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### Transcript of Lecture 7 Gases and Gas Exchange Composition of the atmosphere Gas solubility Gas Exchange Fluxes...

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Lecture 7 Gases and Gas Exchange Composition of the atmosphere Gas solubility Gas Exchange Fluxes Effect of wind Global CO 2 fluxes by gas exchange Emerson and Hedges: Chpts 3 and 10 Slide 2 Sarmiento and Gruber (2002) Sinks for Anthropogenic Carbon Physics Today August 2002 30-36 Q. Was pre-industrial (PI) CO 2 at steady state? Q. Is atm CO 2 at steady state today? Slide 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, CO 2 and H 2 O. GasMole Fraction in Dry Air (f G ) molar volume at STP (l mol -1 ) where f G = moles gas i/total moles 22.414 for an ideal gas (0C) N 2 0.7808022.391 O 2 0.2095222.385 Ar9.34 x 10 -3 22.386 CO 2 3.9 x 10 -4 22.296 Ne1.82 x 10 -5 22.421 He5.24 x 10 -6 22.436 H 2 O~0.013 Q. Why is dry air used? Slide 4 Some comments about units of gases: In AirIn Water Pressure - Atmospheres Volume - liters gas at STP / kg sw 1 Atm = 760 mm Hg STP = standard temperature and pressure Partial Pressure of Gas i = P (i) /760 = 1 atm and 0 C (= 273K) Volume - liters gas / liters air Moles - moles gas / kg sw (ppmv = l / l, etc) Conversion: l gas /kg sw / l gas / mole = moles/kg sw (~22.4 l/mol) Slide 5 Dalton's Law Gas concentrations are expressed in terms of pressures. Total Pressure = P i = Dalton's Law of Partial Pressures P T = P N2 + P O2 + P H2O + P Ar +......... 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 (P i ) result from: 1) variations in P T (atmospheric pressure highs and lows) range = 32 to 25 inch Hg 2) variations in water vapor ( P H2O ) We can express the partial pressure (P i ) of a specific gas on a dry air basis as follows: P i = [ P T - h/100 P o ] f g where P i = partial pressure of gas i P T = Total atmospheric pressure h = % relative humidity P o = vapor pressure of water at ambient T f g = mole fraction of gas in dry air (see table above) Slide 6 Humidity Example: Say we have a humidity of 80% today and the temperature is 15 C Vapor pressure of H 2 O at 15 C = P o = 12.75 mm Hg (from reference books) Then, P H2O = 0.80 x 12.75 = 10.2 mm Hg If P T = 758.0 mm Hg P T Dry = (758.0 - 10.2) mm Hg = 747.8 mm Hg Then: f H2O = P H2O / P T = 10.2 / 758.0 = 0.013 So for these conditions H 2 O 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! Slide 7 Example: Units for CO 2 Atmospheric CO 2 has increased from 280 (pre-industrial) to 398 (present) ppm. In the table of atmospheric concentrations (see slide 3) f G,CO2 = 3.9 x 10 -4 moles CO 2 /total moles = 390 x 10 -6 moles CO 2 /total moles = 390 ppm This can also be expressed in log form as: = 10 0.59 x 10 -4 = 10 -3.41 Slide 8 Example: Units for Oxygen Conversion from volume to moles Use O 2 = 22.385 L / mol at standard temperature and pressure (STP) if O 2 = 5.0 ml O 2 /L SW then 5.0 ml O 2 / L sw x mol O 2 / 22,385 ml = 0.000223 mol O 2 / L sw = 223 mol O 2 / L sw Slide 9 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 ration using an equilibrium constant: K = [A(aq)] / [A(g)] There are two main ways to express solubility (Henry s Law and Bunsen Coefficients). Slide 10 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 K so that the number of moles n divided by the volume is equal to [A(g)] n/V = [A(g)] = P A / RT where P A is the partial pressure of A Then K = [A(aq)] / P A /RT or [A(aq)] = (K/RT) P A [A(aq)] = K H P A units for K are mol kg -1 atm -1 ; in mol kg -1 for P A are atm Henry's Law states that the concentration of a gas in water is proportional to its overlying partial pressure. K H is mainly a function of temperature with a small impact by salinity. Slide 11 Example (Solubility at 0 C): Partial Pressure = P i = f G x 1atm total pressure GasP i K H (0 C, S = 35) C i (0 C, S = 35; P = 760 mm Hg) N 2 0.78080.80 x 10 -3 624 x 10 -6 mol kg -1 O 2 0.20951.69 x 10 -3 354 x 10 -6 Ar0.00931.83 x 10 -3 17 x 10 -6 CO 2 0.00033 63 x 10 -3 21 x 10 -6 Example The value of K H for CO 2 at 25 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 