1. Copyright © R. R. Dickerson 201511 Professor Russell Dickerson Room 2413, Computer & Space Sciences Building Phone(301) 405-5364 russ@atmos.umd.edu web site www.meto.umd.edu/~russ AOSC 620 PHYSICS AND CHEMISTRY OF THE ATMOSPHERE, I

2. 22 Objectives of AOSC 620 & 621 Present the basics of atmospheric chemistry and physics. Teach you experimental and theoretical methods. Show you tools that will help you solve problems that have never been solved before. Prepare you for a career that pushes back the frontiers of atmospheric or oceanic science.

3. Copyright © R. R. Dickerson & Z.Q. Li33 Logistics Office Hours: Tuesdays 3:30 – 4:30 pm (except today) Wednesdays 1:00 – 2:00 pm Worst time is 1- 2 pm Tues or Thrs. Exam Dates: October 13, November 24, 2015 Final Examination: Thursday, Dec. 17, 2015 10:30am- 12:30pm www/atmos.umd.edu/~russ/SYLLABUS_620_2015.html

4. 44 Changes to Syllabus, 2015/16 Basically all of atmospheric chemistry will be taught in AOSC 620. The remainder of cloud physics and radiation will be taught in AOSC 621.

5. 55 Experiment: Room temperature Measure, or estimate if you have no thermometer, the current room temperature. Do not discuss your results with your colleagues. Write the temperature on a piece of paper and hand it in.

6. 66 Homework #1 HW problems 1.1, 1.2, 1.3, 1.6, from Rogers and Yao; repeat 1.1 for the atmosphere of another planet or moon.

7. Copyright © R. R. Dickerson77 Lecture 1. Thermodynamics of Dry Air. Objective: To find some useful relationships among air temperature (T), volume (V), and pressure (P), and to apply these relationships to a parcel of air. Ideal Gas Law: PV = nRT See R&Y Chapter 1 Salby Chapter 1.2 and 2.2-2.3 W&H Chapter 3.

8. Copyright © R. R. Dickerson & Z.Q. Li88 Lecture 1. Thermodynamics of Dry Air. Objective: To find some useful relationships among air temperature (T), volume (V), and pressure (P), and to apply these relationships to a parcel of air. Ideal Gas Law: PV = nRT Where: n is the number of moles of an ideal gas. m = molecular weight (g/mole) M = mass of gas (g) R = Universal gas constant = 8.314 J K -1 mole -1 = 0.08206 L atm K -1 mole -1 = 287 J K -1 kg -1 (for air)

9. Copyright © R. R. Dickerson & Z.Q. Li99 Dalton’s law of partial pressures P =  i p i PV =  i p i RT = RT  i p i The mixing ratios of the major constituents of dry air do not change in the troposphere and stratosphere.

10. Copyright © 2013 R. R. Dickerson & Z.Q. Li 10 Definition of Specific Volume  = V/m = 1/  PV/M = nRT/m P  = R’T Where R’ = R/m Specific volume,  is the volume occupied by 1.0 g (sometimes 1 kg) of air.

11. Copyright © 2015 R. R. Dickerson & Z.Q. Li 11 Definition of gas constant for dry air p  = R’T Upper case refers to absolute pressure or volume while lower case refers to specific volume or pressure of a unit (g) mass. p  = R d T Where R d = R/m d and m d = 28.9 g/mole. R d = 287 J kg -1 K -1 (For convenience we usually drop the subscript)

12. Copyright © 2015 R. R. Dickerson & Z.Q. Li 12 First Law of Thermodynamics The sum of heat and work in a system is constant, or heat is a form of energy (Joules Law). 1.0 calorie = 4.1868 J Q =  U +  W Where Q is the heat flow into the system,  U is the change in internal energy, and W is the work done. In general, for a unit mass: đq = du + đw Note đq and đw are not exact differential, as they are not the functions of state variables.

13. Copyright © R. R. Dickerson & Z.Q. Li13 Work done by an ideal gas. Consider a volume of air with a surface area A. Let the gas expand by a uniform distance of dl. The gas exerts a force on its surroundings F, where: F = pA (pressure is force per unit area) W = force x distance = F x dl = pA x dl = pdV For a unit mass đw = pd 

14. Copyright © 2015 R. R. Dickerson & Z.Q. Li 14 Expanding gas parcel. A dldl

15. Copyright © 2015 R. R. Dickerson & Z.Q. Li 15 In general the specific work done by the expansion of an ideal gas from state a to b is W = ∫ a b pdα p↑p↑ α→α→ a b α1α1 α2α2

16. Copyright © 2015 R. R. Dickerson & Z.Q. Li 16 W = ∮ pdα p↑p↑ α→α→ a b α1α1 α2α2

17. Copyright © 2015 R. R. Dickerson & Z.Q. Li 17 Definition Heat Capacity Internal energy change, du, is usually seen as a change in temperature. The temperature change is proportional to the amount of heat added. dT = đq/c Where c is the specific heat capacity.

