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AT 620 – Notes
These notes were prepared by
Prof. Steven A. Rutledge
(and adapted slightly for the Fall 2009 course,
and again slightly for this year)
You may access Prof. Cotton’s notes, password “cloud9”
http://rams.atmos.colostate.edu/AT620old/AT620Notes.html
DO NOT DISTRIBUTE OUTSIDE CSU
1
Motivation:
understanding energy and its
transformations in the atmosphere
Toutflow
3
4
2
1
Tsea surface
Hurricanes are giant engines that convert heat into wind energy
http://apollo.lsc.vsc.edu/classes/met130/notes/chapter15/vertical_circ.html
Understanding the phase changes of water and associated energy changes
2
Motivation, continued
Thermodynamics provides a basis for understanding nucleation, the
process that forms small water droplets and ice crystals, and thus is the
foundation for the study of cloud and precipitation processes.
• Particles (aerosols) and Nucleation of Water and Ice
– Aerosol indirect effects on climate (those occurring through the link to clouds and
precipitation)
3
Thermodynamics
• …is the study of equilibrium states
• We can use thermodynamics to predict the
direction of transitions, but NOT the rate
– Kinetics, a separate field of study
• We observe equilibrium states via “primitive”
variables – those that can be measured
– Temperature, pressure, composition
• But equilibrium (and the laws governing it) is
really about energy, expressed in terms of
derived properties (“enthalpy”, “entropy”,…)
4
Lecture 1: quick overview of topics
• Open vs. closed system
• What does “adiabatic” mean?
• Intensive vs. extensive properties
• What does “specific” mean (e.g., “specific volume”)?
• Ideal Gas Law
• Equilibria:
1. Thermal
2. Mechanical
3. Chemical
– And: stable, unstable, metastable
• State function vs. work and heat
5
REVIEW AND BASIC CONCEPTS
a. Definitions
Define a “system” (often can be chosen for convenience) – A portion of
matter, whose study focuses on its properties, interaction and evolution,
from a thermodynamic point of view. “That part of the universe under
consideration” (Wikipedia)
All the material environment which may eventually interact with the system is
known as the surroundings
OPEN System – exchange energy (heat and work) and matter with
surroundings
CLOSED System – can exchange energy, but no exchange of matter with
surroundings
A system is described by its properties – or physical variables that express
these properties.
6
A more stringent definition of a closed system (isolated system) is that
there is no exchange of energy, volume or chemical constituents with the
surroundings.
For a closed system:
mass & chemical composition define the system.
properties like temperature & pressure define the system’s state
Define EXTENSIVE PROPERTIES
Take two identical systems each having volume V and chemical
constituents given by N1, N2, Nr where N represents the mole number
Now “merge” these two systems.
New volume = 2 X V
(# molecules of N r )
NA
Nnew = 2 X Nold
7
Parameters that have values in a composite system equal to the sum of the
values in each subsystem are called Extensive Parameters.
Extensive Variables depend on the size of the system!
Intensive Property: does not depend on mass (or size)
Intensive and extensive properties are usually denoted by lowercase and
uppercase symbols, respectively.
p, T- Intensive properties
V- Extensive properties
Defining an intensive property from an extensive property;
z
Example:
is referred to as a “specific” property
vV
m
Z
z
m
Specific Volume
To convert an extensive variable into an intensive variable, simply divide by
mass
V, m are Extensive Variables
1 V density is an Intensive Variable
 m
8
More Definitions:
Homogeneous System- every intensive variable (e.g. T) has the same value at every
point in the system. EXAMPLE: small air parcel if just gas phase.
Heterogeneous System- A collection of homogeneous systems, each different from
one another. Each homogeneous system is referred to as a phase.
Properties of extensive variables for a heterogeneous system;
Z
m
z


m
mass in  phase
z
specific value of extensive variable
Z
e.g.
U
m
u Internal Energy

t
o
t
a
l


Inhomogeneous system- a system in which the intensive variables change in a
continuous manner from one point to another.
Atmosphere: T decreases linearly with height.
9
More Definitions:
Homogeneous System- every intensive variable (e.g. T) has the same value at every
point in the system. EXAMPLE: small air parcel if just gas phase.
Heterogeneous System- A collection of homogeneous systems, each different from
one another. Each homogeneous system is referred to as a phase.
Properties of extensive variables for a heterogeneous system;
Z   m z

