IH Chemistry and Math Review

Transcription

1 IH Chemistry and Math Review Lecture 1 By W. Myers with revisions by S. Guffey 1 Equations Μass Vol SpGr liquid water R 460 R + F K 73 C + C PV nr PV PV Vol cont NoMoles AvVol NoMoles Mass MW C 760mmHg AvVol 4.45 L x x C P bar Vol cont Concppm Vol 6 10 ppm air Volume NoMoles x AvVol Conc MW x Conc AvVol 3 mg / m ppm Conc ppm AvVol x Conc MW 3 mg / m Volcont 6 Concppm x 10 ppm Volair By Warren Myers; modified by S. Guffey 1

2 Reference emperatures and Pressures SP - Standard emperature and Pressure Commonly used for chemical and physical processes. 0 C (73.15 K, 3F) and 1 atm ( kn/m, kpa, 14.7 psia, 0 psig, 9.9 in Hg, 760 torr, Ft.HO, 407. In.W.G, Lbs./Sq.Ft.) NP - Normal emperature and Pressure (OSHA/NIOSH/ACGIH) 5 C (98.15 K) and 760 mm Hg ( kpa), molar vol 4.46 L NP - Normal emperature and Pressure (European and U.S. Ventilation) 0 C (93.15 K, 68 F) and 1 atm ( kn/m, kpa, 14.7 psia, 0 psig, 9.9 in Hg, 760 torr. Density 1.04 kg/m3 (0.075 pounds per cubic foot) SAP - Standard Ambient emperature and Pressure Also used in chemistry as a reference with temperature of 5 degc (98.15 K) and pressure of 101 kpa. ISA - International Standard Atmosphere Reference for aircraft performance kpa, 15 C and 0% humidity. 3 Adjusting LVs to non-np conditions Application of LVs to Unusual Ambient Conditions. When workers are exposed to air contaminants at temperatures and pressures substantially different than those at normal temperature and pressure (NP) conditions (5 C and 760 torr), care should be taken in comparing sampling results to the applicable LVs. For aerosols, the WA exposure concentration (calculated using sample volumes not adjusted to NP conditions) should be compared directly to the applicable LVs published in the LVs and BEIs book. For gases and vapors, there are a number of options for comparing air-sampling results to the LV, and these are discussed in detail by Stephenson and Lillquist (001). One method that is simple in its conceptual approach is 1) to determine the exposure concentration, expressed in terms of mass per volume, at the sampling site using the sample volume not adjusted to NP conditions, ) if required, to convert the LV to mg/m3 (or other mass per volume measure) using a molar volume of 4.4 L/mole, and 3) to compare the exposure concentration to the LV, both in units of mass per volume. 4 By Warren Myers; modified by S. Guffey

3 Avogadro s Volume: molar volume of any gas At 0C, 760 mmhg, AvVol.414 L At 0C, 760 mmhg, AvVol L At 5C, 760 mmhg, AvVol L At 70F, 760 mmhg, AvVol L 5 Density Density () is the amount of matter contained in a unit volume of a substance Mass/Volume Reference density: water English lb/gal Metric - 1 g/ml liq 6 By Warren Myers; modified by S. Guffey 3

4 Specific gravity he ratio of the weight or mass of a given substance, at a specified temperature, to that of an equal volume of another reference substance Reference of liquid or solid is water Reference of gases is air or H SpGr liquid water 7 Problem 1 Find the specific gravity of acetone if the density is 6.54 lb/gal. 8 By Warren Myers; modified by S. Guffey 4

5 Problem 1 solution SpGr liquid water 6.54 lb / gal lb / gal 9 Problem Find the density in lb/gal of carbon disulfide if the SpGr is By Warren Myers; modified by S. Guffey 5

6 Problem solution SpGr liquid water Find the density in lb/gal of carbon disulfide if the SpGr is 1.6. CS SpGr CS x H O 1.6 x 8.33 lb/gal 10.5 lb/gal 11 Problem 3 Given the specific gravity of carbon disulfide is 1.6, what is the mass of 75 ml of CS? 1 By Warren Myers; modified by S. Guffey 6

