LECTURE 2 AOSC 637 Atmospheric Chemistry R. Dickerson Copyright R. R. Dickerson 2011 1 Solubility of CO2 0.065 0.06 H (M/atm) 0.055 0.05 0.045 0.04 0.035 0.03 275 280 285 290 295 300 305 Temp (K) Copyright R. R. Dickerson 2011 2 Solubility of CO2 near mean surface ocean temperature Solubility of CO2 0.048 H (M/atm) 0.0475 0.047 0.0465 0.046 0.0455 0.045 287 287.5 288 288.5 289 Temp (K) Copyright R. R. Dickerson 2011 3 ATMOSPHERIC PHYSICS Seinfeld & Pandis: Chapter 1 Finlayson-Pitts: Chapter 2 1.Pressure: exponential decay 2.Composition 3.Temperature The motion of the atmosphere is caused by differential heating, that is some parts of the atmosphere receive more radiation than others and become unstable. Copyright R. R. Dickerson 2011 4 VERTICAL PROFILES OF PRESSURE AND TEMPERATURE Mean values for 30oN, March Stratopause Tropopause Composition of the Earth’s Troposphere H2 O2 CH4 N2 N2O PM CO O3 ←SO2, NO2, CFC’s, etc Ar CO2 Copyright R. R. Dickerson 2011 Inert gases 6 Banded iron formation. The great Oxygen Catastrophe Copyright R. R. Dickerson 2011 7 Units of pressure: 1.00 atm. = 14.7 psi = 1,013 mb = 760 mm Hg (torr) = 33.9 ft H₂O = 29.9” Hg = 101325 Pa (a Pascal is a Nm⁻² thus hPa = mb) Units of Volume: liter, cc, ml, m³ Units of temp: K, °C Units of R: 0.08206 1 atm mole-1 K-1 8.31 J mole-1 K-1 R’ = R/Mwt = 0.287 J g-1K-1 For a mole of dry air which has the mass 29 g. Problem for the student: calc Mwt. Wet (2% H₂O vapor by volume) air. Copyright R. R. Dickerson 2011 8 Derive the Hypsometric Equation Start with The Ideal Gas Law PV = nRT or P = ρR’T or P = R’T/α Where R’ = R/Mwt Mwt = MOLE WT. AIR ρ = DENSITY AIR (g/l) α = SPECIFIC VOL AIR = 1/ρ We assume that the pressure at any given altitude is due to the weight of the atmosphere above that altitude. The weight is mass times acceleration. P = W = mg But m = Vρ For a unit area V = Z P = Zρg Copyright R. R. Dickerson 2011 9 For a second, higher layer the difference in pressure can be related to the difference in height. dP = − g ρ dZ But ρ = P/R’T dP = − Pg/R’T * dZ For an isothermal atmosphere g/R’T is a constant. By integrating both sides of the equation from the ground (Z = 0.0) to altitude Z we obtain: PZ Z Pg P dP 0 R' T dz 0 PZ Z 1 g dp P p 0 R' T dZ 0 ln(P/P 0 ) Z/H 0 H T R' / g Copyright0R. R. Dickerson 2011 10 Where H₀ = R’T/g we can rewrite this as: PZ P0 exp( Z / H 0 ) *HYPSOMETRIC EQUATION* Note: Scale Height: H₀ ~ 8 km for T = 273K For each 8 km of altitude the pressure is down by e⁻¹ or one “e-fold.” Copyright R. R. Dickerson 2011 11 Problems left to the student. 1. Show that the altitudes at which the pressure drops by a factor of ten and two are 18 and 5.5km. Rewrite the hypsometric Eq. for base 2 and 10. 2. Calculate the scale-height for the atmospheres of Venus and Mars. 3. Derive an expression for pressure as a function of altitude for an atmosphere with a temperature that varies linearly with altitude. 4. Calculate the depth of a constant density atmosphere. 5. Calculate, for an isothermal atmosphere, the fraction of the mass of the atmosphere between 200 and 700 hPa. Copyright R. R. Dickerson 2011 12 Temperature Lapse Rate Going to the mountains in Shenandoah National Park the summer is a nice way to escape Washington’s heat. Why? Consider a parcel of air. If it rises it will expand and cool. If we assume it exchanges no heat with the surroundings (a pretty good assumption, because air is a very poor conductor of heat) it will cool “adiabatically.” CALCULATE: ADIABATIC LAPSE RATE First Law Thermodynamics: dU = DQ + DW Copyright R. R. Dickerson 2011 13 WHERE U = Energy of system (also written E) Q = Heat across boundaries W = Work done by the system on the surroundings H = Internal heat or Enthalpy ASSUME: a) Adiabatic (dH = 0.0) b) All work PdV work (remember α = 1/ρ) dH = Cp dT – α dP CpdT = α dP dT = (α/Cp) dP Copyright R. R. Dickerson 2011 14 Remember the Hydrostatic Equation OR Ideal Gas Law dP ( g )dZ dP (gP/R'T)dZ R'T / P R ' T ( gP ) dT dZ PC p R ' T Result: dT / dZ g / C p This quantity, -g/Cp, is a constant in a dry atmosphere. It is called the dry adiabatic lapse rate and is given the symbol