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Lecture 7 Gases and Gas Exchange

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The atmosphere controls the oceans gas contents for all gases except radon, CO2 and H2O. ... The partial pressure of CO2 in the atmosphere is increasing every day ... – PowerPoint PPT presentation

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Title: Lecture 7 Gases and Gas Exchange


1
Lecture 7 Gases and Gas Exchange
2009
Gas Exchange Fluxes Effect of wind Global CO2
fluxes by gas exchange
Composition of the atmosphere Gas solubility
2
Sarmiento and Gruber (2002) Sinks for
Anthropogenic Carbon Physics Today August 2002
30-36
3
Composition of the Atmosphere More than 95 of
all gases except radon reside in the atmosphere.
The atmosphere controls the oceans gas contents
for all gases except radon, CO2 and
H2O. Gas Mole Fraction in Dry Air (fG)
molar volume at STP (l mol-1 )
where fG moles gas i/total moles
N2 0.78080 22.391 O2 0.20952 22.385 Ar 9
.34 x 10-3 22.386 CO2 3.3 x 10-4
22.296 Ne 1.82 x 10-5 22.421 He 5.24 x
10-6 22.436 H2O 0.012
Why is dry air used?
4
Solubility The exchange or chemical equilibrium
of a gas between gaseous and liquid phases can
be written as A (g) ? A (aq) At
equilibrium we can define the familiar value K
A(aq) / A(g) There are two main ways
to express solubility (Henrys Law and Bunsen
Coefficients).
5
1. Henry's Law We can express the gas
concentration in terms of partial pressure using
the ideal gas law
PV nRT P pressure, V
volume, n moles R gas constant
8.314 J K-1 mol-1, T temp so that the number
of moles n divided by the volume is equal to
A(g) n/V A(g) PA / RT where PA
is the partial pressure of A Then K
A(aq) / PA/RT or
A(aq) (K/RT) PA A(aq)
KH PA units for K are mol kg-1 atm-1
in mol kg-1
for PA are atm Henry's Law states that
the concentration of a gas in water is
proportional to its overlying partial pressure.
KH is mainly a function of temperature with a
small impact by salinity.
6
Example (Solubility at 0?C) Partial Pressure
Pi fG x 1atm total pressure Gas Pi KH (0?C ,
S 35) Ci (0?C, S 35 P 760 mm
Hg) N2 0.7808 0.80 x 10-3 624 x 10-6 mol
kg-1 O2 0.2095 1.69 x 10-3 354 x
10-6 Ar 0.0093 1.83 x 10-3 17 x
10-6 CO2 0.00033 63 x 10-3 21 x 10-6
Example The value of KH for CO2 at 24?C is 29 x
10-3 moles kg-1 atm-1 or 2.9 x 10-2 or
10-1.53. The partial pressure of CO2 in the
atmosphere is increasing every day but if we
assume that at some time in the recent past it
was 350 ppm that is equal to 10-3.456 atm.
7
  • Summary of trends in solubility
  • Type of gas
  • KH goes up as molecular weight
  • goes up (note that CO2 is anomalous)
  • 2. Temperature
  • Solubility goes up as T goes down
  • 3. Salinity
  • Solubility goes up as S goes down

