ATMOSPHERIC STUCTURE

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TOPIC I ATMOSPHERIC STUCTURE AND BULK COMPOSITION
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Density, Temperature, and Pressure
The physical properties of a gas are
characterized by 3 state variables State
variable Units Temperature (T) Measure of the
random kinetic energy of oK, oC, oF individual
gas molecules Pressure (P) Force per unit are
exerted by a gas F L-2 M L T-2 L-2 M L-1
T-2 e.g. g cm-1 s-2, kg m-1 s-2 Density
(?) Mass of gas per unit volume ML-3 e.g.
g cm-3, g lit-1
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Units of temperature and density are familiar.
However, various units are used to express
pressure due to tradition, convenience, etc .
We will often multiply pressure by other
quantities, and so it is useful to become
familiar with the various units of pressure.
Take extra care to with units in calculations 1
dyne cm-2 1 g cm-1 s-2 1 N m-2 1 kg m-1 s-2 1
atm (the average pressure at sea-level) 1.013 x
106 g cm-1 s-2 1 Pa (pascal, unit of pressure in
SI units) 1 kg m-1 s-2 1 hPa (hectopascal)
100 Pa 100 kg m-1 s-2 1 mbar (millibar) 1 hPa
100 kg m-1 s-2
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Example of units conversion (1 atm ? Pa ? hPa
? mbar) 1 atm 1.013 x 106 g cm-1 s-2
1.013 x 106 g cm-1 s-2 x (10-3 kg g-1) x
(100 cm m-1) 1.013 x 105 kg m-1 s-2
1.013 x 105 N m-2 1.013 x
105 Pa 1.013 x 105 Pa x (10-2 hPa
Pa-1) 1.013 x 103 hPa
1.013 x 103 mbar
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IDEAL GAS LAW
To a good approximation, the atmosphere can be
assumed to have the properties of an ideal
gas p V ? Rg T (1) where, p pressure
of an air parcel V volume occupied by the air
parcel ? number of moles of air mass of
air/molecular weight of air T temperature of
air Rg universal gas constant Note Rg has
units of (p V)/(? T) When, p is in atm, V is in
lit, ? is in mole, and T is in K Rg 0.082 atm
lit mole-1 K-1 Aside Remember 1 mole of each gas
contains 6.023 x 1023 molecules (Avogadros )
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Equation (1) can be rewritten in terms of the 3
state variables (p, T, and ?) as follows p V ?
Rg T (1) p (? Rg T)/V (2) Noting
that ? mass/Ma, where mass is the mass of the
air and Ma is the molecular weight of the gas,
equation (2) can be rewritten as p (mass Rg
T)/(Ma V) (3) Noting that mass/V ?, where
? is the density, equation (3) can be rewritten
as p (? Rg T)/Ma (4) Equation (4) is the
ideal gas law in terms of the 3 state variables
(p, T, and ?)
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Calculation of Ma Since air is a mixture of
gases, the molecular weight of air is computed as
follows Ma ??iMi/ ??i ??iMi/? (5) where,
?i is the number of moles of constituent i, Mi is
the molecular weight of constituent i, ? is the
total number of moles of air, and the symbol ?
denotes summation from 1 to i. Equation (5) can
be rewitten as Ma ?(?i/ ?) Mi
?miMi, (6) where, mi ?i/ ?, is the mole
fraction of constituent i. So, if dry air is
assumed to be composed of N2, O2, and 40Ar, with
mole fractions of 0.78, 0.21, and 0.01,
respectively, Ma (0.78 x 28 0.21 x 32 0.01
x 40) /(0.78 0.21 0.01) 28.96 g mole-1
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Calculation of density of dry air at 25C and 1
atm pressure From equation (4), p (? Rg
T)/Ma Rearranging the equation, we get ? (p
Ma)/(Rg T) (1 atm x 28.96 g
mole-1)/(0.082 atm lit mole-1 K-1 x 298 K)
1.2 g lit-1 This value can be converted to
other volume units. For example ? 1.2 g lit-1
1.2 g lit-1 x 10-3 lit cm-3 1.2 x 10-3 g
cm-3 This value can also be converted to other
mass units. For example ? 1.2 x 10-3 g cm-3
1.2 x 10-3 g cm-3/ 28.96 g mole-1 4.14 x
10-5 mole cm-3 4.14 x 10-5 mole cm-3 x
6.023 x 1023 molecules mole-1 2.5 x 1019
molecules cm-3
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The abundance of a particular constituent i in
the atmosphere is expressed in a variety of ways.
