Atmospheric Thermodynamics Part -ILeila M. V. Carvalho

Outline:• A Brief introduction • Gas Laws: applications for the atmosphere• The Hydrostatic Equation • Definition of geopotential• Scale Height and Hypsometric Equation : applications• First Law of Thermodynamics• Adiabatic Processes• Water vapor in the air• Stability• The second Law of Thermodynamics : Application of the

concept of Entropy to the atmosphere

N2: 78%

Water vapor H2O : 0-4% (Variable)Carbon dioxide: CO2 : ~ 0.03% (Variable)

O2:21%

% per volume

Vertical profile of Density and pressure in the Atmosphere

• Determined by the GRAVITY FORCE!

FG = Gravitational force

mgr

mGMF EG

2

The ME mass of the Earth = 5.98*1024 kgG=gravitational constant=6.67x10-11 Nm2kg2

Vertical profile of temperature in the atmosphere

Gas Laws

• Equation of State: is valid for all gases and can be applied to the atmosphere• In this case we have to consider that the atmosphere is a mixture of gases• The ideal gas equation may be written as: pV=mRT

The Gas Law for the dry air

• The ideal gas equation may be written as:pV=mRT (1)

Where: p=pressure (Pa=N/m2); V=volume (m3); m =mass (kg) ; T=absolute temperature (K- Kelvin). R is a constant for 1kg of the gas and DEPENDS ON THE GAS.

Since density ρ = m/V, (1) can be rewritten as:p= ρRT (2)

• The specific volume of the gas, that is the volume occupied by 1kg of gas at pressure p and temperature T is given by α= 1/ρ thus:

p= ρRT -> p α =RT (3)Conclusions: if Temperature of a fixed mass of

gas is held constant during a given process, the volume of the gas is inversely proportional to its pressure (Charles First Law)

Charles’ first law • Remember: temperature is a measure of the mean kinetic energy,

which is a measure of the mean speed at which its molecules move.

Pressure: force in the wall larger volume => decrease in the number of shocks in the walls -. Decrease in pressures

Chances of collision in a given moment decrease

Charles’ Second Law

• Keep volume and mass constant: The pressure of the gas is proportional to its temperature

Pressure: force in the wall

Gram-molecular weight or mole (mol) M

• M is the molecular weight of any substance expressed in grams.

• Example: Water (H2O) ~ 2*1.008 + 16.000 ~ 18g (18.016g)• Given some mass (m) of a substance, the number of moles is

given by:n=m/M (4)

If I say that I have one mol of a given substance it means that I have a mass (g) of the molecular weight of that substance

(whatever it is). Therefore, it make sense to say that 1 mol has the same number or molecules for any substance (universal

constant) known as Avogadro Number NA = 6.022x1023

Implications of the Avogadro Hypothesis

• Gases containing the same number of molecules occupy the same volumes at the same temperature and pressure

• The Law of gases states that : pV=mRT (1) (R depends on the gas).

• If instead of mass of a given gas we express this law in terms of number of molecules (or number of moles), we can generalize the equation:

pV=nR*T (5) R* is also called the Universal Gas Constant = 8.3145JK-1mol-

1

Other important relationships…

• The gas constant for one molecule of any gas is also a universal constant known as Boltzmann’s constant, k

k= R*/NA (7)Hence, for a gas containing no molecules per

unity of volume, the ideal gas equation is:p=nokT (8)

The dry air

• Suppose a mixture of gases (exclude water vapor) exerts a pressure pd and has specific volume αd . The ideal gas equation becomes:

• p dαd = R d T (9)

• Where Rd is the gas constant for 1kg of dry air. The “apparent” molecular weight of the dry air Md is the total mass (in grams) of the constituent gases in dry air divided by the total number o moles of the constituent gases

i i

i

ii

d

M

m

mM

• Md can be also obtained by considering the main constituents of the atmosphere and their proportions:

N2 x 0.78 + O2 * 0.21 + Ar * 0.01=

28.013*0.78+32.000*0.21 + 0.01*39.948= 28.969 or ~28.97 grams

R* is the gas constant for 1 mol of ANY substance, the gas constant for 1 g of dry air is R*/Md, and for 1kg of dry air it is:

Rd=1000 (R*/Md) (11)

= 1000 (8.3145/28.97)=287.0 J K-1kg-1

Water vapor in the atmosphere

• The ideal gas equation may be applied to individual gaseous components. For the water vapor:

• Where e and αv are the pressure and specific volume of water vapor and Rv is the gas constant for 1kg of water vapor.

