Chapter 14.3, Problem 28SP

Interpretation Introduction

Interpretation: The mass in grams of air a child's lungs hold is to be determined.

Concept introduction: Ideal gas, or a perfect gas, is a gas whose physical behavior corresponds to a specific idealized relation between pressure, volume, and temperature, known as the ideal or general gas law.

Answer to Problem 28SP

The mass in grams of air a child's lungs hold is $2.523\text{ g}$ .

Explanation of Solution

The volume of a child's lungs holds is $2.20\text{ L}$ .

The pressure of air is $102\text{ kPa}$ .

The temperature of a child's lungs is ${37}^{\circ }\text{C}$ .

The conversion of degree Celcius into kelvin is done as follows:

  $T\left(\text{K}\right)=T\left({}^{\circ }\text{C}\right)+273.15$

Convert ${37}^{\circ }\text{C}$ into K as follows:

  $\begin{array}{c}T\left(\text{K}\right)=T\left({}^{\circ }\text{C}\right)+273.15\\ \text{=}37+273.15\\ =310.15\text{ K}\end{array}$

Therefore, the temperature in kelvin is $310.15\text{ K}$ .

The value of a gas constant is $8.314{\text{ kPa L K}}^{-1}{\text{ mol}}^{-1}$ .

The ideal gas is as follows:

  $PV=nRT$   ...... (1)

Where,

• $P$ is the pressure.
• $n$ is the number of moles.
• $T$ is the temperature.
• $V$ is the volume.
• $R$ is a gas constant.

Substitute the values of $P$ , $R$ , $T$ and $V$ in equation (1) to calculate the number of moles of air in a sphere as follows:

  $\begin{array}{c}PV=nRT\\ n=\frac{PV}{RT}\\ =\frac{102\bcancel{\text{kPa}}\times 2.20\bcancel{\text{L}}}{8.314\bcancel{\text{kPa}}\bcancel{\text{L }}\bcancel{{\text{K}}^{-1}}{\text{ mol}}^{-1}\times 310.15\bcancel{\text{K}}}\\ =0.0870\text{ mol}\end{array}$

Therefore, the number of moles of air is $0.0870\text{ mol}$ .

If $29\text{ g}$ of air contains $1\text{ mol}$ of air.

The mass of air contains $0.0870\text{ mol}$ of air is as follows:

  $\begin{array}{c}\text{Mass of air}=\frac{29\text{ g}}{1\bcancel{\text{mol}}}\times 0.0870\bcancel{\text{mol}}\\ =2.523\text{ g}\end{array}$

Therefore, the mass of air contains $0.0870\text{ mol}$ is $2.523\text{ g}$ .

Conclusion

The mass in grams of air a child's lungs hold is $2.523\text{ g}$ .