Two of the most widely used in atmospheric
chmeistry are Mole fraction (mi) ?i/? moles
of i per mole of air molecules of I per
molecule of air (units mole i/ mole air
molecules i/molecules air) Since the atmospheric
abundance of a number of gases that we are
interested in is very small, their mole fraction
is very small. So, the mole fraction is often
expressed in as moles of i per million (ppm),
billion (ppb), or trillion moles (ppt) of
air. So, 0.001 mole i/mole air 0.001 x 106 ppm
0.001 x 109 ppb 0.001 x 1012 ppt Number
concentration (ni) number of molecules of i per
unit volume of air (units molecules cm-3) Note
ni mi x (air density in molecules cm-3) Other
ways to express abundance are mass mixing ratio
(i.e. mass of i per mass of air) and mass
concentration (i.e. mass of i per unit volume of
air)
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Occassionally, the abundance of a gas in the
atmosphere is expressed in terms of its partial
pressure( pi). Relationship between pi and
mi Ideal gas law applies to individual gases as
well. From equation (1) for air p V ? Rg T for
constituent i pi V ?i Rg T where, pi is the
partial pressure of constituent i (i.e., it is
the pressure if only constituent i occupied the
whole volume) Thus, pi/p ?i/? mi
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Variation of pressure with altitude Hydrostatic
and isothermal atmosphere Assuming that the
atmosphere is hydrostatic (i.e., upward force on
a parcel due to air below the parcel is balanced
by the downward force on the parcel due to air
above it and due to gravity), results in dp/dz
-?g (7) This equation can be integrated
by using the ideal gas law and assuming that the
atmosphere is isothermal yielding pz
poexp(-z/H), (8) where, pz is the
atmospheric pressure at height z, po is the
atmospheric pressure at the surface, and H is the
scale height defined as H (Rg T)/(Ma
g) (9)
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Hydrostatic and adiabatic atmosphere Assuming
that the atmosphere temperature lapse rate is dry
adiabatic dT/dz -g/Cp, (10) where, Cp is
the heat capacity at constant pressure per unit
mass of air. Cp for dry air is 1004 J kg-1 K-1.
Thus, dT/dz corresponding to a dry adiabatic
lapse rate is -9.81/1004 ? -10K km-1. If Cp is
constant with altitude, then integrating equation
(10) we get Tz To - gz/Cp (11) where, Tz
is the temperature at altitude z and To is the
temperature at the surface. Combining, equation
(7) with equation (10) and integrating, we
get (Tz/To) (pz/po)(Rg/CpMa) (12)
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Mass of the atmosphere Assuming hydrostatic
atmosphere, the mass of the atmosphere (av.
surface pressure) x (area of earths
surface)/(acceleration due to gravity) Assuming
that the average surface pressure is 1 atm
1.013 x 106 g cm-1 s-2, the mass of the
atmosphere (1.013 x 106 g cm-1 s-2) x (5 x 1018
cm2)/(981 cm s-2)
5.1 x 1021 g
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Chemical Composition of the Earths
Atmosphere(mole fraction)
J. C. G. Walker, Evolution of the Atmosphere,
1977
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• Due to abundance of O2, earths atmosphere is
  oxidizing (e.g., SO2 --gt SO42- CH4 --gt CO CO
  --gt CO2 NO2 --gt HNO3)
• After O2, O3 is most abundant oxidant
• But direct oxidation of non-radical species by
  reaction with O2 and O3 is too slow to be of
  consequence
• The OH radical is the most important oxidant in
  the troposphere
• Other atmospheric oxidants include NO3, halogen
  radicals, and H2O2 (in the aqueous-phase)

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BIOGEOCHEMICAL CYCLE OF O2 Study provides
insights into how O2 content of stmosphere is
maintained Stoichiometric representation of O2
production by photosynthesis and loss by
respiration and decay nCO2 nH2O --gt nCH2O
nO2 n CH2O nO2 --gt nCO2 nH2O Here,
CH2O represents organic matter (most common
form is carbohydrate with CHO of 121
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Exchanges between atmosphere and terrestrial
biosphere No net O2 production
ATMOSPHERE O2 36000 (O2)
Mass kTmole (1015 moles) Fluxes kTmole year-1
OCEAN BIOSPHERE C 0.15 (C)
4 (O2)
TERRESTRIAL BIOSPHERE C 200 (C)
decay
npp
4 (C)
4 (CO2)
ATMOSPHERE CO2 50 (CO2)
OCEAN CO2 3253 (CO2)
OCEAN BOTTOM C 1700000 (C)
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Add exchanges with ocean Now there is net O2
production
3.32 (O2)
ATMOSPHERE O2 36000 (O2)
OCEAN BIOSPHERE C 0.15 (C)
3.33 (O2)
4 (O2)
TERRESTRIAL BIOSPHERE C 200 (C)
3.33 (C)
decay
photosynth
npp
4 (C)
3.32 (C)
resp. decay
4 (CO2)
3.33 (CO2)
3.32 (CO2)
org. C burial
ATMOSPHERE CO2 50 (CO2)
OCEAN CO2 3253 (CO2)
6.67 (CO2)
6.66 (CO2)
OCEAN BOTTOM C 1700000 (C)
0.01 (C)
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This is a simple picture of atmospheric O2 Main
features O2 produced whenever C is transferred
from atmospheric and ocean CO2 reservoirs to
terrestrial and ocean biosphere C reservoirs.
