Chapter 12, Problem 69AE

(a)

Interpretation Introduction

Interpretation:

Total moles of gas in mixturehave to be determined.

Concept Introduction:

Ideal gas law represented as equation to relate its volume and pressure with its temperature. Expression for ideal gas equation is as follows:

  $PV=nRT$        (1)

Here,

$P$ is pressure of the gas.

$V$ is volume of gas.

$n$ denotes moles of gas.

$R$ is gas constant.

$T$ is temperature of gas.

Answer to Problem 69AE

Total moles of gas in mixture are $0.085\text{ mol}$.

Explanation of Solution

Rearrange expression (1) to calculate value of $n$ as follows:

  $n=\frac{PV}{RT}$        (2)

Conversion factor to convert $25°\text{C}$ to Kelvin is as follows:

  $\begin{array}{c}T\left(\text{K}\right)=T\left(°\text{C}\right)+273\\ =25°\text{C}+273\\ =298\text{K}\end{array}$

Substitute $\text{2}\text{.0 L}$ for $V$, $790\text{ torr}$ for $P$, $62.364\text{ L}\cdot \text{torr}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$ for $R$, and $298\text{K}$ for $T$ in equation (2).

  $\begin{array}{c}n=\frac{\left(790\text{ torr}\right)\left(\text{2}\text{.0 L}\right)}{\left(62.364\text{ L}\cdot \text{torr}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(298\text{K}\right)}\\ =0.085\text{ mol}\end{array}$

Hence, the total moles of gas in mixture are $0.085\text{ mol}$.

(b)

Interpretation Introduction

Interpretation:

The number of grams of nitrogen in mixture of gas in mixture has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 69AE

The number of grams of nitrogen in mixture of gas in mixture is $1.5\text{ g}$.

Explanation of Solution

Moles of oxygen gas can be calculated as follows:

  $\begin{array}{c}\text{Moles}=\frac{0.65\text{ g}}{32.0\text{ g/mol}}\\ =0.020\text{ mol}\end{array}$

Moles of carbon dioxide can be calculated as follows:

  $\begin{array}{c}\text{Moles}=\frac{0.58\text{ g}}{44.01\text{ g/mol}}\\ =0.013\text{ mol}\end{array}$

Moles of nitrogen in mixture can be calculated as follows:

  $\begin{array}{c}\text{Moles of N}=\left(\text{Total moles}\right)-\left({\text{Moles of O}}_2\right)-\left({\text{Moles of CO}}_2\right)\\ =0.085\text{ mol}-0.020\text{ mol}-0.013\text{ mol}\\ =0.052\text{ mol}\end{array}$

Mass of $0.052\text{ mol}$ nitrogen gas can be calculated as follows:

  $\begin{array}{c}\text{Mass}=\left(\text{moles}\right)\left(\text{molar mass}\right)\\ =\left(0.052\text{ mol}\right)\left(28.0\text{ g/mol}\right)\\ =1.5\text{ g}\end{array}$

Hence, the grams of nitrogen in mixture of gas in mixture are $1.5\text{ g}$.

(c)

Interpretation Introduction

Interpretation:

Partial pressure of each gas in mixture has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 69AE

Values of $p_{{\text{O}}_2}$, $p_{{\text{CO}}_2}$, and $p_{{\text{N}}_2}$ are $19\text{ torr}$, $12\text{ torr}$, and $790\text{ torr}$ respectively.

Explanation of Solution

Mole fraction of ${\text{O}}_2$ in mixture is can be calculated as follows:

  $\begin{array}{c}{x}_{{\text{O}}_2}=\frac{{\text{moles of O}}_2}{\text{Total moles}}\\ =\frac{0.020\text{ mol}}{0.85}\\ =0.024\end{array}$

Therefore, partial pressure of ${\text{O}}_2$ gas can be calculated as follows:

  $\begin{array}{c}{p}_{{\text{O}}_2}=x_{{\text{O}}_2}\cdot P_{\text{T}}\\ =\left(0.024\right)\left(790\text{ torr}\right)\\ =19\text{ torr}\end{array}$

Mole fraction of ${\text{CO}}_2$ in mixture can be calculated as follows:

  $\begin{array}{c}{x}_{{\text{CO}}_2}=\frac{{\text{moles of CO}}_2}{\text{Total moles}}\\ =\frac{0.013\text{ mol}}{0.85}\\ =0.015\end{array}$

Therefore, partial pressure of ${\text{CO}}_2$ gas can be calculated as follows:

  $\begin{array}{c}{p}_{{\text{CO}}_2}=x_{{\text{CO}}_2}\cdot P_{\text{T}}\\ =\left(0.015\right)\left(790\text{ torr}\right)\\ =12\text{ torr}\end{array}$

Partial pressure of ${\text{N}}_2$ gas can be calculated as follows:

  $\begin{array}{c}{p}_{{\text{N}}_2}=P_{\text{T}}-\left(p_{{\text{O}}_2}+p_{{\text{CO}}_2}\right)\\ =790\text{ torr}-\left(19\text{ torr}+12\text{ torr}\right)\\ =759\text{ torr}\end{array}$

Hence, the value of $p_{{\text{O}}_2}$, $p_{{\text{CO}}_2}$, and $p_{{\text{N}}_2}$ are $19\text{ torr}$, $12\text{ torr}$, and $790\text{ torr}$ respectively.