Midterm
ATMO 451B/551B
Spring ‘09
SOLUTIONS
This exam covers chemistry and aerosol size distributions. (HW #1-3), and uses basic
concepts covered in aerosol mechanics and cloud physics. Please turn it into Sonya or
myself 24 hours from the time you pick it up. All exams must be in by class on Friday,
April 17. You must answer all questions. Show all work. You may use class-notes and
textbooks, but do not use any online resources. You must all work independently – do not
discuss the exam with anyone until all exams have been turned in on Friday.
1. A 2001 mass budget of methane (CH4) is given by
dQ
 PR
dt
where Q = is the global amount and P and R are the source and sink rates (598 and 576
Tg CH4 yr-1), respectively. If the current methane mixing ratio is 1745 ppb, calculate the
annual percent rate of increase in methane. Show all equations used and all work.
Answer: The instantaneous % rate of increase is given by:
d ln Q P  R

dt
Q
Whe have P and R, we just need Q. The current mixing ratio of CH4 is 1745 ppb
which is equivalent to a number fraction of CH4 = 1.745 x 10-6. Globally, the
total number of molecules of methane will equal the total number of air molecules
timesCH4. We represent the total # of molecules by dividing the total mass by
the molecular weight,
Q
M

 CH 4
M CH 4 M air
If we have an average molecular weight of air of 29 g/mol and an a molecular
weight of CH4 of 16 g/mol, and M is the total mass of the atmosphere, given by
p
=5.1x1018 kg = 5.1 x109 Tg
g
Thus Q = 4,910 Tg. The % rate of increase is (P-R)/Q = 22/4910 = 0.45% yr-1.
This is small enough to say it is also the annual rate of increase [i.e. ert – 1 = rt
because rt << 1.]
M  4Re2
2. Suppose you have three gas species, A, B and C in a mixing chamber. You know the
following:
A + C  W + X ; kA is known
B + C  Y + Z ; kB is unknown
In your chamber, you can monitor the concentrations of [A], and [B], but not [C]. Show
how you can use the time series of [A](t), and [B](t) to determine the unknown rate
constant kB. Assume there are no other relevant reactions for A and B. However, you
don’t know what the relevant reactions for C are – there may be both sources and sinks.
Hint: Think about the plot of ln[A](t) vs. ln[B](t). If you are having trouble, assume [C]
is constant at first – then convince yourself that you don’t have to assume [C] is constant.
Answer:
d [ A]
  k A [ A][C ]
dt
d [ B]
  k B [ B ][C ]
dt
d ln[ A]
  k A [C ]
dt
d ln[ B ]
  k B [C ]
dt
d ln[ A] k A

d ln[ B ] k B
k B ln
[ A]
[ B]
 k A ln
[ A]0
[ B ]0
So if you plot [B](t) vs. [A](t) on a log-log plot, kB, the desired quantity, will
equal the slope times kA.
3. The sun undergoes a 22-year cycle, which is perceived as an 11-year cycle due to its
symmetry. The amplitude of the fluctuation in total solar output is about 0.1%.
However, the amplitude of the fluctuation in UV (200 nm – 300 nm) is about 1.5% due to
the presence of hot features associated with sunspots. What precisely do you need to
know to diagnose the effect of this change in UV on steady state ozone concentrations?
Assume that the Chapman mechanism is all you have to worry about.
Answer: All you need to know here is the equilibrium Ozone concentration for
the Chapman mechanism (5.13) in S+P.
1/ 2
k j 
[O3 ] SS  0.21 2 O 2  [ M ]3 / 2
k j 
 4 j O3 
You precisely need to know the change in (jO2/jO3)1/2. We know that O2
dissociates due to absorption of photons < 242 nm, and O3 dissociates due to
photons > 240 nm. So in more specific terms, we need to know whether the
amplification of 200-240 nm light is greater than the amplification of 240 nm –
300+ nm light. This was not given in the problem, so you can’t for sure say from
the information provided that O3 will increase (or decrease) in the active phase of
the solar cycle.
4. Suppose you have an NaCl aerosol population whose dry size distribution is
lognormally distributed with a geometric mean dry diameter of <dp> and geometric mean
standard deviation of g. Consider that the wet size distribution depends on the relative
humidity RH. Assume NaCl behaves as an ideal solution (i.e. activity coefficient = 1).
a. Show why we can neglect the Kelvin effect if Dp << Dpc.
Answer: We need to represent the Köhler curve for small Dp in terms of Dpc.
s
A
B
 3
Dp Dp
D pc  3B / A
2
A  3B / D pc
s
3B
B
 3
2
D p D pc D p
  D 2 
3 p   1
  D pc 



