Chapter 20, Problem 70E

Interpretation Introduction

Interpretation:

The temperature, pressure and dimension of a room is given. The mass and number of argon atoms in the room is to be calculated. Also, the number of argon atoms inhaled in one breath is to be calculated. The difference in the health risk between argon gas and radon gas is to be explained.

Concept introduction:

The expression to calculate the number of moles of an atom is,

$\begin{array}{c}\text{PV=nRT}\\ \text{n=}\frac{\text{PV}}{\text{RT}}\end{array}$

Answer to Problem 70E

• The mass of argon atom in the given volume of room is $\underset{\_}{1.5\times 10^{4}g}$.
• The number of argon atoms in the given volume of room is $\underset{\_}{2.2\times 10^{26}atoms}$.
• The number of argon atoms inhaled in one breath is $\underset{\_}{4.5\times 10^{20}atoms}$.
• The main difference between argon and radon is that, radon is a radioactive element whereas argon is not a radioactive element.

The mass and number of argon atoms in the room at $25°\text{C}$ and $1\text{atm}$; the number of argon atoms inhaled in one breath; the explanation of the difference in the health risk between argon gas and radon gas.

Explanation of Solution

Explanation

Given

Dimension of room is $10.0\text{m}\times 10.0\text{m}\times 10.0\text{m}$.

Temperature is $25°\text{C}$.

Pressure is $1.0\text{}\text{}\text{atm}$.

Refer to Table $\text{19-22}$.

Atmospheric abundance of argon is $9.0\times {10}^{-1}$.

Volume of room is equal to the dimension. The volume of room is,

$10.0\text{m}\times 10.0\text{m}\times 10.0\text{m}=1.0\times {10}^3{\text{m}}^3$

The conversion of cubic meter $\left({\text{m}}^3\right)$ into liter $\left(\text{L}\right)$ is done as,

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

Hence, the conversion of $1.0\times {10}^3{\text{m}}^3$ into liter is,

$\begin{array}{c}\text{1}\text{.0}\times {\text{10}}^{\text{3}}{\text{m}}^3=\left(\text{1}\text{.0}\times {\text{10}}^{\text{3}}\times {\text{10}}^{\text{3}}\right)\text{L}\\ =\text{1}\text{.0}\times {\text{10}}^{\text{6}}\text{L}\end{array}$

Volume of argon in the room is $\text{1}\text{.0}\times {\text{10}}^{\text{6}}\text{L}$. But, the percentage of atmospheric abundance of argon is $9.0\times {10}^{-1}$.

Hence, the overall volume of argon is,

$\text{1}\text{.0}\times {\text{10}}^{\text{6}}\text{L}\left(\frac{\text{9}\text{.0}\times {\text{10}}^{-\text{1}}}{\text{100}}\right)=\text{9}\text{.0}\times {\text{10}}^{\text{3}}\text{L}$

The conversion of degree Celsius $\left(°\text{C}\right)$ into Kelvin $\left(\text{K}\right)$ is done as,

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

Hence, the conversion of $\text{25}°\text{C}$ into Kelvin is,

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

Formula

Number of moles of argon is calculated using the formula,

$\begin{array}{c}\text{PV=nRT}\\ \text{n=}\frac{\text{PV}}{\text{RT}}\end{array}$

Where,

• $P$ is the total pressure.
• $V$ is the volume.
• $n$ is the total moles.
• $R$ is the universal gas constant $(\text{0}\text{.08206}\text{L}\cdot \text{atm}/\text{K}\cdot \text{mol})$.
• $T$ is the absolute temperature.

Substitute the values of $P,V,R$ and $T$ in the above equation.

$\begin{array}{c}n=\frac{PV}{RT}\\ =\frac{\text{9}\text{.0}\times {\text{10}}^{\text{3}}\text{L}\left(\text{1}\text{.0}\text{atm}\right)}{(\text{0}\text{.08206}\text{L}\cdot \text{atm}/\text{K}\cdot \text{mol})\left(\text{298}\text{K}\right)}\\ =\underset{\_}{368mol}\end{array}$

The number of moles of argon is $368\text{mol}$.

The atomic mass of argon is $39.95\text{g/mol}$.