18. Copyright © 2015 R. R. Dickerson & Z.Q. Li 18 If no work is done, and for a constant specific volume: đq = c v dT = du or c v = du/dT = Δu/ΔT for an ideal gas At a constant pressure: đq = c p dT = du + pdα = c v dT + pdα or c p = c v + p dα/dT But pα = R’T and p dα/dT = R’ thus c p = c v + R’

19. Copyright © 2015 R. R. Dickerson & Z.Q. Li 19 pα = R’T Differentiating d(pα) = pdα + αdp = R’dT or pdα = R’dT − αdp From the First Law of Thermo for an ideal gas: đq = c v dT + pdα = c v dT + R’dT − αdp But c p = c v + R’ đq = c p dT − αdp This turns out to be a powerful relation for ideal gases.

20. Copyright © 2015 R. R. Dickerson & Z.Q. Li 20 Let us consider four special cases. 1. If a process is conducted at constant pressure (lab bench) then dp = 0. For an isobaric process: đq = c p dT − αdp becomes đq = c p dT 2. If the temperature is held constant, dT = 0. For an isothermal process: đq = c p dT − αdp becomes đq = − αdp = pdα = đw

21. Copyright © 2015 R. R. Dickerson & Z.Q. Li 21 Next two special cases. 3. If a process is conducted at constant density then dρ = dα = 0. For an isosteric process: đq = c v dT = du 4. If the process proceeds without exchange of heat with the surroundings dq = 0. For an adiabatic process: c v dT = − pdα and c p dT = αdp

22. Copyright © 2015 R. R. Dickerson & Z.Q. Li 22 The adiabatic case is powerful. Most atmospheric temperature changes, esp. those associated with rising or sinking motions are adiabatic (or pseudoadiabatic, defined later). For an adiabatic process: c v dT = − pdα and c p dT = αdp du is the same as đw Remember α = R’T/p thus đq = c p dT = R’T/p dp Separating the variables and integrating c p /R’ ∫ dT/T = ∫ dp/p

23. Copyright © 2015 R. R. Dickerson & Z.Q. Li 23 c p /R’ ∫ dT/T = ∫ dp/p (T/T 0 ) = (p/p 0 ) K Where K = R’/c p = 0.286 This allows you to calculate, for an adiabatic process, the temperature change for a given pressure change. The sub zeros usually refer to the 1000 hPa level in meteorology.

24. Copyright © 2015 R. R. Dickerson & Z.Q. Li 24 If we define a reference pressure of 1000 hPa (mb) then: (T/θ) = (p/1000) K Where θ is defined as the potential temperature, or the temperature a parcel would have if moved to the 1000 hPa level in a dry adiabatic process. θ = T (1000/p) K Potential temperature, θ, is a conserved quantity in an adiabatic process.

25. Copyright © 2015 R. R. Dickerson & Z.Q. Li 25 Weather Symbols http://www.ametsoc.org/amsedu/dstreme/ extras/wxsym2.html

26. Copyright © 2014 R. R. Dickerson & Z.Q. Li 26 The Second Law of Thermodynamics dφ ≡ đq/T Where φ is defined as entropy. dφ = c v dT/T + pdα/T = c v dT/T + R’/α dα ∫dφ = ∫ đq/T = ∫ c v /TdT + ∫ R’/α dα For a cyclic process ∮ đq/T = ∮ c v /TdT + ∮ R’/α dα

27. Copyright © 2015 R. R. Dickerson & Z.Q. Li 27 ∮ đq/T = ∮ c v /TdT + ∮ R’/α dα But ∮ c v /T dT = 0 and ∮ R’/α dα = 0 because T and α are state variables; thus ∮ đq/T = 0 ∮ dφ = 0 Entropy is a state variable.

28. Copyright © 2015 R. R. Dickerson & Z.Q. Li 28 Remember đq = c p dT − αdp đq/T = c p /T dT − α/T dp dφ = c p /T dT − α/T dp Remember α/T = R’/p Therefore In a dry, adiabatic process potential temperature doesn’t change thus entropy is conserved.

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