Environmental Physics for Freshman Geography Students
Professor David Faiman. Lecture 12. v.1.5 (January 18, 2007)
Atmospheric phenomena
1. Change of atmospheric pressure with height
We first examine why it is that atmospheric pressure decreases with altitude in the earth’s
atmosphere. In order to see that this must be so, imagine a thin horizontal slab of air with
thickness z and area A = xy. By "thin" I simply mean that z is very much smaller than x or y.
Suppose that the air beneath the slab has a pressure Pup and that the air above it has a
pressure Pdown. Since pressure = force x area, the air beneath the slab pushes up on the slab
with a force Fup = Pupxy and the air above it pushes down on the slab with a force Fdown =
Pdownxy. Now there is one more force at work, namely, the weight of the air in the slab W =
xyzg, where xyz is the volume of the slab,  is the density of the enclosed air and g is the
acceleration due to gravity. These three forces are shown in Fig. 1.
Fdo wn = xyP do wn
x
y
z
W = xyz g
Fup = xyP up
Figure 1: Three forces act on an imaginary slab of air in order to keep it in position:
Its weight, together with air pressure from above, must balance the force of air pressure from below
Since the slab does not go anywhere, these forces must exactly balance one another. I.e.
Fup = Fdown + W
(12.1)
Inserting the values of these forces as shown in the diagram, we see that:
Pup = Pdown +  g z
(12.2)
This implies that Pup is greater than Pdown, i.e., atmospheric pressure must decrease as we go
up in altitude.
If the slab thickness z is very small, eq.(12.2) may be re-written as the so-called equation of
atmospheres:
P' = -  g
(12.3)
64
where P' represents the rate of change of pressure with height. That is to say, (Pdown - Pup) is
the change in pressure as we pass from below the slab to above the slab, and this difference
divided by the thickness of the slab is the rate at which the pressure changes with height as
we pass upward through the slab by an amount z. The right hand side of eq. (12.3) tells us
that this rate is negative, i.e., the pressure decreases and that the rate of decrease is precisely
equal to the air density multiplied by the acceleration of gravity.
Thus far, the equation of atmospheres is exact in that no approximations have been made.
However, in order for the equation to be useful, i.e. "soluble" we must assume a relationship
between pressure and density. But we know of such a relationship for the case of an ideal gas,
i.e.
PV = NkT
(12.4)
where P = pressure (measured in N m-2 = Pascals), V = volume (measured in m3), T =
temperature (measured in K), k = Boltzmann's constant = 1.38 x 10-23 J K-1, and N = the
number of molecules in our volume of interest.
Now the total mass of gas M is simply equal to the number of molecules N multiplied by the
mass of one molecule m, hence we have:
N = M/m
(12.5)
Inserting this into eq. (12.4) we obtain
P = (M/V) (k/m) T
(12.6)
where M/V is simply , the density of the gas. Hence, eq. (12.6) may be re-written:
 = (m/kT) P
(12.7)
If we now assume that air behaves as an ideal gas, we may insert eq.(12.7) into eq.(12.3),
giving:
P' = - (gm/kT) P
(12.8)
Now if the temperature of the air would have a constant value T at all altitudes, eq. (12.8)
would tell us that the rate of decrease of pressure is some constant number (-gm/kT) times the
value of the pressure at that altitude.
This is a the kind of mathematical law that recurs in very many branches of nature, i.e. that
the rate at which something decreases is proportional to the amount of that something that
there is at any given place or time. For example, radioactive decay follows such a law since
the rate at which radioactive atoms decay in time is proportional to the number of atoms we
have at any given time. This fact enables archaeologists to measure the age of various ancient
artifacts (via the amount of the radioactive isotope of carbon, 13C, they contain), and
geologists to date the rocks which comprise the earth. This kind of mathematical relationship
is called a decreasing exponential relationship, for reasons that I shall clarify below.
65
In the present situation, this exponential law says that, for a constant temperature atmosphere,
the rate at which atmospheric pressure decreases with height is proportional to the pressure at
any given height. That is to say, at ground level, the pressure is very large (Patm = 1.01 x 10 5
Pa at sea level), therefore the pressure decreases at a large rate. At higher altitudes, where the
pressure will be lower, the rate of decrease will also be lower. The mathematical function that
has this unique property is the negative exponential function.
