Chapter 13, Problem 13.31P

Interpretation Introduction

Interpretation:

Partial pressure of ${\text{H}}_{\text{2}}$ and ${\text{N}}_{\text{2}}$ has to be calculated.

Concept Introduction:

The net pressure for gaseous mixture is equal to the algebraic sum of partial pressures of its constituent gases. This is referred to 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 13.31P

The partial pressure of ${\text{H}}_{\text{2}}$ is $0.61187\text{ atm}$ and ${\text{N}}_{\text{2}}$ is $1.3818atm$.

Explanation of Solution

The formula to convert mass in grams to moles is as follows:

  $\text{Number of moles}=\frac{\text{Given mass}}{\text{molar mass}}$        (1)

Substitute $\text{0}\text{.513 g}$ for mass and $\text{2}\text{.0158 g/mol}$ for molar mass of ${\text{H}}_{\text{2}}$ in equation (1) to calculate the moles of ${\text{H}}_{\text{2}}$.

  $\begin{array}{c}\text{Number of moles}=\frac{\text{0}\text{.513 g}}{\text{2}\text{.0158 g/mol}}\\ =0.254489\text{ mol}\end{array}$

Substitute $\text{16}\text{.1 g}$ for mass and $\text{28}\text{.0134 g/mol}$ for molar mass of ${\text{N}}_{\text{2}}$ to calculate the moles of ${\text{N}}_{\text{2}}$ in equation (1).

  $\begin{array}{c}\text{Number of moles}=\frac{\text{16}\text{.1 g}}{\text{28}\text{.0134 g/mol}}\\ =0.57473\text{ mol}\end{array}$

The formula to calculate mole fraction is as follows:

  $x_{{\text{H}}_{\text{2}}}=\frac{n_{{\text{H}}_{\text{2}}}}{n_{{\text{N}}_2}+n_{{\text{H}}_{\text{2}}}}$        (2)

Here,

$n_{{\text{H}}_{\text{2}}}$ denotes the number of moles of ${\text{H}}_{\text{2}}$.

$n_{{\text{N}}_2}$ denotes the number of moles of ${\text{N}}_{\text{2}}$.

$x_{{\text{H}}_{\text{2}}}$ denotes the mole fraction of ${\text{H}}_{\text{2}}$.

Substitute $0.254489\text{ mol}$ for $n_{{\text{H}}_{\text{2}}}$, $0.57473\text{ mol}$ for $n_{{\text{N}}_2}$ in equation (2).

  $\begin{array}{c}{x}_{{\text{H}}_{\text{2}}}=\frac{0.254489\text{ mol}}{0.254489\text{ mol}+0.57473\text{ mol}}\\ =0.30690\end{array}$

The total number of moles in mixture is calculated as follows:

  $n_{\text{total}}=n_{{\text{H}}_{\text{2}}}+n_{{\text{N}}_2}$        (3)

Substitute $0.254489\text{ mol}$ for $n_{{\text{H}}_{\text{2}}}$, $0.57473\text{ mol}$ for $n_{{\text{N}}_2}$ in equation (3).

  $\begin{array}{c}{n}_{\text{total}}=0.254489\text{ mol}+0.57473\text{ mol}\\ =\text{0}\text{.829219 mol}\end{array}$

The formula to convert Celsius to kelvin is as follows:

  $T\left(\text{K}\right)=T\left(°\text{C}\right)+273\text{ K}$        (4)

Substitute $\text{20 °C}$ for $T\left(°\text{C}\right)$ in equation (4).

  $\begin{array}{c}T\left(\text{K}\right)=\text{20 °C}+273\text{ K}\\ =293\text{ K}\end{array}$

The expression to calculate pressure as per the ideal gas equation is as follows:

  $P=\frac{nRT}{V}$        (5)

Substitute $293\text{ K}$ for $T$, $\text{10 L}$ for $V$, $\text{0}\text{.829219 mol}$ for $n$, and $0.08206\text{ L}\cdot \text{atm/mol}\cdot \text{K}$ for $R$ in equation (5) to calculate total pressure.

  $\begin{array}{c}P=\frac{\left(\text{0}\text{.829219 mol}\right)\left(0.08206\text{ L}\cdot \text{atm/mol}\cdot \text{K}\right)\left(293\text{ K}\right)}{\left(\text{10 L}\right)}\\ =1.9937\text{ atm}\end{array}$

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

  $P_{{\text{H}}_{\text{2}}}=x_{{\text{H}}_{\text{2}}}{P}_{\text{total}}$        (6)

Here,

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

$x_{{\text{H}}_{\text{2}}}$ denotes the mole fraction of ${\text{H}}_{\text{2}}$.

$P_{{\text{H}}_{\text{2}}}$ denotes the partial pressure exerted by ${\text{H}}_{\text{2}}$.

Substitute 0.30690 for $x_{{\text{H}}_{\text{2}}}$ and $1.9937\text{ atm}$ for $P_{\text{total}}$ in equation (6).

  $\begin{array}{c}{P}_{{\text{H}}_{\text{2}}}=\left(0.30690\right)\left(\text{1}\text{.9937 atm}\right)\\ =0.61187\text{ atm}\end{array}$

The relation of mole fraction that is true is given as follows:

  $x_{{\text{N}}_{\text{2}}}=1-x_{{\text{H}}_{\text{2}}}$        (7)

Substitute 0.30690 for $x_{{\text{H}}_{\text{2}}}$ in equation (7).

  $\begin{array}{c}{x}_{{\text{N}}_{\text{2}}}=1-x_{{\text{H}}_{\text{2}}}\\ =1-0.30690\\ =0.6931\end{array}$

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

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

Here,

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

$x_{{\text{N}}_{\text{2}}}$ denotes the mole fraction of ${\text{N}}_{\text{2}}$.

$P_{{\text{N}}_{\text{2}}}$ denotes the partial pressure exerted by ${\text{N}}_{\text{2}}$.

Substitute 0.6931 for $x_{{\text{N}}_{\text{2}}}$ and $1.9937\text{ atm}$ for $P_{\text{total}}$ in equation (8).

  $\begin{array}{c}{P}_{N_2}=\left(0.6931\right)\left(1.9937atm\right)\\ =1.3818atm\end{array}$