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Unformatted text preview: Chapter 10: Gases and the Atmosphere 450 EndofChapter Solutions for Chapter 10 Summary Problem Answer: (a) 154 mm Hg (b) X N 2 =0.777, X O 2 = 0.208, X Ar = 0.0093, X CO 2 = 0.0003, X H 2 O = 0.0053 (c) 77.7% N 2 , 20.8% O 2 , 0.93% Ar, 0.03% CO 2 , 0.54% H 2 O; slight difference, since this sample is wet Strategy and Explanation: (a) Given the total atmospheric pressure, determine the partial pressure of N 2 and O 2 , the mole fraction of each gas, and the percent by volume. Use Table 10.1 to get the percentages by volume of N 2 and O 2 in a sample of the atmosphere: N 2 is 78.084% by volume; O 2 is 20.948% by volume. According to Avogadro’s law, moles and gas volumes are proportional, so the mole fraction is equal to the volume fraction. X N 2 = % N 2 100.000% = 78.084% 100.000% = 0.78084 Similarly, X O 2 = 0.20948. Partial pressure p i = X i P tot p N 2 = X N 2 P tot = (0.78084) × (734 mm Hg) = 573 mm Hg p O 2 = X O 2 P tot = (0.20948) × (734 mm Hg) = 154 mm Hg (b) When the pressure is increased by a factor of 7 p N 2 = 7 × (573 mm Hg) = 4012 mm Hg p O 2 = 7 × (154 mm Hg) = 1076 mm Hg (c) Given temperature, cylinder volume, fuel volume, and fuel density, calculate the partial pressure of the fuel in the cylinder. First, calculate the moles of fuel: 0.050 mL C 8 H 18 " 0.760 g C 8 H 18 1 mL C 8 H 18 " 1 mol C 8 H 18 114.2 g C 8 H 18 = 3.3 " 10 # 4 mol C 8 H 18 T = 150 °C + 273.15 = 423.15 ≈ 420 K (rounded to tens place) 485 mL " 1 L 1000 mL = 0.485 L P C 8 H 18 = n C 8 H 18 RT V = (3.3 " 10 # 4 mol C 8 H 18 ) " 0.08206 L $ atm mol $ K % & ’ ( ) * " (420 K) (0.485 L) = 0.024 atmC 8 H 18 0.024 atm " 760. mmHg 1 atm = 18 mmHg (d) The balanced equation for the complete combustion of is C 8 H 18 : 2 C 8 H 18 ( l ) + 25 O 2 (g) 16 CO 2 (g) + 18 H 2 O(g) Use the calculated moles of C 8 H 18 and the stoichiometry to find the moles of O 2 . 3.3 " 10 # 4 mol C 8 H 18 " 25 mol O 2 2 mol C 8 H 18 = 0.30 mol O 2 (e) The partial pressure of the N 2 (g) in the cylinder, the volume, and the temperature are calculated in (b). Use the ideal gas law to determine the number of moles in the reaction cylinder: 4012 mmHg " 1 atm 760.0 mmHg = 5.279 atm Chapter 10: Gases and the Atmosphere 451 n N 2 = PV RT = (5.279 atm) " (0.485 L) 0.08206 L # atm mol # K $ % & ’ ( ) " (420 K) = 7.4 " 10 * 2 mol N 2 The reaction that N 2 (g) undergoes to form NO has this equation: N 2 (g) + O 2 (g) 2 NO(g) We are told to assume 10.% of the N 2 (g) reacts to form NO. 7.4 " 10 # 2 mol N 2 " (0.10) " 2 mol NO 1 mol N 2 = 0.015 mol NO (f) The reaction that produces photochemically active NO 2 has the following equation: NO(g) + 1 2 O 2 (g) 2 NO 2 (g) (g) Using stoichiometry: 0.015 mol N 2 " 1 mol NO 2 1 mol NO " 46.006 g NO 2 1 mol NO 2 = 0.68 g NO 2 (h) There is a variety of information needed to calculate the NO emissions for an entire city for a year, including: population of the city, fraction of the population using vehicles, the number of vehicles used, the number of...
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This note was uploaded on 12/17/2009 for the course CHE 129/130 taught by Professor Hanson during the Spring '09 term at SUNY Stony Brook.
 Spring '09
 HANSON
 Chemistry, pH

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