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ACCTG418 (5)
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Department
Accounting
Course
ACCTG418
Professor
Gail Amort- Larson
Semester
Winter

Description
1 2 The upcoming midterm exam next Wednesday is The following were pumped into a 5.00 L tank at 25C: cumulative. Anything we covered in class, problem sets 46 L of O at 1.00 atm 2 (g) and handouts, from day one up to and including the 12 L of He (g)t 2.00 atm lecture before the exam, can be on it. Calculate the partial pressures o2 O and He, the total pressure and the mole fractions of O2and He. Bizarro by Dan Piraro The two gases don't react and therefore for each gas, n and T are constant. P1V1= nRT = const = P V2 2 hence P V1= 1 V 2 2 P 1 1 1 atm 46L O2 P 2 V = 5L = 9.2 atm 2 He P = P 1V1 = 2 atm 12L = 4.8 atm 2 V 2 5L Ptotal9.2 + 4.8 = 14.0 atm XO = 9.2 = 0.66 Unitless ! 2 14.0 4.8 XHe = 1 – X O2 = 1 – 0.66 = 0.34 =14.0 3 4 Applications of Dalton’s law See figure 5.12 Collecting gas over water Total pressure = pressure of collected gas + pressure of water vapor above the water liquid. The pressure of water vapor above liquid depends only on the temperature and is called saturated vapor pressure (SVP) at the temperature of the experiment. At 22C the SVP is 19.8 torr (see Table 9 CDS). So, if the total pressure is 700.0 torr (typical for Edmonton), then the pressure of the gas collected is 700.0 – 19.8 = 680.2 torr 5 6 A sample of SO o2cupies 37.5 L at 22C and 475 torr Example involving a chemical reaction total pressure, when collected over water. Calculate the N O dissociates partially into NO 2 4 (g) 2 (g) weight of SO 2ollected. N 2 4 (g)s put into a 55.0 L container at 75C. Before P = 475 torr = P + P 22 = P + 19.8 the reaction starts, the pressure is 0.686 atm. After the T SO 2 H 2 SO 2 reaction occurs the final pressure is 1.14 atm. What are 455.2 PSO = 475 – 19.8 = 455.2 torr = atm the final partial pressures of2N 4nd NO ? 2 2 760
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