m
mass in  phase
z
specific value of extensive variable
Z
e.g.
U total   m u

Internal Energy
This gradient would imply
energy transfer 
a RATE of change
Inhomogeneous system- a system in which the intensive variables change in a
continuous manner from one point to another.
Atmosphere: T decreases linearly with height.
10
Equilibrium (brief look; details later)
Most of our work in thermodynamics of the atmosphere will be dealing with
thermodynamic systems moving from an “unstable” state to a “stable” state.
Equilibrium: The state of a system remaining constant over time (an ok
definition for now); a minimum in energy (or maximum in entropy)
UNSTABLE equilibrium: a small change in the state of a system causes large
departures from the original state.
EXAMPLE: small droplet within a field of water vapor
RHcrit
RH
-small changes in drop size lead to changes in the
vapor pressure, relative to the environment
droplet
Rcritr
-what are accompanying changes in total energy?
METASTABLE EQUILIBRIUM
Example: supercooled droplet
“local” minimum, not global minimum; stable to small changes in temperature
11
& pressure; sudden freezing if droplet contacts “freezing nucleus”
A system in “stable equilibrium” is not only invariant in time, but small changes in
the intensive properties of the system always finds the system referring to its
equilibrium position (state).
There are 3 components to equilibrium we will consider:
THERMAL ( T )
MECHANICAL ( p )
CHEMICAL ( µ, chemical potential; related to Gibbs Free Energy)
-> we can sometimes use vapor pressure here
Aside: we assume the environment seeks equilibrium. We often formulate
models this way, where the rate of change is proportional to the departure from
equilibrium.
Example: latent heat transfer at sea-air interface is modelled by
Qc  CD L(qw  qa ) v
Saturation specific
humidity of water (ocean)
Saturation specific humidity
in air (boundary layer) 12
TEMPERATURE
Consider the following system:
adiabatic wall- no heat flow
A
B
diathermic wall- heat flow
Wait for a period of time: A&B are said to be in thermal equilibrium. All bodies in
thermal equilibrium have the same temperature.
Temperature is defined relative to a reference body that attains thermal equilibrium
with the substance (body).
(This reference body is of course a thermometer, which is any device with a
thermometric substance that can be related to the thermal state of the body)
Substance
Gas, constant volume
Gas, constant pressure
Thermocouple
Platinum wire
Hg, const p
T

a
xb
Property X
Pressure
Specific volume
EMF (voltage)
Resistance
Specific volume
= a property of a thermometric substance
13
System of Units:
adopt SI/MKS
Kelvin scale of temperature is defined by assigning the number 273.16 to the
temperature of a mixture of pure ice, water and water vapor in equilibrium.
T
(

C
)

T
(

K
)

2
7
3
.
1
6
Acceleration (g)
m s-2
9.8 m s-2
Density (  )
kg
Force (N, Newton)
kg m s-2
Pressure (Pa, pascal)
N m-2 =kg m-1 s-2
m-3
9
T(ÞF)  T(ÞC)  32
5
2 = 10 baryes= 10-5 bars
1 Pa= 10 dynes/ cm
1 mb= 102 Pa (surface pressure; 1013 mb or about 105 Pa)
Energy (J Joule)
kg m2 s-2
1J=107 ergs=0.2389 calories
Specific energy
J kg-1 ; m2 s-2
Calorie
energy required to raise temperature of 1g of H2O at 15°C
by 1 °K
14
1 calorie= 4.1868 J
Another important unit/term is the MOLE (mol)
The amount of a substance which contains as many elementary units as there
are C atoms in 12 g of 12C. A mole of any substance contains the same
number of molecules, which is NA (Avogadro’s number)
NA = 6.023 X 1023 molecules / mole
One (1) atomic weight of any element contains NA particles.
15
EQUATION OF STATE (“GAS LAW”)
Relationships that express intensive parameters (e.g., p, T) in terms of
extensive parameters are called “equations of state.”
Early lab studies determined that p, V, and T could be related by an “equation
of state.”
Trivia: Boyle’s Law, 1662! (P and V inversely proportional)
Charles’ Law, 1787: V proportional to T
Avogadro’s Law, 1811: equal volumes of ideal gases have same # of
molecules
Ideal Gas Law: from combining gas laws, proposed by Clapeyron, 1834
IDEAL GAS LAW: Equation of state for ideal gases (gases where molecules
have no interactions; no molecular attractive forces; and molecules are
assumed to occupy zero volume).
pV= nR*T= mR*T/M= mRT
n= number of moles= m/M; m = mass, g; M= molecular weight, g/mole
R* = universal gas constant, 8.3143 J mol-1 K-1
R = R*/M, specific gas constant, J g-1 K-1
16
p, V and T are often referred to as STATE VARIABLES