You can easily explore its properties if you have a scientific calculator. In order to see what
an “exponential” decrease looks like, use your scientific calculator to calculate the values:
EXP(0) = 1, EXP(-0.5) = 0.6065, EXP(-1) = 0.3679, EXP(-1.5) = 0.2231, EXP(-2) = 0.1353,
etc. and plot a graph of the results. (Some calculators label the function “ex”. Others call it
“EXP”).
Fig. 2 shows an exponential decrease for the pressure of an ideal gas having the molecular
weight of air and an assumed constant average temperature of -43 oC. Since one mole (i.e.
6.02 x 1023 molecules) of air weighs 29 gm, k/m for air takes the value 287 J kg-1 K-1. Fig. 2
also shows typical actual values of the atmospheric pressure observed in the so-called
troposphere, i.e. the part of the atmosphere closest to ground level. Note that in this graph I
have used pressure units that are more familiar to meteorologists, namely, millibars (mb). [1
standard atmosphere = 1013 mb = 1.01 x 105 Pa = 760 mm Hg].
12 00
y = 1012 .7 * 1 0^(-6.43 04e-2 x) R^2 = 1.00 0
10 00
P [m b]
80 0
P [mb]
60 0
exp fit
40 0
20 0
0
0
10
20
30
he ight [km]
Figure 2: A negative exponential curve corresponding to sea level air pressure = 1013 mb and a constant
average air temperature of - 43 oC. Also shown (squares) are typical average atmospheric pressures at various
altitudes
You will notice that the exponential curve does not fit the data precisely because the air
temperature is not constant (we have assumed an average value) and because air is not a truly
ideal gas. But the shape is not too bad! We see how the pressure really does decrease sharply
at first and then more gently as we get to higher altitudes.
66
Now since eq. (12.7) must be true at any given altitude, it is clear that the rate at which the
two variables P and  are changing must also be related by the same equation. Therefore, an
alternative form of eq. (12.8), which describes the rate of change of density with height, is:
' = - (gm/kT) 



(12.9)
where ' represents the rate of increase of air density with increasing altitude.
2. Variation of Air Temperature with Height
In plotting eq. (12.8) I have assumed an average value of -43 oC for the air temperature. This
average is quite reasonable for the troposphere, where air temperatures vary with height from
typically about +20 oC at ground level, to about -60 oC at an altitude of about 12 km. But for
the rest of the atmosphere the situation is much more complicated, as shown in Fig. 3.
Io nizatio n by extreme UV,
solar X-rays an d
co smic rays
UV-C absorption
UV-B absorption
0
Meso sph ere
Stra tosphere
Th ermo sph ere
-50
Tro posphere
Temperature [degC]
50
-10 0
0
20
40
60
80
10 0
12 0
Altitude [km]
Figure 3: Typical temperature variations with altitude in the troposphere,
stratosphere, mesosphere and thermosphere of the earth’s atmosphere
Fig. 3 shows that we may distinguish 4 distinct parts of the atmosphere as regards changes in
temperature: the troposphere, in which the temperature decreases with altitude; the
stratosphere, in which it rises again; the mesosphere in which it falls again; and the
thermosphere in which the temperature rises once more with increasing altitude. Let us
examine each of these regions in more detail.
2.1 The troposphere
AS previously mentioned, the part of the atmosphere that is in contact with the ground is
called the troposphere. Here the temperature decreases with increasing altitude, from
ambient values at ground level, to approximately - 60 oC at a height of about 12 km. At this
altitude we encounter a transition region known as the tropopause, where the air temperature
stops falling. The precise values of the tropospheric temperature vary from summer to winter,
67
and the height of the tropopause is different at the equator compared to the poles. But the
values given in Fig. 3 are typical representative values.