8
Temperature control on gas concentrations
O2 versus temperature in surface ocean solid
line equals saturation for S 35 at different
temperatures average supersaturation 7
mmol/kg (3)
9
Rates of Gas Exchange Stagnant Boundary Layer
Model.
well mixed atmosphere
Cg KH Pgas equil. with atm
ATM
0
OCN
Stagnant Boundary Layer transport by
molecular diffusion
ZFilm
Depth (Z)
CSW
well mixed surface SW
Z is positive downward ?C/ ?Z F (flux
into ocean)
10
Flux of Gas The rate of transfer across this
stagnant film occurs by molecular diffusion from
the region of high concentration to the region
of low concentration. Transport is described by
Fick's First Law which states simply that flux
is proportional to the concentration
gradient.. F - D dA / dZ where D
molecular diffusion coefficient ( f (gas and T))
(cm2sec-1) dZ is the thickness of the stagnant
film (Zfilm)(cm) dA is the concentration
difference across the film (mol cm-3) The water
at the top of the stagnant film (Cg) is assumed
to be in equilibrium with the atmosphere. We can
calculate this value using the Henry's Law
equation for gas solubility. The bottom of the
film has the same concentration as the
mixed-layer (CSW). Thus Flux F -
D/Zfilm (Cg - CSW) - D/Zfilm (KHPg - CSW)
11
Because D/Zfilm has velocity units, it has been
called the Piston Velocity (k) e.g., D cm2
sec-1 Z film cm Typical values are D
1 x 10-5 cm2 sec-1 at 15ºC
Zfilm 10 to 60 mm Example D 1 x 10-5 cm2
sec-1 Zfilm 17 mm determined for the average
global ocean using 14C data Thus Zfilm 1.7 x
10-3 cm The piston velocity D/Z k 1 x
10-5 cm s-1 /1.7 x 10-3 cm 0.59 x
10-2 cm/sec ? 5 m / d note 1 day
8.64 x 104 sec Each day a 5 m thick layer of
water will exchange its gas with the
atmosphere. For a 100m thick mixed layer the
exchange will be completed every 20 days.
12
Gas Exchange and Environmental Forcing Wind
Wanninkhof, 1992 from 14C
Liss and Merlivat,1986 from wind tunnel exp.
5 m d-1
20 cm hr-1 20 x 24 / 102 0.48 m d-1
13
U-Th Series Tracers
14
222Rn Example Profile from North Atlantic
Does Secular Equilibrium Apply? t1/2 222Rn ltlt
t1/2 226Ra (3.8 d) (1600
yrs) YES! A226Ra A222Rn
222Rn
226Ra
Why is 222Rn activity less than 226Ra?
15
222Rn is a gas and the 222Rn concentration in the
atmosphere is much less than in the ocean mixed
layer (? mixed layer). Thus there is a net
evasion of 222Rn out of the ocean.
The 222Rn balance for the mixed layer, ignoring
horizontal advection and vertical exchange with
deeper water, is
d222Rn/dt sources sinks decay of 226Ra
decay of 222Rn - gas exchange to atmosphere
?ml l222Rn d222Rn/dt ? ml l226Ra 226Ra ?
l 222Rn 222RnML
D/Zfilm 222Rnatm
222RnML
Knowns l222Rn, l226Ra, DRn Measure ? ml,
A226Ra, A222Rn, d222Rn/dt Solve for Zfilm
16
?ml l222Rn d222Rn/dt ? ml l226Ra 226Ra
?ml l222Rn 222Rn
D/Zfilm 222Rnatm
222RnML ?ml dA222Rn/ dt ?ml (A226Ra
A222Rn) D/Z (CRn, atm CRn,ML)
atm Rn 0
for SS 0
Then -D/Z ( CRn,ml) ?ml (A226Ra A222Rn)
D/Z (ARn,ml/lRn) ?ml (A226Ra A222Rn)
D/Z (ARn,ml) ?ml lRn (A226Ra
A222Rn) ZFILM D (A222Rn,ml) / ?ml lRn (A226Ra
A222Rn) ZFILM (D / ?ml lRn) (
)
17
Stagnant Boundary Layer Film Thickness
Z DRn / ? l 222Rn
(1/A226Ra/A222Rn) ) - 1
Histogram showing results of film
thickness calculations from many
stations. Organized by Ocean and by Latitude
Average Zfilm 28 mm
  • Q. What are limitations of
  • this approach?
  • unrealistic physical model
  • steady state assumption

18
One of main goals of JGOFS was to calculate the
CO2 flux across the air-sea interface
Flux F - D/Zfilm (Cg - CSW) - D/Zfilm
(KHPg - CSW)
- D/Zfilm (KHPo KHPSW)

-D/Zfilm KH (Po PSW)
19
Expression of Air -Sea CO2 Flux
  • Magnitude
  • Mechanism
  • Apply over larger space time domain

k-transfer velocity From Sc wind speed
S Solubility From SST Salinity
F k s (pCO2w- pCO2a) K ? pCO2
pCO2a
pCO2w
From CMDL CCGG network
From measurements and proxies
20
Global Map of Piston Velocity (k in m yr-1) times
CO2 solubility (mol m-3) K from satellite
observations (Nightingale and Liss, 2004 from
Boutin).
21
?pCO2 fields
Overall trends known Outgassing at low
latitudes (e.g. equatorial) Influx at high
latitudes (e.g. circumpolar) Spring blooms
draw down pCO2 (N. Atl) El Niños decrease
efflux
22
JGOFS Gas Exchange Highlight 4 -
?pCO2 fieldsTakahashi climatology
Monthly changes in pCO2w
23
Fluxes JGOFS- Global monthly fluxes
Combining pCO2 fields with k F k s (pCO2w-
pCO2a)
  • On first order flux and ?pCO2 maps do not look
    that different