Introducing density,

m
V
p
V

m
R
T

V
R
T

p

R
T
The SPECIFIC VOLUME is the inverse of density:
 1/
*
The universal gas constant R follows from Avogadro’s hypothesis, which says, gases
containing the same number of molecules occupy the same volume at the same p and T
and
*

(
p
V

n
R
T
)
Ideal Gas law can also be written in the form for individual molecules using Boltzmann’s
constant

k  R / N A  1.38110
p  n kT
p  kT
n 
23
J
deg K  molecule
# molecules
unit volume
17
units
Check the units for the gas constants:
P V = n R* T
N
3
m
 moles ?? K
m2
then
R * must have units :
But we had a value of

Does it work out?
----------------------
Recall
Force (N, Newton)
Pressure (Pa, pascal)
Energy (J, Joule)
R *  8.3143
N m
moles K
J
mole K
kg m s-2
N m-2 =kg m-1 s-2
kg m2 s-2
18
units, continued
Check the units for the gas constants:
PV=mRT
N
3
m
 kg ?? K
2
m
then
R * must have units :
N m
kg K
We found a value for DRY AIR (MW ~ 29 g mol-1) of

Does it work out?
----------------------Recall
Force (N, Newton)

Pressure (Pa, pascal)
Energy (J, Joule)
Rd  287
J
kg K
kg m s-2
N m-2 =kg m-1 s-2
kg m2 s-2
19
Aside: REAL GASES
Behavior of real gases (molecular attractive forces; molecules have non-zero volume)
can be approximated by the Van der Waals equation (other equations also proposed):
For 1 mole
*
RT
p
v
R T a
p
 2
v b v
 =specific volume
a,b constants for each particular gas
Follows from a model of point molecules moving independently from
one another
Two simple ‘corrections’ were developed by Van der Waals
1. Molecules have non-zero volume
(
Vn
b
)
or
(vb
)
2.
nb is volume “occupied” by molecules
n=# of moles
V=total volume
2
a
v
Pressure is reduced by
; decrease of pressure is proportional to number of
interacting pairs of molecules per unit volume (1v2)
Molecule approaching wall experiences a net backward
attraction due to remaining molecules in volume; thus
pressure exerted on wall is reduced.
20
(See notes at end for more)
Mixture of IDEAL GASES
Dalton’s Laws of partial pressures
In a mixture of gases, the pressure Pi of the ith
constituent is defined as the pressure that the gas
would exert if the ith gas occupied the volume alone.
Total pressure
For each gas,
Pt   Pi
i

PV

n
R
T  mi RiT
i
i
Applying Dalton’s Law:
 
with

mi R
Mi 

ni Ri


V Pi  PV


n
R
T   mi Ri T
t
i
i
i
Since our goal is to represent properties of a mixture of ideal gases as those
of a single gas, we define a mean gas constant and molecular weight for the
mixture as R  m R / m Gas constant
i i
i
M  ni M i / ni
Hence
Molecular weight
Mair ~ 29 g mole-1 (useful)