The physical reason for this fall in temperature with increasing altitude in the troposphere is
that air is basically transparent to solar radiation. The incoming solar radiation is absorbed in
the ground and heats it. That is to say, the incoming photons impart kinetic energy to the
molecules that comprise the solid earth. The nearby air is then heated by conduction from the
warm ground. What this means is that the vibrating earth molecules cause the nearby air
molecules to vibrate. These molecules in turn transfer some kinetic energy to their neighbors,
and so on. Clearly, the further we get away from ground level, the weaker will be this
conductive heating mechanism. Hence, the air temperature will fall as we proceed higher in
altitude.
2.2 The stratosphere
At the height of the tropopause we notice that the temperature of the air stops falling, and
then starts to rise. Here, we have entered the stratosphere, where the temperature now rises
the higher we go, reaching almost 0 oC at about 50 km above ground. At this altitude we
encounter a second transition region, known as the stratopause, where the air temperature
stops rising. The agent responsible for heating in the stratosphere is the kinetic energy
generated by solar ultraviolet light. As we shall discuss in detail below, these photons are so
energetic that they can break up oxygen molecules into free oxygen atoms. This process, as
we shall also see, is responsible for the creation of the stratospheric ozone layer.
2.3 The mesosphere
By the time we reach the stratopause, where the stratosphere ends, there are very few oxygen
molecules left. So solar UV radiation ceases to be an effective heating agent. We have
reached the mesosphere, where the atmosphere once more starts to cool off with increasing
altitude. Here the temperature drops to nearly -100 oC at an altitude of about 80 km, where a
third transition region, the mesopause, is encountered.
2.4 The thermosphere
After the mesopause we enter the thermosphere, where the air pressure has dropped to 10-12
mb. In fact there are so few gas molecules per unit volume that "temperature" loses its normal
meaning. However, there are now several agents that can increase the energy of these residual
gas molecules; namely, extremely short-wave solar UV radiation, X-rays and cosmic
radiation. If one translates the resultant kinetic energy of the gas molecules into
"temperatures", using (3/2)kT as the kinetic energy per molecule, one obtains values up to a
thousand degrees Celsius, or so! In this region the air molecules become ionized (i.e.
stripped of their electrons) by the incoming radiation, resulting in the nocturnal phenomenon
of Aurora. This phenomenon is observed close to the two poles of the earth because the
ionizing radiation becomes trapped by the earth's magnetic field, and near the poles (where
the field lines enter and leave the earth’s surface) the density of this radiation is highest. More
generally, the layers of ionized molecules - the so-called ionosphere - cause disturbances to
radio transmissions.
68
Beyond the thermosphere there is no thermopause: the residual atmosphere gradually merges
with the vacuum of space in a region known as the exosphere. This region has not been
included in Fig. 3.
3. The stratospheric ozone layer
3.1. Formation of the ozone layer
Ozone is a gas that is formed when solar ultraviolet light interacts with oxygen molecules. If
the incoming solar photons have sufficiently high energy (so-called UV-C radiation) they tear
apart the oxygen molecules, forming oxygen atoms, which, as we have previously discussed,
are highly reactive. The chemical reaction is:
 (< 290 nm) + O2 -> O + O
(12.10)
where  represents a photon with wavelength less than 290 nm (this is the UV-C range of
photon energies). These free oxygen atoms quickly attach themselves to neighboring oxygen
molecules, forming molecules of ozone (O3). The chemical reaction is:
O + O2 + M -> O3 + M
(12.11)
In eq. (12.11), M represents a neighboring molecule (such as nitrogen) which must be present
in order for the collision process to conserve both momentum and energy.
This is the way ozone is formed in the earth’s atmosphere. Now, unlike the principal gases
we have discussed, ozone is not thoroughly mixed with the rest of the atmospheric
constituents. Instead, it forms a layer in the stratosphere. It is easy to see why this is so. The
formation of ozone requires two conditions: First, a plentiful supply of UV-C photons, and
second, a plentiful supply of oxygen targets. Far from the earth, there are plenty of UV-C
photons arriving from the sun, but there are very few oxygen molecules to stop them.
