CHEM 4800
Atmospheric Chemistry
Assignment #1
Due Friday September 24, 2010 by 3 PM
1. Find appropriate data and calculate the mass of the atmosphere of Callisto, the second
largest moon of Jupiter (in kg).
g0(Callisot) = 1.24 m s-2
Rp(Callisto) = 2400 km = 2.400 x 106 m
P0(Callisto) = 0.75 x 10-6 Pa
M atm 
4 R p 2 P0
g0

4 (2.40 106 m) 2 (0.75 10 6 Pa)
 4.4 107 kg
2
1.24 m s
2. Venus and the Earth have about the same size and mass as one another (assume they
are the same). Venus’ atmosphere is essentially just CO2. Calculate the scale height of
CO2 on Earth and on Venus. Note that Venus’ atmosphere is rather hot (perhaps 200oC
on average). Comment on the effects that such different scale heights may have on the
chemistries of Earth’s and Venus’ atmospheres.
H s (CO2 , Earth )
RT
8.314 J K 1mol 1 (223K )


 4,308 m  4.3 km
Mg 0.044 kg mol 1 (9.78 m s 2 )
H s (CO2 ,Venus )
RT
8.314 J K 1mol 1 (473K )


 10, 008 m  10.0 km
Mg 0.044 kg mol 1 (8.93 m s 2 )
Both the different scale heights and the different partial pressure of CO2 on Venus and
Earth will make tremendous differences in their atmospheric chemistries. Even at the
surface of Earth, PCO2 is very small (perhaps 377 x 10-6 atm), whereas on Venus it is
about 90 atm, since the atmosphere is almost all CO2. At some arbitrary altitude, PCO2 on
Earth will have dropped more than at the same altitude on Venus since Hs,Earth < HS,Venus,
and this will only increase the difference in partial pressures. The result therefore is that
at high altitudes on Earth, PCO2 will become even smaller than it is at the surface and so
CO2 does not play much of a role in Earth’s upper atmospheric chemistry. However on
Venus, the atmospheric chemistry is dominated by CO2 from the surface to the top of the
atmosphere.
3. Calculate the total concentration of air molecules (in molecules cm-3) at z = 0, 2, and 5
km. Use an appropriate average temperature in each case.
pV  nRT
n
P
thus, 
V RT
Which would be expressed in mol L-1 if using P in atm and R = 0.082 L atm K-1 mol-1.
Multiplying by Avogadro’s number and dividing by 1000 converts this to
molecules mL-1:
n( N Av )
PN Av

1000V 1000 RT
And we can use the hydrostatic equation for the pressure, giving us:
molecules mL1 
  Mgz 
N Av P0 exp 
 RT 
molecules mL1 
1000 RT
For example, at the surface (z = 0), the concentration of air molecules is:
 0.0288 kg mol 1 (9.78 m s 2 )(0 m) 
6.02 1023 mol 1 (1.00 atm) exp 

8.314 J K 1mol 1 (288K )

  2.55 1019 mL1
n0 
1
1
1000(0.082 L atm K mol )(288K )
(Note that in this calculation, I have used the SI units for R inside the exponential, since I
have also used SI units for M and g. Outside the exponential in the denominator, I used
non-SI units for R, since I used atm in the numerator. I could have instead used P0 =
101325 bar and R = 8.314 J K-1mol-1)
Similarly at the other altitudes,
 0.0288 kg mol 1 (9.78 m s 2 )(2000 m) 
6.02 10 mol (1.00 atm) exp 

8.314 J K 1mol 1 (281K )

  2.05 1019 mL1

1000(0.082 L atm K 1mol 1 )(281K )
23
n2 km
1
(I used the lapse rate of -6.5 K km-1 to calculate an approximate average temperature
from 0 to 2 km)
n5 km
 0.0288 kg mol 1 (9.78 m s 2 )(5000 m) 
6.02 1023 mol 1 (1.00 atm) exp 

8.314 J K 1mol 1 (272 K )

  1.45 1019 mL1

1
1
1000(0.082 L atm K mol )(272 K )
4. Re–derive the hydrostatic equation for the troposphere assuming that the temperature
is not constant. In the troposphere you can assume a lapse rate of –6.5 K/km. Use this
new expression to calculate the pressure at z = 5 km. Compare to the value found
assuming T is a constant.
At one point in the derivation, we have:
 z   Mg  
p  p0 exp   
dz 
 0  RT  
and to solve this, we assumed T is not a function of z. But in this problem, T = T0 – 6.5 z
(where T and T0 are in K and z is in km).
z
 
