Subfield: Thermodynamics

CHANGE OF AIR TEMPERATURE WITH ALTITUDE, ATMOSPHERIC STABILITY AND AIR POLLUTION 

Vertical motion of air governs many atmospheric processes, such as the formation of clouds and precipitation and the dispersal of air pollutants. If the atmosphere is stable, vertical motion is restricted and air pollutants tend to be accumulated around the emission site rather than dispersed and diluted. Meanwhile, in an unstable atmosphere, vertical motion of air encourages the vertical dispersal of air pollutants. Therefore, the pollutants' concentrations depend not only on the strength of emission sources but also on the stability of the atmosphere.

We shall determine the atmospheric stability by using the concept of air parcel in meteorology and compare the temperature of the air parcel rising or sinking adiabatically in the atmosphere to that of the surrounding air. We will see that in many cases an air parcel containing air pollutants and rising from the ground will come to rest at a certain altitude, called a mixing height. The greater the mixing height, the lower the air pollutant concentration. We will evaluate the mixing height and the concentration of carbon monoxide emitted by motorbikes in the Hanoi metropolitan area for a morning rush hour scenario, in which the vertical mixing is restricted due to a temperature inversion (air temperature increases with altitude) at elevations above $119 \mathrm{~m}$.

Let us consider the air as an ideal diatomic gas, with molar mass $\mu=29 \mathrm{~g} / \mathrm{mol}$.

Quasi equilibrium adiabatic transformation obey the equation $p V^{\gamma}=$ const, where $\gamma=\frac{c_{p}}{c_{V}}$ is the ratio between isobaric and isochoric heat capacities of the gas.

The student may use the following data if necessary:

The universal gas constant is $R=8.31 \mathrm{~J} /($ mol.K).

The atmospheric pressure on ground is $p_{0}=101.3 \mathrm{kPa}$

The acceleration due to gravity is constant, $g=9.81 \mathrm{~m} / \mathrm{s}^{2}$

The molar isobaric heat capacity is $c_{p}=\frac{7}{2} R$ for air.

The molar isochoric heat capacity is $c_{V}=\frac{5}{2} R$ for air.



Mathematical hints 

a. $\int \frac{d x}{A+B x}=\frac{1}{B} \int \frac{d(A+B x)}{A+B x}=\frac{1}{B} \ln (A+B x)$

b. The solution of the differential equation $\frac{d x}{d t}+A x=B \quad$ (with $\quad A$ and $B$ constant) is $x(t)=x_{1}(t)+\frac{B}{A}$ where $x_{1}(t)$ is the solution of the differential equation $\frac{d x}{d t}+A x=0$.

c. $\lim _{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^{x}=e$

1. Change of pressure with altitude.

Question: 1.1. Assume that the temperature of the atmosphere is uniform and equal to $T_{0}$. Write down the expression giving the atmospheric pressure $p$ as a function of the altitude $z$.

Solution:
For an altitude change $d z$, the atmospheric pressure change is :

$$
d p=-\rho g d z
\tag{1}
$$

where $g$ is the acceleration of gravity, considered constant, $\rho$ is the specific mass of air, which is considered as an ideal gas:

$$
\rho=\frac{m}{V}=\frac{p \mu}{R T}
$$

Put this expression in (1) :

$$
\frac{d p}{p}=-\frac{\mu g}{R T} d z
$$

If the air temperature is uniform and equals $T_{0}$, then

$$
\frac{d p}{p}=-\frac{\mu g}{R T_{0}} d z
$$

After integration, we have :

$$
p(z)=p(0) \mathbf{e}^{-\frac{\mu g}{R T_{0}} z}
\tag{2}
$$

Final answer: $p(z)=p(0) e^{-\frac{\mu g}{R T_{0}} z}$