Chapter 3, Problem 3C.4E

(a)

Interpretation Introduction

Interpretation:

Total pressure of mixture has to be determined.

Concept Introduction:

Dolton’s law of partial pressure gives relation between total pressure of mixture of gases and partial pressure of individual gases. The expression of relation can be represented as follows:

  $P=P_{\text{A}}+P_{\text{B}}+\cdot \cdot \cdot$

Here,

A and B are individual gases.

$P_{\text{A}}$ is partial pressure of gas A.

$P_{\text{B}}$ is partial pressure of gas B.

$P$ is total pressure of mixture of gases.

Mole fraction is a relation between moles of constituent and total moles of all constituent in mixture. The expression of relation can be represented as follows:

  $x_{\text{A}}=\frac{n_{\text{A}}}{n}$

Here,

$x_{\text{A}}$ is mole fraction of gas A.

$n_{\text{A}}$ are moles of gas A.

$n$ is sum moles of all gases.

Answer to Problem 3C.4E

Partial pressure of ${\text{N}}_2$ and ${\text{H}}_2$ is $1.0\text{ atm}$ and $2.0\text{ atm}$ respectively.

Explanation of Solution

The expression to calculate number of moles of ${\text{CH}}_4$ is as follows:

  $\text{Mole}\left({\text{CH}}_4\right)=\frac{\text{Mass}\left(\text{g}\right){\text{ CH}}_4}{\text{Molar mass}\left(\text{g/mol}\right){\text{ CH}}_4}$        (1)

Substitute $376\text{ mg}$ for mass of ${\text{CH}}_4$ and $\text{16}\text{.04 g/mol}$ for molar mass in equation (1).

  $\begin{array}{c}\text{Mole}\left({\text{CH}}_4\right)=\left(\frac{376\text{ mg}}{\text{16}\text{.04 g/mol}}\right)\left(\frac{{10}^{-3}\text{ g}}{1\text{ mg}}\right)\\ =2.34\times {10}^{-2}\text{ mol}\end{array}$

The expression to calculate number of moles of $\text{Ar}$ is as follows:

  $\text{Mole}\left(\text{Ar}\right)=\frac{\text{Mass}\left(\text{g}\right)\text{ Ar}}{\text{Molar mass}\left(\text{g/mol}\right)\text{ Ar}}$        (2)

Substitute $154\text{ mg}$ for mass of $\text{Ar}$ and $39.95\text{ g/mol}$ for molar mass in equation (2).

  $\begin{array}{c}\text{Mole}\left(\text{Ar}\right)=\left(\frac{154\text{ mg}}{39.95\text{ g/mol}}\right)\left(\frac{{10}^{-3}\text{ g}}{1\text{ mg}}\right)\\ =3.85\times {10}^{-3}\text{ mol}\end{array}$

The expression to calculate number of moles of ${\text{N}}_2$ is as follows:

  $\text{Mole}\left({\text{N}}_2\right)=\frac{\text{Mass}\left(\text{g}\right){\text{ N}}_2}{\text{Molar mass}\left(\text{g/mol}\right){\text{ N}}_2}$        (3)

Substitute $252\text{ mg}$ for mass of ${\text{N}}_2$ and $28.02\text{ g/mol}$ for molar mass in equation (3).

  $\begin{array}{c}\text{Mole}\left({\text{N}}_2\right)=\left(\frac{252\text{ mg}}{28.02\text{ g/mol}}\right)\left(\frac{{10}^{-3}\text{ g}}{1\text{ mg}}\right)\\ =8.99\times {10}^{-3}\text{ mol}\end{array}$

Therefore, the mole fraction for ${\text{N}}_2$ gas can be calculated as follows:

  $\begin{array}{c}\text{Mole fraction}\left({\text{N}}_2\right)=\frac{\left({\text{mol N}}_2\right)}{\left({\text{mol CH}}_4\right)+\left(\text{mol Ar}\right)+\left({\text{mol N}}_2\right)}\\ =\frac{\left(8.99\times {10}^{-3}\text{ mol}\right)}{\left(2.34\times {10}^{-2}\text{ mol}\right)+\left(3.85\times {10}^{-3}\text{ mol}\right)+\left(8.99\times {10}^{-3}\text{ mol}\right)}\\ =0.688\end{array}$

The expression to calculate total pressure form partial pressure nitrogen gas is as follows:

  $P=\frac{P_{{\text{N}}_2}}{x_{{\text{N}}_2}}$        (4)

Here,

$x_{{\text{N}}_2}$ is mole fraction of ${\text{N}}_2$ gas.

$P_{{\text{N}}_2}$ is partial pressure of ${\text{N}}_2$ gas.

$P$ is total pressure of mixture of gases.

Substitute $0.688$ for $x_{{\text{N}}_2}$ and $21.3\text{ kPa}$ for $P_{{\text{N}}_2}$ in equation (3).

  $\begin{array}{c}P=\frac{21.3\text{ kPa}}{0.688}\\ =31.0\text{ kPa}\end{array}$

(b)

Interpretation Introduction

Interpretation:

Volume of sample mixture has to be determined.

Concept Introduction:

An ideal gas contains a large number of randomly moving particles that are supposed to have perfectly elastic collisions among themselves. It is a theoretical concept. Gases that show perfect elastic collision are practically not possible. At higher $T$ and lower $P$ gases behave almost ideally.

Ideal gas law is an equation of state for any hypothetical gas. Expression for ideal gas equation is as follows:

  $PV=nRT$

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 3C.4E

Volume of sample mixture is $1.05\text{ L}$.

Explanation of Solution

Total number of moles of sample can be calculated as follows:

  $\begin{array}{c}\text{Mole}=\text{Mole}\left({\text{CH}}_4\right)+\text{Mole}\left(\text{Ar}\right)+\text{Mole}\left({\text{N}}_2\right)\\ =\left(2.34\times {10}^{-2}\text{ mol}\right)+\left(3.85\times {10}^{-3}\text{ mol}\right)+\left(8.99\times {10}^{-3}\text{ mol}\right)\\ =1.31\times {10}^{-2}\text{ mol}\end{array}$

The expression to calculate volume of sample is as follows:

  $V=\frac{nRT}{P}$        (5)

Substitute $1.31\times {10}^{-2}\text{ mol}$ for $n$, $31.0\text{ kPa}$ for $P$, $8.31446\text{ L}\cdot \text{kPa}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$ for $R$ and $300\text{ K}$ for $T$ in equation (5).

  $\begin{array}{c}V=\frac{\left(1.31\times {10}^{-2}\text{ mol}\right)\left(8.31446\text{ L}\cdot \text{kPa}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(300\text{ K}\right)}{\left(31.0\text{ kPa}\right)}\\ =1.05\text{ L}\end{array}$