Download Chapter 3: Atmospheric pressure and temperature

Chapter 3: Atmospheric pressure and temperature
3.1 Distribution of pressure with altitude
The barometric law
Atmospheric pressure declines with altitude, a fact familiar to everyone who has flown in
an airplane and felt pressure changes in their ears, or climbed a mountain and struggled to
breathe. The decrease of pressure with altitude is a simple consequence of the
relationship between weight of overlying fluid and pressure: each increment in altitude
leaves below a slab of atmosphere with a certain weight due to its mass and the force of
gravity. As we go up, pressure must decline by exactly the weight of the slab with unit
area, we leave behind. An equivalent relationship holds descending into the ocean, apart
from a change in the convention of measuring depth vs. altitude (see Figure 3.1).
Figure 3.1. Changes in pressure with
altitude in the atmosphere (left) and
depth in the ocean (right). Pressure
always increases as the observer moves
downward because the weight of the
overlying column of fluid (air or water)
increases. Altitude is conventionally
measured increasing upwards from the
surface of the earth, and depth
increasing downwards. Therefore
pressure decreases with increasing
altitude in the atmosphere and pressure
increases with increasing depth in the
ocean.
Air and water are both fluids, but they differ with respect to compressibility. We know
about compressibility from everyday experience. For example, if we take a bicycle pump
and seal the outlet, the plunger can be pressed in, decreasing the volume of the air trapped
inside. The trapped air exerts a spring-like force on the handle, the smaller the volume
into which the air is squeezed, the stronger the force. This is Boyle's Law. If we place
water in the pump, we cannot significantly compress the volume.
This simple thought experiment illustrates that air is compressible, density depends on
pressure, following the Perfect Gas Law. Water is almost incompressible. This
difference makes the change in pressure with height very different between the
atmosphere and the ocean. The relationship between density, pressure and altitude is
illustrated in Figures 3.2 and 3.3. In Figure 3.2, the weight of the slab of fluid between
Z1 and Z2 is given by the density, ρ, multiplied by (volume of the slab) ×g = ρ×(area x
height) ×g. If we let the area be 1 m2, the weight is therefore ρ g × (Z2 −Z1). If the
atmosphere is not being accelerated, there must be a difference in pressure (P2 - P1)
across the slab that balances the force of gravity. Therefore,
− (P2 −P1) = weight = ρ g ×(Z2 −Z1).
(3.1)
(Note the minus sign, pressure is lower at P2.) We can insert the Perfect Gas Law into
Eq. 3.1 (Chapter 2, use the form P = ρR' T) to calculate ρ,
ρ = Pav/(R'T),
(3.2)
where Pav = (P1+P2)/2. If we put together (3.1) and (3.2) and use the value for the
constant R' (=k/m) from Chapter 2, we obtain the barometric law,
∆P = -Pav mg / (kT) ∆Z.
(3.3a)
Here the notation ∆P represents P2−P1, ∆Z is Z2−Z1, and P is the pressure in the middle
of the slab (approximating Pav).
P2
P2
↓
↓
Z2============================ Z2
↑
↑
ρg (Z2 − Z1) = P1 − P2
weight of slab = pressure difference
P1
P1
↓
↓
Z1============================ Z1
↑
↑
P1
P1
Figure 3.2. The balance of pressure
forces across a slab of fluid. The
pressure P1 is greater P2 by −∆P,
corresponding to the weight of the slab
over unit area. This weight is given by
the mass (mass density (ρ) times the
volume, 1m2×∆Z×g.
Because air is compressible the change in pressure across the slab (Eq. 3.3a) is
proportional to both the pressure itself and to the thickness of the slab. Therefore, for a
given altitude change, the atmospheric pressure changes by a fixed percentage that
depends on temperature but not on altitude or pressure.
The scale height, H, and the simple form of the barometric law
We make reference to Eq. 3.3 to define the scale height H ≡ kT/(mg). H represents a
key length in the atmosphere, as it is the only parameter in the barometric law,
∆P = -P ∆Z/H.
(3.3b)
The value of the scale height, H, at room temperature (298 K) is approximately 8.7 km.
