explanation of Solution

Given Information:

Pressure of ${\text{CH}}_{\text{4}}$ is $\text{0}\text{.175 atm}$ .

Pressure of ${\text{O}}_{\text{2}}$ is $\text{0}\text{.250 atm}$ .

The total pressure as per the Dalton’s law of partial pressure is calculated as follows:

  $P_{\text{total}}=P_{{\text{CH}}_{\text{4}}}+P_{{\text{O}}_2}$

The value of $P_{{\text{CH}}_{\text{4}}}$ is $\text{0}\text{.175 atm}$ .

The value of $P_{{\text{O}}_2}$ is $\text{0}\text{.250 atm}$ .

Substitute the value in above formula.

  $\begin{array}{c}{P}_{\text{total}}=P_{{\text{CH}}_{\text{4}}}+P_{{\text{O}}_2}\\ =\text{0}\text{.175 atm}+\text{0}\text{.250 atm}\\ =\text{0}\text{.425 atm}\end{array}$

The formula to calculate mole fraction of ${\text{CH}}_{\text{4}}$ is as follows:

  ${\chi }_{{\text{CH}}_{\text{4}}}=\frac{P_{{\text{CH}}_{\text{4}}}}{P_{\text{Total}}}$

The value of $P_{{\text{CH}}_{\text{4}}}$ is $\text{0}\text{.175 atm}$ .

The value of $P_{\text{total}}$ is $\text{0}\text{.425 atm}$ .

Substitute the value in above formula.

  $\begin{array}{c}{\chi }_{{\text{CH}}_{\text{4}}}=\frac{P_{{\text{CH}}_{\text{4}}}}{P_{\text{Total}}}\\ =\frac{\text{0}\text{.175 atm}}{\text{0}\text{.425 atm}}\\ =0.4117\end{array}$

The formula to calculate mole fraction of ${\text{O}}_{\text{2}}$ is as follows:

  ${\chi }_{{\text{O}}_2}=\frac{P_{{\text{O}}_2}}{P_{\text{Total}}}$

The value of $P_{{\text{CH}}_{\text{4}}}$ is $\text{0}\text{.250 atm}$ .

The value of $P_{\text{total}}$ is $\text{0}\text{.425 atm}$ .

Substitute the value in above formula.

  $\begin{array}{c}{\chi }_{{\text{O}}_2}=\frac{P_{{\text{O}}_2}}{P_{\text{Total}}}\\ =\frac{\text{0}\text{.250 atm}}{\text{0}\text{.425 atm}}\\ =0.588\end{array}$

Hence, mole fraction of ${\text{CH}}_{\text{4}}$ and ${\text{O}}_{\text{2}}$ is 0.4117 and 0.588 respectively.

(b)

Interpretation Introduction

Interpretation:Total moles of gas in mixture are to be calculated.

Concept introduction: The assimilation of various gas laws namely Boyle’s law, Charles law and Avogadro’s results in formulation of ideal gas equation as given below:

  $PV=nRT$

Where,

• $R$ is gas constant.
• $V$ denotes the volume.
• $n$ denotes number of moles.
• $T$ denotes temperature.
• $P$ denotes pressure.

The formula to convert degree Celsius to kelvin is as follows:

  $T\left(\text{K}\right)=T\left(°\text{C}\right)+273\text{ K}$

Answer to Problem 49E

Total moles of gas in mixture are $0.16089\text{ mol}$ .

Explanation of Solution

The formula to convert degree Celsius to kelvin is as follows:

  $T\left(\text{K}\right)=T\left(°\text{C}\right)+273\text{ K}$

The temperature in Celsius is $\text{65 }°\text{C}$ .

Substitute the value in above formula.

  $\begin{array}{c}T\left(\text{K}\right)=T\left(°\text{C}\right)+273\text{ K}\\ =6\text{5 }°\text{C}+273\text{ K}\\ =338\text{ K}\end{array}$

The formula to calculate moles as per ideal gas law is as follows:

  $n=\frac{PV}{RT}$

The value of $P$ is $\text{0}\text{.425 atm}$ .

The value of $V$ is $\text{10}\text{.5 L}$ .

The value of $T$ is $338\text{ K}$ .

The value of $R$ is $0.08206\text{ L}\cdot \text{atm/mol}\cdot \text{K}$ .

Substitute the values in above equation.

  $\begin{array}{c}n=\frac{PV}{RT}\\ =\frac{\left(\text{0}\text{.425 atm}\right)\left(\text{10}\text{.5 L}\right)}{\left(0.08206\text{ L}\cdot \text{atm/mol}\cdot \text{K}\right)\left(338\text{ K}\right)}\\ =0.16089\text{ mol}\end{array}$

Thus, total moles of gas in mixture is $0.16089\text{ mol}$ .

(c)

Interpretation Introduction

Interpretation:Masses of ${\text{CH}}_{\text{4}}$ and ${\text{O}}_{\text{2}}$ gas in mixture are to be calculated.

Concept introduction:The formula to convert degree Celsius to kelvin is as follows:

  $T\left(\text{K}\right)=T\left(°\text{C}\right)+273\text{ K}$

The formula to calculate mass from moles is given as follows:

  $\text{Mass}=\left(\text{Number of moles}\right)\left(\text{Molar mass}\right)$

Answer to Problem 49E

Masses of ${\text{CH}}_{\text{4}}$ and ${\text{O}}_{\text{2}}$ is $1.0627\text{ g}$ and $3.027\text{ g}$ respectively.

Explanation of Solution

Since mole fraction of methane is 0.4118 so moles of methane is calculated as follows:

  $\begin{array}{c}{\text{Moles of CH}}_4=\left(0.4118\right)\left(0.16089\text{ mol}\right)\\ =0.06625\text{ mol}\end{array}$

Since mole fraction of methane is 0.588 so moles of methane is calculated as follows:

  $\begin{array}{c}{\text{Moles of O}}_{\text{2}}=\left(0.588\right)\left(0.16089\text{ mol}\right)\\ =0.0946033\text{ mol}\end{array}$

The formula to calculate mass from moles is given as follows:

  $\text{Mass}=\left(\text{Number of moles}\right)\left(\text{Molar mass}\right)$

Number of moles is $0.06625\text{ mol}$ .

Molar mass is $16.04\text{ g/mol}$ .

Substitute the values in above formula.

  $\begin{array}{c}\text{Mass}=\left(0.06625\text{ mol}\right)\left(16.04\text{ g/mol}\right)\\ =1.0627\text{ g}\end{array}$

For ${\text{O}}_{\text{2}}$

Number of moles is $0.0946033\text{ mol}$ .

Molar mass is $\text{32 g/mol}$ .

Substitute the values in above formula.

  $\begin{array}{c}\text{Mass}=\left(0.0946033\text{ mol}\right)\left(\text{32 g/mol}\right)\\ =3.027\text{ g}\end{array}$