24
CO2 Fluxes Status
Do different parameterizations between gas
exchange and wind matter?
Global uptakes Liss and Merlivat-83 1 Pg C
yr-1 Wanninkhof-92 1.85 Pg C
yr-1 WanninkhofMcGillis-98 2.33 Pg C
yr-1 Zemmelink-03 2.45 Pg C yr-1
Yes!
Global average k (21.4 cm/hr) 2.3 Pg C yr-1
We might not know exact parameterization with
forcing but forcing is clearly important
Compare with net flux of 1.3 PgCy-1 (1.9 -
0.6) in Sarmiento and Gruber (2002), Figure 1
25
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26
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27
Some comments about units of gases In
Air In Water Pressure - Atmospheres
Volume - liters gas at STP / kgsw 1 Atm 760
mm Hg STP standard temperature and
pressure Partial Pressure of Gasi P (i) /760
1 atm and 0?C (
273ºK) Volume - liters gas / liters air
Moles - moles gas / kgsw (ppmv ml / l,
etc) Conversion lgas/kgsw / lgas / mole
moles/kgsw (22.4
l/mol)
28
Dalton's Law Gas concentrations are expressed in
terms of pressures. Total Pressure SPi
Dalton's Law of Partial Pressures PT PN2
PO2 PH2O PAr ......... Dalton's Law
implies ideal behavior -- i.e. all gases behave
independently on one another (same idea as ideal
liquid solutions with no electrostatic
interactions). Gases are dilute enough that this
is a good assumption. Variations in partial
pressure (Pi) result from 1) variations in PT
(atmospheric pressure highs and lows) 2)
variations in water vapor ( PH2O) We can express
the partial pressure (Pi) of a specific gas on a
dry air basis as follows Pi PT - h/100 Po
fg where Pi partial pressure of gas i
PT Total atmospheric pressure
h relative humidity Po vapor
pressure of water at ambient T fg
mole fraction of gas in dry air (see table above)
29
Example Say we have a humidity of 80 today and
the temperature is 15?C Vapor pressure of H2O at
15?C Po 12.75 mm Hg (from reference
books) Then, PH2O 0.80 x 12.75 10.2 mm
Hg If PT 758.0 mm Hg PTDry (758.0 - 10.2)
mm Hg 747.8 mm Hg Then fH2O
PH2O / PT 10.2 / 758.0 0.013 So for
these conditions H2O is 1.3 of the total gas in
the atmosphere. That means that water has a
higher concentration than Argon (Ar). This is
important because water is the most important
greenhouse gas!
30
Example Units for CO2 Atmospheric CO2 has
increased from 280 (pre-industrial) to 380
(present) ppm. In the table of atmospheric
concentrations (page 1 this Lecture) fG,CO2 3.3
x 10-4 moles CO2/total moles 330 x
10-6 moles CO2/total moles 330
ppm This can also be expressed in log form as
100.52 x 10-4 10-3.48
31
Example Units for Oxygen Conversion from volume
to moles Use O2 22,385 L / mol at standard
temperature and pressure (STP) if O2 5.0 ml
O2/LSW then 5.0 ml O2 / Lsw x mole O2 /
22,385 ml 0.000223 ml O2 / Lsw 223
mmol O2 / Lsw
32
  • 2. Bunson Coefficients
  • Since oceanographers frequently deal with gas
    concentrations not only in molar units but also
    in ml / l, we can also define
  • A(aq) a PA
  • where a 22,400 x KH (e.g., one mol of gas
    occupies 22,400 cm3 at STP)
  • is called the Bunsen solubility coefficient.
    Its units are cm3 mol-1.

33
Solubilities of Gases in Seawater
from Broecker and Peng, (1982)
Bunson Coefficient
Henrys Law
Solubility increases with mole weight and
decreasing temperature
Concentration ratio for equal volumes of air
and water.
KH 29 x 10-3 2.9 x 10-2 10-1.53
34
Example What is the concentration of CO2 (aq) in
equilibrium with the atmosphere? For PCO2 350
ppm 10-3.456 For CO2 KH 29 x 10-3 2.9 x
10-2 10-1.53 moles /kg atm then CO2 (aq) KH
x PCO2 10-1.53 x 10-3.456 10-4.986 mol kg-1
100.014 10-5 1.03 x 10-5
10.3 x 10-6 mol/l at 25?C The concentration of
CO2(aq) will be dependent only on PCO2 and
temperature. It is independent of pH.
35
Gas Solubility - CFCs
36
Causes of deviations from Equilibrium Causes of
deviation from saturation can be caused
by 1. nonconservative behavior (e.g.
photosynthesis () or respiration (-) or
denitrification ()) 2. bubble or air injection
() 3. subsurface mixing - possible
supersaturation due to non linearity of KH
or a vs. T. 4. change in atmospheric pressure -
if this happens quickly, surface waters
cannot respond quickly enough to reequilibrate.
37
Is beer carbonated? Calculate the flux of CO2
(in mol m-2 s-1) out of your favorite, frosty,
carbonated beverage. The PCO2 in the beverage
0.125atm Assume the surface of the beverage is
in equilbrium with the PCO2 of the atmosphere
(375 x 10-6 atm.). Let DCO2 2 x 10-9 m2 s-1
and let the stagnant boundary layer thickness be
Zfilm 5 x 10-5 m. What is the flux? (ans
0.144 x 10-1 mol m-2 s-1) Which way does the CO2
flux go? (ans out of the beer)
38
Effect of El Nino on ?pCO2 fields High resolution
pCO2 measurements in the Pacific since Eq. Pac-92
Eq Pac-92 process study
PCO2sw
Always greater than atmospheric
Cosca et al. in press
El Nino Index
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