Temperatures decline with altitude (as discussed later), and an approximate mean scale
height is about 7 km (corresponding to 240 K) for the whole atmosphere. If we had an
atmosphere where the temperature did not change with altitude, the barometric law would
have a very simple form in terms of the exponential function exp(), which appears on
most hand calculators, exp(x)≡ ex, where e=2.718282⋅⋅⋅⋅,
P(Z) = Po e-Z/H.
(3.4)
Here Po is the pressure at the ground (1000 mb, see Chapter 2). You may readily check
that the exponential function in (3.4) has the property defined in the barometric law, i.e.
that the pressure declines by a fixed fraction or percentage for a fixed increase in altitude
(14% per km in the example). Although temperatures do vary with altitude, the changes
are modest (H varies from about 6 to 8.7 km), thus Eq. 3.4 is useful as an approximation
to the barometric law giving variation of pressure with altitude.
15.0
0
12.5
1
Altitude (km)
Depth (km)
10.0
2
7.5
5.0
3
2.5
4
0.0
0
100
200
300
Pressure (bar)
oceans
400
0
0.25
0.5
0.75
Pressure (bar)
atmosphere
Figure 3.3 The distribution of pressure with depth in the oceans (left) and with altitude in the
atmosphere (right, scale height H = 7km). Pressure increases linearly with depth in the ocean because
water is incompressible, and therefore density (ρ) is constant. Pressure decreases exponentially with
altitude in the atmosphere because density not constant, but it proportional to pressure according to the
Perfect Gas Law (Chapter 2).
Eq. (3.4), with H=7 km, is illustrated in Figure 3.3 (right panel), which shows the
dependence of pressure on altitude. We have placed pressure on the x-axis and altitude
on the y-axis, a convention often used in atmospheric science because it illustrates
altitude as the vertical dimension.
3.2 Pressure vs depth in the ocean
The weight of a slab of ocean with unit area (1 m2) is the [mass of the slab]×g= ρ g (D1 D2)⋅which gives the pressure difference between the top and bottom of the slab,
1
∆P = ρ g ∆D.
(3.5)
Since ρ is essentially constant for water (water is incompressible), the change in pressure
across the slab is proportional to the thickness of the slab but not proportional to pressure
itself (contrast to atmosphere). The pressure changes by the same increment for a given
depth change, and pressure increases linearly, not exponentially, with depth in the ocean,
P = Po + ρw g D,
(3.6)
where ρw is the mean mass density of seawater (density of seawater changes slightly with
salinity and temperature) and Po is the pressure at the surface (1 atm). Since the mass
density of liquid water is about 1000 times greater than the density of air, the pressure
becomes very large in the deep ocean is quite large (see Figure 3.3).
3.3 Buoyancy
Buoyancy is the tendency for less dense fluids to be forced upwards by more dense fluids
under the influence of gravity. Buoyancy is extremely significant as a driving force for
motions in the atmosphere and oceans, and hence we will examine the concept very
carefully here.
The mass density of air ρ is given by mn, where m is the mean mass of an air molecule
(4.81×10-26 kg molecule-1 for dry air), and n is the number density of air (n =2.69×1025
molecules m-3 at T=0o C, or 273.15 K). Therefore the density of dry air at 0 C is ρ = 1.29
kg m-3. If we raise the temperature to 10° C (285.15 K), the density is about 4% less, or
1.24 kg m-3. This seemingly small difference in density would cause air to move in the
atmosphere, i.e. to cause winds.
Example 1a: Buoyant block in a tank of water
To understand buoyancy, let's first consider the familiar example of a solid object with
density ρb held below the surface in a tank of water with density ρw (see Figure 3.4a).
Figure 3.4a. Buoyancy forece Forces on a
solid body immersed in a tank of water. The
solid is assumed less dense than water and to
area A (m2 ) on all sides. P1 is the fluid
pressure at level 1, and P1x is the downward
pressure exerted by the weight of overlying
atmosphere, plus fluid between the top of
the tank and level 2, plus the object. The
buoyancy force is P1 – P1x (up ↑) per unit
area of the submerged block.
The pressure at depth D1in the water is the weight of the overlying atmosphere plus the
weight of a column of water, per unit area. At depth D1, the pressure P1 is uniform
everywhere at that depth in the water, and the force pushing up on the block is the
pressure×area of the block = ρw ×g× A×D1, since pressure P1 = ρw ×g ×D1.
