Solutions: Environmental Chemistry - a global perspective 4th Edition
Chapter 6

Atmospheric aerosols

PROBLEMS/SOLUTIONS
1.

Which of the two—cloud droplets or raindrops—would you expect to be more effective in scavenging
gases from the atmosphere? Explain your reasoning.

Solution
It would be expected that cloud droplets would be more effective than rain drops in scavenging gases from
the atmosphere. The smaller size of the cloud droplets (aerosol), compared to the rain drops, means greater
total surface area, giving more opportunity for the interaction with gases. The longer residence times
experienced by the cloud droplets would also lead to more efficient scavenging compared to rain drops.
Rain drops are larger particles with a smaller total surface area and therefore have less contact with the
gases. As well the heavier particle would fall quickly through the atmosphere allowing less time for gases
that could be scavenged from the atmosphere. Counteracting this, to a small extent, is the fact that the
falling drops would move through a larger portion of the air mass during their descent. It is likely the
capacity of individual rain drops would be higher but, it would not be fully utilised, while the capacity of
the cloud droplets would be much smaller but better utilised.
2.

A fog consists of 10 000 droplets of water per cm3. The average diameter of the drops is 1.5 μm.
Compare the mass of water in the liquid phase to that in the gaseous form if the temperature is 35ºC
and the relative humidity is 100%.

Solution
The fog has 10 000 droplets of water per cm3, where T = 35ºC (308 K). The average diameter is 1.5 µm
(1.5 x 10-6 m) and the density of water at 308 K is 0.99406 g mL-1. Appendix A.2 gives the density of water

for 20ºC as 0.99823 g mL-1.
100% relative humidity indicates that the atmosphere is saturated with moisture at the temperature
indicated (308 K). Assume total pressure = P o = 101 325 Pa. Also assume that the droplet of water is a
perfect sphere (radius = 7.5 x 10-7 m).
Appendix C.2 has geometric relations, i.e. volume of a sphere:
volume of a sphere = 4/3πr3
or 1.77 x 10-15 L,

= (4/3) x 3.14 x (7.5 x 10-7 m)3
= 1.77 x 10-18 m3
or 1.767 x 10-12 mL

The mass of one droplet of water at 308 K:
1.767 x 10-12 mL x 0.99406 g mL-1 = 1.757 x 10-12 g (per drop)
1.757 x 10-12 g (per drop) x 10 000 drops (per cm3) = 1.757 x 10-8 g
The mass of liquid water in 1 mL of the fog is 1.76 x 10 -8 g = 0.0176 µg.
The mass of water in the vapour phase is determined as follows:

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Solutions: Environmental Chemistry - a global perspective 4th Edition
Chapter 6

Atmospheric aerosols

Using information like that provided in Figure 8.2, at 35ºC the vapour pressure of water is 5.5 kPa. Since
there is 100% relative humidity, the total vapour pressure for water remains at:
1.00 x 5.5 kPa = 5.5 kPa
The mixing ratio of water vapour is therefore 5.5 kPa

Using PV = nRT, the number of moles of water vapour can be determined.
V = 10-6 m3, P = 5500 Pa, R = 8.314 J mol-1 K-1, T = 308 K
n = 2.15 x 10-6 mol (of water molecules)
2.15 x 10-6 mol x 18.015 g mol-1 = 3.87 x 10-5 g (of water)
There is 38.7 µg of water in the vapour phase in one cm3 (1 mL) at 35ºC at 100% humidity. The ratio of
water in the vapour phase to water in the liquid phase is:
38.7 µg
---------------- = 2200
0.0176 µg
There is about 2200 times as much water in the vapour phase as there is in the liquid phase under the given
set of conditions.
3.

Compare the number of moles of nitrogen–oxygen species in air containing 300 ppbv nitric oxide,
with fog consisting of 10 000 droplets per cm3, having an average diameter of 2 μm, and containing
nitrate ion at a concentration of 3 × 105 mol L–1.

Solution
Data: 300 ppbv NO, 10 000 droplets per cm3, 2 µm = 2 x 10-6 m diameter, [NO3-] = 3 x 10-5 mol L-1.
Calculate the volume of water in one litre of fog. Assume that the droplet of water is a perfect sphere
(radius = 1.0 x 10-6 m).
volume of a sphere = 4/3πr3 = (4/3) x 3.14 x (1.0 x 10-6 m)3
= 4.19 x 10-18 m3

(or 4.19 x 10-15 L) per drop

4.19 x 10-15 L per drop x 10 000 drops per cm3
= 4.19 x 10-11 L water per cm3 air
= 4.19 x 10-8 L water per L air


or

moles of NO3- in water droplets in one L of air
3 x 10-5 mol L-1 x 4.19 x 10-8 L = 1.26 x 10-12 mol
There are approximately 1.3 x 10-12 mol of NO3- in 1 L of fog water drops for the conditions described.

