AOSC 620
PHYSICS AND CHEMISTRY
OF THE ATMOSPHERE, I
Professor Russell Dickerson
Room 2413, Computer & Space Sciences Building
Phone(301) 405-5364
russ@atmos.umd.edu
web site www.meto.umd.edu/~russ
Copyright © R. R. Dickerson 2015
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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.
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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 10:30am12:30pm
www/atmos.umd.edu/~russ/SYLLABUS_620_2015.html
Copyright © R. R. Dickerson & Z.Q. Li
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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.
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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.
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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.
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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.
Copyright © R. R. Dickerson
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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)
Copyright © R. R. Dickerson & Z.Q. Li
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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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Definition of Specific Volume
 = V/m = 1/r
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.
Copyright © 2013 R. R. Dickerson
& Z.Q. Li
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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 = 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)
Copyright © 2015 R. R. Dickerson
& Z.Q. Li
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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 = 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.
Copyright © 2015 R. R. Dickerson
& Z.Q. Li
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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
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Expanding gas parcel.
dl
A
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& Z.Q. Li
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In general the specific work done
by the expansion of an ideal gas
from state a to b is W = ∫ab pdα
a
b
p↑
α1
α→
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& Z.Q. Li
α2
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W = ∮ pdα
a
b
p↑
α1
α→
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& Z.Q. Li
α2
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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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& Z.Q. Li
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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’
Copyright © 2015 R. R. Dickerson
& Z.Q. Li
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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.
Copyright © 2015 R. R. Dickerson
& Z.Q. Li
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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 = cpdT − αdp becomes
đq = − αdp = pdα = đw
Copyright © 2015 R. R. Dickerson
& Z.Q. Li
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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
Copyright © 2015 R. R. Dickerson
& Z.Q. Li
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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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& Z.Q. Li
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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.
Copyright © 2015 R. R. Dickerson
& Z.Q. Li
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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.
Copyright © 2015 R. R. Dickerson
& Z.Q. Li
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Weather Symbols
http://www.ametsoc.org/amsedu/dstreme/
extras/wxsym2.html
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& Z.Q. Li
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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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& Z.Q. Li
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∮ 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.
Copyright © 2015 R. R. Dickerson
& Z.Q. Li
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Remember
Remember
Therefore
đq = cpdT − αdp
đq/T = cp/T dT − α/T dp
dφ = cp/T dT − α/T dp
α/T = R’/p
dT
dp
dj = c p - R'
T
p
= c p dq
Dj = c p ln (q / q 0 )
In a dry, adiabatic process potential temperature
doesn’t change thus
entropy is conserved.
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& Z.Q. Li
7am
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& Z.Q. Li
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10 am
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& Z.Q. Li
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