P1 A = A 105 + ρw g D1 A
(3.7a)
where 105 is the pressure of the overlying atmosphere. However, the pressure P1x in the
downward direction, and thus the force on the block, is different that the upward force at
the base of the object. We already have the tools to calculate the magnitude of this
difference by computing the weight of (fluid + object):
P1x = 105 + ρw g D2 + ρb g (D1-D2),
(3.7b)
where the height of the block is D1 − D2. Thus there is a net force on the object, the
difference in pressure× A, that will accelerate the object, from Eq. 3.7
Fnet =A×(P1x − P1) = (ρb − ρw) g A (D1-D2) = (ρb − ρw) g Vblock.
(3.8)
The difference in pressure at the bottom of the object creates an upward (negative)
buoyant force given by Equation 3.8, if ρw > ρb, as in the familiar case of wood (ρb =
700 kg m-3) and water (ρw = 1000 kg m-3). In Eq. 3.8, we identify (ρb g Vblock) as the
weight of the block, and ρw g Vblock is the buoyancy force, an upward force on the object
that we see is equal to the weight of the water displaced by the block.
Eq. 3.8 is therefore a form of Archimedes' principle: the buoyancy force is equal to the
weight of fluid displaced by the object. Note that the total pressure of overlying water
and atmosphere above level 2 is irrelevant, since only pressure differences lead to
unbalanced forces, and thus to acceleration of fluids or objects.
Example 1b: Floating block in a tank of water, add mineral oil
Figure 3.4b shows the behavior of a block of Teflon immersed in water and in cooking
oil, and addresses the question as to what will happen if the block is first placed into a
bucket of water, then oil place on top.
?
?
A
B
C
Fig. 3.4b. Buoyancy experiment: A block of Teflon has slightly lower density than water, and therefore
it floats [A] in water. Teflon is more dense than oil, and therefore it sinks [B] in oil. What will happen if
we have the block floating in water, and then add oil on top [C] (will it move up, down, or stay put)?
In panel A, about 1/8 of the volume of the block is surrounded by air, and 7/8 by water.
The block is floating, so the net force on it is zero (no acceleration, it stays put),
F = 0 = (ρb - ρair) f Vb g + (ρb - ρw) (1 –f ) Vb g
(ρB=880; ρAIR =1.3; ρW 1000; ρOIL=830 kg/m3)
3.8'
where f =1/8. The weight of the block is ρb Vb g, the sum of the terms in f and (1 – f ),
and buoyancy force, which just balances the weight, is the sum of the other two terms,
(-1) (ρair f + ρw ( 1 – f ) ).
The density of air is about 800 times smaller than the density of water, so almost all of
the buoyancy comes from the water.
When oil is added it forms a separate layer that floats on the surface of the water,
surrounding the top of the block. The magnitude of the buoyancy force on the upper part
of the block, initially ρair f, increases to ρoil f, since the density of oil is about 700 times
that of air. There is a larger upward force and the block therefore rises into the oil.
Eventually it reaches a new position where only a small bit is supported by water, and
most of the block is immersed in the oil, and none is in the air.
Example 2. Buoyancy due to density variations in a fluid
The next example illustrates how buoyancy forces tend to push less dense fluids (e.g.
warm air) upwards in the presence of more dense fluids (Fig. 3.5).
Figure 3.5. Tank with connected reservoirs, filled with dense fluid (light blue, e.g. cold air, density ρ
d) and less-dense fluid (dark red, e.g. warm air, density ρl)). We assume that the fluids cannot mix. If
fluid heights are equal (left panel, height = h), the pressure at the bottom of the left reservoir (ρ
ρdh) will be
higher than the right reservoir (ρ
ρlh) because ρd > ρl. The denser fluid will flow under the less-dense fluid
(right panel), until the pressures are equalized. The tendency for the less-dense fluid to be displaced
upward by the more-dense fluid is a consequence of gravity acting on density difference to produce
pressure gradients and buoyancy.
As discussed above, the pressure in each tank of a pair is equal to the weight of the
column of liquid per unit area, i.e. at the bottom of the light blue tank, Pblue = ρblue ×
(height of liquid), and in the dark red tank, Pred = ρred x (height of liquid). Since ρblue>ρ
red, it follows that Pblue > Pred. This puts a higher pressure on the left side of the tube
connecting the two columns, and the liquid will flow from left to right until the pressure
is the same on each side of the pipe connecting the tanks.