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Solutions: Environmental Chemistry - a global perspective 4th Edition
Chapter 6

Atmospheric aerosols

For the gaseous nitrogen oxide, the total moles of gas in 1 L (at 101 325 Pa and 298 K) is:
PV = nRT

n = 0.04088 mol

Number of moles of NO (based on 300 ppbv = 3 x 10-7 mole fraction) is:
Moles NO
----------------- = 3 x 10-7
0.04088
Moles NO = 1.23 x 10-8 mol
Ratio of nitrogen species in gaseous and dissolved forms is:
1.23 x 10-8 mol NO ÷ 1.26 x 10-12 mol NO3- = 9760
There is about 9800 times more moles of nitric oxide in the gaseous forms in the air (300 ppbv NO) than in
dissolved form in the fog with 10 000 droplets per cm3 (average diameter of 2 µm) with a nitrate
concentration of 3 x 10-5 mol L-1.
4.


A fly ash (ρ = 1.8 g mL–1) aerosol consists of particles averaging 13 μm in diameter and with a
concentration of 800 μg m–3. Use the average diameter to calculate the settling velocity (cm s–1) and
settling rate (g m–2 s –1) of the particles in still air.

Solution
Data: ρ = 1.8 g mL-1 (= 1.8 x 106 g m-3), dp = 13 µm (= 1.3 x 10-5 m) and assume standard conditions for
pressure and temperature, Po (101 325 Pa) and 25ºC (298 K)
Settling velocity (vt): (see section 6.2 for explanations of terms)
(ρp - ρa)Cgdp2
vt = -------------------------- =
18η

(1.8 x 106 – 1.2 x 103) 1.015 x 9.8 x (1.3 x 10-5)2
----------------------------------------------------------------18 x 1.9 x 10-2

vt = 8.8 x 10-3 m s-1
The settling velocity in unit of cm s-1 is:

vt = 8.8 x 10-3 m s-1 x 100 cm m-1 = 0.88 cm s-1

Units are not shown in the calculation above for convenience, but are:
ρ = g m-3 ; C = dimensionless;

g = m s-2 ;

dp = m ;

η = g m-1 s-1


The value of C was obtained by interpolation using data provided in Table 6.4 (p. 143).
(50 – 37)
1.003 + --------------- (1.016 – 1.003) = 1.015
(50 – 10)
The settling rate can be calculated by multiplying the settling velocity by the concentration (in this case 800
µg m-3).
8.8 x 10-3 m s-1 x 800 µg m-3 = 7.0 µg m-2 s-1 (settling rate)

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Solutions: Environmental Chemistry - a global perspective 4th Edition
Chapter 6

5.

Atmospheric aerosols

Suggest a possible sequence of reactions (beginning with oxidation via the hydroxyl radical) by which
dimethyl sulfide ((CH3)2S) can be oxidized to produce a sulfuric acid aerosol.

Solution
One possible mechanism is as follows:
(CH3)2S +

•OH → H2O +

•CH2 – S – CH3 + O2 →

•CH2 – S – CH3


•O – O – CH2SCH3

•O – O – CH2SCH3 + NO → NO2 + •O – CH2SCH3
•O – CH2SCH3 → CH2O + CH3S•
CH3S• + O2 →

•CH3 + SO2

SO2 + •OH + M → HO •SO2 + M
O2 + HO •SO2 + M → HOO• + SO3 + M
SO3 + H2O → H2SO4
6.

Polyaromatic hydrocarbons are commonly associated with soot particles, as adsorbates on the surface.
Calculate the relative and actual surface areas (actual values in m2 g–1) of soot particles (ρ = 0.6 g
mL–1) having diameters 20 μm and 2 μm, respectively, assuming that the particles are spheres.
Speculate on the health implications with regard to human intake of PAH compounds.

