Question

Section 1.7 showed that in 1997 los angeles county air had carbon monoxide (co) levels of 15.0 ppm. an average human inhales about 0.50 l of air per breath and takes about 20 breaths per minute. part a how many milligrams of carbon monoxide does the average person inhale in an 6 hour period for this level of carbon monoxide pollution? assume that the carbon monoxide has a density of 1.2 g/l. (hint: 15.0 ppm co means 15.0 lco per 106l air.)

Answer 1

Given information:

Concentration of CO in air = 15 ppm

Volume of air inhaled per breath = 0.50 l

Number of breaths taken per min = 20

Density of CO = 1.2 g/L

To determine:

mg of CO inhaled in 6 hours

Explanation:

15 ppm of CO (parts per million) implies that there is 15 L of CO in 10⁶ L of air

Therefore, volume of CO inhaled through 0.50 L of air is

= 15 L CO * 0.50 L air/10⁶ L air = 7.5 *10⁻⁶ L CO/breath

Now, there are 20 breaths per minute

Therefore, the number of breaths in a 360 min (i.e. 6 hr) period would be

=360 min * 20 breaths/1 min = 7200 breaths

Thus, the volume of CO inhaled during this time span would be

= 7200 breaths * 7.50*10⁻⁶L/1 breath = 0.054 L

Since the density of CO = 1.2 g/L

The corresponding mass of CO inhaled = Density*Volume

= 0.054 * 1.2 = 0.0648 g = 64.8 mg

Ans: Mass of CO inhaled in a 6 h period is 64.8 mg

Answer 2

First, let's determine the amount of air a human intakes per minute:

0.5 L/breath * (20 breaths/min) = 10 L/min

So, for 6 hours, the amount of volume would be:

6 hrs = 360 min
10 L/min ( 360 min = 3,600 L of total air inhaled

Next, we convert the concentration to g/L. The solution is as follows:
15 L CO/10⁶ L air * 1.2 g CO/L = 18 g CO/L air
So, we multiply to this the volume of air inhaled.

Amount of CO inhaled = (18 g CO/L air)(3,600 L of total air)(1,000 mg/1 g)
Amount of CO inhaled = 64.8×10⁶ mg of CO inhaled