Chapter 1, Problem 118AP

(a) Referring to Problem 1.119. calculate the total volume (in liters) of air an adult breathes in a day. (b) In a city with heavy traffic, the air contains $\text{2}{\text{.1 × 10}}^{\text{-6}}\text{ L}$ of carbon monoxide (a poisonous gas) per liter. Calculate the average daily intake of carbon monoxide in liters by a person.

Interpretation Introduction

Interpretation:

The volume of air and carbon monoxide intake of an adult, in a day, is to be calculated.

Concept introduction:

A conversion factor is used to convert one unit into another.

The appropriate conversion factor for any equality is selected in such a way that it results in the proper unit cancellation.

The conversion factors for this are $\frac{24\text{hrs}}{\text{1\#x2009;day}}$ and $\frac{\text{1\#x2009;day}}{24\text{hrs}}$.

1hr = 60 min. The conversion factors for this are $\frac{1\text{hr}}{60\min }$ and $\frac{60\min }{1\text{hr}}$.

$\text{1\#x2009;L=1000\#x2009;mL}$. The conversion factors for this are $\frac{\text{1\#x2009;L}}{\text{1000\#x2009;mL}}$ and $\frac{\text{1000\#x2009;mL}}{\text{1L}}$.

Answer to Problem 118AP

Solution:

a) $\text{8}\text{.6}\times {\text{10}}^3\text{L}/\text{day}$

b) $\text{0}\text{.018 L}/day$

Explanation of Solution

a)

Given information:

Number of breaths per minute is 12. The volume of air per breath of a resting adult is 500 mL.

Explanation:

An adult takes in 500mL of air per breath. The conversion factor for this is $\frac{\text{500 mL}}{\text{1 breath}}$.

Number of breaths per minute is 12.

1 min = 12 breath

The conversion factor for this is $\frac{12\text{breath}}{1\min }$.

Multiply $\frac{\text{500mL}}{\text{1breath}}$ with $\frac{12\text{breath}}{1\min }$, and cancel the units to find the volume of air taken per minute.

$\left(\frac{\text{500mL}}{\text{1}\cancel{\text{breath}}}\right)\left(\frac{12\cancel{\text{breath}}}{1\min }\right)=\frac{\text{6000mL}}{\text{1\#x2009;min}}$

Multiply $\frac{\text{6000mL}}{\text{1\#x2009;min}}$ by the conversion factor $\frac{60\min }{1\text{hr}}$, and $\frac{24\text{hrs}}{\text{1\#x2009;day}}$, and cancel the units to get the volume of air breathed in a day.

$\left(\frac{\text{6000mL}}{\text{1\#x2009;}\cancel{\text{min}}}\right)\left(\frac{60\cancel{\min }}{1\cancel{\text{hr}}}\right)\left(\frac{24\cancel{\text{hrs}}}{\text{1\#x2009;day}}\right)=8640\times {10}^3\text{mL}/\text{day}$

Convert volume to liters by multiplying the calculated volume by the conversion factor $\frac{\text{1\#x2009;L}}{\text{1000\#x2009;mL}}$.

$\left(\frac{8640\times {10}^3\cancel{\text{mL}}}{\text{1 day}}\right)\left(\frac{\text{1 L}}{\text{1000 }\cancel{\text{mL}}}\right)=\text{8}\text{.6}\times {\text{10}}^3\text{L}/\text{day}$

Volume of air breathed in a day is $\text{8}\text{.6}\times {\text{10}}^3\text{L}/\text{day}$.

b)

Given information:

Air contains $2.1\times {10}^{-6}\text{L}$ of carbon monoxide per liter.

Explanation:

An adult breathes 8640 liters of air in a day.

$1\text{day}=\text{8640 L air}$

One liter of air contains $2.1\times {10}^{-6}\text{L\#x2009;CO}$.

$1L\text{air}=2.1\times {10}^{-6}\text{L\#x2009;CO}$.

The conversion factor to find the volume of carbon monoxide is $\frac{2.1\times {10}^{-6}\text{L\#x2009;CO}}{1L\text{air}}$.

Multiply $\frac{2.1\times {10}^{-6}\text{L\#x2009;CO}}{1L\text{air}}$ with $\frac{\text{8640 L air}}{1\text{day}}$, and cancel the units to find the volume of carbon monoxide intake in a day.

$\begin{array}{c}\left(\frac{2.1\times {10}^{-6}\text{L CO}}{1\cancel{\mathrm{L}\text{air}}}\right)\left(\frac{\text{8640}\cancel{\text{Lair}}}{1\text{day}}\right)=18144\times {10}^{-6}\text{L CO}/day\\ \text{=0}\text{.018 L CO}/day\end{array}$

Volume of carbon monoxide is $\text{0}\text{.018 L}/day$.