1
Appendix 3: The pressure in the earth’s atmosphere
Does the pressure inside the earth’s atmosphere depend on the altitude h measured from
the sea level? Using the ideal gas law, we can show below that the pressure does
exponentially decrease as we climb up to a location with a higher altitude.
Consider an air mass inside an imaginary box whose center is at an altitude of h in the
earth’s atmosphere. The volume of the box is V = Adh, where A is the area of the base of
the box and dh is its height. In static mechanical equilibrium, a net force on the box must
vanish so that
!
#
#
1 &
1 &
AP% h " dh ( " AP% h + dh ( " m( h ) g = 0,
$
$
2 '
2 '
where P ( h ± dh 2) is the pressure at the bottom (–) and the top (+) of the box,
!
respectively, and m( h ) is the mass of the air inside the box, which is related to the
!
density of the air "( h ) inside the box by
!
m( h ) = " ( h ) Adh.
!
The above force-balance equation becomes a differential equation for the pressure as a
!
function of the altitude:
"
"
1 %
1 %
P$ h + dh ' ( P$ h ( dh '
dP
#
#
2 &
2 &
=
= () ( h ) g .
dh
dh
To express the density as a function of the pressure and temperature of the air in the
!
box, we use the ideal gas law,
PV = nRT ,
!
2
which is an approximate equation of state for air. n is the mole number of the air in the
box and T is the temperature of the air in the box. Note that the use of the ideal gas law
here implies that we are assuming local thermal equilibrium inside the box, which means
that the temperature inside the box is well defined and constant.
Of course, the
atmosphere as a whole is not in thermal equilibrium, as we know that the temperature
depends on the altitude and heat is being transferred from the warmer ground to the
colder top of the atmosphere. We can express the density as a function of the pressure
and temperature of the air in the box by
"( h ) =
Mn ( h ) MP ( h )
=
,
V
RT ( h )
where M is the molar mass or the mass of 1 mole of air:
!
M = (0.78)(28 g mol) + (0.21)( 32 g mol) + (0.01)( 40 g mol) = 29 g mol,
!
where the molar mass of nitrogen ( N 2 ) gas is 28 g mol; the molar mass of oxygen ( O 2 )
gas is 32 g mol; the molar mass of argon (Ar) gas is 40 g mol.
The equation for the pressure now becomes
!
!
!
!
!
MgP
dP
(h )
="
dh
RT ( h )
or
!
dP
Mg
="
dh
P
RT ( h )
so that
" P h % P (h)
! ln$ ( ) ' = ( dP = ) Mg
R
# P (0) & P ( 0) P
h
dh *
( T (h*) .
0
In the troposphere ( 0 " h " 11 km), the temperature is found to decrease with the altitude
!
roughly at rate of a = 6.5 K km:
!
!
3
#
ah &
T ( h ) = T (0) " ah = T (0)$1"
',
% T (0) (
where T (0) " 300 K . As ah T (0) " 0.24 < 1 in the troposphere, we neglect the decrease
!
in temperature with the altitude to obtain
!
!
!
" P (h )%
Mg
ln$
'=(
R
# P (0) &
h
*
0
dh )
Mg
+(
T (0) ( ah )
RT (0)
h
Mg
* dh) = ( RT (0) h
0
so that
$ Mg '
P ( h ) " P (0) exp&#
h),
% RT (0) (
where
!
29 #10$3 kg mol)(9.8 m s2 )
(
Mg
"
" 1#10$4 m$1 .
RT (0)
(8.3 J (mol% K))(300 K)
The density is then given by
!
"( h ) #
!
% Mg (
% Mg (
MP ( h ) MP (0)
=
exp'$
h* = "(0) exp'$
h* .
RT (0)
RT (0)
& RT (0) )
& RT (0) )
The characteristic length scale or “thickness” of the atmosphere can be defined as
)1
# Mg &
4
ht " %
( * 10 m = 10 km
$ RT (0) '
so that
$ h'
P ( h )!" P (0) exp&# )
% ht (
!
and
!
% h(
"( h ) # " (0) exp'$ * .
& ht )
4
The total mass of the air directly above a base area A on the ground is then given by
M total
$
.
( h# +
( h # +1
= % "( h #) Adh # & A"(0) % exp*' - dh # = A" (0)0'ht exp* ' -3 = A"(0) ht
) ht ,
) ht ,20
/
0
0
$
$
The atmospheric pressure at the ground level is then
!
P (0) =
!
M total g
" # (0) ght " (1 kg m3 )(10 m s2 )(10 4 m) = 10 5 Pa ,
A
which justifies a quick order-of-magnitude estimate of the atmospheric pressure by
P (0) " # (0) ght .
A math aside: it’s good to remember the following:
!
)
# x&
* exp%$" a ('dx = a,
0
which means that the area under the exponential function exp(" x a) is equal to the
!
area of a rectangle with width a and height 1.
!