ESPM/EPS c180/CEE 106
Problems Set #6 - due March 3, 2016.
1) Atmospheric Radiation
a) Is ammonia (NH3) a greenhouse gas? Why or why not?
b) Three gases X, Y, and Z are being considered as CFC substitutes. Each has
strong absorption bands in the infrared.

ANSWER KEY Week 11
Oxidizing power of the atmosphere
The reactions used in this problem can be drawn as a tree diagram as below:
Rxn. 1 CH4 + OH + O2
CH3OO + H2O
Rxn. 2
a.67%
b.33%
NO NO2
HO2
CH3OOH
h
a.33%
Rxn. 3
HO2 + OH
+ HCHO
b.33% OH
HCHO + HO2
OH
c.

ANSWER KEY Week 5
1) Role of molecular diffusion in atmospheric transport
D = Do (Po/P),
where D = molecular diffusion
P = pressure
Do = 0.2 cm2/s
Po = 1 atm
1a) (20 PTS) time it takes for molecule to travel 1 m (or 100 cm)
- at sea level:
D = (0.2 cm2/s)

ESPM C180/EPS C180/CEE 106 Air Pollution
Homework #2 - due February 4, 2016.
1) Geochemical Box Models:
Consider an element X exchanging between two geochemical reservoirs A and B. Let
MA and MB be the masses of X in reservoirs A and B, respectively; let

ANSWER KEY Problem Set 7
1.) Flux at TOA = I (this is the flux before any attenuation, therefore I0)
a) (25 PTS)
Show that:
R( z ) O2 O2 fna (0) I e
z
O2 fhna ( z )
h
O2 = quantum yield for O2 photolysis
O2 = absorption cross-section for O2
f = O2 mole f

ANSWER KEY Week 4
1a) (15 PTS) From RH (%) 100
PH 2O
PSat , H 2O
we can find that:
RH (%)
PSat , H 2O
100
From the plot we can read RH(%) and temperature, which gives us saturated vapor
pressure, PSat, H2O, (careful about units!) and we can therefore calc

ANSWER KEY Week 1
1) (20 PTS)
Prove: m = A(0)h
Given: m = mass of the species in atmosphere (kg)
A = surface area of Earth (m2) = 4R2 (R = radius of Earth = 6400 km)
exp
Proof:
The area slice of the atmosphere can be defined by the equation
4
Assume A

ANSWER KEY Week 12
1.
(15 PTS) PAN as a reservoir for NOx
To determine NOx ratios, start with steady-state for NO or NO2:
PNO = LNO
J 2 NO2 k1 O3 NO
where k1 = 2.2x10-12e(1420/293)=1.67x10-14 cm3 molec-1 s-1
NO J 2 NO2
k1 O3
Substitute into NOx:
NO2 NO

ANSWER KEY Week 3
1) (15 PTS (10 for calculation, 5 for explanation)
Fin = evap.
MH2O
(1.3x1016 kg)
Fout = precip.
(0.2 cm/day)
Convert precipitation rate to mass flux:
Fout = precipitation rate x area of Earth
area of Earth, Ae = 4r2 = 4 x x (6.4 x 108 c

ANSWER KEY Week 6
1a) (15 PTS) Ammonia is
.
H-N-H
H
It is asymmetric and so we can expect it to have a dipole moment and to be a greenhouse gas (in
fact, it absorbs at approximately 10 m). While ammonia is a greenhouse gas, it does not have
any significan

ANSWER KEY Week 9
1) Polar stratospheric ozone loss
Given: air parcel at 16 km (at max ozone density)
na of air parcel = [M] = 3 x 1018 molecules cm-3
[O3] = 5 x 1012 molecules cm-3
T = 190 K
mean ozone loss rate = LO3 1x10 6 molec cm 3 s 1
a) (15 PTS) No

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ESPM c180
Problems Set #5 - due February 25, 2016.
1) Role of molecular diffusion in atmospheric transport:
The molecular diffusion coefficient D of air varies inversely with pressure:
D = D0 (P0/P)
where P0 is the pressure at sea level and D0

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ESPM/EPS c180
Problems Set #8 due March 31, 2016.
Chapman Mechanism:
We examine here some features of the detailed Chapman mechanism presented in class.
Consider an air parcel at 44 km altitude, 30N latitude, overhead sun, T = 263 K, na = 5.0

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ESPM/EPS c180
Problems Set #9 due April 7, 2016.
1) Polar stratospheric ozone loss
We examine here the origin of the ozone hole in the Antarctic spring stratosphere.
Consider an air parcel at 16 km, the altitude of maximum ozone density, befor

ESPM/EPS c180/CEE 106
Problems Set #4 - due February 18, 2016.
1) The diurnal variation of temperature and relative humidity on a spring day in
Washington, D.C. is shown on the attached plot. Read the values from the plot and
the attached table of saturat

ESPM c180
Problems Set #7 - due March 17, 2016.
1) Analytical Model of the Stratospheric Ozone Layer
Solar UV radiation below 210 nm wavelength is absorbed in the atmosphere mainly by
O2 (not O3). Consider a radiation beam of wavelength < 210 nm with a fl

ANSWER KEY Week 2
1) Box Model:
A
B
FA-B
MA
MB
M = MA + MB
FB-A
i) (15 PTS) at steady state: FA-B = FB-A
given F = k[M] and FA-B = FB-A :
kA-B MA = kB-A MB
so:
MA
and
k B A M B
k A B
MA
A
MB
B
where k B A B
1
B 1 M B A
MB
B
A 1
given: M = MA + MB => MB

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ESPM c180/EPS c180/CEE 106 Air Pollution
Problems Set #12 - due April 28, 2016.
PAN as a reservoir for NOx
1) Consider an atmosphere containing 10 ppb NOx and 100 ppb O3 at T=293K and P =
1000 mb. Calculate the steady state concentrations of N

ESPM/EPS C180/CEE 106 Air Pollution
Problems Set #3 - due February 11, 2016.
1) The main source of water vapor to the atmosphere is evaporation. The main sink for
water vapor is precipitation. If the total mass of water vapor in the atmosphere is
1.3x1016

ESPM/EPS C180/CEE 106 Air Pollution
Problems Set #1
Due: Thursday January 28, 2016
NOTE: IN ORDER TO HELP YOURSELVES, AND THE GRADER, PLEASE SHOW
ALL UNITS IN CALCULATIONS ON YOUR HOMEWORKS AND EXAMS!
Knowing the scale height of a chemical species in the