Chapter 1, Problem 1.76P

(a)

Interpretation Introduction

Interpretation:

The number of particles present in the air for the given space is to be determined.

Concept Introduction:

The air in the atmosphere is a mixture of different gases and particles. Usually, air contains a balanced composition of gases and particles. However, emissions from industries and automobiles can increase the amount of some gases and particles in air. It causes an imbalance of the atmosphere which can lead to some harmful outcomes.

Answer to Problem 1.76P

The number of particles present in the air for the given space is $7.15\times {10}^7$ .

Explanation of Solution

Assuming the particles in air are spherical structures with a dimeter of $2.5\mu m$ , the volume of a single particle can be calculated. The equation for the calculation of the volume of a sphere is as follows.

  $V=\frac{4}{3}\pi r^3$

Where, $V$ is the volume of the sphere, and $r$ is the radius of the sphere.The diameter of the sphere is given in the question.

  $\begin{array}{l}r=\frac{d}{2}=\frac{2.5}{2}=1.25\mu m\\ V=\frac{4}{3}\pi r^3\\ $ V=\frac{4}{3}\times 3.14\times {(1.25)}^3=8.17\mu m^3$\end{array}$

Thereafter, the total volume of the room has to be calculated. The dimensions of the room are given in the units of feet, thus, the units should be converted into meters.

  $V_{room}=\frac{10}{3.28}\times \frac{8.25}{3.28}\times \frac{12.5}{3.28}{m}^3=29.21m^3$

The equation can be used to determine the number of particles in the room.

  $\text{Standard density=}\frac{\text{mass of a particle}\times \text{number of particles}}{\text{total volume of the room}}$

Therefore, first, the mass of a particles should be calculated.

  $\begin{array}{l}\text{Mass of a particle=density}\times \text{volume of a particle}\\ \text{Mass of a particle=2}{\text{.5gcm}}^{-3}\times 8.17\mu m^3=2.5\times {10}^{6}g{m}^{-3}\times 8.17\times {10}^{-18}{m}^3=20.425\times {10}^{-12}g\end{array}$

Then the obtained values can be used in the above mentioned equation to get the number of particles in the room.

  $\begin{array}{l}\text{50}\text{.0}\times {\text{10}}^{-6}{\text{gm}}^{-3}\text{=}\frac{\text{20}\text{.425}\times {\text{10}}^{-12}\text{g}\times \text{number of particles}}{29.21m^3}\\ \text{number of particles}=\frac{\text{50}\text{.0}\times {\text{10}}^{-6}\times 29.21}{\text{20}\text{.425}\times {\text{10}}^{-12}}=7.15\times {10}^7\end{array}$

(b)

Interpretation Introduction

Interpretation:

The amount of particles in air per breath a human takes is to be calculated.

Concept Introduction:

The air in the atmosphere is a mixture of different gases and particles. Usually, air contains a balanced composition of gases and particles. However, emissions from industries and automobiles can increase the amount of some gases and particles in air. It causes an imbalance of the atmosphere which can lead to some harmful outcomes.

Answer to Problem 1.76P

The allowable amount of particles in air per breath a human takes is $1.22\times {10}^3$ .

Explanation of Solution

Assuming the particles in air are spherical structures with a dimeter of $2.5\mu m$ , the volume of a single particle can be calculated. The equation for the calculation of the volume of a sphere is as follows.

  $V=\frac{4}{3}\pi r^3$

Where, $V$ is the volume of the sphere, and $r$ is the radius of the sphere. The diameter of the sphere is given in the question.

  $\begin{array}{l}r=\frac{d}{2}=\frac{2.5}{2}=1.25\mu m\\ V=\frac{4}{3}\pi r^3\\ $ V=\frac{4}{3}\times 3.14\times {(1.25)}^3=8.17\mu m^3$\end{array}$

Mass of a particle can be calculated using this volume.

  $\begin{array}{l}\text{Mass of a particle=density}\times \text{volume of a particle}\\ \text{Mass of a particle=2}{\text{.5gcm}}^{-3}\times 8.17\mu m^3=2.5\times {10}^{6}g{m}^{-3}\times 8.17\times {10}^{-18}{m}^3=20.425\times {10}^{-12}g\end{array}$

The allowable amount of particles in the human breath can be calculated as follows:

  $\begin{array}{l}\text{Standard density=}\frac{\text{mass of a particle }\times \text{ number of particles}}{\text{volume of human breath}}\\ \text{50}\text{.0}\times {\text{10}}^{-6}{\text{gm}}^{-3}\text{=}\frac{\text{20}\text{.425}\times {\text{10}}^{-12}\text{g}\times \text{number of particles}}{0.500\times {10}^{-3}{m}^3}\\ \text{number of particles}=\frac{\text{50}\text{.0}\times {\text{10}}^{-6}\times 0.500\times {10}^{-3}}{\text{20}\text{.425}\times {\text{10}}^{-12}}=1.22\times {10}^3\end{array}$