PV

n
R
T  mRT
t
t
IDEAL GAS LAW FOR A MIXTURE OF GASES (LIKE OUR ATMOSPHERE)
21
Mole fraction for each constituent:
For 2 gases
Of course
N1 
n1
n1  n2
Ni  1
N2 
Ni 
ni
ni
n2
n1  n2
(sum of mole fractions is equal to 1)
Atmospheric Composition
Mixture of ‘Ideal’ gases
Water in 3 phases
Liquid
Cloud Formation
Aerosols
Solid
Atmospheric Chemistry
Radiation
Optics
Climate
Atmospheric Electricity ( J  gE )
22
TABLE I-4
Main components of dry atmospheric air
Gas
Molecular
Weight
Molar (or volume)
fraction
Mass
Fraction
(J kg-1 K-1)
Specific gas
constant
(J kg-1 K-1)
mi Ri / m
(Jkg-1K-1)
Nitrogen (N2)
28.013
0.7809
0.7552
296.80
224.15
Oxygen (O2)
31.999
0.2095
0.2315
259.83
60.15
Argon (Ar)
39.948
0.0093
0.0128
208.13
2.66
Carbon dioxide (CO2)
44.010
0.0003
0.0005
188.92
0.09
1.0000
H2O(v)
18.02 g/mole
1.0000
mi Ri
 287.05
m
Volume Fraction
0-0.04
23
TABLE I-5
Minor gas components of atmospheric air.
Gas
Molar(or volume) Fraction
Neon
(Ne)
1.8x10-5
Helium
(HE)
5.2x10-6
Methane
(CH4)
1.5x10-6
Krypton
(Kr)
1.1x10-6
Hydrogen
(H2)
5.0x10-7
Nitrous oxide
(N2O)
2.5x10-7
Carbon monoxide
(CO)
1.0x10-7
Xeon
(Xe)
8.6x10-8
Ozone
(O3)
Variable. Up to 10-5 in stratosphere.
Sulfur dioxide
(SO2)
Hydrogen sulfide and
other reduced
sulfur compounds
H2S, ect.)
Nitric oxide
(NO)
Nitrogen dioxide
(NO2)
Ammonia
(NH3)
Formaldehyde
(CH2O)
Variable, under 10-8
24
TABLE 4.1 Fractional Composition of Planetary Atmospheres
Earth
Venus
Mars
Titan
1013
92,000
6
1496
-
~200
-
~ 10
0.03b
96.5
95.3
(2-10)
N2
78
3.5
2.7
65-98
O2
21
<0.1ppm
0.13
-
Pressure (mbar)
Surface
Cloud Top
TRACE
GASES; BUT
IMPORTANT!
Major Gases, %
CO2, (CH4)
 100ppm
 100ppm
H2O
0-2c
Ar
0.9
70ppm
1.6
0-25?
H2
0.5
?
~ 10
2000
He
5
~ 12
-
-
CO
0.2
50
700
~ 100
NO
0.0005
-
~3
-
O3
0.4
-
~ 0.01
-
HCl
-
0.6
-
-
HF
-
0.005
-
-
Ne
18.2
7
2.5
-
Kr
1.14
1
0.3
-
Xe
0.087
< 0.1
0.08
-
-
Trace Gases (ppm)
25
Tables I-4 and I-5 show 4 major constituents and numerous minor constituents for
Earth’s Atmosphere. These gases compose ‘dry air’ or ‘moist air’ if H2O is included.
These gases exist in these proportions in roughly the lowest 100km of our atmosphere
due to turbulent mixing:
100-120km
Diffusive equilibrium
occurs above the turbo
pause (homopause)
Heterosphere
Homosphere
Turbulent mixing in the
homosphere
In the absence of other forces acting on them, gases diffuse at different rates according
to their individual molecular weights
R
From solving hydrostatic
RiT
R

