Chapter 13, Problem 13.33P

Interpretation Introduction

Interpretation:

Partial pressure of $\text{Ar}$, $\text{He}$ and ${\text{F}}_2$ has to be calculated.

Concept Introduction:

The net pressure for gaseous mixture is equal to algebraic sum of partial pressures of its constituent gases. This is referred as Dalton’s law.

For a mixture of two gases, 1 and 2, $P_{\text{total}}$ is calculated as follows:

  $P_{\text{total}}=P_1+P_2$

In terms of mole fraction the partial pressure is calculated as follows:

  $P_{\text{i}}=x_{\text{i}}{P}_{\text{total}}$

Here,

$P_{\text{total}}$ denotes total pressure exerted by whole mixture of gases.

$x_{\text{i}}$ denotes the mole fraction.

$P_{\text{i}}$ represents the partial pressure.

Answer to Problem 13.33P

Partial pressure of $\text{Ar}$, $\text{He}$ and ${\text{F}}_2$ is $0.05912\text{ atm}$, $1.8227\text{ atm}$ and $0.02604\text{ atm}$ respectively.

Explanation of Solution

The formula to obtain mass from mass percent of given mixture is as follows:

  $\text{Mass}=\left(\text{Mass \%}\right)\left(\text{Mass of mixture}\right)$        (1)

Substitute $\text{0}\text{.622 g}$ for mass of mixture and $22.2\text{ \%}$ for mass percent of $\text{Ar}$ in equation (1).

  $\begin{array}{c}\text{Mass}=\left(\text{22}\text{.2 \%}\right)\left(\text{0}\text{.622 g}\right)\\ =0.138084\text{ g}\end{array}$

Substitute $\text{0}\text{.622 g}$ for mass of mixture and $\text{68}\text{.5 \%}$ for mass percent of $\text{He}$ in equation (1).

  $\begin{array}{c}\text{Mass}=\left(\text{68}\text{.5 \%}\right)\left(\text{0}\text{.622 g}\right)\\ =0.42607\text{ g}\end{array}$

Substitute $\text{0}\text{.622 g}$ for mass of mixture and $\text{9}\text{.3 \%}$ for mass percent of ${\text{F}}_2$ in equation (1).

  $\begin{array}{c}\text{Mass}=\left(\text{9}\text{.3 \%}\right)\left(\text{0}\text{.622 g}\right)\\ =0.057846\text{ g}\end{array}$

The expression to evaluate moles is as follows:

  $\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar Mass}}$        (2)

Substitute $0.138084\text{ g}$ for mass and $\text{39}\text{.948 g/mol}$ for molar mass in equation (2) to calculate moles of $\text{Ar}$.

  $\begin{array}{c}\text{Number of moles}=\frac{0.138084\text{ g}}{\text{39}\text{.948 g/mol}}\\ =0.003456\text{ mol}\end{array}$

Substitute $0.42607\text{ g}$ for mass and $\text{4 g/mol}$ for molar mass in equation (2) to evaluate moles of $\text{He}$.

  $\begin{array}{c}\text{Number of moles}=\frac{0.42607\text{ g}}{\text{4 g/mol}}\\ =0.1065175\text{ mol}\end{array}$

Substitute $0.057846\text{ g}$ for mass and $\text{37}\text{.99681 g/mol}$ for molar mass of ${\text{F}}_2$ in equation (2) to calculate the moles of ${\text{F}}_2$.

  $\begin{array}{c}\text{Number of moles}=\frac{0.057846\text{ g}}{\text{37}\text{.99681 g/mol}}\\ =0.001522\text{ mol}\end{array}$

The conversion factor to convert $\text{atm}$ to $\text{torr}$ is as follows:

  $1\text{ atm}=\text{760 torr}$

Thus, $\text{1450 torr}$ is converted to $\text{atm}$ as follows:

  $\begin{array}{c}\text{Pressure}=\left(\text{1450 torr}\right)\left(\frac{1\text{ atm}}{760\text{ torr}}\right)\\ =1.90789\text{ atm}\end{array}$

The formula to calculate mole fraction of argon is as follows:

  $x_{\text{Ar}}=\frac{n_{\text{Ar}}}{n_{\text{Ar}}+n_{\text{He}}+n_{{\text{F}}_{\text{2}}}}$        (3)

Substitute $0.003456\text{ mol}$ for $n_{\text{Ar}}$, $0.1065175\text{ mol}$ for $n_{\text{He}}$ and $0.001522\text{ mol}$ for $n_{{\text{F}}_{\text{2}}}$ in equation (3).