The less dense liquid is buoyant relative to the blue liquid; the difference in density
causes the less dense fluid to float on top of the more dense liquid, or or to rise if it is
inserted into the denser liquid.
Example 3. Buoyancy: warm and cold air
The Perfect Gas Law tells us that, if a gas is kept at a constant pressure, increasing the
temperature will cause it to expand, reducing its number density and its mass density
(Charles' Law). The liquids in the previous example could be cold air (more dense, light
blue) and warm air (less dense, dark red). Therefore, if an air parcel is heated it will
become less dense than the surrounding air and it will rise, as illustrated in Fig. 3.6.
BUOYANCY AND MASS DENSITY
Hot and cold gas, both at P = 1
-->>
add heat
=> molecules move faster
hit the piston harder
PRESSURE STAYS FIXED
the number density is lower for higher T
the mass density is lower for higher T
{n, P, T}
⇒
{n/2,P,Tx2}
-->>
the dense (cold) air
displaces the less dense (warmer)
air under the force of gravity
cold
warm
Figure 3.6. (upper) If two air
parcels are at the same pressure, the
warmer parcel is less dense than the
colder air, following the Perfect Gas
Law: P=ρ1R'T1=ρ2R'T2 the
product ρT=constant. (lower) If a
parcel of warm air is surrounded by a
cooler volume of air, the pressures
are the same in both parcels, thus the
density of the warmer parcel is lower
than the surroundings and it
experiences a buoyancy force just as
the fluids in Figure 3.5.
warm air rises
above the cold air
We now understand the real meaning of the well-known idea that “warm air rises”. Air
rises if it is warmer than surrounding air at the same pressure, because it is less dense,
and therefore buoyant. (Hence the phrase is not accurate: it's temperature differences, not
"warmth", that counts.) Curiously, the force that pushes buoyant air parcels upwards is
the force of gravity, acting in a fluid as illustrated in Figures 3.5 and 3.6.
Buoyancy is responsible for the over-turning of the troposphere, as we discuss in the next
chapter. The sun heats the earth’s surface and air near the surface is warmed, becoming
buoyant. As they rise, they lose their heat to the surrounding air or by radiating heat to
space, and then eventually sink. The buoyant rising motion is called convection and the
over-turning moves heat from the equator to higher latitudes, representing the driving
force for climate.
3.4 Water vapor in the atmosphere
It takes energy to evaporate liquid water, and thus when water vapor condenses, energy is
released, warming the air. Thus wet air carries latent energy, the potential for
condensation of water vapor to heat the air, a key factor in storms and rain. In addition,
water molecules have a lower mass than air molecules, and wet air at a given pressure has
lower density than dry air. If water vapor is added to dry air, buoyancy will be created,
causing wet air at the same temperature and pressure to be lifted over dry air. In the
following section we study evaporation and condensation of water vapor in air, the effect
of water vapor on atmospheric density, and the energy content associated with
condensation.
An air parcel is a useful hypothetical construct, defined as an unconfined volume of air
that may be followed around as a unit for some period of time. It must therefore be small
enough that it has a single temperature, pressure, and composition, but large enough that
it can be treated as a fluid rather than individual molecules. This concept is used to aid
the discussion in this and following chapters.
Vapor pressure of water and the Clausius-Clapeyron Equation
If we place liquid water or ice in a closed vessel with dry air and maintain a constant
temperature, we will find that, over time, the amount of water vapor in the air becomes
steady. This amount depends only on the temperature of the liquid or solid. Chemists
summarize the results of this experiment by plotting the partial pressure Ps of water
vapor as a function of absolute temperature T of the condensed phase. Partial pressure is
defined as the pressure due to water molecules hitting the walls of the vessel, a concept
readily visualized in terms of the ping-pong-ball atmosphere (paint the water “molecules”
a distinctive color).
The vapor pressure (partial pressure) of water over liquid and solid water increases
steeply with temperature (Fig. 3.7). This relationship (the Clausius-Clapeyron Equation,
named for the scientists who measured it in the 19th century), has the approximate form
(Eq. 3.9)
Ps = A exp(B(1/273.15 - 1/T),
with A= 6.11 mbar (the vapor pressure at 0°C) and B=5308 (°K).