Gradually, as the UV-C radiation gets closer to the earth, the density of the atmosphere
increases. Therefore, more and more UV-C photons are absorbed until none are left. We thus
have a situation in which at low altitudes there are plenty of oxygen molecules but no UV-C
radiation to break them up (because it has all been absorbed at higher altitudes). On the other
hand, at the highest altitudes there are plenty of UV-C photons but no oxygen molecules to
stop them. There will accordingly be an optimal altitude where ozone is produced. This is the
so-called stratospheric ozone layer. It is the collisions between UV-C photons and oxygen
molecules that is responsible for the heating of the stratosphere, as we have seen, to
temperatures that are much higher than those at the top of the troposphere.
How thick is this layer? Well, clearly, it will not have well-defined boundaries. It will start
off with extremely low density at the top of the stratosphere. It will then become
progressively denser as we proceed to lower altitudes, reaching a maximum and then
gradually disappearing altogether. The maximum density of the ozone layer occurs at
altitudes between 20 km and 30 km above sea level. Now, if we could take all of that ozone
and compress it into a spherical shell at standard temperature and pressure (i.e. 273 K and 1
atmosphere), the shell would have a thickness of about 3 mm. Atmospheric physicists have
defined a more convenient unit for discussing the thickness of the ozone layer. It is the
Dobson unit (DU): 3 mm = 300 DU.
69
3.2 Seasonal changes in the thickness of the ozone layer
Now, ozone can also be broken apart by photons. For this purpose their energy need not be as
high as that of the UV-C photons we discussed above. Wavelengths in the range 290 nm 320 nm (so-called UV-B radiation) are sufficient. The chemical reaction is:
 (290 nm < < 320 nm) + O3 -> O2 + O (12.12)
The free oxygen atom that is released in reaction (12.12) quickly attaches itself to a nearby
molecule of air, forming O3 or N2O. However, the important point about this reaction is that
it absorbs out most (but not all!) of the harmful UV-B radiation that reaches us from the sun.
Unlike UV-C radiation, which is all absorbed out by the plentiful supply of atmospheric
oxygen, some UV-B radiation does reach ground level because the ozone layer is so thin.
But the thickness of the ozone layer varies with the seasons of the year. In summer time,
when the flux of destructive UV-B photons is highest, the ozone layer becomes thinner. On
the other hand, in winter, the flux of UV-B is smaller and the ozone layer recovers. It reaches
its maximum thickness in the springtime and its minimum thickness in the autumn. Naturally,
when the ozone layer is thickest in the northern hemisphere, it is thinnest in the southern
hemisphere, and vice-versa.
In recent years, it has been discovered that many industrially-produced gases (such as the
CFCs used in air-conditioners and spray cans) can break up ozone by chemical means. These
agents are believed to be responsible for the so-called hole in the ozone layer that has been
observed in the southern hemisphere. Clearly, any artificially induced thinning of the
stratospheric ozone layer will have the effect of allowing more harmful UVB radiation to
reach ground level. It is this concern that has led to the world-wide ban on the use of certain
kinds of refrigerant gases.
Problem set 12 (atmospheres)
1. For an average air temperature of –43 0C, calculate the numerical value of the constant
(gm/kT) in eq. (12.8).
2. The exponential curve drawn in Fig. 2 is the function P = Po EXP[-(gm/kT) z], where Po
= 1012.7 mb is the atmospheric pressure at ground level (z = 0), g = 9.81 m s-2 is the
acceleration of gravity at the earth’s surface, k/m = 287 J kg-1 K-1 is the gas constant for air,
and T = 230 K (= -43 oC) is the assumed average value of air temperature in the troposphere.
For values of the altitude z = 0, 2,500 m, 5,000 m, 10,000 m, 15,000 m, 20,000 m, 25,000 m
and 30,000 m, use a scientific calculator to evaluate the corresponding values of the pressure
P and plot a graph of P against z. You should obtain the same curve as Fig. 2.
3. Repeat these calculations for a warmer atmosphere (T = 330 K) and a colder atmosphere
(T = 130 K), and plot the results on the same graph you made for question 2. For each of
these situations, would the atmosphere be more spread out or more drawn in? Explain why
these results make good intuitive sense.
70