Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Slide 16Slide 17Slide 18Slide 19Slide 20Slide 21Slide 22Slide 23Slide 24Slide 25Slide 26Slide 27Slide 28Gas Solubility - CFCsSlide 30Lecture 7 Gases and Gas ExchangeComposition of the atmosphereGas solubilityGas Exchange FluxesEffect of windGlobal CO2 fluxes by gas exchangeEmerson and Hedges: Chpts 3 and 10Sarmiento and Gruber (2002) Sinks for Anthropogenic CarbonPhysics Today August 2002 30-36Q. Was pre-industrial (PI) CO2 at steady state?Q. Is atm CO2 at steady state today?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) N20.78080 22.391O20.20952 22.385Ar 9.34 x 10-322.386CO23.9 x 10-4 22.296Ne 1.82 x 10-522.421He 5.24 x 10-622.436H2O ~0.013Q. Why is dry air used?Some comments about units of gases:In Air In WaterPressure - Atmospheres Volume - liters gas at STP / kgsw1 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)Dalton's LawGas concentrations are expressed in terms of pressures.Total Pressure = SPi = 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) range = 32 to 25 inch Hg2) 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 ] fgwhere 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)Humidity Example:Say we have a humidity of 80% today and the temperature is 15CVapor pressure of H2O at 15C = Po = 12.75 mm Hg (from reference books)Then, PH2O = 0.80 x 12.75 = 10.2 mm HgIf PT = 758.0 mm HgPTDry = (758.0 - 10.2) mm Hg = 747.8 mm HgThen: fH2O = PH2O / PT = 10.2 / 758.0 = 0.013So 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 CO2Atmospheric CO2 has increased from 280 (pre-industrial) to 398 (present) ppm.In the table of atmospheric concentrations (see slide 3)fG,CO2 = 3.9 x 10-4 moles CO2/total moles = 390 x 10-6 moles CO2/total moles = 390 ppmThis can also be expressed in log form as: = 100.59 x 10-4 = 10-3.41Example: Units for OxygenConversion from volume to molesUse O2 = 22.385 L / mol at standard temperature and pressure (STP)if O2 = 5.0 ml O2/LSWthen5.0 ml O2 / Lsw x mol O2 / 22,385 ml = 0.000223 mol O2 / Lsw = 223 mol O2 / LswSolubilityThe 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).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 Kso 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 AThen K = [A(aq)] / PA/RT or [A(aq)] = (K/RT) PA [A(aq)] = KH PAunits 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 0C):Partial Pressure = Pi = fG x 1atm total pressureGas PiKH (0C , S = 35) Ci (0C, S = 35; P = 760 mm Hg)N20.7808 0.80 x 10-3 624 x 10-6 mol kg-1 O20.2095 1.69 x 10-3 354 x 10-6Ar 0.0093 1.83 x 10-3 17 x 10-6CO20.00033 63 x 10-3 21 x 10-6 ExampleThe value of KH for CO2 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 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 for all T and SExample: What is the concentration of CO2 (aq) in equilibrium with the atmosphere?For PCO2 = 350 ppm = 10-3.456 and T = 25°CFor CO2 KH = 29 x 10-3 = 2.9 x 10-2 = 10-1.53 moles /kg atmthenCO2 (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 25C (0.5% of DIC)The equilibrium concentration of CO2(aq) will be dependent only on PCO2 and temperature.CO2(aq) is independent of pH.But this is only the first step!Summary of trends in solubility:1. Type of gas: KH goes up as molecular weight goes up (note that CO2 is anomalous)See solubility table.2. Temperature: Solubility goes up as T goes downMajor effect3. Salinity: Solubility goes up as S goes downMinor effectControls on SolubilityO2 versus temperaturein surface oceansolid line equals saturationfor S = 35 at different temperaturesaverage supersaturation≈ 7 mmol/kg (~3%)Temperature controlon gas concentrationsCauses 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.Rates of Gas ExchangeStagnant Boundary Layer