Chapter 18, Problem 121AE

a.

Interpretation Introduction

Interpretation:The mass of Xe in the given room needs to be calculated.

Concept introduction: The expression relating the pressure, P, volume, V with number of moles, n and temperature, T of an ideal gas is:

  $\text{PV = nRT}$ - (1)

Where R is Universal gas constant.

Answer to Problem 121AE

  $\text{0}\text{.177 g}$ is the mass of Xe in the given room.

Explanation of Solution

Given:

The volume of Xe in atmosphere = $\text{9}\text{.0}\times {\text{10}}^{-6}\text{ \%}$ at ${25}^o\text{ C}$ and $1\text{ atm}$ .

The dimensions of room is 7.26 m by 8.80 m by 5.67 m.

The volume of the room is:

  $\begin{array}{l}\text{Volume = 7}\text{.26 m}\times \text{8}\text{.80 m}\times \text{5}\text{.26 m}\\ \text{Volume = = 336}{\text{.05 m}}^3\end{array}$

Converting volume from m³to L as:

  ${\text{1 m}}^3={\text{10}}^3\text{ L}$

  $\text{= 336}{\text{.05 m}}^3\text{ = 336}\text{.05 }\times {\text{ 10}}^3\text{ L}$

The amount of Xe present in the atmosphere is:

  $\begin{array}{l}\text{9}\text{.0 }\times {\text{ 10}}^{-6}\text{ \% of Xe = }\left(\frac{\text{9}\text{.0 }\times {\text{ 10}}^{-6}}{100}\right)\\ \text{= 9}\text{.0 }\times {\text{ 10}}^{-8}\end{array}$

So, $\text{9}\text{.0 }\times {\text{ 10}}^{-8}$ is the amount of Xe present at the given conditions (temperature and pressure).

The amount of Xe present in the room is:

  $\begin{array}{l}\text{Volume of Xenon = volume of room }\times \text{ amount of Xenon in air}\\ \text{Volume of Xenon = 336}\text{.05 }\times {\text{ 10}}^3\text{ L }\times \text{ 9}\text{.0 }\times {\text{ 10}}^{-8}\\ \text{Volume of Xenon = 3}\text{.02 }\times {\text{ 10}}^{-2}\text{ L}\end{array}$

Now, the number of moles of Xe present in the room is determined using equation (1):

  $\text{PV = nRT}$

The values to be substituted are:

  $\text{P = 1}\text{.0 atm}$ , $\text{V = }3.3\times {\text{ 10}}^{-2}\text{ L}$ , $\text{R = 0}{\text{.08206 Latm K}}^{-1}{\text{mol}}^{-1}$ and $\text{T = 25+273 = 298 K}$

  $\begin{array}{l}\text{n = }\frac{\text{PV}}{\text{RT}}\\ \text{n = }\frac{\text{(1}\text{.0 atm) (}3.3\times {\text{10}}^{-2}\text{ L) }}{\text{(0}{\text{.08206 Latm K}}^{-1}{\text{mol}}^{-1}\text{) (}298\text{ K)}}\\ \text{n}=\frac{3.3\times {\text{10}}^{-2}}{24.44}\text{ mol}\\ \text{n = 1}\text{.35}\times {\text{10}}^{-3}\text{ mol}\end{array}$

Now, the mass of Xe is determined using formula:

  $\text{number of moles, n = }\frac{\text{mass, m}}{\text{molar mass, M}}$

Rearranging as:

  $\text{Mass = n }\times \text{ M}$

Molar mass of Xe is $\text{131}\text{.3 g/mol}$ .

Substituting the values:

  $\begin{array}{l}\text{Mass = 1}\text{.35 }\times {\text{ 10}}^{-3}\text{ mol }\times \text{ 131}\text{.3 g /mol}\\ \text{Mass = 0}\text{.177 g}\end{array}$

Thus, the mass of Xe in the given room is $\text{0}\text{.177 g}$ .

b.

Interpretation Introduction

Interpretation: The number of Xe atoms inhaled in each breath needs to be calculated when the volume of air taken by a person is about 2 L.

Concept introduction: The expression relating the pressure, P, volume, V with number of moles, n and temperature, T of an ideal gas is:

  $\text{PV = nRT}$ - (1)

Where R is Universal gas constant.

Answer to Problem 121AE

  $\text{5}\times {\text{10}}^{15}\text{ atoms}$ is the number of Xe atoms inhaled in each breath.

Explanation of Solution

The amount of Xe in $2\text{ L}$ air is:

  $\begin{array}{l}\text{Volume of Xe in 2 L air = Xe per L }\times \text{ given amount}\\ \text{Volume of Xe in 2 L air = 9}\text{.0 }\times {\text{ 10}}^{-8}\times \text{ 2 L}\\ \text{Volume of Xe in 2 L air = 1}\text{.8 }\times {\text{ 10}}^{-7}\text{ L}\end{array}$

Now, the number of moles of Xe present in the breath is determined using equation (1):

  $\text{PV = nRT}$

The values to be substituted are:

  $\text{P = 1}\text{.0 atm}$ , $\text{V = 1}\text{.8 }\times {\text{ 10}}^{-7}\text{ L}$ , $\text{R = 0}{\text{.08206 Latm K}}^{-1}{\text{mol}}^{-1}$ and $\text{T = 25+273 = 298 K}$

  $\begin{array}{l}\text{n = }\frac{\text{PV}}{\text{RT}}\\ \text{n = }\frac{\text{(1}\text{.0 atm) (1}\text{.8 }\times {\text{ 10}}^{-7}\text{ L) }}{\text{(0}{\text{.08206 Latm K}}^{-1}{\text{mol}}^{-1}\text{) (}298\text{ K)}}\\ \text{n = 7}\text{.4 }\times {\text{ 10}}^{-9}\text{ mol}\end{array}$

Now, the number of atoms ofXe is determined using formula:

  $\text{Number of atoms = n }\times {\text{ N}}_{\text{A}}$

Where ${\text{N}}_{\text{A}}$ is Avogadro’s constant having value $\text{6}\text{.022 }\times {\text{ 10}}^{23}\text{ atoms/mol}$ .

  $\begin{array}{l}\text{Number of atoms = 7}\text{.4}\times {\text{10}}^{-9}\text{ mol }\times \text{ 6}\text{.022}\times {\text{10}}^{23}\text{ atoms/mol}\\ \text{Number of atoms = 4}\text{.5}\times {\text{10}}^{15}\text{ atoms}\\ \text{Number of atoms }\simeq \text{ 5}\times {\text{10}}^{15}\text{ atoms}\end{array}$

Thus, the number of Xe atoms inhaled in each breath is $\text{5}\times {\text{10}}^{15}\text{ atoms}$ .