1. AOSC 620 Physics and Chemistry of the Atmosphere, I
Professor Russell Dickerson Room 2413, Computer & Space Sciences Building Phone(301)
web site
Copyright © R. R. Dickerson 2016 1

2. 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. 2

3. Copyright © R. R. Dickerson
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 22, 2016
Final Examination: Saturday Dec. 17, :30am-12:30pm
www/atmos.umd.edu/~russ/SYLLABUS_620_2016.html
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4. New Textbook for 2016 Rogers and Yau is outdated.
Salby is good for atmospheric physics – prone to errors in chemistry e.g., page 32. 4

5. Experiment: we will launch a rawinsonde
You will plot the data on a skew-T.
Identify PBL and inversions.
Identify the LCL.
Calculate CAPE.
Forecast weather. 5

6. 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. You may substitute Salby's problem 1.6 for R&Y 1.1. 6

7. 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 1.2 (Chapter 1Section 2), and
W&H Chapter 3.
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8. 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
= J K-1 mole-1
= L atm K-1 mole-1
= 287 J K-1 kg-1 (for air)
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9. Dalton’s law of partial pressures
P = Si pi
PV = Si piRT = RT Si pi
The mixing ratios of the major constituents of dry air do not change in the troposphere and stratosphere.
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10. Definition of Specific Volume
= V/m = 1/r
PV/M = nRT/m
Pa = R’T
Where R’ = R/m
Specific volume, a, is the volume occupied by 1.0 g (sometimes 1 kg) of air. It is intensive rather than extensive like V.
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11. Definition of gas constant for dry air
pa = R’T
Upper case refers to absolute pressure or volume while lower case refers to specific volume or pressure of a unit (g) mass.
pa = RdT
Where Rd = R/md and md = 28.9 g/mole.
Rd = 287 J kg-1 K-1
(For convenience we usually drop the subscript)
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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 = J
Q = DU + DW
Where Q is the heat flow into the system, DU 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.
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13. 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 = pda
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14. Copyright © 2015 R. R. Dickerson & Z.Q. Li
Expanding gas parcel. dl A
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15. Copyright © 2016 R. R. Dickerson
In general, the specific work done by the expansion of an ideal gas from state a to b is a b p↑ α1 α→ α2 15
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16. Copyright © 2016 R. R. Dickerson
W = ∮ pdα a b p↑ α1 α2 α→ 16
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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.
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18. If no work is done, and for a constant specific volume:
đq = cvdT = du or
cv = du/dT = Δu/ΔT for an ideal gas
At a constant pressure:
đq = cpdT = du + pdα
= cvdT + pdα or
cp = cv + p dα/dT
But pα = R’T and
p dα/dT = R’ thus
cp = cv + R’
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19. d(pα) = pdα + αdp = R’dT or pdα = R’dT − αdp
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 = cvdT + pdα = cvdT + R’dT − αdp
But cp = cv + R’
đq = cpdT − αdp
This turns out to be a powerful relation for ideal gases.
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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 = cpdT − αdp becomes
đq = cpdT
2. If the temperature is held constant, dT = 0.
For an isothermal process:
đq = − αdp = pdα = đw
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21. 3. If a process is conducted at constant density then dρ = dα = 0.
Next two special cases.
3. If a process is conducted at constant density then dρ = dα = 0.
For an isosteric process:
đq = cvdT = du
4. If the process proceeds without exchange of heat with the surroundings dq = 0.
For an adiabatic process:
cvdT = − pdα and cpdT = αdp
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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:
cvdT = − pdα and cpdT = αdp
du is the same as đw
Remember α = R’T/p thus
đq = cpdT = R’T/p dp
Separating the variables and integrating
cp/R’ ∫dT/T = ∫dp/p
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23. Copyright © 2015 R. R. Dickerson & Z.Q. Li
cp/R’ ∫dT/T = ∫dp/p
(T/T0) = (p/p0)K
Where K = R’/cp = 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.
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24. Copyright © 2015 R. R. Dickerson & Z.Q. Li
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.
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25. Copyright © 2015 R. R. Dickerson & Z.Q. Li
Weather Symbols
extras/wxsym2.html
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26. The Second Law of Thermodynamics
dφ ≡ đq/T
Where φ is defined as entropy.
dφ = cvdT/T + pdα/T
= cvdT/T + R’/α dα
∫dφ = ∫đq/T = ∫cv/TdT + ∫R’/α dα
For a cyclic process
∮ đq/T = ∮ cv/TdT + ∮R’/α dα
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27. ∮ đq/T = ∮ cv/TdT + ∮R’/α dα But ∮ cv/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.
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28. Copyright © 2015 R. R. Dickerson & Z.Q. Li
Remember
đq = cpdT − αdp
đq/T = cp/T dT − α/T dp
dφ = cp/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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29. Copyright © 2015 R. R. Dickerson & Z.Q. Li
7am
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30. Copyright © 2015 R. R. Dickerson & Z.Q. Li
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