Type:
Solution Manual
Resource:
Atmospheric Chemistry and Physics From Air Pollution to Climate Change
Edition:
3rd Edition
Author(s):
John H. Seinfeld Spyros N. Pandis
ATMOSPHERIC CHEMISTRY AND PHYSICS: FROM AIR POLLUTION TO CLIMATE CHANGE Third Edition John H. Seinfeld California Institute of Technology Spyros N. Pandis University of Patras and Carnegie Mellon University John Wiley 2015
1.4
CH3SCH3 concentrations range from 250 to 500 ng m-3 (0.25 to 0.50 μg m-3).
ξ CH SCH (ppm) = 3
3
=
RT pM CH 3SCH 3
[CH3SCH 3 ] (μg m −3 )
(8.314) (298) (0.25) = 9.86 × 10-5 ppm
(1.0133 × 10 ) 5
= 0.0986 ppb At 0.5 μg m-3, ξ CH 3SCH 3 = 0.197 ppb
1.5 ⎛ 3.7879 L ⎞ ⎛ 1000 cm 3 ⎞ ⎛ 0.85 g ⎞ ⎟⎟ ⎜ ⎟⎟ ⎜⎜ 500 gallons ⎜⎜ = 1,609,857.5 g Gasoline 3 ⎟ L ⎝ gallon ⎠ ⎝ ⎠ ⎝ cm ⎠
C: 12.01 g/mol H: 1.01 g/mol O: 16.00 g/mol gasoline (C8H18) = [8(12.01) + 18(1.01)] g/mol = 114.26 g/mol CO2: (12.01 + 2·16.00) g/mol = 44.01 g/mol ⎞ ⎛ 44.01 g CO 2 ⎞ ⎛ 1 kg ⎞ ⎟⎟ ⎜⎜ ⎟⎟ ⎟⎟ ⎜⎜ 1 mol gasoline 1 mol CO 1000 g ⎠ ⎠⎝ ⎠⎝ 2 ⎠⎝ ⎞⎛
⎛
(1,609,857.5 g )⎜⎜ 1 mol gasoline ⎟⎟ ⎜⎜ 8 mol CO 2 ⎝
= 4960 kg CO2
114.26 g
1.6 Mass of the atmosphere = 5.148 × 1018 kg = 5.148 × 1021 g Moles of the atmosphere = 5.148/29 × 1021 = 1.77 × 1020 mol air Moles of 1 ppm CO2 = (10-6) (1.77 × 1020) = 1.77 × 1014 mol CO2 Mass of 1 ppm CO2 = (1.77 × 1014 mol CO2) (44 g/mol) = 7.8 × 1015 Mass of carbon = (7.8 × 1015) (12/44) = 2.1 × 1015 g = 2.1 Pg C Current mass of C in atmosphere = (2.1) (400) = 840 Pg C
1.7
The scale heights for the atmospheres of Venus and Mars are: Venus Mars
15.85 km 10.89 km
Although the temperature on Mars is colder than that of Earth and the molecular weight of its atmosphere is slightly higher than that of our atmosphere, the scale height on Mars is larger than Earth’s owing to the much smaller gravity on Mars.
Chapter 2
2.1 The mean lifetime of CH4 in the atmosphere is estimated as the ratio of the amount of CH4 in the atmosphere (Tg) to its source or sink strength (~ 600 Tg yr-1). To determine the abundance of CH4 we need to convert a mixing ratio of 1745 ppb into total atmospheric mass. The total number of moles of air in the atmosphere = 1.62 × 1020. So Moles of CH4 = (1745 × 10-9) (1.62 × 1020) = 2.83 × 1014 Mass of CH4 = (2.83 × 1014 moles) (16 g mol-1) = 45.3 × 1014 g = 4530 Tg Mean atmospheric lifetime of CH4 = 4530/600 ≅ 7.6 years For CO: Mass of CO = (100 × 10-9) (1.62 × 1020) (28) = 45.4 × 1013 g = 454 Tg Mean atmospheric lifetime of CO = 454/2780 = 0.16 year ≅ 2 months Thus the explanation lies in the much longer atmospheric lifetime of CH4.
2.2 z (km)
ξ O (ppm)
cO3 (molec cm-3)
cair (molec cm-3)
cair (ppm)
5 10 20 25 30 35 40 50
0.0326 0.0581 1.0283 5.7632 8.8874 9.6667 6.0214 4.6858
5.00 × 1011 5.00 × 1011 1.90 × 1012 4.80 × 1012 3.40 × 1012 1.70 × 1012 5.00 × 1011 1.00 × 1011
1.53024 × 1011 8.59253 × 1018 1.84755 × 1018 8.32862 × 1017 3.8256 × 1017 1.75861 × 1017 8.30369 × 1016 2.13411 × 1016
736 413.5 88.91 40.08 18.41 8.463 3.996 1.027
3
Mixing ratio is cO3 /cair , and cair is not a constant with altitude. Absolute O3 concentration peaks
at about 25 km; O3 mixing ratio peaks at about 35 km.
2.3 Total amount of N2O in the atmosphere = (315 × 10-9) (1.62 × 1020 moles) (44 g mol-1) = 2245 Tg N2O = 2245 (28/44) = 1429 Tg N Mean N2O lifetime = 1429 Tg N/12.6 Tg N yr-1 = 113 years.
