lecture 4 gases and gas exchange composition of the atmosphere gas solubility gas exchange fluxes...
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Lecture 4 Gases and Gas ExchangeComposition of the atmosphereGas solubility
Gas Exchange FluxesEffect of windGlobal CO2 fluxes by gas exchange
2009
Emerson and Hedges: Chpts 3 and 10
Sarmiento and Gruber (2002) Sinks for Anthropogenic CarbonPhysics Today August 2002 30-36
Composition of the AtmosphereMore 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 22.414 for an ideal gas (0°C)
N2 0.78080 22.391O2 0.20952 22.385Ar 9.34 x 10-3 22.386CO2 3.3 x 10-4 22.296Ne 1.82 x 10-5 22.421He 5.24 x 10-6 22.436
H2O ~0.013
Why is dry air used?
Some comments about units of gases:
In Air In WaterPressure - 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 0C (= 273ºK)
Volume - liters gas / liters air Moles - moles gas / kgsw
(ppmv = l / l, etc)
Conversion: lgas/kgsw / lgas / mole = moles/kgsw
(~22.4 l/mol)
Dalton's LawGas concentrations are expressed in terms of pressures.
Total Pressure = Pi = Dalton's Law of Partial PressuresPT = 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)
Example:Say we have a humidity of 80% today and the temperature is 15C
Vapor pressure of H2O at 15C = Po = 12.75 mm Hg (from reference books)
Then, PH2O = 0.80 x 12.75 = 10.2 mm HgIf 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!
Example: Units for CO2
Atmospheric CO2 has increased from 280 (pre-industrial) to 380 (present) ppm.
In the table of atmospheric concentrations (see slide 3)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
Example: Units for Oxygen
Conversion from volume to molesUse 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 mol O2 / 22,385 ml = 0.000223 mol O2 / Lsw
= 223 mol O2 / Lsw
SolubilityThe 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 valueK = [A(aq)] / [A(g)]
There are two main ways to express solubility (Henry’s Law and Bunsen Coefficients).
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.
Example (Solubility at 0C):Partial Pressure = Pi = fG x 1atm total pressure
Gas Pi KH (0C , S = 35) Ci (0C, 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
ExampleThe value of KH for CO2 at 24C 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.
See Emerson and Hedges: Table 3.6 for 20°C and Table 3A1.1 for regressions fpr all T and S
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
= 10+0.014 10-5 = 1.03 x 10-5 = 10.3 x 10-6 mol/l at 25C
The concentration of CO2(aq) will be dependent only on PCO2 and temperature.It is independent of pH.
Summary of trends in solubility:1. 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
O2 versus temperaturein surface ocean
solid line equals saturationfor S = 35 at different temperatures
average supersaturation≈ 7 mmol/kg (~3%)
Temperature controlon gas concentrations
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 vs. T.4. change in atmospheric pressure - if this happens quickly, surface waters cannot respond quickly enough to reequilibrate.
Rates of Gas ExchangeStagnant Boundary Layer Model.
Depth (Z)
ATM
OCN
Cg = KH Pgas = equil. with atm
CSW
ZFilm
Stagnant BoundaryLayer – transport by molecular diffusion
well mixed surface SW
well mixed atmosphere
0
Z is positive downward
C/ Z = F = + (flux into ocean)
Flux of GasThe 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 d[A] / dZwhere D = molecular diffusion coefficient in water (= f (gas and T)) (cm2sec-1)dZ is the thickness of the stagnant film on the ocean side (Zfilm)(cm)d[A] is the concentration difference across the stagnant 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)
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 m
Example:D = 1 x 10-5 cm2 sec-1
Zfilm = 17 m determined for the average global ocean using 14C dataThus 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.
0
20
40
60
80
100
0 5 10 15 20
Tracer N-2000Tracer Wat-91Tracer-WannRnC-14bombC-14 natk, 600 W-92k, 600 L&MS-86W-99
k, 6
00
(cm
/hr)
U10
(m/s)
Gas Exchange and Environmental Forcing: Wind
Liss and Merlivat,1986from wind tunnel exp.
Wanninkhof, 1992from 14C
20 cm hr-1 = 20 x 24 / 102 = 4.8 m d-1
~ 5 m d-1
U-Th Series Tracers
222Rn Example Profile fromNorth Atlantic
226Ra
222RnDoes Secular Equilibrium Apply?
t1/2 222Rn << t1/2 226Ra
(3.8 d) (1600 yrs)
YES!A226Ra = A222Rn
Why is 222Rn activity less than 226Ra?
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:
ml 222Rn [222Rn]/t = ml 226Ra [226Ra] – 222Rn [222RnML] + D/Zfilm { [222Rnatm] – [222RnML]}
Knowns: 222Rn, 226Ra, DRn
Measure: ml, A226Ra, A222Rn, d[222Rn]/dt
Solve for Zfilm
222Rn/dt = sources – sinks = decay of 226Ra – decay of 222Rn - gas exchange to atmosphere
ml 222Rn d[222Rn]/dt = ml 226Ra [226Ra] – ml 222Rn [222Rn] + D/Zfilm { [222Rnatm] – [222RnML]}
ml δA222Rn/ δt = ml (A226Ra – A222Rn) + D/Z (CRn, atm – CRn,ML)
for SS = 0 atm Rn = 0
Then
-D/Z ( – CRn,ml) = ml (A226Ra – A222Rn)
+D/Z (ARn,ml/Rn) = ml (A226Ra – A222Rn)
+D/Z (ARn,ml) = ml Rn (A226Ra – A222Rn)
ZFILM = D (A222Rn,ml) / ml Rn (A226Ra – A222Rn)
ZFILM = (D / ml Rn) ( )226
222
1
1Ra
Rn
A
A
Z = DRn / 222Rn (1/A226Ra/A222Rn) ) - 1
Average Zfilm = 28 m
Stagnant Boundary Layer Film Thickness
Histogram showing results of film thicknesscalculations from many stations.
Organized by Ocean and by Latitude
Q. What are limitations of this approach?1. unrealistic physical model2. steady state assumption
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)
Expression of Air -Sea CO2 Flux
k-transfer velocity
From Sc # & wind speed
From CMDLCCGG network
S – Solubility
From SST & Salinity
From measurements and proxies
F = k s (pCO2w- pCO2a) = K ∆ pCO2
pCO2apCO2w
MagnitudeMechanismApply over larger space time domain
Global Map of Piston Velocity (k in m yr-1) times CO2 solubility (mol m-3) = Kfrom satellite observations (Nightingale and Liss, 2004 from Boutin).
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
∆pCO2 fields
Monthly changes in pCO2w
∆pCO2 fields:Takahashi climatology
JGOFS Gas Exchange Highlight #4 -
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
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
Wanninkhof&McGillis-98: 2.33 Pg C yr-1
Zemmelink-03: 2.45 Pg C yr-1
Yes!
CO2 Fluxes: Status
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
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)] = PA
where = 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.
Solubilities of Gases in Seawater
Solubility increases with mole weight and decreasing temperature
KH = 29 x 10-3
= 2.9 x 10-2
= 10-1.53
Concentration ratio for equal volumes of air andwater.
Henry’s LawBunson Coefficient
from Broecker and Peng, (1982)
Gas Solubility - CFCs
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)
-2
-1
0
1
2
3
4
1985 1990 1995 2000
E
NS
O IN
DE
X (M
EI)
year
Effect of El Nino on ∆pCO2 fieldsHigh resolution pCO2 measurements in the Pacific since Eq. Pac-92
Eq Pac-92 process study
Cosca et al. in press
El Nino Index
PCO2sw
Always greater than atmospheric