METO 620 PHYSICS AND CHEMISTRY OF THE ATMOSPHERE I

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METO 620PHYSICS AND CHEMISTRYOF THE
ATMOSPHERE I
Professor Zhanqing Li Room 2114, Computer
Space Sciences Building Phone [phone].
Zli_at_atmos.umd.edu www.atmos.umd.edu/zli
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Schedule

• Office Hours 930 530pm
• Exams 2 mid and one final.
• Final score composition
• Homework 30
• Mid exam 1 and 2 20 each
• Final exam 30.
• Course note
• www/atmos.umd.edu/zli

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Part 1 Thermodynamic Applications in
Understanding of the Atmosphere

• What is thermodynamics?
• Thermodynamics is concerned with the conservation
  and conversion of various forms of energy and
  relationships between energy and changes in the
  properties of mater, such as water. It is a
  fundamental theory to explain observations of
  many physical, chemical and biological systems.
• Applications in atmospheric sciences
• Wind and ocean currents are essentially driven
  by thermodynamic imbalance caused by non-uniform
  solar heating, in attempt to achieve a
  thermodynamic equilibrium.
• Clouds are formed due to condensation/freezing
  of water vapor. Conversation of water phases is
  governed by thermodynamic theories
• Thermodynamic feedbacks in the atmosphere and
  ocean are critical to understanding of climate
  change

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Atmospheric Composition
by volume by mass Molecular Weight Nitrogen
78.08 75.51 28.02 Oxygen 20.95 23.14 32.00
Argon 0.93 1.28 39.95 Carbon
dioxide 0.03 0.05 44.01 Water vapor 0-4 (highly
variable) 18.01 Except for water vapor,
atmospheric composition is very stable around the
globe and up to 90km above the surface. Water is
the only substance that exists in three forms
gas, liquid (droplets and raindrops) and
ice Hydrometeors raindrops and snowflakes
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Thermodynamics of Dry Air.
Objective To find some useful relations among
air temperature, volume, and pressure. Ideal Gas
Law PV nRT (M/m)RT 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
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Definition of Specific Volume

• V/M 1/r
• PV/M RT/m
• Pa RT
• Where R R/m
• Specific volume, a, is the volume occupied by
  1.0 g of air.
• R is a specific gas constant

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Understanding of the Idea Gas Law

• What is P pressure is exerted by the impact of
  moving molecules
• What is T temperature is a form of kinetic
  energy of the molecules
• P depends on
• speed of the molecules (T),
• the mass of molecules (M), and
• The frequency of the impact (1/V)

• P MT/V
• Pa RT

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Daltons law of partial pressures
P Si pi The mixing ratios of the major
constituents of dry air do not change in the
troposphere and stratosphere. Note that each
individual gas obeys the ideal gas law. Following
this law, we can derive the specific gas constant
for the mixture of air gases (dry gas)
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Definition of gas constant for dry air

• pa RT
• 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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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.
• In general, for a unit mass
• dq du dw
• Note dq and dw are not exact differential, as
  they are not the functions of state variables

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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 unit 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 dw pda

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In general the specific work done by the
expansion of an ideal gas from state a to b isW
?ab pda
a
b
p?
a1
a2
a?
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Or, for a cyclic systemW ? pda

• W is not equal to zero for cyclic process and
  thus dw is not an exact differential.

a
b
p?
a1
a2
a?
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Definition Heat Capacity

• Internal energy change, du, is measured as a
  change in temperature.
• The temperature change is proportional to the
  amount of heat added.
• dT dq/c
• where c is the specific heat capacity.
• If no work is done, specific volume is a
  constant
• dq cvdT du or
• cv du/dT ?u/?T

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• At a constant pressure
• dq cpdT du pda
• cvdT pda or
• cp cv p da/dT
• But pa RT and
• p da/dT R thus
• cp cv R
• For dry air,
• cp 1005 J kg-1 K-1
• cv 718 J kg-1 K-1

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• pa RT
• Differentiating
• d(pa) pda adp RdT or
• pda RdT - adp
• From the First Law of Thermo for an ideal gas
• dq cvdT pda cvdT RdT - adp
• But cp cv R
• dq cpdT - adp
• This turns out to be a powerful relation for
  ideal gases.

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• Let us consider four special cases.
• If a process is conducted at constant pressure,
  then dp 0.
• For an isobaric process
• dq cpdT - adp becomes
• dq cpdT
• If the temperature is held constant, dT 0.
• For an isothermal process
• dq - adp pda dw

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• Next two special cases.
• If a process is conducted at constant density
  then d? da 0.
• For an isochoric process
• dq cvdT du
• If the process proceeds without exchange of heat
  with the surroundings dq 0.
• For an adiabatic process
• cvdT - pda and cpdT adp

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• The adiabatic case is powerful.
• Most atmospheric temperature changes are
  adiabatic.
• For an adiabatic process
• cvdT - pda and cpdT adp
• du dw
• Remember a RT/p thus
• cpdT RT/p dp
• Separating the variables and integrating
• cp/R ?dT/T ?dp/p

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• cp/R ?dT/T ?dp/p
• (T/T0) (p/p0)K
• Where K cp/R 0.286
• This allows you to calculate, for an adiabatic
  process, the temperature change for a given
  pressure change.

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• If we define a reference pressure of 1000 hPa
  (mb) then
• (T/?) (p/1000)K
• Where ? 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.
• ? T (1000/p)K
• Potential temperature, ?, is a conserved quantity
  in an adiabatic process.

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Entropy

• df dq/T
• Where f is defined as entropy.
• df cvdT/T pda/T
• cvdT/T R da /a
• ? dq/T ? cvdT/T ?R da /a

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• ? dq/T ? cvdT/T ?R da /a
• But ? cvdT /T 0 and ?Rda /a 0
• Because T and a are state variables thus
• ? df 0
• Entropy is a state variable, and df is an exact
  differential.
• It is worthnoting that dq is not an exact
  differential, but its ratio over T becomes an
  exact differential.

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• Remember
• dq cpdT - adp
• dq/T cpdT /T - adp /T
• df cpdT /T - a/T dp
• Remember a/T R/p
• therefore
• df cpdT/T - Rdp/p
• cpd? /?
• ?f cpdln(?/?0)
• ?f cpln(?/?0) constant
• In an adiabatic process potential temperature
  doesnt change thus entropy is conserved.

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Homework 1

• HW problems 1.1, 1.2, 1.3, 1.6, from Rogers and
  Yao.
• Due 9/6/07