Solution
PAHs are associated with soot particles whose density is ρ = 0.6 g mL-1. Assume volume of 1 mL and
therefore 0.6 g of soot.
For the 20 µm particles, the volume of the sphere is determined by:
4/3πr3 = 4/3 x 3.14 x (10 x 10-6 m)3
= 4.19 x 10-15 m3 (or 4.19 x 10-9 mL) per sphere
The mass of the sphere is:

0.6 g mL-1 x 4.19 x 10-9 mL = 2.51 x 10-9 g

In 1 g of solid soot particles there are 1 ÷ 2.51 x 10-9 = 3.98 x 108 spheres

The surface area of each sphere is:

4πr2 = 4 x 3.14 x (10 x 10-6 m)2 = 1.3 x 10-9 m2

The total surface area of all spheres weighing 1 g is:
1.3 x 10-9 m2 x 3.98 x 108 = 0.50 m2
For the 2 µm particle the volume of the sphere is:

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Solutions: Environmental Chemistry - a global perspective 4th Edition
Chapter 6

Atmospheric aerosols
4/3πr3 = 4/3 x 3.14 x (1.0 x 10-6 m)3
= 4.19 x 10-18 m3 (or 4.19 x 10-12 mL) per sphere
0.6 g mL-1 x 4.19 x 10-12 mL = 2.51 x 10-12 g

The mass of the sphere is:

In 1 g of solid soot particles there are 1 ÷ 2.51 x 10-12 = 3.98 x 1011 spheres
The surface area of each sphere is:

4πr2 = 4 x 3.14 x (1.0 x 10-6 m)2 = 1.3 x 10-11 m2

Total surface area of all spheres weighing 1 g is:
1.3 x 10-11 m2 x 3.98 x 1011 = 5.0 m2
The surface area of the 2 µm soot particles is 10 times greater than that of the 20 µm particles.
0.50 m2 g-1)


(5.0 vs.

The smaller particles (2 µm) have a greater surface area, providing a greater area on which to adsorb PAHs.
They will also have a longer residence time (and possibly remain suspended until removed by a process
such as washout in rain). The larger soot particles (20 µm) will settle fairly rapidly and will not have as
great an ability to adsorb PAHs. Furthermore, the smaller particles are less likely to be retained in the nasal
passages, than are the larger particles. The smaller particles therefore could reach the lungs. Overall, the
health implications are that the smaller soot particles could carry high concentrations of PAHs and are
likely to be inhaled, exposing the body to a source of potentially dangerous PAH compounds.
7.

In fly ash, the lead and chromium concentrations are measured for four particle diameter ranges.
Results are as follows:
Particle diameter / μm

Concentration / μg g–1
Lead

Chromium

> 10

870

330

6–10

990


760

2–6

1100

1800

<2

1300

2700

Suggest reasons for these trends.
Solution
As the particle size increases total surface area decreases and less metal (Pb or Cr) is adsorbed on the
surface of the particle. This would suggest that both metals are being adsorbed onto the surface of the fly
ash. The metals may have been present in trace amounts in the fuel, were volatilized during combustion of
the fuel, and then deposited on the surface of the fly ash particles as the particles cooled. The smaller sized
fractions end up with a larger concentration, while the larger particles with the greater mass have a smaller
concentration, despite possibly sorbing the same amount of Pb or Cr per unit surface area. The Pb results
also suggest that the particles themselves may contain some Pb, not just that absorbed on the surface, since
the concentration, even for the large particles, is still relatively high.

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Solutions: Environmental Chemistry - a global perspective 4th Edition

Chapter 6

8.

Atmospheric aerosols

Refer to the NASA NEO website given in Additional Resources 7. From the main page, select the
‘Atmosphere’ link at the top left hand side of the page, then choose aerosol particle radius or aerosol
optical thickness. Explore the data and compare monthly changes over the years of the state of aerosols
around the globe.

For classroom discussion, no solution provided. What is the aerosol condition where you live and how
does it compare? What can you do to reduce any anthropogenic contribution to aerosol issues?

9. A sample of solid aerosol is collected from a rural area and found to have as composition Al = 2400 mg
kg-1 and Pb = 3.4 mg kg-1. Using data for crustal abundance of elements (Appendix B.1), calculate the
chemical concentration factor for lead. What might be the reason for a value such as you have
obtained?
Solution
From Appendix B.1:

Pb 14 mg kg-1

(CPb / CAl)aerosol
CCF = --------------------------(CPb / CAl)crustal

=

Al


81 300 mg kg-1

3.4 / 2400
------------------------ =
14 / 81 300

0.0014166
------------------- = 8.2
0.0001722

One reason for this high CCF for lead found in an aerosol collected in a rural area could be due to our past use
of leaded fuel.

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