AOSC 620 Lecture 2
PHYSICS AND CHEMISTRY
OF THE ATMOSPHERE I
Professor Russell Dickerson
Room 2413, Computer & Space Sciences Building
Phone(301) 405-5391
russ@atmos.umd.edu
web site www.meto.umd.edu/~russ
Copyright © 2015 R. R. Dickerson
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Experiment: Room temperature
What is the temperature of the room?
We’ll get 17 answers.
How well do they agree?
Does location matter?
Does observer bias play a role?
Copyright © 2015 R. R. Dickerson
& Z.Q. Li
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Experiment: Room temperature
Last class
Average oC
Stdev oC
Max oC
Min oC
TMean = +/- °C (+/- °F)
Measured with uncalibrated or no thermometers at n
locations.
Copyright © 2010 R. R. Dickerson
& Z.Q. Li
3
What is the right answer?
What is the uncertainty?
Are the data Gaussian?
How can we improve the measurement?
Copyright © 2015 R. R. Dickerson
& Z.Q. Li
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Experiment: Room temperature
Let’s repeat the experiment with calibrated
thermometers.
Copyright © 2015 R. R. Dickerson
& Z.Q. Li
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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
Pα = R’T
First Law of Thermodynamics:
đq = du + đw
W = ∫ pdα
Copyright © 2015 R. R. Dickerson
& Z.Q. Li
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Review (cont.)
Definition of heat capacity:
cv = du/dT = Δu/ΔT
cp = cv + R
Reformulation of first law for unit mass of an
ideal gas:
đq = cvdT + pdα
đq = cpdT − αdp
Copyright © 2015 R. R. Dickerson
& Z.Q. Li
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Review (cont.)
For an isobaric process:
đq = cpdT
For an isothermal process:
đq = − αdp = pdα = đw
For an isosteric process:
đq = cvdT = du
For an adiabatic process:
cvdT = − pdα and cpdT = αdp
Copyright © 2015 R. R. Dickerson
& Z.Q. Li
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Review (cont.)
For an adiabatic process:
cvdT = − pdα and cpdT = αdp
du = đw
(T/T0) = (p/p0)K
Where K = R’/cp = 0.286
(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 © 2015 R. R. Dickerson
& Z.Q. Li
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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 = cpln(θ/θ0)
In a dry, adiabatic, process potential temperature
doesn’t change, thus entropy is conserved.
Copyright © 2015 R. R. Dickerson
& Z.Q. Li
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Useful idea - a perfect or exact differential:
If z = f(x,y), dz is a perfect differential iff:
∂2f/∂x∂y = ∂2f/∂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 © 2015 R. R. Dickerson
& Z.Q. Li
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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 © 2015 R. R. Dickerson
& Z.Q. Li
12
Virtual Temperature: Tv 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 © 2015 R. R. Dickerson
& Z.Q. Li
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Consider a mixture of dry air and water vapor.
Let
Md = mass of dry air
Mv = mass of water vapor
md = molecular weight of dry air
mv = molecular weight of water.
Copyright © 2015 R. R. Dickerson
& Z.Q. Li
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Dalton’s law: P = Spi
*
*
R
R
P  pd  e   d
T  v
T
md
mv
 Md Mv 



 md mv 
M Md  Mv


V
V
*
RT
P
V
(V  volume)
Copyright © 2015 R. R. Dickerson
& Z.Q. Li
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Combine P and  to eliminate V:
 Md Mv 
1

P  R T 

 md mv  M d  M v
w 1
*  1

 R T 

 md mv  1  w
*
*
R
1

T (1  w /  )
md
1 w
(1  w /  )
P  RT
(1  w)
Copyright © 2015 R. R. Dickerson
& Z.Q. Li
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Since P = RT*
(1  w /  )
and P  RT
(1  w)
(1+ w / e )
T = virtual temperature = T
(1+ w)
*
T > T because e = 0.622
*
Copyright © 2015 R. R. Dickerson
& Z.Q. Li
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Alternate derivation:
Since  proportional to Mwt
( md )
T  virtual temperature  T
( mw )
md  29 g / mole
*
mw  29(1  x )  18 x
Where w is the mass mixing ratio and x (molar or volume
mixing ratio) =
[H2O] = w/0.62
T* = T (29/(29-11[H2O]))
e.g., [H2O] = 1% then T* = T(1.004)
If , [H2O] = 1% then w = 0.01*.62 = 0.062 T* = 1.01/1.0062 =
Copyright © 2015 R. R. Dickerson
& Z.Q. Li
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Where w is the mass mixing ratio and x (molar or volume
mixing ratio) =
[H2O] = w/0.62
T* = T (29/(29-11[H2O]))
e.g., [H2O] = 1% then T* = T(1.004)
Test: if [H2O] = 1% then w(18/29) = 0.01*.62 = 0.0062
T* = T (1 + w/)/(1+w) =
T (1 + 0.01)/(1+0.0062) = T(1.004)
Copyright © 2015 R. R. Dickerson
& Z.Q. Li
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Unsaturated Moist Air
Equation of state: Pa = RT*
(1+ w / e )
T =T
(1+ w)
(1+ w / e )
Pa = R
T = RmT
(1+ w)
(1+ w / e )
where Rm = R
@ R(1+ 0.6w)
(1+ w)
*
because w £ 10 -1 so (1+ w)-1 @ 1- w
Copyright © 2015 R. R. Dickerson
& Z.Q. Li
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Specific Heats for Moist Air
Let mv = mass of water vapor
md = mass of dry air
To find the heat flow at constant volume:
d Q  (mv  md )d q  md cv dT  mv cvv dT
 mv 
 mv cvv 
  md cv dT 1 

 md 1 
 md 
 md cv 
(1  w)d q  cv dT (1  rw)
with r 
cvv
cv
 1.96
Copyright © 2015 R. R. Dickerson
& Z.Q. Li
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For constant pressure
c pm  c p (1  0.9 w)
So Poisson’s equation becomes
k
 1000mb 
  T

 P 
Rm
R
k
 (1  0.2 w)
c pm c p
(1+0.6w)/(1+0.9w) => (1-0.2w) due to rounding error.
Copyright © 2015 R. R. Dickerson
& Z.Q. Li
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Water Vapor Pressure
Equation of state for water vapor: ev = v Rv T
where ev is the partial pressure of water vapor
R*
Rv 
mv
R * md
R * md
eva  RvT 
T
T
mv md
md m v
 R dT / 
mv

 0.622
md
Copyright © 2015 R. R. Dickerson
& Z.Q. Li
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