Question

1.a. (3 pts) On 8/28 during our class the concentration of PM2.5 particles was 8.6 μg m- 3. Assuming you take 15 breaths in a minute and an average breath volume of 500 mL (at rest). How much PM2.5 entered our lungs during the 50 min class?

Compound	Mass Fraction	Density (g/mL)	Density (μg/μm3)
Sulfate	0.20	1.76	1.76 x 10-6
Crustal	0.13	2.50	2.5 x 10-6
Black Carbon	0.12	1.80	1.80 x 10-6
Nitrate	0.05	1.72	1.72 x 10-6
Residual (trace organics)	0.50	1.40	1.40 x 10-6

1.b. (3 pts) Assuming all those particles (in question 1.a.) are perfect spheres 2.5 μm in diameter and are of the composition reported in table 1, how many particles of black carbon entered our lungs? (round to nearest thousand)

1.c. (3 pts) Using the same assumptions (as question 1.a.), how much PM10 entered our lungs during class considering the concentration of PM10 was 145 μg m-3?

Answer 1

1.a We take average

15 breath =1 min So at 50 min 50*15= 750 breaths in whole class

Volume of breath 750*500 ml = 375000 ml = 375 litre. = 0.375 m³

Density was 8.6*10^-6 gm/m³ . So total amount of PM2.5 Particles = 8.6*10^-6 gm/m³ * 0.375 m³ = 3.225*10^-6 gm = 3.225 .

b) Mass fraction of black carbon is 0.12.

We took 0.375 m³ breath in whole class. So amount of black carbon present there 0.375 m³ * 0.12 =0.045 m³

All of them are spherical and diameter 2.5 . So radius 1.25 ​. So volume of one particle

4/3 * * (1.25)³ = 8.18 * 10^-6 m³​​​​​​

So no of particles ( 0.045/ 8.18 * 10 ^-6) = 5500

c) It is same as a) So again

volume of breath taken in class 0.375 m³ Now the density is 145*10^-6 gm/m³ .

So amount 0.375*145*10^-6 = 54.375