Copyright R. R. Dickerson 201511 Professor Russell Dickerson Room 2413, Computer Space Sciences...

Copyright © R. R. Dickerson 2015 11

Professor Russell Dickerson Room 2413, Computer & Space Sciences Building Phone(301) [email protected] web site www.meto.umd.edu/~russ

AOSC 620PHYSICS AND CHEMISTRY

OF THE ATMOSPHERE, I

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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.

Copyright © R. R. Dickerson & Z.Q. Li 33

Logistics

Office Hours: Tuesdays 3:30 – 4:30 pm(except today)Wednesdays 1:00 – 2:00 pmWorst time is 1- 2 pm Tues or Thrs.Exam Dates: October 13, November 24, 2015Final Examination: Thursday, Dec. 17, 2015 10:30am-12:30pmwww/atmos.umd.edu/~russ/SYLLABUS_620_2015.html

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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.

Copyright © R. R. Dickerson 77

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 1Salby Chapter 1.2 and 2.2-2.3W&H Chapter 3.


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)


Dalton’s law of partial pressures

P = i pi

PV = i piRT = RT i pi

The mixing ratios of the major constituents of dry air do not change in the troposphere and stratosphere.

Copyright © 2013 R. R. Dickerson & Z.Q. Li

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Definition of Specific Volume

= V/m = 1/PV/M = nRT/m

P = R’TWhere R’ = R/m

Specific volume, is the volume occupied by 1.0 g (sometimes 1 kg) of air.


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Definition of gas constant for dry air

p = R’TUpper 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)


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First Law of ThermodynamicsThe 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 = U + WWhere Q is the heat flow into the system, U is the

change in internal energy, and W is the work done.In general, for a unit mass:

đq = du + đwNote đq and đw are not exact differential, as they are

not the functions of state variables.


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.

A

dl


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In general the specific work done by the expansion of an ideal gas from state a to b is W = ∫a

b pdα

p↑

α→

ab

α1 α2


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W = ∮ pdα

p↑

α→

ab

α1 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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If no work is done, and for a constant specific volume:

đq = cvdT = du orcv = du/dT = Δu/ΔT for an ideal gas

At a constant pressure:đq = cpdT = du + pdα

= cvdT + pdα orcp = cv + p dα/dT

But pα = R’T and p dα/dT = R’ thus

cp = cv + R’


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pα = R’TDifferentiating

d(pα) = pdα + αdp = R’dT orpdα = 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.


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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 becomesđq = cpdT

2. If the temperature is held constant, dT = 0.For an isothermal process:

đq = cpdT − αdp becomesđq = − αdp = pdα = đw


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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


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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 đwRemember α = R’T/p thus

đq = cpdT = R’T/p dpSeparating 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 = 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.


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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.


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Weather Symbolshttp://www.ametsoc.org/amsedu/

dstreme/extras/wxsym2.html


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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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∮ 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 = 0Entropy is a state variable.


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Rememberđq = cpdT − αdp

đq/T = cp/T dT − α/T dpdφ = cp/T dT − α/T dp

Remember α/T = R’/p Therefore

In a dry, adiabatic process potential temperature doesn’t change thus entropy is conserved.


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7am


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Copyright R. R. Dickerson 201511 Professor Russell Dickerson Room 2413, Computer Space Sciences...

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