  $\begin{array}{c}{x}_{\text{Ar}}=\frac{0.003456\text{ mol}}{0.003456\text{ mol}+0.1065175\text{ mol}+0.001522\text{ mol}}\\ =0.03099\end{array}$

The formula to calculate mole fraction of argon is as follows:

  $x_{\text{He}}=\frac{n_{\text{He}}}{n_{\text{Ar}}+n_{\text{He}}+n_{{\text{F}}_{\text{2}}}}$        (4)

Substitute $0.003456\text{ mol}$ for $n_{\text{Ar}}$, $0.1065175\text{ mol}$ for $n_{\text{He}}$ and $0.001522\text{ mol}$ for $n_{{\text{F}}_{\text{2}}}$  in equation (4).

  $\begin{array}{c}{x}_{\text{He}}=\frac{0.1065175\text{ mol}}{0.003456\text{ mol}+0.1065175\text{ mol}+0.001522\text{ mol}}\\ =0.95535\end{array}$

The formula to calculate mole fraction of argon is as follows:

  $x_{{\text{F}}_{\text{2}}}=\frac{n_{{\text{F}}_{\text{2}}}}{n_{\text{Ar}}+n_{\text{He}}+n_{{\text{F}}_{\text{2}}}}$        (5)

Substitute $0.003456\text{ mol}$ for $n_{\text{Ar}}$, $0.1065175\text{ mol}$ for $n_{\text{He}}$ and $0.001522\text{ mol}$ for $n_{{\text{F}}_{\text{2}}}$  in equation (5).

  $\begin{array}{c}{x}_{{\text{F}}_{\text{2}}}=\frac{0.001522\text{ mol}}{0.003456\text{ mol}+0.1065175\text{ mol}+0.001522\text{ mol}}\\ =0.01365\end{array}$

In terms of mole fraction $P_{\text{Ar}}$ is calculated as follows:

  $P_{\text{Ar}}=x_{\text{Ar}}{P}_{\text{total}}$        (6)

Substitute 0.03099 for $x_{\text{Ar}}$ and $1.90789\text{ atm}$ for $P_{\text{total}}$ in equation (6).

  $\begin{array}{c}{P}_{\text{Ar}}=\left(0.03099\right)\left(1.90789\text{ atm}\right)\\ =0.05912\text{ atm}\end{array}$

In terms of mole fraction $P_{\text{Ar}}$ is calculated as follows:

  $P_{\text{He}}=x_{\text{He}}{P}_{\text{total}}$        (7)

Substitute 0.95535 for $x_{\text{He}}$  and $1.90789\text{ atm}$ for $P_{\text{total}}$ in equation (7).

  $\begin{array}{c}{P}_{\text{He}}=\left(0.95535\right)\left(1.90789\text{ atm}\right)\\ =1.8227\text{ atm}\end{array}$

Partial pressure of ${\text{F}}_2$ is calculated as follows:

  $P_{{\text{F}}_{\text{2}}}=x_{{\text{F}}_{\text{2}}}{P}_{\text{total}}$        (8)

Substitute 0.01365 for $x_{{\text{F}}_{\text{2}}}$ and $1.90789\text{ atm}$ for $P_{\text{total}}$ in equation (8).

  $\begin{array}{c}{P}_{{\text{F}}_{\text{2}}}=\left(0.01365\right)\left(1.90789\text{ atm}\right)\\ =0.02604\text{ atm}\end{array}$