Dew point, frost point, and relative humidity
The Clausius-Clapeyron Equation (3.9)(Figure 3.7) describes the maximum content of
H2O in air at a given temperature. An air parcel can have less if the air is dry, but no
more: if we try to add more than this amount of water, liquid or solid particles will
condense, removing the excess water until the partial pressure is given by Eq. 3.9. For
example, suppose we have an air parcel at 16°C (289.15°K) with 10.2 mbar of water
vapor. If I inject a drop of liquid water into this air, the vapor pressure over the liquid is
higher than in the air, and the droplet will evaporate, moistening the parcel. If we cool
that air to 7.48°C and try again, the vapor pressure over the drop is equal to that in the air,
and the drop will remain as it was. The liquid and vapor are said to be in equilibrium, or
equivalently, the air at 7.48°C with 10.2 mbar of water vapor is said to be saturated with
water vapor (see Figure 3.8). If we cool this air parcel further, water will condense onto
the droplet. There are always particles around to help water to condense, so we never
observe significant excess vapor pressure over the saturation vapor pressure.
T (Celcius)
T (Celcius)
-10
60
70
80
90
1000
900
800
10
20
30
40
0.06
0.05
0.04
0.03
0.02
0.01
700
0.0
260
270
280
290
300
310
T (Kelvin)
600
T (Celcius)
500
200
100
-6
-4
-2
0
2
5
liquid
3
4
ice
2
300
-8
1
400
-10
6
-12
Vapor pressure of water (mbar)
Vapor pressure of water (mbar)
0
100
260
265
270
275
0
T (Kelvin)
270
280
290
300
310
320
330
340
350
360
T (Kelvin)
Figure 3.7a. Vapor pressure of water (mbar) over
liquid, plotted as a function of temperature (K).
370
Figure 3.7b (upper). Vapor pressure of water on an
expanded scale. The scale on the right shows the
fraction of air molecules that are H2O at 1 atm.
pressure for air in equilibrium with water at the
given temperature. (lower). Vapor pressures of water
(——) and ice (---) at temperatures below 0°C
(273.15°K).
Figures 3.7b and 3.8b shows what happens when liquid water is cooled below 0°C: The
vapor pressure over ice becomes lower than that over liquid water. This makes ice more
stable than liquid, as may be visualized in the following thought experiment. Imagine ice
and water both present in a closed vessel at a temperature below 0°. The liquid will tend
to saturate the air with vapor, but the corresponding partial pressure of water is higher
than the saturation pressure over ice (see Fig. 3.7b, lower panel), so the excess vapor will
condense on the ice. The process will continue until all the liquid has disappeared and
only ice remains.
fraction, mole/mole
50
60
40
40
30
20
20
0
10
Vapor pressure of water (mbar)
0
10
5
Saturated Vapor Pressure (mb)
15
Air
Parcel
10
Saturated Vapor Pressure (mb)
15
Relative Humidity and Frost Point
20
Relative Humidity and Dew Point
Dew Point
7.48 C
Air
Parcel
5
0
Frost Point
-8.3 C
0
5
10
15
Temperature (C)
RH = 10.2 mb / 17.9 mb x 100% = 57%
Figure 3.8a Diagram illustrating condensation of
water vapor from an air parcel initially at 16°C with
10.2 mb water. Liquid will not condense from this
parcel, because the vapor pressure over liquid water
at 16°C (18 mb, see diagram) is higher than in the
air. As we cool the air (follow horizontal dotted
line), condensation will start at 7.5°C, where the
vapor pressure over liquid water equals that in the
air. If the parcel is cooled further, condensation will
occur and the partial pressure of water will fall
following the Clausius-Clapeyron equation (solid
line).
20
-15
-10
-5
0
5
10
15
Temperature (C)
RH = 3 mb / 12.1 mb x 100% = 25%
Figure 3.8b Same as for Figure 3.8a, except the
parcel is initially at 10°C and contains 3 mb of H2O
vapor. Condensation starts at -8.2°C, and ice may
form rather than liquid water.