Formula

The mass of argon is calculated using the formula,

$\text{m = nM}$

Where,

• $\text{m}$ is the mass of argon.
• $\text{M}$ is the atomic mass of argon.
• $\text{n}$ is the number of moles of argon.

Substitute the values of $\text{M}$ and $\text{n}$ in the above equation.

$\begin{array}{c}\text{m = nM}\\ \text{=386}\text{mol×39}\text{.95}\text{g/mol}\\ \text{=}\underset{\_}{\text{1}{\text{.5×10}}^{\text{4}}\text{g}}\end{array}$

The number of moles of argon is $368\text{mol}$.

Formula

The number of argon atoms is calculated using the formula,

$\text{N = nA}$

Where,

• $N$ is the number of argon atoms.
• $A$ is the Avogadro’s number $(6.023\times {10}^{23})$.
• $n$ is the number of moles of argon.

Substitute the values of $\text{A}$ and $\text{n}$ in the above equation.

$\begin{array}{c}\text{N = nA}\\ \text{= (6}{\text{.023×10}}^{\text{23}}\text{)}\left(\text{386}\right)\\ \text{=}\underset{\_}{\text{ 2}{\text{.2×10}}^{\text{26}}\text{atoms}}\end{array}$

Given

Volume of air in the room is $\text{2}\text{.0}\text{L}$.

Temperature is $25°\text{C}$.

Pressure is $1.0\text{}\text{}\text{atm}$.

The conversion of degree Celsius $\left(°\text{C}\right)$ into Kelvin $\left(\text{K}\right)$ is done as,

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

Hence, the conversion of $\text{25}°\text{C}$ into Kelvin is,

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

Given volume of air in the room is $\text{2}\text{.0}\text{L}$. Since, the percentage atmospheric abundance of argon is $9.0\times {10}^{-1}$.

Hence, the overall volume of argon breathed is,

$\text{2}\text{.0}\text{L}\left(\frac{\text{9}\text{.0}\times {\text{10}}^{-\text{1}}}{\text{100}}\right)=\text{0}\text{.018}\text{L}$

Formula

Number of moles of argon is calculated using the formula,

$\begin{array}{c}\text{PV=nRT}\\ \text{n=}\frac{\text{PV}}{\text{RT}}\end{array}$

Where,

• $P$ is the total pressure.
• $V$ is the volume.
• $n$ is the total moles.
• $R$ is the universal gas constant $(\text{0}\text{.08206}\text{L}\cdot \text{atm}/\text{K}\cdot \text{mol})$.
• $T$ is the absolute temperature.

Substitute the values of $P,V,R$ and $T$ in the above equation.

$\begin{array}{c}\text{n = }\frac{\text{PV}}{\text{RT}}\\ \text{=}\frac{\text{0}\text{.018}\text{L}\left(\text{1}\text{.0}\text{atm}\right)}{\text{(0}\text{.08206}\text{L×atm/K×mol)}\left(\text{298}\text{K}\right)}\\ \text{=}\underset{\_}{\text{7}{\text{.4×10}}^{\text{-4}}\text{mol}}\end{array}$

Number of moles of argon taken in one breath is $7.4\times {10}^{-4}\text{mol}$.

Formula

The number of argon atoms is calculated using the formula,

$N=nA$

Where,

• $N$ is the number of argon atoms.
• $A$ is the Avogadro’s number $(6.023\times {10}^{23})$.
• $n$ is the number of moles of argon.

Substitute the values of $A$ and $n$ in the above equation.

$\begin{array}{c}N=nA\\ =(6.023\times {10}^{23})\left(7.4\times {10}^{-4}\right)\\ =\underset{\_}{4.5\times 10^{20}atoms}\end{array}$

Both argon and radon are inert gas.  But radon is a radioactive element.  The prolonged exposure to this element may cause lung cancer.

Conclusion

The required value of mass of argon atom in the given volume of room is $\underset{\_}{1.5\times 10^{4}g}$. The number of argon atoms in the given volume of room is $\underset{\_}{2.2\times 10^{26}atoms}$.  The number of argon atoms inhaled in one breath is $\underset{\_}{4.5\times 10^{20}atoms}$.