B
  3 for D p  D pc
Dp
B
 3
Dp
b. At what relative humidity is Dp/Dpc = 0.1?
Answer: To make this simple, we must continue to assume Dp >> dp, even when
Dp = 0.1 Dpc. First let’s explore what this means for NaCl.
d p  (0.1) D pc
D pc  3B / A
1/ 2

S MW i 3 
3T
 
d p 
 0.66 μm K W M S

3
For NaCl, S = 2.2 g/cm ; MS=58 g/mol; i = 2. Assume T = 298 K.

S MW i 3 
3T
d p  0.1
d p 
 0.66 μm K W M S

 0.66 μm K W M S 

d p  100
3T
 S M W i 

1/ 2
 0.66 μm K (3.2) 

 100
3
(
298
K
)
(
2
.
2
)(
2
)


 50 nm
As sub-steps we have found B = 0.72dp3. and Dpc = (31 m-1/2) dp3/2. To get
relative humidity (saturation is more convenient analytically), we use
  D 2 
3 p   1
  D pc 



B
s  970 3
D pc
B
s 3
Dp
S  1 s


 1  2.3  10  2 m 3/2 d p3 / 2
Note that in this form of the Köhler equation, we have assumed that |s| << 1,
which is only satisfied here if dp >> 87 nm. Here, we see that there isn’t one
relative humidity for all dp. So we choose an appropriate example of dp = 1 m,
and see that S = 1 - .023 = .977, and thus RH = 97.7%. The lowest dp that
satisfies the |s| << 1 and Dp = 0.1Dpc criteria is about 0.25 m, for which RH ~=
80%.
c. Show that, with these assumptions, the equilibrium size distribution at relative
humidity RH is still lognormally distributed with the same geometric standard deviation
as the dry size distribution. (This is the main part of the problem – do this even if you
don’t get the others).
Answer: Here, we are considering a much simpler situation. We consider a
lognormal aerosol distribution in equilibrium at a fixed Relative humidity If the
solution is ideal, then we have the Raoult effect telling us that
S
nw
ns  nw
 S 
nw  ns 

1 S 
We note that ns ~ dp3 and will continue to assume that nw ~ Dp3. (i.e. dp << Dp)
Since at fixed relative humidity we have ns ~ nw , we can conclude that dp3 ~ Dp3,
which is the same as saying Dp = cdp where c is a constant related to terms
discussed above (density, molar weight, relative humidity, vant Hoff factor). The
size distribution is given by
n(ln d p )d ln d p 
 1  ln d  ln d
p
pg
exp  

ln  g
 2
2 ln  g

Nt




2

 d ln d p


The same particles that fall within the size range lndp  lndp +dlndp will fall
within the size range lnDp  lnDp + dlnDp. But since Dp = cdp, we can say dlnDp
= dlncdp = dlndp. So we can convert the dry size distribution to a size distribution
at S using the relation n(lndp)dlndp = n(lnDp)dlnDp, which yields
n(ln D p )d ln D p 
 1  ln D  ln cd
p
pg
exp  

ln  g
 2
2 ln  g

Nt




2

 d ln D p


Clearly, the geometric standard deviation hasn’t changed. The whole thing has
just shifted in log-diameter space.
d. If RH is slowly increasing, such that the distribution remains in equilibrium, then the
width of the distribution is constant. Describe what happens to the width of the
distribution if RH increases too rapidly to maintain equilibrium?
Answer: The key here is to remember that
dD p
dt
~
d ln D p
dt
1
s
Dp
~
1
e
2 ln D p
s
This says that the larger particles will grow substantially more slowly than the
small ones as long as the difference between ambient supersaturation and particle
supersaturation is much greater than the difference between the smaller and larger
particles’ equilibrium supersaturations. This growth kinetics has the effect of
narrowing the distribution, and decreasing g. as long as the large supersaturation
difference is maintained.
Another way to put it is that the timescale for a particle to reach its equilibrium
size (with respect to an ambient saturation below the particle’s critical
supersaturation) is much faster for small particles than large ones. So if you
impulsively increase relative humidity, the small ones will grow quickly to their
new equilibrium size, while the larger ones will take a bit more time. This has the
effect of temporarily narrowing the distribution during the relaxation process.

5. Using the table in S+P problem 8.7, show what the standard deviation of a lognormal
size distribution is in terms of the geometric mean diameter and geometric mean standard
deviation. What is the relative dispersion, defined as the ratio of the standard deviation to
the mean diameter?
  D p2  D p
2



2
2
 D pg
exp 2 ln 2  g  D pg
exp ln 2  g



1

 D pg exp  ln 2  g  exp ln 2  g  1
2

 D
p


exp ln 2  g  1



RD  exp ln 2  g  1
The slope of the line I plotted is ¾. Is it surprising that such a non-linear function
would yield a nearly linear relationship with its independent variable over such a
wide range of values? It’s not obvious to me…