O2 removed whenever biospheric C is returned to
ocean and atmospheric reservoirs. Net O2
production involves only ocean biosphere and not
terrestrial atmosphere. Net O2 production rate
corresponds to organic carbon burial rate in
ocean sediments. Inferences If all the
biosphere is oxidized (C O2 --gt CO2),
atmospheric O2 consumed 200 0.15 200.15
kTmole. This is a small fraction (200.15/36000)
of the atmospheric O2 content.
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Problem with cycle Cycle is not in balance.
Atmospheric CO2 reservoir is being depleted , and
atmospheric O2 and ocean bottom C reservoirs are
increasing, at a rate of 0.01 kTmole year-1 If
this were the case, all the C would ultimately be
drawn into the ocean bottom with a concomittant
small increase in atmospheric O2. The fix is in
the rock cycle - sediments brought to the earths
surface are weathered (C O2 --gt CO2).
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Add weathering to balance cycle
0.01 (O2)
3.32 (O2)
ATMOSPHERE O2 36000 (O2)
OCEAN BIOSPHERE C 0.15 (C)
3.33 (O2)
4 (O2)
TERRESTRIAL BIOSPHERE C 200 (C)
3.33 (C)
decay
photosynth
npp
4 (C)
3.32 (C)
resp. decay
4 (CO2)
3.33 (CO2)
3.32 (CO2)
org. C burial
ATMOSPHERE CO2 50 (CO2)
weathering
OCEAN CO2 3253 (CO2)
6.67 (CO2)
0.01 (CO2)
6.66 (CO2)
OCEAN BOTTOM C 1700000 (C)
0.01 (C)
0.01 (C)
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Inferences If all the biosphere is oxidized (C
O2 --gt CO2), atmospheric O2 consumed 200
0.15 200.15 kTmole. This is a small fraction
(200.15/36000) of the atmospheric O2 content
Timescales of variation The concept of
lifetime (?) is useful in this regard - rough
measure of timescale on which variations in
process rates can cause significant variations
in amount in reservoir Defining ? mass in
reservoir/loss rate,we get ?O2 due to wethering
(36600 kTmole)/(0.01 kTmole year-1) 3.66
million years ?O2 due to interactions with
biosphere (36600 kTmole)/(43.32 kTmole year-1)
? 5000 years Thus, fluctuations in interactions
with the biosphere can cause significant changes
in atmospheric O2 on timescales of about 5000
years -gt this is still long enough to make
detection of seasonal changes difficult.
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If production of O2 halted, it would take about 4
million years to significantly deplete
atmospheric O2. Imbalances in weathering rates
could cause significant fluctuation rates on
timescales of 4 million years. But atmospheric
O2 content has been relatively stable for much
longer (mammals have been around for about 65
million years) and multicelled animals with
oxidative metabolisms have been around for 10
times as long. This is due to feedbacks in the
system (see reading). Another problem with cycle
is that if each atmospheric O2 produced
corresponds to an organic C in sediments, why is
the amount of C in sediments so much greater than
atmospheric O2 content. Answer is that most of
the O2 produced is buried in the form of sulfur
(SO4) and iron oxides (Fe2O3). (see section 9.3
of reading for detailed cycle).
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Fossil-fuel combustion can be though of as
enhanced weathering of C about 40 times
normal weathering rate (40 x 0.01 kTmole C
year-1) 0.4 kTmole C year-1 But fossoil-fuels
comprise only about 0.06 of org C sediments -
thus , total fossil-fuels 0.0006 x 1700000
1020 kTmoles C. At present rate, all
fossil-fuels woulb be gone is 1020/0.4 2550
years At this were to happen, O2 would decrease
only by 1020 kTmole (i.e. by 2-3)