Name Examination #1, OSH 420 February 23, 2016 This examination consists of 100
ID: 701472 • Letter: N
Question
Name Examination #1, OSH 420 February 23, 2016 This examination consists of 100 total points Answer the following questions using the information provided from the 2015 TLV booklet in the table below. Concentrations are expressed at Standard Conditions (760 mm Hg, 250) show your work to receive full or partial credit!! No work shown-No credit. Substance Isoamyl alcohol 100 ppm 125 ppm STEL/C Notation Mo. TLV Basis TWA Wt 88.15 Eye & URT irr 74.12 Skin & eye irr 138.21 Eye&URT; irr | lso butanol 50 ppm A3 Isophorone N-Isopropylaniline 2 Skin; BEI 135.21 MeHb-emia Skin; A3 Varies Skin&URT; ir, Kerosene 300 mg/m CNS impair 1. Express the Ceiling for Isophorone in mg/m. (5 points total) 2. Express the TLV-TWA for Isoamyl alcohol in mg/m.(5 points) 3. Express the TLV-TWA for Isobutanol in mg/m' at 635 mm Hg and 27 °C. (10 points) A worker is exposed to Kerosene for 2.5 hours at 250 ppm, 1.5 hours at 336 ppm, 1 ho 287 ppm and the remaining 3 hours of the shift at 177 ppm. What is their 8-hour TWA osure to Kerosene in percent? (10 points)Explanation / Answer
Ans 1
At 760 mmHg (1atm) and 25°C(298K)
From the ideal gas equation
V = RT/P = 0.0821 L-atm/mol·K *298K /1atm
= 24.46 L/mol
Ceiling for isophorone in mg/m3
= concentration in ppm x molecular weight /V
= 5 x 138.21/24.46
= 28.252 mg/m3
Ans 2
TLV-TWA for isoamyl alcohol in mg/m3
=concentration in ppm x molecular weight /V
= 100 x 88.15/24.46
= 360.384 mg/m3
Ans 3
P= 635 mmHg * 1atm/760mmHg = 0.8355 atm
Temperature T = 27 + 273 = 300 K
From the ideal gas equation
V = RT/P = 0.0821 L-atm/mol·K *300K /0.8355atm
= 29.478 L/mol
TLV-TWA for isobutanol in mg/m3
= concentration in ppm x molecular weight /V
= 50 x 74.12/29.478
= 125.721 mg/m3
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.