Chapter 12, Problem 34SP

At a height of 10 km (33 000 ft) above sea level, atmospheric pressure is about 210 mm of mercury. What is the net resultant normal force on a 600 cm2 window of an airplane flying at this height? Assume the pressure inside the plane is 760 mm of mercury. The density of mercury is 13 600 kg.

To determine

The mass of the dust in a $20\text{ m}\times 15\text{ m}\times 8.0\text{ m}$ room in an unhealthy, dusty cement mill, where there were $2.6\times {10}^9$ dust particles in $1{\text{ m}}^3$ air. The specific gravity of the dust is $3.0$, and assume the diameter of the spherical particle to be $2.0\text{μm}$.

Answer to Problem 34SP

Solution:

$78\text{ g}$

Explanation of Solution

Given data:

The specific gravity of the dust particle is $3.0$.

The diameter of the spherical particle is $2.0\text{μm}$.

The size of the room is $20\text{ m}\times 15\text{ m}\times 8.0\text{ m}$.

The number of dust particles in $1{\text{ m}}^3$ air is $2.6\times {10}^9$.

Formula used:

Write the expression of the specific gravity of the material:

$S{G}_{material}=\frac{{\rho }_{material}}{{\rho }_{water}}$

Here, $S{G}_{material}$ is the specific gravity of the material, ${\rho }_{material}$ is the density of the material, and ${\rho }_{water}$ is the density of the water.

Write the expression for density of the material:

$\rho =\frac{m}{V}$

Here, $m$ is the mass and $V$ is the volume.

Explanation:

Calculation for the one dust particle:

Calculate the radius of the dust particle:

$r_{dust}=\frac{d}{2}$

Here, $d$ is the diameter of the dust particle.

Substitute $2.0\text{μm}$ for $d$

$\begin{array}{c}{r}_{dust}=\frac{2.0\text{μm}}{2}\\ =1.0\text{μm}\end{array}$

Recall the expression of the specific gravity of the dust:

$S{G}_{dust}=\frac{{\rho }_{dust}}{{\rho }_{water}}$

Here, $S{G}_{dust}$ is the specific gravity of the dust and ${\rho }_{dust}$ is the density of the dust.

Rearrange for ${\rho }_{dust}$

${\rho }_{dust}=\left(S{G}_{dust}\right)\left({\rho }_{water}\right)$

Understand that the specific density of the water is $1000{\text{kg/m}}^3$.

Substitute $1000{\text{kg/m}}^3$ for ${\rho }_{water}$ and $3.0$ for $S{G}_{dust}$

$\begin{array}{c}{\rho }_{dust}=\left(3.0\right)\left(1000{\text{kg/m}}^3\right)\\ =3000{\text{kg/m}}^3\end{array}$

Recall the expression of density of the dust.:

${\rho }_{dust}=\frac{m_{dust}}{V_{dust}}$

Here, $m_{dust}$ is the mass of the dust and $V_{dust}$ is the volume of the dust.

Rearrange for $m_{dust}$

$m_{dust}=\left({\rho }_{dust}\right)\left(V_{dust}\right)$

Understand that the volume is spherical. Hence, calculate the volume of the dust particle by using the formula:

$V_{dust}=\frac{4}{3}\pi {\left(r_{dust}\right)}^3$

Substitute $1.0\text{μm}$ for $r_{dust}$

$\begin{array}{c}{V}_{dust}=\frac{4}{3}\pi {\left(1.0\text{μm}\right)}^3\\ =\frac{4}{3}\pi {\left(1.0\text{μm}\left(\frac{{10}^{-6}\text{ m}}{\text{1 μm}}\right)\right)}^3\\ =4.2\times {10}^{-18}{\text{ m}}^3\end{array}$

Substitute $3000{\text{kg/m}}^3$ for ${\rho }_{dust}$ and $4.2\times {10}^{-18}{\text{ m}}^3$ for $V_{dust}$

$\begin{array}{c}{m}_{dust}=\left(3000{\text{kg/m}}^3\right)\left(4.2\times {10}^{-18}{\text{ m}}^3\right)\\ =1.26\times {10}^{-14}\text{kg}\end{array}$

This is the mass of one dust particle.

Understand that the total number of dust particle in $1{\text{ m}}^3$ volume is $2.6\times {10}^9$.

Therefore, the mass in $1{\text{ m}}^3$ volume is

$m=n\left(m_{dust}\right)$

Substitute $2.6\times {10}^9$ for $n$ and $1.26\times {10}^{-14}\text{kg}$ for $m_{dust}$

$\begin{array}{c}m=\left(2.6\times {10}^9\right)\left(1.26\times {10}^{-14}\text{kg}\right)\\ =3.27\times {10}^{-5}\text{kg}\end{array}$

This is the mass of the dust particle in $1{\text{ m}}^3$ volume.

The mass of the dust particle in $20\text{ m}\times 15\text{ m}\times 8.0\text{ m}$ volume room is

$M=m\left(V_{room}\right)$

Here, $V_{room}$ is the volume of the room.

$\begin{array}{c}M=\left(3.27\times {10}^{-5}\text{kg}\right)\left(\left(20\text{ m}\right)\left(15\text{ m}\right)\left(8.0\text{ m}\right)\right)\\ =0.078\text{kg}\\ =\text{78 g}\end{array}$

This is the total mass of the dust particle in $20\text{ m}\times 15\text{ m}\times 8.0\text{ m}$ room.

Conclusion:

Therefore, the mass of the dust particle in $20\text{ m}\times 15\text{ m}\times 8.0\text{ m}$ room is $\text{78 g}$.

To determine

The mass of the dust in each average breath of $400{\text{ cm}}^3$ volume in an unhealthy, dusty cement mill where there were $2.6\times {10}^9$ dust particles in $1{\text{ m}}^3$ air. The specific gravity of the dust is $3.0$, and assume the diameter of the spherical particle to be $2.0\text{μm}$.

Answer to Problem 34SP

Solution: $13\text{ μg}$

Explanation of Solution

Introduction:

If the number of the particle in the volume $V$ is $n$, then the number of the particle in some part of its volume $v$ is $\frac{V}{v}\left(n\right)$.

Here, $n$ is the number of the particle in volume $V$ and $v$ is the some part of the volume.

Explanation:

Understand that the mass of the dust particle in $20\text{ m}\times 15\text{ m}\times 8.0\text{ m}$ room is $\text{78 g}$.

From the previous part, the mass of the dust particle in $1{\text{ m}}^3$ volume is $3.27\times {10}^{-5}\text{kg}$.

The volume of averagebreadth of the air is $400{\text{ cm}}^3$.

Then, mass of the dust particle in $400{\text{ cm}}^3$ volume is

$\begin{array}{c}{M}^{\prime }=\left(3.27\times {10}^{-5}\text{kg}\right)\left(400{\text{ cm}}^3\right)\\ =\left(3.27\times {10}^{-5}\text{kg}\right)\left(400{\text{ cm}}^3\left(\frac{{10}^{-6}{\text{ m}}^3}{{\text{cm}}^3}\right)\right)\\ =1.308\times {10}^{-8}\text{ kg}\end{array}$

Convert it into small unit:

$\begin{array}{c}{M}^{\prime }=1.308\times {10}^{-8}\text{ kg}\left(\frac{{10}^9\mu \text{g}}{\text{1 kg}}\right)\\ =13\text{ μg}\end{array}$

Conclusion:

Therefore, the mass of the dust in each average breath of $400{\text{ cm}}^3$ volume is $13\text{ μg}$.