# Atmospheric Thermodynamics Part -I

Mechanics

Oct 27, 2013 (4 years and 6 months ago)

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Atmospheric Thermodynamics
Part
-
I

Leila 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

Water vapor in the air

Stability

The second Law of Thermodynamics : Application of
the concept of Entropy to the atmosphere

N
2
: 78%

Water vapor H
2
O : 0
-
4% (Variable)

Carbon dioxide: CO
2
: ~
0.03%
(Variable)

O
2
:21%

% per volume

Vertical profile of Density and pressure in the
Atmosphere

Determined by the GRAVITY FORCE!

F
G

= Gravitational

force

The ME mass of the Earth = 5.98*10
24

kg

G=gravitational constant=6.67x10
-
11

Nm
2
kg
2

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/m
2
)
;

V=volume

(m
3
)
;

m

=mass

(kg)

;

T=absolute

temperature

(K
-

Kelvin)
.

R

is

a

constant

for

1
kg

of

the

gas

and

DEPENDS

ON

THE

GAS
.

Since

density

ρ

=

m/V,

(
1
)

can

be

rewritten

as
:

p=

ρ
RT (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=

ρ
RT
-
> 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 (H
2
O) ~ 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

=
6.022x10
23

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
-
1
mol
-
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*/N
A

(7)

Hence, for a gas containing n
o

molecules per
unity of volume, the ideal gas equation is:

p=
n
o
kT

(8)

The dry air

Suppose a mixture of gases (exclude water vapor) exerts a
pressure p
d

and has specific volume
α
d
. The ideal gas
equation becomes:

p

d
α
d

= R

d
T

(9)

Where

R
d

is

the

gas

constant

for

1
kg

of

dry

air
.

The

“apparent”

molecular

weight

of

the

dry

air

M
d

is

the

total

mass

(in

grams)

of

the

constituent

gases

in

dry

air

divided

by

the

total

number

o

moles

of

the

constituent

gases

M
d

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

N
2

x 0.78 + O
2

* 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:

R
d
=1000 (R*/M
d
) (11)

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

Water vapor in the atmosphere

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

Where
e

and
α
v

are the pressure and specific volume
of water vapor and R
v

is the gas constant for 1kg of
water vapor.

Remember M
w

water vapor: H
2
O ~ 18.016g =

18.016 x 10
-
3

kg

R
v

= 1000 R*/M
w
(13)

=1000 (8.3145)/18.016 = 461.51

J K
-
1
kg
-
1

Easy to conclude that…

If R
v

= 1000 R*/M
w
and R
d
=1000 (R*/M
d
)

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 to

water 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

p
tot
=p
O2

+ p
N2

+ p
H20v
+…

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
p
H20

Problem 3.1: If at 0
o
C 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 0
o
C?

Hints : application of Dalton’s law of the partial pressures

How to find partial pressures? By applying the gas equation separately for each gas

Dry air:

The density of the dry air
ρ
d

=
1,275kgm
-
3
and R
d
=287.0 J K
-
1
kg
-
1

Substituting in the above equation: p

d

= 9.997x10
4

Pa = 999.7 10
2
Pa or 999.7hPa

Partial pressure exerted by the water vapor:

The density of the dry air
ρ
v

=
4.770x10
-
3

kgm
-
3
and R
v
=461.5 J K
-
1
kg
-
1

Substituting in the above equation:
e
= 601.4 Pa = 6.014hPa

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 10
23

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 m
d

of dry
air and mass m
v

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???)

Applying the ideal gas equation (2) for each component
of the air we obtain
:

From Dalton’s Law of partial pressures:

Substituting in the equation of density:

Remember that Rd/Rv =
ε

=>
Rv=

ε

Rd. Therefore, substituting in the above equation

M

Now, what does
Tv

mean??

Note that we are using
only Rd in this equation

Interpretation of Virtual Temperature

Since
ε

is positive and < 1 (=0.622), and
e

≤ p, the denominator is
positive and <1 and, therefore,
T
v

> 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

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 +
δ
p

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

Z

δ
z

Gravity acceleration

Consider “g” the gravity acceleration

(m/s
2
)

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

In a hydrostatic equilibrium, and
considering the limit as
δ
z → 0

or:

pressure p

Hydrostatic Equation

A few thoughts

The Hydrostatic equation is one of the most
useful relationships in Meteorology.
Understanding (and ultimately memorizing)
problems in the future:

How to calculate pressure at a given height z?

(17) Hydrostatic Equation

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

and

or, because

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
–Šƒ–?•—•–?„‡?†‘•‡?ƒ‰ƒ‹••–?–Š‡??ƒ”–Š?ï•?
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

Using that

(20)

(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

dz

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

Geopotential Height Z

Defined as

where g
o

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 g
o

~ g

(22)

z(km)

Z(km)

g(ms
-
2
)

0

0

9.81

1

1.00

9.80

10

9.99

9.77

100

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

and

For the mixture of gases

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…

Rearranging the last expression and using (20)

Hydrostatic Equation

Combined with Variation in the Geopotential Equation

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:

Φ
2
,
Z
2
, p
2

Φ
2,
Z
1
, p
1

A few important things to think about:

Note the negative sign

Definition of Geopotential Height

It means that if we divide the right hand equation by g
o

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!

The difference Z
2

Z
1

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?

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
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)

Constant : can move outside the integral

And can be treated as a
Constant
H

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

?

Which calculated between p1 and
p2 will be

(
ln

is =Natural Logarithm
)

Now, going back to the thickness:

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

(
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:

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
(
R
gas

= 1000 R*/
M
gas

)

0
2
4
6
8
10
12
14
16
0
20
40
60
80
100
120
This figure shows the behavior of an
arbitrary exponential function

y=
yo

exp (
-
constant)

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
Z
m

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

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

(a)

(b)

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”

In this case we have
to obtain a

data
and resolve the
integral

Tv

data will
provide the temperature
profile and we can obtain
the mean
Tv

for that
layer (Eq. 3.28)

ln
(p1)

ln
(p2)

(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

Considering Z
1

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

Z
1000hPa

~ 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 15
o
C and (b) at a point in the
polar regions where the corresponding mean virtual temperature is
-
40
o
C

For
Tv
=15
o
C or
288K=>
Δ
Z=5846m

=585dam
(decameters)

For
Tv
=
-
40
o
C or
233K=>
Δ
Z=4730m
=473 dam

Geopotential Height 500hPa

Surface pressure:

It is clear that the terrain varies
considerably in most regions of our
planet

It is important to compare surface
pressure from place to place

This 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:

Which can be resolved to obtain the sea
-
level pressure

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)

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
100hPa/km ~ 10hPa/100m
(
10hPa/328
ft)

Δ
p/

Δ
z =
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)

Δ
p/

Δ
z =
50hPa/km

Δ
z=1000m