• Remember Mw water vapor: H2O ~ 18.016g =

18.016 x 10-3 kg• Rv = 1000 R*/Mw (13)

=1000 (8.3145)/18.016 = 461.51 J K-1kg-1

TRe vv

Easy to conclude that…

• If Rv = 1000 R*/Mw and Rd=1000 (R*/Md) Therefore, Rd/Rv = Mw/Md = ε = 0.622

• Because air is a mixture of gases, it Obeys Dalton’s law of partial pressures, which states the total pressure exerted by a mixture of gases that do not interact chemically is equal to the sum of the partial pressure of the gases.

The part of the total atmospheric pressure due towater vapor is referred to as the vapor pressure.

Dalton's Law of Partial Pressures Total pressure of gas mixture is sum of pressures of individual gases

ptot=pO2 + pN2 + pH20v+…

The vapor pressure of a volume of air depends on both the temperature and the density of water vapor molecules.

For the same T, more Wv molecules pH20

Problem 3.1: If at 0oC the density of dry air alone is 1,275kgm-3 and the density of water vapor alone is 4.770x10-3 kgm-3 (for T=273.2K) what is the total pressure exerted by a mixture of the dry air and water vapor at 0oC?

Hints : application of Dalton’s law of the partial pressuresHow to find partial pressures? By applying the gas equation separately for each gas

Dry air:

TRTRpTRp dddd

dddd

1

The density of the dry air ρd = 1,275kgm-3 and Rd=287.0 J K-1kg-1 Substituting in the above equation: p d = 9.997x104 Pa = 999.7 102Pa or 999.7hPa

Partial pressure exerted by the water vapor:

The density of the dry air ρv = 4.770x10-3 kgm-3 and Rv=461.5 J K-1kg-1 Substituting in the above equation: e= 601.4 Pa = 6.014hPa

TRTReTRe vvv

vvv

1

The total pressure exerted by the mixture of dry air and water vapor is =(999.7 + 6.014) hPa or 1006hPa

Water vapor and virtual temperature

• Moist air has a smaller apparent molecular weight than dry air (why??)

Answer: The molecular weight of volume with 1 mole (6.022x 1023 molecules) of dry air is ~ 28g. The same volume filled with 1 mole of water vapor has Molecular weight ~ 18g. One mole of a mixture of water vapor + dry air will have less apparent weight (depending on how much water vapor it contains). This is a very important concept that will be useful to understand the dynamics of hurricanes, dry fronts, formation of storms and many others issues…

• However, rather than use a gas constant for moist air, the exact value of which would depend on the amount of water vapor in the air (which varies considerably), it is convenient to retain the gas constant for dry air (which we know how to measure) and use a ‘FICTITIOUS TEMPERATURE’ called VIRTUAL TEMPERATURE in the ideal gas equation.

How to derive a ‘Virtual Temperature’?

• Consider a volume V of moist air at temperature t and total pressure p that contains a mass md of dry air and mass mv of water vapor. The density of the moist air can be given by:

• Where prime ‘ indicates partial densities.• By looking at this equation you may think that the

total density of the moist air is greater than that of the dry air: this is a big mistake (why???)

vd

vd

V

mm

Applying the ideal gas equation (2) for each

component of the air we obtain: TRe vv TRp dd

From Dalton’s Law of partial pressures:

vd

vd

V

mm

eppepp dd

Substituting in the equation of density:

TR

e

TR

ep

vd

Remember that Rd/Rv = ε => Rv= ε Rd. Therefore, substituting in the above equation

p

e

p

e

TR

p

d

1

)1(1

p

e

TR

p

d

M

M

TR

p

d

M

TRp d

vdTRp

)1(1

p

eT

Tv Now, what does Tv mean??