γ₀, defined as −dT/dZ. 0 9.8K / km For a parcel of air moving adiabatically in the atmosphere: T2 T1 0 ( Z2 Z1 ) Copyright R. R. Dickerson 2011 15 Where Z₂ is higher than Z₁, but this presupposes that no heat is added to or lost from the parcel, and condensation, evaporation, and radiative heating can all produce a non-adiabatic system. The dry adiabatic lapse rate is a general, thermodynamic property of the atmosphere and expresses the way a parcel of air will cool on rising or warm on falling, provided there is no exchange of heat with the surroundings and no water condensing or evaporating. The environmental lapse rate is seldom equal to exactly the dry adiabatic lapse rate because radiative processes and phase changes constantly redistribute heat. The mean lapse rate is about 6.5 K/km. Problem left to the students: Derive a new expression for the change in pressure with height for an atmosphere with a constant lapse rate, TZ T0 Z Copyright R. R. Dickerson 2011 16 Stability and Thermodynamic Diagrams (Handout) Solid lines – thermodynamic property Dashed (colored) lines – measurements or soundings day γa > γ₀ unstable γb = γ₀ neutrally stable γc < γ₀ stable On day a a parcel will cool more slowly than surroundings – air will be warmer and rise. On day b a parcel will always have same temperature as surroundings – no force of buoyancy. On day c a parcel will cool more quickly than surroundings – air will be cooler and return to original altitude. Copyright R. R. Dickerson 2011 17 This equation is easily corrected for a wet air parcel. The heat capacity must be modified: Cp' = (1-a)Cp + aCp(water), where "a" is the mass of water per mass of dry air. If the parcel becomes saturated, then the process is no longer adiabatic. Condensation adds heat. DQ = -Lda Where L is the latent heat of condensation -dT/dZ = g/Cp + (L/Cp) da/dZ Because the amount of water decreases with altitude the last term is negative, and the rate of cooling with altitude is slower in a wet parcel. The lapse rate becomes the wet adiabat or the pseudoadiabat. Copyright R. R. Dickerson 2011 18 (HANDOUT) Very important for air pollution and mixing of emissions with free troposphere. Formation of thermal inversions. (VIEWGRAPH) In the stratosphere the temperature increases with altitude, thus stable or stratified. DFN: Potential temperature, θ : The temperature that a parcel of air would have if it were brought to the 1000 hPa level (near the surface) in a dry adiabatic process. You can approximate it quickly, T Z z Z 0 or a proper derivation yields: ( ) θz = T 1000 hPa/Pz ** R’/Cp = T (1000 hPa/Pz) ** 0.286 Copyright R. R. Dickerson 2011 19 DIURNAL CYCLE OF SURFACE HEATING/COOLING: z Subsidence inversion MIDDAY 1 km Mixing depth NIGHT 0 MORNING T NIGHT MORNING AFTERNOON Plume looping, Baltimore ~2pm. Copyright R. R. Dickerson 2011 21 Copyright R. R. Dickerson 2011 22 Copyright R. R. Dickerson 2011 23 Plume Lofting, Beijing in Winter ~7am. Copyright R. R. Dickerson 2011 24 Atmospheric Circulation and Winds Copyright R. R. Dickerson 2011 25 ITCZ Copyright R. R. Dickerson 2011 26 Mid latitude cyclone with fronts Copyright R. R. Dickerson 2011 27 Warm Front Fig. 9.13 Copyright R. R. Dickerson 2011 28 Cold front Fig. 9.15 Copyright R. R. Dickerson 2011 29 Copyright R. R. Dickerson 2011 Fig. 30 1-17, p. 21 Detailed weather symbols Copyright R. R. Dickerson 2011 31 Copyright R. R. Dickerson 2011 Fig. 32 1-16, p. 19 Electromagnetic Radiation Copyright R. R. Dickerson 2011 33 Electromagnetic spectrum Copyright R. R. Dickerson 2011 34 Planck’s Law for blackbody radiation: E ( , T ) 2hc 5 2 1 e hc / kT 1 ( Wm -2 m -1 ) Take first derivative wrt T and set to zero: Wien’s Law: max 3 a 2.9 10 m K (m) T T -1 Integrate over all wavelength: Stephan-Boltzmann Law: ET T ( W/m ) 4 Copyright R. R. Dickerson 2011 2 35 UV-VISIBLE SOLAR SPECTRUM Copyright R. R. Dickerson 2011 36 Planck’s Law says that the energy of a photon is proportional to its frequency. E h (J) Where Planck’s Const., h = 6.6x10-34 Js The energy associated with a mole of photons at a given wavelength in nm is: E 1.2 x10 / (kJ/mole ) 5 Copyright R. R. Dickerson 2011 37