CO 2 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 20C and Table 3A1.1 for regressions for all T and S Slide 12 Example: What is the concentration of CO 2 (aq) in equilibrium with the atmosphere? For P CO2 = 350 ppm = 10 -3.456 and T = 25C For CO 2 K H = 29 x 10 -3 = 2.9 x 10 -2 = 10 -1.53 moles /kg atm then CO 2 (aq) = K H x P CO2 = 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 25 C (0.5% of DIC) The equilibrium concentration of CO 2 (aq) will be dependent only on P CO2 and temperature. CO 2 (aq) is independent of pH. But this is only the first step! Slide 13 Summary of trends in solubility: 1.Type of gas: K H goes up as molecular weight goes up (note that CO 2 is anomalous) See solubility table. 2. Temperature: Solubility goes up as T goes down Major effect 3. Salinity: Solubility goes up as S goes down Minor effect Controls on Solubility Slide 14 O 2 versus temperature in surface ocean solid line equals saturation for S = 35 at different temperatures average supersaturation 7 mol/kg (~3%) Temperature control on gas concentrations Slide 15 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 K H or vs. T. 4. change in atmospheric pressure - if this happens quickly, surface waters cannot respond quickly enough to reequilibrate. Slide 16 Rates of Gas Exchange Stagnant Boundary Layer Model. Depth (Z) ATM OCN C g = K H P gas = equil. with atm C SW Z Film Stagnant Boundary Layer transport by molecular diffusion well mixed surface SW well mixed atmosphere 0 Z is positive downward C/ Z = F = + (flux into ocean) Slide 17 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 d[C g ] / dZ where D = molecular diffusion coefficient in water (= f (gas and T)) (cm 2 sec -1 ) dZ is the thickness of the stagnant film on the ocean side (Z film )(cm) D[C g ] is the concentration difference across the stagnant film (mol cm -3 ) The water at the top of the stagnant film (C g ) 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 (C SW ). Thus: Flux = F = - D/Z film (C g - C SW ) = - D/Z film (K H P g - C SW ) Slide 18 Because D/Z film has velocity units, it has been called the Piston Velocity (k) e.g., D = cm 2 sec -1 Z film = cm Typical values are D = 1 x 10 -5 cm 2 sec -1 at 15C Z film = 10 to 60 m (from 222 Rn and 14 C) Example: D = 1 x 10 -5 cm 2 sec -1 Z film = 17 m determined for the average global ocean using 14 C data Thus Z film = 1.7 x 10 -3 cm The piston velocity = D/Z = k = 1 x 10 -5 cm 2 s -1 /1.7 x 10 -3 cm = 0.59 x 10 -2 cm/sec 5 m / d note: 1 day = 8.64 x 10 4 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. Slide 19 Gas Exchange and Environmental Forcing: Wind Liss and Merlivat,1986 from wind tunnel exp. Wanninkhof, 1992 from 14 C 20 cm hr -1 = 20 x 24 / 10 2 = 4.8 m d -1 ~ 5 m d -1 Slide 20 One of main goals of JGOFS was to calculate the CO 2 flux across the air-sea interface Flux = F = - D/Z film (C atm - C SW ) = - D/Z film (K H P atm - C SW ) = - D/Z film (K H P atm K H P SW ) = -D/Z film K H (P atm P SW ) See Boutin Figure Slide 21 Expression of Air -Sea CO 2 Flux k-transfer velocity From Sc # & wind speed From CMDL CCGG network S Solubility From SST & Salinity From measurements and proxies F = k s (pCO 2w - pCO 2a ) = K pCO 2 pCO 2a pCO 2w Magnitude Mechanism Apply over larger space time domain Slide 22 Global Map of Piston Velocity (k in m yr -1 ) times CO 2 solubility (mol m -3 ) = K from satellite observations (Nightingale and Liss, 2004 from Boutin). Slide 23 Overall trends known: * Outgassing at low latitudes (e.g. equatorial) * Influx at high latitudes (e.g. circumpolar) * Spring blooms draw down pCO 2 (N. Atl) * El Nios decrease efflux pCO 2 fields Slide 24 Monthly changes in pCO 2w pCO 2 fields:Takahashi climatology JGOFS Gas Exchange Highlight #4 - Slide 25 Fluxes: JGOFS- Global monthly fluxes Combining pCO 2 fields with k: F = k s (pCO 2w - pCO 2a ) On first order flux and pCO 2 maps do not look that different http://www.ldeo.columbia.edu/res/pi/CO2/carbondioxide/pages/pco2_flux_rate_maps.html Slide 26 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! CO 2 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