7 Problem 3 solution Given the specific gravity of carbon disulfide is 1.6, what is the mass of 75 ml of CS? Mass of CS Vol liq SpGr CS x Η Ο Vol liq Mass of CS g g 75ml ml 13 Problem 4 If 94.5 g of CS were vaporized, how much vapor would be produced? 14 By Warren Myers; modified by S. Guffey 7

8 Problem 4 solution Volume NoMoles x Vol / mole Mass NoMoles x MassPerMole If 94.5 g of CS were vaporized, how much vapor would be and sea level? Solution: Determine number of moles Based on the molar volume determine the total volume C 760mmHg AvVol 4.45 x x 4.45 L / mole C P Mass NoMoles MW bar 94.5g 94.5moles 1.43moles (1 + x3) g/ mole 76 Volume NoMoles x AvVol 1.34 moles x 4.45 L / mole 30.4 L 15 Problem 5 If that volume of vapor were to be diluted with 10 m 3 of clean air, what would the concentration be in PPM? 16 By Warren Myers; modified by S. Guffey 8

9 Problem 5 solution If that volume of vapor were to be diluted with 10 m 3 of clean air, what would the concentration be in PPM? Vol cont 6 Concppm x10 ppm Vol air L+ 30.3L m x L m 3 6 x 10 ppm 3030 ppm 17 Problem 6 What is the mass, in mg, of CS in a 10 Liter sample of air collected at 5 C, if the concentration of CS in the sample is 500 PPM? 18 By Warren Myers; modified by S. Guffey 9

10 Problem 6 solution What is the mass, in mg, of CS in a 10 Liter sample of air collected at 5 C, if the concentration of CS in the sample is 500 PPM? 1) Calculate the volume of pure CS needed to produce 500 PPM in 10 L. ) Determine number of moles in that volume. 3) Determine mass from the number of moles and the gram molar (GMW) weight. 19 Vol Problem 6 solution - continued 10 cont 6 Concppm x ppm Volair What is the mass in mg of CS in a 10 Liter sample of air collected at 5 C, if the concentration of CS in the sample is 500 PPM? 1) Calculate the volume of pure CS needed to produce 500 PPM in 10 L. Vol cont Conc ppm Vol 6 10 ppm 500 ppm 10L 10 ppm air 6 5x10-3 L 0 By Warren Myers; modified by S. Guffey 10

11 Problem 6 solution - continued What is the mass in mg of CS in a 10 Liter sample of air collected at 5 C, if the concentration of CS in the sample is 500 PPM? 1) Calculate the volume of pure CS needed to produce 500 PPM in 10 L. Vol 5x10-3 L ) Determine number of moles in that volume. Vol NoMoles Vol cont one mole x L.05x L/ mole 4 1 Problem 6 solution - continued What is the mass in mg of CS in a 10 Liter sample of air collected at 5 C, if the concentration of CS in the sample is 500 PPM? ) Number of moles in that volume.05x10-4 moles 3) Determine mass from the number of moles and the MW. MW CS 1+x3 76g/mole78000 mg/mole Mass NoMoles x WtPerMole NoMoles x MW / mole 4.05 x10 moles x mg / mole 15.6mg By Warren Myers; modified by S. Guffey 11

12 C 760mmHg AvVol 4.45 L x x C P Equations to convert from ppm to mg/m 3 bar Conc ppm AvVol MW x Conc 3 mg / m Conc 3 / ppm mg m MW x Conc AvVol When converting values expressed as an element (e.g., as Fe, as Ni), the molecular weight of the element should be used, not that of the entire compound. 3 Absolute emperature Degree Rankin ( R)- English R 460 R + F Degree Kelvin ( K) metric K 73 C + C 4 By Warren Myers; modified by S. Guffey 1

13 General Gas Law PV nr PV nr PV PV Permits the determination of the value of any of the three basis characteristics of the gas/vapor being evaluated that result from changes to one or both of the other two characteristics. 5 SP volume correction for NP P1 V 1 P V 1 V P1 V 1 P 1 V P1 V 1 1 P V V 98 K 760 mmhg.5 L 73 K 760 mmhg 4.5L at 5 C and 760 mmhg 6 By Warren Myers; modified by S. Guffey 13

14 he End 7 By Warren Myers; modified by S. Guffey 14