Since the water vapor content of air is a key property for weather and climate, it is often
reported as part of weather forecasts. However, most people are unfamiliar with partial
pressures, and therefore alternative measures are used, as illustrated in Fig. 3.8:
• partial pressure of water, also called absolute humidity Pw, is the pressure in
Newtons m-2 or mbar due to water molecules only, i.e. equal to nwaterkT.
• relative humidity is the ratio of actual water vapor partial pressure Pw to the
saturation pressure Ps. The air parcel in our first example had relative humidity of
57%.
• specific humidity is the number of grams of water vapor per kilogram of dry air.
• dew point or frost point refer to the temperature to which an air parcel must be cooled
to first make condensation occur.
-->>
add water keep P & T FIXED
(P = n kT, both sides)
n in dry piston (unit vol) = n in wet unit vol
*BUT <
mass air molecule > mass water molecule
mass in dry piston > mass in wet piston
=>>
mass density dry air > mass density wet air
******************************************
dry air {n,P,T}
Fig 3.9. Density of wet and dry
air at fixed pressure P and
temperature T. Diagram
showing why dry air has higher
mass density than wet air when
both parcels at the same
temperature and pressure.
Molecular weights:
air=29, water=18 (grams/mole).
wet air {n,P,T}
Adding water vapor to air reduces its density. We know from the Perfect Gas Law, P =
nkT, that pressure depends on the number density n (number of molecules per cubic
meter) and temperature, but not on the mass of the molecules. (Don’t be confused by the
form of the Perfect Gas Law P=ρR’T, the dependence of ρ and R’ on molecular mass
cancel out!) Therefore if we have two air parcels at the same P and T, n must also be the
same. However the average molecular mass, and hence the density ρ, is lower in the wet
parcel (see Fig. 3.9). Thus addition of moisture to dry air may create buoyancy.
Discussion.
The rapid rise of Pw with increasing temperature is very important for climate. Powerful
storms occur in the tropics, and at midlatitudes in summer, such as huge thunderstorms or
hurricanes; these storms draw their energy from release of latent heat. The ClausiusClapeyron equation tells us that warm air can contain much more water vapor than cooler
air, and thus potentially much more latent heat can be released to produce buoyancy,
convection, and rain or snow. Thus convective storms involving warm, moist air are
often more intense than storms that develop in cooler regions. Water vapor is also an
important contributor to warming of the surface of the earth through the greenhouse
effect, the absorption and re-emission of radiant heat by the atmosphere. Through the
Clausius-Clapeyron equation, warming the atmosphere allows higher amounts of water
vapor, increasing the efficiency of the greenhouse effect. This phenomenon will be
discussed in detail in a later chapter.
3.5: Demonstrations for this chapter
These laboratory demonstrations illustrate phenomena discussed in this chapter.
1)
Mercury barometer: The height of mercury in the column is supported by air
pressure pushing down on the surface of the fluid reservoir. Atmospheric
pressure changes produce changes in the height of the column. We use mercury
instead of water because the density of mercury is 13.6 × the density of water:
Height of a mercury barometer = 0.76m (760 mm Hg).
Height of a water barometer = 10 m (33 feet!) Difficult to use.
2)
3)
Buoyancy of fluids and density. Colored water vs. paint thinner: The two fluids
are placed in a glass U-tube. The paint thinner is less dense than the water, and
the column of the less dense fluid is push upwards as in Figure 3.5. The pressure
at the bottom of the tube with water (neglecting the pressure of the atmosphere,
which is the same for both tubes) is Pwater = ρwater × gh, where h = height of the
water column. Initially h was the same in the tube with paint thinner, and since ρ
thinner < ρwater, Pthinner = ρthinner × gh was less than Pwater. After the valve is opened
the less dense liquid rises higher in the tube, equalizing the pressure in both tubes.
This demo shows that the driving force for buoyancy, and for stratification (light
fluid floating on top of denser fluid) is the difference in mass density (ρthinner <ρ
water) under the influence of gravity.
Latent heat of vaporization. We showed that water can be made to boil at room
temperature by lowering the pressure to the point at which atmospheric pressure
over the water is less than its vapor pressure. As soon as the water begins to boil,
its temperature begins to drop--the latent heat of vaporization is supplied by
cooling the water itself (and by heat transferred from air around the flask). We
can actually make the water freeze if we have a good vacuum line, a phenomenon
exploited in freezing of foods and proteins.
