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Solutions: Environmental Chemistry - a Global Perspective 4Th Edition

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Solutions: Environmental Chemistry - a Global Perspective 4Th Edition Solutions: Environmental Chemistry - a global perspective 4th Edition Chapter 2: The Earth’s atmosphere PROBLEMS/SOLUTIONS 1. The mixing ratio of oxygen in the atmosphere is 20.95%. Calculate the concentration in mol L–1 and in g m–3 at Pº (101 325 Pa, 1.00 atm) and 25ºC. Solution The mixing ratio for O2 (g) is 20.95% (Table 2.1). Use the ‘Ideal Gas’ law (PV= nRT) to calculate the total number of moles of gas in 1.00 L for the given conditions. Constants are given in Appendices B.1 and C.1. T = 25ºC (298 K), P = 1.01325 x 105 Pa, R = 8.315 J K-1 mol-1, and V = 1.00 L = 1.00 x 10-3 m3 The total number of moles of gas in 1.00 L is n = PV/RT n = 0.04089 mol The number of moles of oxygen can be calculated from the mole fraction (which is the same as the mixing ratio). % O2 = 20.95, mole fraction = 0.2095 The number of moles of O2 in 1.00 L is: 0.2095 x 0.04089 = 0.00857 mol -3 -1 The concentration of O2 (g) is: 8.57 x 10 mol L . -3 The concentration of O2 (g) in units of g m is determined as follows: 3 -1 1 m = 1000 L, therefore, 1000 L contains 8.57 moles of O2 (g). (M.M. of O2 is 31.9988 g mol ) -1 -3 -3 31.9988 g mol x 8.57 mol m = 274 g m -3 -1 -3 The O2 (g) concentration can be expressed as either 8.57 x 10 mol L or 274 g m . 2. Calculate the atmospheric pressure at the stratopause. What are the concentrations (mol m–3) of dioxygen and dinitrogen at this altitude? How do these concentrations compare with the corresponding values at sea level? Solution Use Equation 2.3 to calculate the pressure at 50 km (stratopause). P Poe Ma g h / R T h _ -1 -1 -2 -1 R = 8.315 J K mol , Pº = 101 325 Pa, g = 9.81 m s , Ma = 0.02896 kg mol (average molar mass of air), h = 50 km (50 000 m) and T = –2ºC (271 K). Note: these last two parameters are the approximate altitude and temperature at the stratopause. 1 1 Solutions: Environmental Chemistry - a global perspective 4th Edition Chapter 2: The Earth’s atmosphere -(0.029 kg mol-1 x 9.81 m s-2 x 50 000 m) (8.315 J K-1 mol-1 x 271 K) Ph 101 325 Pa x e The units in the inverse-natural logarithm term completely cancel. (recall that J = kg m2 s-2) Ph = 101 325 Pa x 0.001812 = 184 Pa (about 0.2 % of atmospheric pressure at sea level) -3 Next, calculate the concentration of O2 and N2 in units of mol m using PV = nRT. Begin by choosing a reasonable volume for the question. In this case assume the volume = 1 m3. The mixing ratios of oxygen gas and nitrogen gas at the stratopause are unchanged from the values at sea level, that is 0.2095 and 0.7808, respectively. The partial pressures are then 38.5 and 143.7 Pa, respectively. The concentration of oxygen is then calculated using n/V = P/RT n/V = 38.5 Pa / 8.315 J mol-1 K-1 x 271 K = 0.0171 mol m-3 The concentration of nitrogen is similarly found to be = 0.0638 mol m-3 Comparison to sea level conditions: -3 -1 The answer to Problem 1 gave the O2 concentration as 8.56 x 10 mol L . Therefore concentration in units -3 -3 -3 of mol m at sea level would be 8.56 mol m . A similar calculation for N2 yields 31.9 mol m at sea level. Comparison of sea level concentrations to stratopause concentrations: -3 -3 8.56 mol m 31.9 mol m for O2 --------------------- = 500 for N2 --------------------- = 500 0.0171 mol m-3 0.0638 mol m-3 The concentrations for both species are about 500 times greater at sea level than they are at the stratopause. 3. What is the total mass of the stratosphere (the region between 15 and 50 km above the Earth’s surface)? What mass fraction of the atmosphere does this make up? Solution The radius of Earth = 6.3782 x 106 m and the average molar mass of air = 0.02896 kg mol-1 Using equation 2.3: o Ma g h / R T Ph P e we can calculate the pressure at the ‘top and bottom’ of the stratosphere. 