Chapter 6, Problem 6.100QP

Interpretation Introduction

Interpretation: The mole fraction of ${\text{H}}_{\text{2}}$, pressure of gaseous mixture and the pressure of each constituent gas is to be calculated.

Concept introduction: The total pressure of a gaseous mixture is sum of the partial pressure of the constituent gases.

To determine: The mole fraction of ${\text{H}}_{\text{2}}$, pressure of gaseous mixture and the pressure of each constituent gas.

Answer to Problem 6.100QP

Solution

The pressure of ${\text{H}}_{\text{2}}$ is $\underset{\_}{22.18atm}$, the pressure of ${\text{N}}_{\text{2}}$ is $\underset{\_}{5.59atm}$, the pressure of ${\text{CH}}_{\text{4}}$ is $\underset{\_}{22.34atm}$, the total pressure is $\underset{\_}{50.11atm}$ and the mole fraction of ${\text{H}}_{\text{2}}$ is $\underset{\_}{0.44}$.

Explanation of Solution

Explanation

Given

The volume of the container is $\text{1}\text{.0 L}$.

The temperature is ${\text{0 }}^{\text{ο}}\text{C}$.

The mass of ${\text{H}}_{\text{2}}$ is $\text{2}\text{.00 g}$.

The mass of ${\text{N}}_{\text{2}}$ is $\text{7}\text{.00 g}$.

The mass of ${\text{CH}}_{\text{4}}$ is $\text{16}\text{.00 g}$.

The conversion of Celsius to Kelvin is done as,

$\text{0}{}^{\text{ο}}\text{C}=273\text{K}$

The partial pressure of ${\text{H}}_{\text{2}}$ ( $P_{{\text{H}}_{\text{2}}}$) is calculated by the formula,

$P_{{\text{H}}_{\text{2}}}=\frac{mRT}{MV}$

Where,

• $m$ is the mass of ${\text{H}}_{\text{2}}$.
• $V$ is the volume.
• $R$ is gas constant ( $0.08206\text{L}\cdot \text{atm/mol}\cdot \text{K}$).
• $M$ is the molar mass of ${\text{H}}_{\text{2}}$.
• $T$ is temperature.

The molar mass of ${\text{H}}_{\text{2}}$ is equal to $\text{2}\text{.02 g/mol}$.

Substitute the value of $m$, $V$, $R$, $M$ and $T$ in the above equation.

$\begin{array}{c}{P}_{{\text{H}}_{\text{2}}}=\frac{\text{2}\text{.00 g}\times 0.08206\text{L}\cdot \text{atm/mol}\cdot \text{K}\times 273\text{K}}{\text{2}\text{.02 g/mol}\times \text{1}\text{.0 L}}\\ =\underset{\_}{22.18atm}\end{array}$

The partial pressure of ${\text{N}}_{\text{2}}$ ( $P_{{\text{N}}_{\text{2}}}$) is calculated by the formula,

$P_{{\text{N}}_{\text{2}}}=\frac{m_{1}RT}{M_{1}V}$

Where,

• $m_1$ is the mass of ${\text{N}}_{\text{2}}$.
• $V$ is the volume.
• $R$ is gas constant ( $0.08206\text{L}\cdot \text{atm/mol}\cdot \text{K}$).
• $M_1$ is the molar mass of ${\text{N}}_{\text{2}}$.
• $T$ is temperature.

The molar mass of ${\text{N}}_{\text{2}}$ is equal to $\text{28}\text{.02 g/mol}$.

Substitute the value of $m_1$, $V$, $R$, $M_1$ and $T$ in the above equation.

$\begin{array}{c}{P}_{{\text{N}}_{\text{2}}}=\frac{\text{7}\text{.00 g}\times 0.08206\text{L}\cdot \text{atm/mol}\cdot \text{K}\times 273\text{K}}{\text{28}\text{.02 g/mol}\times \text{1}\text{.0 L}}\\ =\underset{\_}{5.59atm}\end{array}$

The partial pressure of ${\text{CH}}_{\text{4}}$ ( $P_{{\text{CH}}_{\text{4}}}$) is calculated by the formula,

$P_{{\text{CH}}_{\text{4}}}=\frac{m_{2}RT}{M_{2}V}$

Where,

• $m_2$ is the mass of ${\text{CH}}_{\text{4}}$.
• $V$ is the volume.
• $R$ is gas constant ( $0.08206\text{L}\cdot \text{atm/mol}\cdot \text{K}$).
• $M_2$ is the molar mass of ${\text{CH}}_{\text{4}}$.
• $T$ is temperature.

The molar mass of ${\text{CH}}_{\text{4}}$ is equal to $\text{16}\text{.04 g/mol}$.

Substitute the value of $m_2$, $V$, $R$, $M_2$ and $T$ in the above equation.

$\begin{array}{c}{P}_{{\text{CH}}_{\text{4}}}=\frac{\text{16}\text{.00 g}\times 0.08206\text{L}\cdot \text{atm/mol}\cdot \text{K}\times 273\text{K}}{\text{16}\text{.04 g/mol}\times \text{1}\text{.0 L}}\\ =\underset{\_}{22.34atm}\end{array}$

The pressure of the gas mixture ( $P_{\text{total}}$)is calculated as,

$P_{\text{total}}=P_{{\text{H}}_{\text{2}}}+P_{{\text{N}}_{\text{2}}}+P_{{\text{CH}}_{\text{4}}}$

Substitute the value of $P_{{\text{H}}_{\text{2}}}$, $P_{{\text{N}}_{\text{2}}}$ and $P_{{\text{CH}}_{\text{4}}}$ in the above equation.

$\begin{array}{c}{P}_{\text{total}}=22.1\text{8 atm}+5.5\text{9 atm}+22.34\text{ atm}\\ =\underset{\_}{50.11atm}\end{array}$

The mole fraction of ${\text{H}}_{\text{2}}$ ( ${\chi }_{H_2}$)is calculated as,

${\chi }_{H_2}=\frac{P_{{\text{H}}_{\text{2}}}}{P_{\text{total}}}$

Substitute the value of $P_{{\text{H}}_{\text{2}}}$ and $P_{\text{total}}$ in the above equation.

$\begin{array}{c}{\chi }_{H_2}=\frac{22.1\text{8 atm}}{50.\text{11 atm}}\\ =\underset{\_}{0.44}\end{array}$

Conclusion

The pressure of ${\text{H}}_{\text{2}}$ is $\underset{\_}{22.18atm}$, the pressure of ${\text{N}}_{\text{2}}$ is $\underset{\_}{5.59atm}$, the pressure of ${\text{CH}}_{\text{4}}$ is $\underset{\_}{22.34atm}$, the total pressure is $\underset{\_}{50.11atm}$ and the mole fraction of ${\text{H}}_{\text{2}}$ is $\underset{\_}{0.44}$