Copyright © 2014 R. R. Dickerson1 Professor Russell Dickerson Room 2413, Computer & Space Sciences Building Phone(301) web site AOSC 620 Lecture 2 PHYSICS AND CHEMISTRY OF THE ATMOSPHERE I

Copyright © 2013 R. R. Dickerson & Z.Q. Li 22 Experiment: Room temperature What is the temperature of the room? We’ll get 22 answers. How well do they agree? Does location matter? Does observer bias play a role?

Copyright © 2010 R. R. Dickerson & Z.Q. Li 3 Experiment: Room temperature Last class Average o C Stdev o C Max o C Min o C T Mean = +/- °C (+/- °F) Measured with uncalibrated or no thermometers at n locations.

Copyright © 2013 R. R. Dickerson & Z.Q. Li 4 What is the right answer? What is the uncertainty? Are the data Gaussian? How can we improve the measurement?

Copyright © 2013 R. R. Dickerson & Z.Q. Li 5 Experiment: Room temperature Let ’ s repeat the experiment with calibrated thermometers.

Copyright © 2013 R. R. Dickerson & Z.Q. Li 6 Lecture 2. Thermodynamics of Air, continued – water vapor. Objective: To find some useful relationships among air temperature, volume, and pressure. Review Ideal Gas Law: PV = nRT P α = R ’ T First Law of Thermodynamics: đq = du + đw W = ∫ pdα

Copyright © 2013 R. R. Dickerson & Z.Q. Li 7 Review (cont.) Definition of heat capacity: c v = du/dT = Δu/ΔT c p = c v + R Reformulation of first law for unit mass of an ideal gas: đq = c v dT + pdα đq = c p dT − αdp

Copyright © 2013 R. R. Dickerson & Z.Q. Li 8 Review (cont.) For an isobaric process: đq = c p dT For an isothermal process: đq = − αdp = pdα = đw For an isosteric process: đq = c v dT = du For an adiabatic process: c v dT = − pdα and c p dT = αdp

Copyright © 2013 R. R. Dickerson & Z.Q. Li 9 Review (cont.) For an adiabatic process: c v dT = − pdα and c p dT = αdp du = đw (T/T 0 ) = (p/p 0 ) K Where K = R ’ /c p = (T/θ) = (p/1000) K Define potential temperature: θ = T(1000/p) K Potential temperature, θ, is a conserved quantity in an adiabatic process.

Copyright © 2013 R. R. Dickerson & Z.Q. Li 10 Review (cont.) The Second Law of Thermodynamics is the definition of φ as entropy. dφ ≡ đq/T ჶ dφ = 0 Entropy is a state variable. Δφ = c p ln(θ/θ 0 ) In a dry, adiabatic process potential temperature doesn ’ t change, thus entropy is conserved.

Copyright © 2013 R. R. Dickerson & Z.Q. Li 11 Useful idea - a perfect or exact differential: If z = f(x,y), dz is a perfect differential iff: ∂ 2 f / ∂x∂y = ∂ 2 f / ∂y∂x ჶ dz = 0 For example, v = f(T,p) dv = (∂v / ∂p) T dp + (∂v / ∂T) p dT This is true for dU, dH, dG, but not đw or đq.

Copyright © 2013 R. R. Dickerson & Z.Q. Li 12 Various Measures of Water Vapor Content Vapor pressure Vapor density – absolute humidity Mixing ratio, w (g/kg) Specific humidity Relative humidity Virtual temperature (density temp) Dew point temperature Wet bulb temperature Equivalent temperature Isentropic Condensation Temperature Potential temperature Wet-bulb potential temperature Equivalent potential temperature

Copyright © 2013 R. R. Dickerson & Z.Q. Li 13 Virtual Temperature: T v or T * Temperature dry air would have if it had the same density as a sample of moist air at the same pressure. Question: should the virtual temperature be higher or lower than the actual temperature?

Copyright © 2013 R. R. Dickerson & Z.Q. Li 14 Consider a mixture of dry air and water vapor. Let M d = mass of dry air M v = mass of water vapor m d = molecular weight of dry air m v = molecular weight of water.

Copyright © 2013 R. R. Dickerson & Z.Q. Li 15 Dalton ’ s law: P =  p i

Copyright © 2013 R. R. Dickerson & Z.Q. Li 16 Combine P and  to eliminate V:

Copyright © 2013 R. R. Dickerson & Z.Q. Li 17 Since P =  RT *

Copyright © 2013 R. R. Dickerson & Z.Q. Li 18 Alternate derivation: Since  proportional to  Mwt Where w is the mass mixing ratio and x (molar or volume mixing ratio) = [H 2 O] = w/0.62 T* = T (29/(29-11[H 2 O])) e.g., [H 2 O] = 1% then T* = T(1.004) If, [H 2 O] = 1% then w = 0.01*.62 = T* = 1.01/ =

Copyright © 2013 R. R. Dickerson & Z.Q. Li 19 Where w is the mass mixing ratio and x (molar or volume mixing ratio) = [H 2 O] = w/0.62 T* = T (29/(29-11[H 2 O])) e.g., [H 2 O] = 1% then T* = T(1.004) Test: if [H 2 O] = 1% then w(18/29) = 0.01*.62 = T* = T (1 + w/  )/(1+w) = T ( )/( ) = T(1.004)

Copyright © 2013 R. R. Dickerson & Z.Q. Li 20 Unsaturated Moist Air Equation of state: P  = RT *

Copyright © 2013 R. R. Dickerson & Z.Q. Li 21 Specific Heats for Moist Air Let m v = mass of water vapor m d = mass of dry air To find the heat flow at constant volume:

Copyright © 2013 R. R. Dickerson & Z.Q. Li 22 For constant pressure So Poisson ’ s equation becomes (1+0.6w)/(1+0.9w) => (1-0.2w) due to rounding error.

Copyright © 2013 R. R. Dickerson & Z.Q. Li 23 Water Vapor Pressure Equation of state for water vapor: e v =  v R v T where e v is the partial pressure of water vapor