 Mg
Thus, p  p0 exp   
dz 
 0  R (T0  6.5 z )  
dx
1
This is an integral of the form 
, for which the solution is ln( a  bx ) . In this
a  bx
b
case, a = T0 and b = – 6.5.
Therefore,
z
 
 Mg
p  p0 exp   
dz 
R
(
T

6.5
z
)
0

 
0

z
  Mg 1

 Mg
 p0 exp 
ln(T0  6.5 z )   p0 exp 
 ln(T0  6.5 z    ln(T0   

 R 6.5

 6.5 R

0

 0.0288 kg mol 1 (9.78 m s 2 )

 (1.00 atm) exp 
((ln(288  6.5(5))  ln(288)) 
1
1
1
 0.0065 K m (8.314 J K mol )

 0.536 atm
If we instead assume the temperature in constant, we need to choose an appropriate
temperature to do the calculation. I took the average value of the temperature between the
ground (T = 288 K) and at 5 km (T = 255.5); average = 272 K:
 0.0288 kg mol 1 (9.81 m s 2 )(5000 m) 
 Mgz 
p  p0 exp 
  0.535 atm
  1.00 atm exp 
8.315 J K 1mol 1 (272 K )
 RT 


which is negligibly different from the value calculated using temperature in the integral.
5. Some values for spatial and temporal scales of various atmospheric constituents are
presented in the table below.
Species
OH
HO2
H2O2
CO
CH4
CFC’s
Spatial Scale
2m
10 m
1 km
30 km
3000 km
10,000 km
Lifetime
10 s
90 s
5 days
0.3 y
20 y
100 y
SO2 has a lifetime in the atmosphere of 3.0 days. Use the data to calculate the spatial
scale of SO2. You must use regression to solve this problem. Eye-ball-o-metric methods
are not acceptable. Show a plot and the regression coefficients.
log(lifetime) vs log(distance)
6
5
y = 1.2526x - 3.9386
4
log(lifetime, days)
3
2
1
0
-1
0
1
2
3
4
5
6
7
8
-2
-3
-4
-5
log(distance, m)
I used Excel to plot log() versus log(d) (where  is in days and d is in m) and obtained
the regression equation log  = 1.253 log d – 3.939. Rearranging this equation gives
 log  3.939 



log   3.939
log d 
or d  10 1.253
1.253
If the lifetime of SO2 is 3.0 days, then
d  10
 log (3.0)  3.939 


1.253


 103.52  3,345 m  3.35 km
6. Benzene (a known carcinogen) is put into the atmosphere at a rate of approximately
6.3 x 1012 kg y-1. The average concentration of benzene in the atmosphere is 4.6 ppbv.
Calculate the lifetime of benzene (days). Be careful with your units.
Q
P
We have the value of P (6.3 x 1012 kg y-1), so we need to calculate Q (in kg).

4.6 ppbv = 4.6 x 10-9 moles benzene per mole of air, or L benzene per L of air.
As shown in class, the volume of the atmosphere is Vatm = 4.2 x 1018 m3. Thus, the
volume of pure gaseous benzene is VB = 4.9 x 10-9 x 4.2 x 1018 m3 = 2.06 x 1010 m3.
The density of an ideal gas can be derived from the ideal gas law pV = nRT as follows:
PV  nRT
n
P

V RT
n( MW ) P( MW )

V
RT
Where MW is the molecular weight of the gas (78.0 g/mol in this case). Note that the
units of the left side of this equation are therefore g/L, which are the units of density.
Thus  
P( MW )
RT
Most of the mass of the atmosphere is contained within the troposphere, as will be most
of the benzene. I therefore used an average temperature between 0 and 20 km (223K) and
an average pressure (0.5 atm)

P( MW ) 0.5 atm(101325 Pa atm1 )(0.078 kg mol 1 )

 2.13 kg m3
1
1
RT
8.314 J K mol (223K )
Thus, the mass of benzene in the atmosphere = 2.13 kg m-3 x 2.06 x 1010 m3 = 4.39 x 1010
kg.
Thus,
Q
4.39 1010 kg
 
 0.00696 y
P 6.30 1012 kg y 1
365 d y 1  2.5 d