Note that we are using only Rd in this equation

Interpretation of Virtual Temperature

)1(1

p

eT

TvSince ε is positive and < 1 (=0.622), and e ≤ p, the denominator is positive and <1 and, therefore, Tv > T (if e ≠ 0)

In conclusion, The virtual temperature is the temperature that dry air would need to attain in order to have the same density as the moist air at the same pressure.

Remember that moist air is less dense than dry air at the same temperature and pressure. Therefore, the virtual temperature is always greater than the actual temperature.

However, even for very warm and moist air, the virtual temperature exceeds the actual temperature by only a few degrees

The Hydrostatic Equation Air pressure at any height in the atmosphere is due to the force per unit area exerted against a surface by the weight of the air molecules above that surface. This “weight” results from the pull of gravity That explains why pressure decreases with increasing height above the ground

P2

P1

P1 > P2

However, pressure gradients cause a force that points from regions with high pressure towards regions with low pressure (vertical Gradient Pressure).

That explains why the atmosphere does not collapses on the ground and does not escape to the space

How to derive this balance of forces?

pressure p + δp

Column with unit cross-sectional area (say 1m-2)

Z

δzVertical Gradient Acceleration=Force/mass

Gravity acceleration

Consider “g” the gravity acceleration (m/s2)

Consider that δp is negative ( that is, z increases p decreases)

In a hydrostatic equilibrium, and considering the limit as δz → 0 or:

pressure p

gz

p

1

Hydrostatic Equationgz

p

A few thoughts

• The Hydrostatic equation is one of the most useful relationships in Meteorology. Understanding (and ultimately memorizing) this equation will help you to solve many problems in the future:

gz

p

How to calculate pressure at a given height z?

(17) Hydrostatic Equationgz

p

If the pressure at height z is p(z), from (17) above a fixed point on the Earth:

/1 dpgdz and

)(

)(

p

zp zdzgdp

zdzgzp )(or, because 0)( p

That is, the pressure at height z is equal to the weight of the air in the vertical column of unit cross-sectional area lying above that level. If the mass of the Earth’s atmosphere were distributed uniformly over the globe, retaining the Earth’s topography in its present form, the pressure at sea level would be 1013 hPa, which is referred to as 1 atmosphere (or 1 atm)

Geopotential• The geopotential Φ at any point in the Earth’s atmosphere is defined as the work that must be done against the Earth’s gravitational field to raise a mass of 1 kg from sea level to that point. In other words, Φ is the gravitational potential per unity of mass Z

gdzd Using that dpgdz dpgdzd (20)

zgdzz

0)( (21)

Where Φ(0) at sea level (z=0) has, by convention, been taken as zero. Note that the geopotential at a particular point in the atmosphere depends ONLY ON THE HEIGHT of that point and NOT ON THE PATH

dzThe geopotential Φ(z) at height z is thus given by:

Geopotential Height Z• Defined as

where go is the globally averaged acceleration due to gravity at the Earth’s

surface (~ 9.81ms-2)Geopotential height is used as the vertical coordinate in most atmospheric

applications in which energy plays an important role (e.g., in large-scale atmospheric motions). Values of z (actual height) and Z are almost the same in the lower atmosphere where go ~ g

z

oo

gdzgg

zZ

0

1)( (22)

z(km) Z(km) g(ms-2)0 0 9.811 1.00 9.8010 9.99 9.77100 98.47 9.50

Nevertheless… In meteorological practice it is not convenient to deal with the density of a gas. So the next step is to re-write the geopotential height equation such that we avoid using ρ

How to do that??? First, remember that among the equations that ρ appears is the Hydrostatic Equation and the Gas Law

gz

p and RTp For the mixture of gases

gRT

p

z

p

For the mixture of gases : we need some adjusts to take into account the eventual presence of water vapor. What to do??? Use the concept of virtual temperature, and life will be easier…

g

TR

p

z

p

vd

Rearranging the last expression and using (20)

g

TR

p

z

p

vd

Hydrostatic Equation Combined with Variation in the Geopotential Equation

p

dpTR

p

dpRTdpgdzd vd

p

dpTRd vd

We see here that infinitesimal variations in geopotential between two pressure levels are related to temperature and pressure. To know the Geopotential difference between Z1 and Z2 we need to integrate the above equation (sum all infinitesimal contributions) between two levels since temperature and pressure changes when we consider large variations:

Φ2,Z2, p2

Φ2,Z1, p1

2

1

2

1

p

p vd p

dpTRd

2

112

p

p vd p

dpTR

A few important things to think about:

2

112

p

p vd p

dpTR

Note the negative sign

z

oo

gdzgg

zZ

0

1)(

Definition of Geopotential Height

It means that if we divide the right hand equation by go and get rid of the minus sign we have an equation for the Geopotential Height at p1 and p2! How to get rid of the minus sign?? Invert the limits of your integral and you are done!

1

212

p

p vo

d

p

dpT

g

RZZ

The difference Z2 – Z1 is referred to as the geopotential thickness of the layer between pressure levels p2 and p1 and is dependent on the virtual temperature

So, why is this so important?

1

212

p

p vo

d

p

dpT

g

RZZ

The answer, my friend, is blowing in the wind… Since thickness is dependent on temperature, we can use this concept to observe warm and cold regions and related that to waves that propagating in the atmosphere and how they change wind patterns… We will see more about these aplications soon.

Exploring these concepts further: Scale height and the Hypsometric Equation

Let’s suppose (for simplification) that an isothermal atmosphere( constant temperature with height and negligible corrections of the virtual temperature (which is an approach as moisture changes with height)

1

212

p

p vo

d

p

dpT

g

RZZ

Constant : can move outside the integral

1

212

p

po

vd

p

dp

g

TRZZ

And can be treated as a Constant H

Now, you need to remember (Calculus-I): what is the solution for the integral

dxx

x

x2

1

1 ? Answer: xln Which calculated between p1 and p2 will be

2121 lnlnln pppp (ln is =Natural Logarithm)

Now, going back to the thickness: )/ln( 21

1

212 ppHp

dpHZZ

p

p

Or you may want to find p2 given p1 and the thickness:

H

ZZpp 1212 exp

Tg

RTH

o

3.29 (H is called the Scale Height of the atmosphere)

Exponential decrease of pressure with height

Vertical structure of the atmosphere

• If we assume that the first Z is at sea level and is equal zero, we can determine p as a function of height as:

H

Zpp o

11 exp

Considerations -1• The atmosphere is well mixed ~

below 105km: the pressures and densities of the individual gases decrease with altitude at the same rate and with a scale of height proportional to the gas constant R (and inversely proportional to the apparent molecular weight of the mixture (Rgas = 1000 R*/Mgas )

0 20 40 60 80 100 1200

2

4

6

8

10

12

14

16

This figure shows the behavior of an arbitrary exponential functiony=yo exp (-constant)

Tg

RTH

o

3.29

Consideration-2

• Below 105km the atmosphere is well mixed• If we take a value for Tv of 255K (the

approximate mean value for the troposphere and stratosphere), H~ 7.5km. Let’s examine some implications of these findings…

Exercise 1.2 – (Chapt. 1): At approximately what height above sea level Zm does half the mass of the atmosphere lie above and the other half lie below? [Hint: Assume an exponential pressure dependence with H=7.5 km and neglect the small vertical variations of g with height

Solution: Let Zm be the level that half the mass of the atmosphere lies above and half lies below. The pressure at the Earth’s surface is equal to the weight (per unit area) of the overlying column of air. The same is true of the pressure at any level in the atmosphere. Hence the problem states that, p1=2p2

Unity of area )/ln( 21 ppHZm

)2ln(5.7)/2ln( 22 ppHZm

km2.5

This explains why 5.2km is approximately the height or altitude of the 500mb pressure level

Implications

Density decreases with height in the same manner as pressure

Exercise 1.3 – (Chapt. 1): Assuming an exponential pressure and density dependence with H=7.5km, estimate the heights in the atmosphere at which (a) the air density is equal to 1kg m-3 and (b) the height at which the pressure is equal to 1hPa. Assume that density at sea level is equal 1.25kg m-3 and pressure is equal 1000hPa

