5. (20 points) Phenanthrene (PNT, CuHo) is an rganic comound foion Phenanthrene

Question

5. (20 points) Phenanthrene (PNT, CuHo) is an rganic comound foion Phenanthrene made an appearance on Exam 1 of this class. In addition to cigarettes may also contain fine particles. Potentially useful information about phenanthrene(picture below is of PNT Vapor Pressure 2.17x 10 atm Henry's Constant = 0.0313 ain-Lino! Density 1.18glcm (5 points) A bar on University Avenue that permits indoor smoking is 60 m' in volume, and several customers are smoking and drinking water. If the partial pressure of phenanthrene in this room is 1.25 x 10° atm at equilibrium, what is the concentration of PNT in the glasses of water (units ppm)? a. Ignore your calculations from Part a. The outside air brings in 1.5 g /ms of (8 points) fine particulates into the 60 m bar from outside, at an air exchange of 1.5 hr (air changes-Q/V). Inside the bar, cigarettes are smoked at a rate of 12 cigarettes/hour (1 cigarette 10 mg fine particles). The bar behaves as a completely mixed, steady state system. What is the concentration of fine particles in the bar? Compare the results of this apartment with NAAQS standard for fine particulates (65 ug/m') b. (4 points) Why are these classes of concern to human health? What are the two classes of fine particles, as established by the NAAQS? c. d. (3 points) Fine particles are of concen for both the AOI and the NAAQS. What major piece of legislation are these components of, and how are the NAAQS and AQI different? Page 7 of 8

Explanation / Answer

a) Given, partial pressure p= 1.25 x 10-9 atm, Kh= 0.0313 atm-L/mol

concentration (C)= ?

according to Henry's law,

p= KhC

C= p/Kh

= 1.25 x 10-9 atm/ 0.0313 atm-L/mol

= 3.99 x 10-8mol/L.

to convert into ppm multiply moles by molecular weight which gives you g/L and convert gm into mg. thus, mg/L is ppm

C= 3.99x10-8 * 178.23(molar mass of phenanthrene)

= 711.14 x 10-8 g/L * 1000

= 711.14 x 10-5 mg/L

C = 7.11 x 10-3 ppm

b) air exchange = Q/V

1.5 hr-1 = Q/ 60m3

Q= 60*1.5 m3/hr

Q= 90 m3/hr

So, 90 m3 of particulate matter enters the room in an hour, now you have 1.5 ug/m3

so in one m3 there is 1.5ug of particulate matter. hence, in 90m3 we have = 90*1.5 =135 ug of particulate matter in one hour entering from outside

adding the contribution of cigarretes we have 12*10= 120mg/hr =120*1000 ug/hr= 120000 ug in one hour

therefore in one hour the room has a concentration of 120000 + 135 ug particulate matter = 120,135 ug/m3  which as compared to NAAQS standard is 1848 times more !!

C) Two classes are PM10 and PM2.5 . These, stand for PM 10 and 2.5 microns and smaller respectively.

when, PM 2.5 levels are increased in the air it causes the air to become hazy and reduces visibility. duee to small size they can travel deep into the respiratory tract and cause eyes, nose or lung irritation.

PM10 increased level can cause respiratory irritation and even cancer.

d) particulate matter come under the air quality legislation. AQI stands for air quality standard used by many government agencies to calculate the level of pollutants in the air. NAAQS is established by the EPA as national ambient air quality standard. an AQI level of 100 corresponds to the NAAQS level for the pollutant.

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