z / H
i
H
balance equation (see
M
n z  ne
g
next slide)
or p
H=SCALE HEIGHT
 
Above ~100 km, heavier gases (large M) have smaller scale heights –e folding distance is less.
For our purposes we well consider ‘dry air’ as a constant mixture which behaves like a pure gas.
mi Ri
Rd 
mi
(.755)(296.8)  .23(259.83)  0.0128(208.13)  .0005(188.92)
mi
26
1
Rd  287 J deg kg
1
GAS CONSTANT
Hydrostatic balance (http://eesc.columbia.edu/courses/ees/climate/lectures/atm_phys.html)
The atmosphere does not collapse under the downward pull of gravity
because of the energy embedded in the movement of the air molecules.
This movement creates the force of pressure which counters the
gravitational pull on the atmosphere. The balance between the force of
pressure and gravity is the hydrostatic balance.
To find the expression for the hydrostatic balance, we first note that
atmospheric surface pressure is due to the weight of the entire atmospheric
column above. As we ascend, there is less of an atmosphere above us, and
hence the pressure drops. Consider a column of gas Δz meters tall
suspended somewhere in the atmosphere. The reason this column of air
does not "fall" down under the pull of gravity is because the pressure acting
on its bottom surface is higher than the pressure acting on its top surface.
The pressure difference Δp exactly balances the weight (per unit area) of
the column. Stated in mathematical terms this balance is written as:
Δp = - ρ g Δz
Write density in terms of T and p to get in form
that yields the scale height
where g is the acceleration due to gravity = 9.8 m/s2.
27
Rd  287.05
J kg-1 °k-1
R
Md 
 28.96 g mole-1
Rd
Useful
number!
The “standard atmosphere” (see depiction to
the right, from Wikipedia) indicates an
average scale height below ~100 km of ~7.64
km, consistent with an average air
temperature over that range of close to 260 K
Note that equation can be rearranged as
ln
p
z

po
H
At higher 
altitudes each gas tends to have its
own scale height (gravity plays more of a role
than mixing)
THE FIRST LAW of THERMODYNAMICS: Conservation of energy
Consider a “body” of molecules
v
Macroscopic velocity=v
Body has 3 types of energies:
macroscopic kinetic energy 1 mv 2
2
macroscopic potential energy
mgh
microscopic internal energy, KE and PE of molecules
Potential energy (PE) at the microscopic level is associated with molecular
attractive forces
Kinetic energy (KE) at the microscopic level is associated with molecular
motions which are of course proportional to the temperature of the gas.
First Law is concerned with conservation of microscopic energy
29
WORK (OF EXPANSION)
Atmosphere is a compressible gas!!
One type of work we will consider in our study of atmospheric thermodynamics is the
work of expansion (e.g., an expanding gas ). If the gas expands, 
. If the gas
W

0
contracts, 
.
W

0
Formulation of work term (mechanical work):
p,V
piston displacement dl
cross-sectional area A
V
2
Frictionless Piston
W

F

d
x

p
A

d
l

d
V


p
a
2
V
Work = area under P-V curve
2
For work done in a cycle,
p
2
1
1
W

pdv

pdv

pdv



1
2
V
V
1
30
More on Van der Waal’s Equation
R*T
a
p
 2
V  nb v
2
For large V, V»nb and p» a v so
we see that Van der Waal’s
Equation takes on the same form
as the ideal gas law.
For n=1 mole
R*T
a
p
 2
v b v
Examine pressure reduction term
a
v2
Consider outermost layer of molecules, and layer of molecules just below this layer.
There is an attractive force between these 2 layers. Molecules at other positions are
subject to a net-zero force.
Point attractive forces between molecules will be proportional
2
to n , where n  N ; molecular density
V
So pressure reduction term is like
2
N

2
 n   

V 
Since N  nN A n=# of moles
We have
NA=Avogadro’s #
n 2 N A2  N A2
N
   
 2
2
V
v
V 
2
31
Volume reduction term
V  nb
Consider atoms or molecules as spheres with diameter D:
molecules
Molecular “sphere of influence” is

6
.  108 cm
D
for
D3
(no other molecule exists within diameter D of another molecule, or volume of
Total volume occupied by N molecules is
Or since N  nN A , moles 
N
#
moles
Volume reduction term is nN  D 3
A
6
or
NA

6

6
D
6
3
D3

D3 )
per mole
This term, N  D 3 would be “b” in the volume reduction term. So volume reduction is
A
nb
6
V  nb
32
Intermolecular forces
“ Van der Waals forces “
Intermolecular forces can be repulsive or attractive
F
In general-
Repulsive
Attractive
r
r  Dmolecule
For r large compared to diameter of the molecule, F is attractive
-
-
+
+
(Correlation in electron positions)
Van der Waals force is ELECTRICAL in nature
positive nucleus
-
+
negative electron cloud
-
For r  Dmolecule
F  Repulsive (electrostatic repulsion of electron clouds). This
33
process is the reason that a liquid is incompressible.