)/ln( 21 ppHZ )/ln( 21 HZ

kmkm 7.1

00.1

25.1ln5.7

(a) (b)

kmkm 5200.1

1000ln5.7

Thickness and temperature

• The temperature of the atmosphere generally varies with height and the virtual temperature correction can not always be neglected. How to in this case, we have to find a way to resolve the integral below, since Tv is not always a “constant”

1

212

p

p vo

d

p

dpT

g

RZZ

In this case we have to obtain a radiosonde data and resolve the integral

Tv

Radiosonde data will provide the temperature profile and we can obtain the mean Tv for that layer (Eq. 3.28)

ln(p1)

ln(p2)

2

112 ln

p

p

g

TRZZ

o

vd (29)

Exercise 3.3 – (Chapt. 3): Calculate the geopotential height of the 1000hPa pressure surface when the pressure at sea level is 1014hPa. The scale height of the atmosphere may be taken as 8km

The obvious solution for this problem is to use the Hypsometric equation

)/ln( 2112 ppHZZ

)/ln( 211000 ppHZZ levelseahPa

Considering Z1 =0 (sea level), p2= 1000hPa and p1=1014 hPa

Z1000hPa ~ 111m

Thickness and Heights of Constant Pressure Surfaces

• Because pressure decreases monotonically with height, pressure surfaces (imaginary surfaces on which pressure is constant) never intersect.

• From the hypsometric Equation we learned that the thickness of the layer between any two pressure surfaces p2 and p1 is proportional to the mean virtual temperature Tv

• Therefore, as Tv increases the air between the two pressure levels expand and the layer becomes thicker

Exercise 3.4 – (Chapt. 3): Calculate the thickness of the layer between the 1000 and 500hPa pressure surfaces (a) at the point in the tropics where the mean virtual temperature of the layer is 15oC and (b) at a point in the polar regions where the corresponding mean virtual temperature is -40oC

2

112 ln

p

p

g

TRZZ

o

vd

500

1000ln1000500

o

vdhPahPa g

TRZZ

For Tv=15oC or 288K=> ΔZ=5846m=585dam (decameters)

For Tv=-40oC or 233K=> ΔZ=4730m =473 dam

Geopotential Height 500hPa

Surface pressure:It is clear that the terrain varies considerably in most regions of our planetIt is important to compare surface pressure from place to placeThis is because differences in pressure in a given level are essential for the horizontal movement of the air (wind)For this purpose, we apply equations to know what should be the pressure if a particular location was at sea level.When we look at a map with sea level pressure we can then compare pressure among locations.

P2

P1

P2<P1

Reduction of pressure to sea level

• Let the subscripts g and 0 refer to conditions at the ground and at sea level (Z=0), respectively. Then, for the layer between the Earth’s surface and sea level, the hypsometric equation assumes the form:

g

o

p

pHZg ln

Which can be resolved to obtain the sea-level pressure

vd

goggo TR

Zgp

H

Zgpp expexp

vd

goggo TR

Zgp

H

Zgpp expexp

Considerations

• If Zg is small, the scale height H can be evaluated from the ground temperature. Also, if Zg/H <<1, the exponential in (31) can be approximated by 1 + Zg/H, in which case (31) becomes

(31)

vd

gog

gggo TR

Zgp

H

Zppp

• Because pg is approximately equal to 100hPa and H ~ 8000m, the pressure correction (in hPa) is roughly equal to Zg(m)/8. In other words, for altitudes a few hundred meters above (or below) sea level, pressure decrease by 1hPa for every 8m of vertical ascent

4,322m

1000m

If you are climbing a mountain from the sea level to about 1000m, you will notice that pressure will decrease about 100hPa/km ~ 10hPa/100m (10hPa/328 ft)

Δp/ Δz = 100hPa/km

If we were to climb between 9-10km (a little bit above the Everest), pressure fall would be only 50hPa/km (Notice that a mountain at this height should have much more snow than this one)

Δp/ Δz = 50hPa/km

Δz=1000m