Answer
a) $8.6\times10^{3}$ L of air per day.
b) $1.8\times10^{-2} $ L of CO per day.
Work Step by Step
We determined from the previous question that an adult inhales 500 mL of air per breath.
a) If an adult breathes 12 times in a minute, his daily intake of air is:
$ \frac{500 mL}{breath} \times\frac{1L}{1000 mL} \times \frac{12 breaths}{1 min} \times\frac{60 min}{1 h} \times\frac{24 h}{1 day} = \frac{8640 L}{day} = 8.6 \times 10^{3} L $
b) If a person breathes $8.6\times10^{3} L$ of air per day and there are $2.1\times10^{-6} L$
carbon monoxide (CO) per liter of that air, then daily CO intake is:
$ \frac{8.6 \times 10^{3} L}{day} \times\frac{2.1\times10^{-6}}{1 L} = 1.8 \times10^{-2} L $
