Copyright R. R. Dickerson 2011 1

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.
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VERTICAL PROFILES OF PRESSURE AND TEMPERATURE
Mean values for 30 o N, March
Stratopause
Tropopause

O
2
Composition of the Earth’s Troposphere
H
2
CH
4
N
2
N
2
O
PM
CO
O
3
←SO
2,
NO
2
,
CFC’s, etc
Ar
CO
2
Inert gases
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Banded iron formation.
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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/M wt
= 0.287 J g -1 K -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.
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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.
But
For a unit area V = Z
P = W = mg m = Vρ
P = Zρg
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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:
P
Z
P
0
 dP

Z
0
 
Pg
R' T dz
P
Z 
P
0
1 p dp
 
Z
0

R' g
T dZ ln(P/P
0
)
H
 
Z/H
0

T R ' / g
8

Where H₀ = R’T/g we can rewrite this as:
P
Z

P
0 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.”
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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.
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student a
Basic Thermo
14
Thermo
Diag 10
14 9 student b
14 10 student c
14 10 student d
12 10 student e pass
14
5/5
2
4/5
Aerosols
(10) Clouds 10
Biogeoche m 11
Photo
Chem 12 Sow ice 10
Cld models 10
Kinetics
20
10 7 7 10 5 8 18
7
6.5
10
10
10
3
12
1
6
6
9
8
20
11
0
3
3/5
3
1
3/5
7
9
4/5
10
0
3/5
4
0
2/5
0
0
3/5
12
7
3/5

2
Solubility of CO2
0.065
0.06
0.055
0.05
0.045
0.04
0.035
0.03
275 280 285 290
Temp (K)
295 300 305
Solubility of CO
2
0.048
0.0475
0.047
0.0465
0.046
0.0455
0.045
287 287.5
288
Temp (K)
288.5
289

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
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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
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Remember the Hydrostatic Equation
OR dP
 
( g

) dZ dP
 
(gP/R'T)dZ
Ideal Gas Law
Result:
 dT


R ' T / P

R ' T ( gP ) dZ
PC p
R ' T dT / dZ
  g / C p
This quantity, –g/C p
, 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 .
8 K / km
For a parcel of air moving adiabatically in the atmosphere:
T
2

T
1
 
0
( Z
2

Z
1
)
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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,
T
Z

T
0
 
Z
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(Handout)
Solid lines – thermodynamic property
Dashed (colored) lines – measurements or soundings day
γ a
γ b
> γ
₀ unstable
= γ₀ neutrally stable
γ c
< 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.
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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.
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(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,
 z

T
Z

Z

0 or a proper derivation yields:
z
(
1000 hPa/P z
)
** R’/Cp
= T
(
1000 hPa/P z
)
** 0.286
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z
Subsidence inversion
MIDDAY
1 km
Mixing depth
NIGHT
0
MORNING
T NIGHT MORNING AFTERNOON

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ITCZ
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Fig. 9.13
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Fig. 9.15
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Planck’s Law for blackbody radiation:
E (

, T )

2
 hc
2

5
1 e hc /
 kT 
1
( Wm
2 m
1
)
Take first derivative wrt T and set to zero: Wien’s Law:
 max
 a
T

2 .
9

10

3
T m K
-1
( m)
Integrate over all wavelength: Stephan-Boltzmann Law:
E
T
 
T
4
( W/m
2
)
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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 x 10
5
/

( kJ /mole )
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