3.6 Summary of main points of this chapter
Barometric law
• The pressure in the atmosphere declines with height according to the barometric
law, which is given approximately by P(Z) = Po e(-Z/H) where Po is the pressure at the
ground, Z is altitude, H is the scale height [H ≡ kT/(mg) ∼ 7 km on average n the
atmosphere, where the mean T ~ 250 K]. Atmospheric pressure declines by factor 1/e
≈ 1/2.7 for each scale height or a factor of about 10 for two scale heights.
• The exact form of the barometric law is ∆ P/ P = - ∆ Z/ H, where ∆ P is the change
in pressure that occurs for vertical displacement ∆Z.
• The barometric law results from hydrostatic balance combined with the Perfect Gas
Law: the pressure of the atmosphere at each altitude supports the weight of the
overlying atmosphere.
Buoyancy
• Buoyancy is the force of gravity acting on fluids with different density (commonly
due to a difference in temperatures); buoyancy may drive atmospheric motions such
as convection("warm air rises", more accurately, air warmer than its surroundings
becomes buoyant).
Water vapor
• The vapor pressure of water increases steeply (exponentially) as temperature
increases.
• The latent heat of vaporization of water is large (due to hydrogen bonding between
water molecules, to be discussed in a later chapter). Transport of water vapor in the
earth's atmosphere by winds provide an efficient way to move energy from one
location to another.
• We defined specific humidity (grams of water vapor per gram of air) and relative
humidity (fraction or % of saturation vapor pressure) and showed how to calculate
these quantities using information about the thermodynamic properties of water, and
atmospheric temperature and pressure.
• Here are some useful numbers about water:
molecular weight of water = 18 g/mole (compared to 29 gm/mole = mean molecular
weight of dry air).
ρw, density of liquid water = 1 g/cm3 = 1000 kg/m3 (compare to air, ρair = 1.3 kg/m3).
1 kg = mass of 1 liter (1 liter = 1000 cm3 = 10-3 m3 ) of water.
latent heat of vaporization = 2.52 106 J/kg to evaporate water.
latent heat of sublimation of ice = 2.86 106 J/kg (to evaporate ice); latent heat of
freezing = 0.34 106 J/kg [=(2.86 - 2.52) 106]
specific heat of water = 4.2 J/g (or 4.2 103/kg), energy required to raise the temperature
of 1 g of liquid water (or 1 kg) by 1° C. The definition of the traditional measure of heat,
the calorie, is the heat required to raise the temperature of 1 cm3 of liquid water by 1°C
thus 4.2 J ≅ 1 calorie.
3.7 Exercises
Problem 1: Compute the temperature rise from condensation of a particular amount of
water vapor. Compare to solar input.
Problem 2: Compute the density difference between wet and dry air at 305K. Show the
effects of this small density difference in the “dry line” in New Mexico in summer.
Problem 3: (a)Find the heat required to raise the temperature of 1 mole of water from 0
to 100° C (the boiling point). (b)By what factor is the latent heat of vaporization larger
than the amount needed to raise the temperature by 1° C?
Problem 4: Hurricane Mitch dropped .7 m of rain over a large area in 2 days. Compute
the amount of latent heat release from the storm and compare to total US energy use for a
year.
Chapter 3: Atmospheric pressure and temperature..............1
3.1 Distribution of pressure with altitude ................................................................. 1
The barometric law ................................................................................................... 1
The scale height, H, and the simple form of the barometric law ................................ 2
3.2 Pressure vs depth in the ocean............................................................................. 3
3.3 Buoyancy .............................................................................................................. 4
Example 1: Buoyant block in a tank of water ......................................................... 4
Example 2. Buoyancy due to density variations in a fluid...................................... 6
Example 3. Buoyancy: warm and cold air.............................................................. 7
3.4 Water vapor in the atmosphere ........................................................................... 8
Vapor pressure of water and the Clausius-Clapeyron Equation.................................. 8
Dew point, frost point, and relative humidity ............................................................ 8
3.5: Demonstrations for this chapter...................................................................... 11
3.6 Summary of main points of this chapter .......................................................... 12
Barometric law ....................................................................................................... 12
Buoyancy................................................................................................................ 12
Water vapor ............................................................................................................ 13