Basic Chemical Thermodynamics - Elements of Environmental Engineering: Thermodynamics and Kinetics - page 19

the density of air (kg/m 3 ) , and M is the average molecular weight of air ( ≈ 29). Consider
a parcel of air rising from the ground (sea) level to the lower atmosphere. It experiences
a change in pressure with altitude given by the well-known relation d P(h) /d h =−ρ g .
Use the ideal gas law to obtain an expression for P(h) , given that P 0 is the atmospheric
pressure at sea level.
Assume that as the parcel rises it undergoes a change in volume in relation to a
decreasing pressure, but that there is no net heat exchange between it and its surround-
ings, that is, it undergoes an adiabatic expansion . This is a reasonable assumption,
since the size of the reservoir (ambient atmosphere) is much larger than the size of the
air parcel, and any changes in the ambient temperature are imperceptible whereas that
within the air parcel will be substantial. Therefore, the volume expansion of the parcel
leads to a decrease in temperature. This variation in temperature of dry air with height
is called the dry adiabatic lapse rate .Apply the first law relation and use the expression
d U = C v d T (as in Section 2.2.5) to obtain an equation for the variation in temperature
of the air parcel with height.
Solution : Considering the air to be an ideal gas at any point in the atmosphere, we have
the following:

P = ρ RT

M .

(2.11)

Since the pressure at any point is due to the weight of the air above, we have d P(h) /d h =
−ρ g and hence

d P(h)
d h =−

P(h)Mg
RT

. (2.12)

Integrating the above equation with P( 0 ) = P 0 , we obtain
P(h) = P 0 e − (Mhg/RT) (2.13)

The above equation gives the pressure variation with height in the atmosphere.
Now we shall obtain the temperature profile in the atmosphere using the concepts
from the first law. For an adiabatic process we know that δ q = 0. Hence d U = δ w .As
will be seen in Section 2.2.5, d U = C v d T . Since it is more convenient to work with P
and T as the variables rather than with P and V , we shall convert the P d V term to a
form involving P and T using the ideal gas law. Thus,

m
M R d T .

d (PV) = P d V + V d P = (2.14)

Hence we have for the first law expression

m

M RT d P

m

M R d T .

C v d T =

P −

(2.15)

By rearranging, we obtain

m
M ·

RT
P

d T
d P =

. (2.16)

m
M · R

C v +

Combining the equation for d P /d h with the above, we obtain

d T
d h =−

g
C p , m ,

(2.17)

continued