2.4
nO3 = (3 × 1012 molecules cm-3) (15 × 105 cm) = 4.5 × 1018 molecules cm-2 Then
h = (3.72 × 10-20) (4.5 × 1018) = 0.167 cm = 1.67 mm = 167 DU
2.5
2.6
a.
4.2 × 1015 and 2.5 × 1015
b.
5.3 months
9.0 months
5.0 months
6.6 months
1 ppb CH4 = 2.78 Tg CH4 46 Tg y-1/2.78 Tg ppb-1 = 16.5 ppb y-1
2.7
Two compartment model for CH3Br in the atmosphere.
mSH
kOH
kT kT
mNH
kOH
S
The mass balances on CH3Br for the two hemispheres are:
dmNH = S + kT mSH − kT mNH − kOH mNH dt dmSH = kT mNH − kT mSH − kOH mSH dt
Assume steady state: dmNH dmSH = 0 = 0 dt dt Obtain O = = S + kT mSH − kT mNH − kOH mNH kT mNH − kT mSH − kOH mSH O = a) To get mNH /mSH consider (2), kT ( mNH − mSH ) = kOH mSH
(1)
(2)
mNH k = 1+ OH mSH kT = 1+ b)
0.667 = 1.74 0.09
The observed mNH /mSH ratio is 1.3. This suggests that there are sources of CH3Br in the SH, possibly from the ocean.
2.8
a.
For the 3 species at 273 K:
CHBr3: CH2Br2: C2H5Br:
j = 1.4 × 10-‐6 s-‐1 j = 1.2 × 10-‐8 s-‐1 j = 2.1 × 10-‐10 s-‐1
τOH 0.21 yr 0.35 yr 0.12 yr
kOH [OH] = (1.49 × 10-‐13) (106) = 1.49 × 10-‐7 s-‐1 kOH [OH] = (9.12 × 10-‐14) (106) = 9.12 × 10-‐8 s-‐1 kOH [OH] = (2.76 × 10-‐13) (106) = 2.76 × 10-‐7 s-‐1
For loss by photolysis: CHBr3: CH2Br2: C2H5Br:
τp 0.023 yr 2.64 yr 151 yr
The overall tropospheric lifetime for each is computed from: 1 1 1 = + τ total τ OH τ p The results for τtotal are: CHBr3 0.021 yr CH2Br2 0.31 yr C2H5Br 8.33 yr Note: 31,536,000 seconds/year b. Assume Q = 1.5 Gmol Br yr-‐1 emission flux into the troposphere τ = 0.021 yr tropospheric lifetime kTS = 0.063 yr -‐1 flux of air from troposphere to stratosphere At steady state, Q – (τ-‐1 + kTS) c = 0 We want to calculate c. Note that Q = 0.5 x 109 mol CHBr3 yr-‐1. Solving for c, C = 0.5 x 109 / (47.7 x 8.14 x 1024) = 1.3 x 10-‐18 mol cm-‐3 = 7.83 x 105 molecules cm-‐3 The mole fraction of CHBr3 is 7.83 x 105/2.7 x 10 19 = 2.9 x 1014 = 0.029 ppt.
This is to be compared with the surface measurements of about 1 ppt. Owing to the relatively short lifetime of CHBr3 in the troposphere, its concentration is not uniform over the entire troposphere. With its source at the surface, the mixing ratio is expected to be highest at the surface. c. The annual flux of CHBr3 into the stratosphere is (0,063 yr-‐1) (1.3 x 10-‐18 mol cm-‐3) (8.14 x 1024 cm3) (3 Br atoms) = 2 x 106 mol Br yr-‐1
2.9 Reaction with OH is the main sink of CH3CCl3 in the troposphere: CH3CCl3 + OH -------> CH2CCl3 + H2O with the reaction rate constant k = 6.11 × 10-15 cm3 molec-1 s-1 at 273 K. The dynamic behavior of CH3CCl3 due to reaction with OH is given by d[CH3 CCl3 ] =k CH3 CCl3 OH dt
(1)
Assuming constant OH radical concentration, Equation (1) becomes d[CH3 CCl3 ] =k' CH3 CCl3 dt
(2)
where k’ = k[OH]. In this way, the time-dependent concentration of CH3CCl3 follows the first order decay [CH3 CCl3 ] = [CH3 CCl3 ]0 exp -k't
(3)
Plot [CH3CCl3] vs. [t] and apply exponential regression to the data: 8
11
x 10
10
8
7
6
3
[CH CCl ] (molecule cm−3)
9
3
5
4
3
2
1
0
0.5
1
1.5
2
time (s)
2.5
3
3.5
4 8
x 10
We get k’ = 5.698 × 10-9 s-1. Therefore, [OH] = k’/k = 9.3 × 105 molecule cm-3 Or plot log([CH3CCl3]/ [CH3CCl3]0) vs. [t] and apply linear regression to the data:
0
−0.5
log(C/C0)
−1
−1.5
−2
−2.5
0
0.5
1
1.5
2
time (h)
2.5
3
3.5
4 8
x 10
We get k’ = 5.698 × 10-9 s-1. Therefore, [OH] = k’/k = 9.3 × 105 molecule cm-3