2 2 Solutions: Environmental Chemistry - a global perspective 4th Edition Chapter 2: The Earth’s atmosphere P at 15 km = 9099 Pa (‘bottom’ of stratosphere) P at 50 km = 184 Pa (‘top’ of stratosphere) Calculate the following three masses using equation 2.2 (p. 24). 1) the total mass of the atmosphere, 2) the mass of the atmosphere down to the tropopause (h = 15 000 m), and 3) the mass of the atmosphere down to the stratopause (h = 50 000 m). Mass (of the defined part of the atmosphere) = P4πr2/g pressure / Pa radius / m 4π g / m s-2 mass / kg 1) 101 325 6 378 200 12.566 9.81 5.28 x 1018 (mass 1) 2) 9099 6 393 200 12.566 9.81 4.76 x 1017 (mass 2) 3) 184 6 428 200 12.566 9.81 9.74 x 1015 (mass 3) The radius used in the calculation of mass 2 and mass 3 (above) reflect the increased height of 15 000 m and 50 000 m above the surface of the Earth. The pressures shown are the corresponding pressures calculated at these altitudes. Mass 1 represents the total mass (mass of entire atmosphere). Mass 2 represents the mass of the atmosphere minus the troposphere. Mass 3 represents the mass of the atmosphere minus the sum of the troposphere plus the stratosphere. The mass of the stratosphere can now be determined by difference. Mass of troposphere = mass 1 – mass 2 = 4.804 x 1018 kg (mass 4) Mass of troposphere + mass of stratosphere = mass 1 – mass 3 = 5.270 x 1018 kg (mass 5) Mass of stratosphere = mass 5 – mass 4 = 4.66 x 1017 kg The stratospheric mass fraction is 4.66 x 1017 kg ÷ 5.28 x 1018 kg = 0.08825 (or ~ 8.8%) It is clear from this calculation that the remaining mass of the atmosphere above the stratosphere (> 50 km) would contribute very little to the overall mass of the atmosphere and that most of the mass is contained in the troposphere. The tropospheric mass fraction is 4.804 x 1018 kg ÷ 5.28 x 1018 kg (x100%) = ~ 91 % of the entire atmosphere. Together the troposphere and stratosphere account for more than 99 % of the mass of the atmosphere from these calculations. 3 3 Solutions: Environmental Chemistry - a global perspective 4th Edition Chapter 2: The Earth’s atmosphere 4. If the mixing ratio of ozone in a polluted urban atmosphere is 50 ppbv, calculate its concentration: (a) in mg m –3; and (b) in molecules cm–3. Solution -3 a) Convert 50 ppbv O3 to units of mg m . Assume 20ºC and 101 325 Pa for a typical polluted urban atmosphere and calculate the total moles of gas for a 1 m3 volume using PV = nRT. 3 -1 -1 101 325 Pa x 1 m = n x 8.315 J mol K x 293 K 3 n = 41.59 mol (for 1 m ) 50 ppbv mixing ratio of O3 represents a mole ratio determined by the formula: mole ratio 9 (x / 41.59 mol) x 1 x 10 = 50 ppbv 3 where x is the number of moles of O3 in 1 m -6 x = 2.08 x 10 mol of O3 The mass of O3 in this volume is -6 -1 -1 2.08 x 10 mol x 48.0 g mol x 1000 mg g = 0.10 mg -3 Therefore the concentration of 50 ppbv O3 is equivalent to 0.10 mg m O3. 3 b) Convert 50 ppbv O3 to molecules per cm . Continuing on from the calculation in part (a) from the value -6 -3 2.08 x 10 mol m O3, simply convert the units mol to molecules and cubic meters to cubic centimetres, as shown by the following calculation: -6 -3 3 6 3 23 -1 2.08 x 10 mol m x (1 m / 1 x 10 cm ) x 6.022 x 10 molecules mol 12 -3 = 1.25 x 10 molecules cm 12 -3 50 ppbv O3 can also be expressed as 1.3 x 10 molecules cm . 5. The gases from a wood-burning stove are found to contain 1.8% carbon monoxide at a temperature of 65ºC. Express the concentration in units of g m –3. Solution Use PV = nRT, assume 101 325 Pa for pressure and use 1 m3 for volume 101 325 Pa x 1 m3 = n x 8.315 J mol-1 K-1 x 338 K 4 4 Solutions: Environmental Chemistry - a global perspective 4th Edition Chapter 2: The Earth’s atmosphere n = 36.05 mol (total moles of gas for 1 m3 at 65ºC) CO (g) moles: 0.018 x 36.05 mol = 0.6490 mol mass of CO (g): 0.6490 mol x 28.0 g mol-1 = 18.2 g of CO (in 1 m3) The concentration of CO gas from the wood-burning stove under these conditions is 18.2 g m-3 ! 6.
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