Atmospheric Properties
R. Wordsworth
September 13, 2016
1
Hydrostatic equation
The derivation here is more traditional than what was done in the lectures.
Consider a column of air with density ρ(z) under gravity g. Pressure at base is p(z), pressure
at top is p(z + δz). Then by Taylor’s theorem
p(z + δz) ∼ p(z) +
dp
δz
dz
(1)
Pressure is force per unit area so net upwards force is simply
δFp = A
dp
δz
dz
(2)
This must balance the gravitation force, which is δFg = −gδm = −gρAδz. Hence
A
dp
δz = −gρAδz
dz
(3)
dp
= −gρ.
(4)
dz
Hydrostatic balance holds to a very good approximation in almost all planetary atmospheres. It is
key to understanding their structure.
Now, if we use the ideal gas law (see below) we can write
dp
pg
=−
dz
RT
(5)
g
d ln[p]
=−
dz
RT
Z z
g
ln[p] − ln[p0 ] = −
dz 0
RT
0
p(z) = p0 e−
Rz
g
0 RT
dz 0
(6)
(7)
(8)
which we can write as
p(z) = p0 e−z/H .
(9)
with scale height H ≡ RT /g in simplest limit of a thin isothermal atmosphere. This is height over
which pressure of the atmosphere decays by e−1 .
1
2
Ideal Gas Law
Planetary atmospheres are often well-treated as ideal gases. The ideal gas law is
pV = N kB T
(10)
p is pressure, V is volume, N is (total) number of molecules or atoms, T is temperature and kB is
Boltzmann’s constant. We can write
p = nkB T
(11)
where n is number density [molecules / m3 ]. Then,
p = ρ/(mp µa )kB T
(12)
where ρ is mass density [kg / m3 ], mp is the proton mass, and µa is the mean molecular weight of
the gas. Defining R = kB /mp µa
p = ρRT
(13)
where R is the specific gas constant [J/kg/K], with a value that depends on the properties of the
gas.
Ideal gas law no longer holds when energy in interactions between molecules (v.s. mean kinetic
energy) is no longer negligible. Theory gets more difficult rapidly! E.g. Venus deep atmosphere,
gas giants.
3
Dry tropospheric convection
The lower atmosphere is extremely turbulent. To understand the temperature variation, consider
a fluid blob / gas parcel moving upwards.
The first law of thermodynamics is
dU = δQ − δW
(14)
Change in internal energy of system = heat gained minus work done by system. The internal energy
is dU = Cv dT , where Cv is the heat capacity at constant volume. Furthermore, the work done is
δW = pdV , where p is pressure and V volume. Divide everything by the parcel mass to get
δQ
= cv dT + pdv.
M
(15)
Here cv = Cv /M and v = V /M is specific volume (1/ρ). First term is increase in internal energy of
the system (δU = cv dT ), second term is work done by the system, δQ is heat supplied to system.
Apply the chain rule to get
δQ
dp
= cv dT + d(p/ρ) −
.
(16)
M
ρ
Now use ideal gas law, set δQ = 0 (adiabatic assumption) and remember that the specific heat
capacity at constant pressure cp = cv + R. Then
0 = cp dT −
2
dp
.
ρ
(17)
Solving this we get
R
d ln p
cp
R/cp
p
T = T0
.
p0
d ln T =
(18)
(19)
R/cp ∼ 0.28 for N2 . T vs. p is useful for climatologists because p determines the atmospheric
physics. The atmospheric lapse rate as a function of z is of most practical importance for engineering
and some scientific applications. To derive it, combine (17) with the hydrostatic balance relation
and get
dT
g
(20)
=− .
dz
cp
4
Mass column
Pressure is force divided by area p = F/A so in a thin atmosphere (small variation of g with height)
F = matm g → p = ug
where u = matm /A is the atmospheric mass column in kg/m2 .
3
(21)