Assignment 1 2010-11 Fall Answers

Assignment 1 2010-11 Fall Answers - Qatar University...

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Qatar University Department of Chemistry: CHEM 241 2010-11 Fall Term Assignment 1: Gases Hand in: Tuesday October 5 in class. ANSWERS 1. (4 points) Benzene is a common industrial petrochemical. It is however a known carcinogen, and there are very strict workplace safety standards for benzene exposure. The generally accepted TLV (Threshold Limit Value) for benzene in air in the workplace is 1 ppm for an 8 hr/day 40 hr/wk exposure. Calculate the amount of benzene (i.e., its total mass in the laboratory air) in a laboratory of 10x20x3m at 1 atm pressure and at 20 o C when the benzene vapour concentration is 1 ppm (parts per million mass concentration). Assume air to be 80% N 2 and 20% O 2 and behaving as an ideal gas. Answer : ppm benzene = (mass benzene/mass air)x10 6 Therefore we first have to calculate the mass of air in the space of 10x20x3=600 m 3 . We could say we have 0.80 atm N 2 and 0.20 atm O 2 n N2 = pV/RT = 0.80x101,325x600/(8.314x293) = 2.00x10 4 moles N 2 mass N 2 = 2.00x10 4 x28 = 5.6x10 5 g (2 sign. figs) n O2 = pV/RT = 0.20x101,325x600/(8.314x293) = 4.99x10 3 moles O 2 mass O 2 = 4.99x10 3 x32 = 1.6x10 5 g (s sign. Figs) mass air total = 7.19x10 5 g Or we could say the average MM of air is 0.8x28 + 0.2x22 = 28.8 g/mole, with n total = 2.50x0 4 . The answer for the mass of air of course will be the same. TLV = 1 ppm, so then 1 ppm benzene = (mass benzene/mass air)10 6 Mass benzene = 1(ppm)xmass air(g)/10 6 = 1x7.19x10 5 /10 6 = 0.719 g benzene Answer: mass of benzene = 0.719 g Notes: 1) Always report units in your answer. 2) Pay attention to the significant figures 3) Imagine what the benzene concentration would be if by accident you left a bottle with just 100 g of benzene open to evaporate in the air, it would be 140x the TLV! 2. Calculate the density of CO 2 gas at 40 o C and p = 1 atm. (This temperature is just above the critical temperature of carbon dioxide, see problem 3 on the following page), a). (3 points) Assuming that CO 2 is an ideal gas under these conditions. Hint: to calculate the density, use the ideal gas equation to calculate the number of moles e.g. in 1 m 3 . Note that the answer requires units! Answer: With V = 1.00 m 3 , T = 313K, p = 1.01x10 5 Pa, we calculate n = pV/RT = 1.01x10 5 x1.00/(8.314x313) = 38.8 moles N 2 (in 1 m 3 ) 38.8 moles = 38.8x44.0 = 1.69x10 3 g d = m/V = 1.69x10 3 (g) /1.00(m 3 ) = 1.69x10 3 g.m –3 or 1.69 kg.m –3 Again, units must be shown, and correct (or reasonable) significant figures b). (3 points) Assuming that CO 2 is a non-ideal gas following the van der Waals equation, with the constants a and b as in the Lecture Notes Table 1.1 (p.6)
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This note was uploaded on 11/26/2010 for the course CHEM 241 taught by Professor Huda during the Spring '10 term at Texas College.

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Assignment 1 2010-11 Fall Answers - Qatar University...

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