The atmospheric pressure (force per unit area) on a surface at an altitude
z
is due to the weight of the column of air situated above the surface. Therefore, the drop in air pressure
ρ
between the top and bottom of a cylindrical volume element of height
Δ
z
and cross-section area
A
equals the weight of the air enclosed (density
ρ
times volume
V
=
A
Δ
z
times gravity g), per unit area:
p
(
z
+
Δ
z
)
−
p
(
z
)
=
−
ρ
(
z
)
(
A
Δ
z
)
g
A
=
−
ρ
(
z
)
g
Δ
z
.
Let
Δ
z
→
0
to derive the differential equation
d
p
/
d
z
=
−
ρ
g
. To analyze this further we must postulate a formula that relates pressure and density. The perfect gas law relates pressure, volume, mass
m
, and absolute temperature
T
according to
p
V
=
m
R
T
/
M
, where
R
is the universal gas constant and
M
is the molar mass (i.e., the mass of one mole) of the air. Therefore density and pressure are related by
ρ
:
=
m
/
V
=
M
p
/
R
T
.
a. Derive the equation
d
p
d
z
=
−
M
g
R
T
p
and solve it for the “isothermal” case where
T
is constant to obtain the barometric pressure equation
p
(
z
)
=
p
(
z
0
)
exp
{
−
M
g
(
z
−
z
0
)
/
R
t
}
.
b. If the temperature also varies with altitude
T
=
T
(
z
)
, derive the solution
p
(
z
)
=
p
(
z
0
)
exp
{
−
M
g
R
∫
z
0
z
d
ζ
T
(
ζ
)
}
.
c. Suppose an engineer measures the barometric pressure at the top of a building to be
99
,
000
Pa
(pascals), and
101
,
000
Pa
at the base
(
z
=
z
0
)
. If the absolute temperature varies as
T
(
z
)
=
288
−
0.0065
(
z
−
z
0
)
, determine the height of the building. Take
R
=
8.31
N-m/mol-K
,
M
=
0.029
kg/mol
, and
g
=
9.8
m/sec
2
. (An amusing story concerning this problem can be found at http://www.snopes.com/college/exam/barometer.asp)