Volatilization is the process where solids and liquids vaporize into the atmosphere. There are three mechanisms: vapors escape the air/liquid interface, diffuse through the boundary layer via molecular and turbulent diffusion, and are transported away by advection and dispersion. The rate of vaporization depends on the mass transfer coefficient, existing contaminant concentration in water, and saturation concentration in air. If the concentration is below saturation, no vaporization occurs; above saturation, vaporization takes place. The loss of contaminants from a water body can be calculated using the rate, surface area, depth, and change in concentration over time.

1. Volatilization
• Process of vaporization of solid and liquid to atmosphere.
Mechanisms of Volatilization:
• Initially Vapour escapes through the air/liquid or air/solid interface
into the atmosphere boundary sublayer.
• Then Gas/Vapour diffuses through the boundary sublayer by molecular
and turbulent diffusion.
• The gaseous compound is transported away from the site by advection
and dispersion.

2. The rate of vaporization:
r = -k (C-Cs)
r = vaporization rate , g/m2h
k = Mass transfer coefficient , m/h
C= existing VOC concentration in water body , g/m3
Cs= saturation concentration of VOC in atmosphere , g/m3
Case 1: C < Cs no vaporization
Case 2: C > Cs vaporization occurs

3. If VOC is assumed to be vaporizing from a water body of finite area,
say a water tank of surface area As and depth h then:
Outflow due to vaporization = r As ,g/h
Loss of VOC to volume Ash =
𝒅𝑪
𝒅𝒕
Ash
Assumption: Vaporization is independent of surface temperature , wind
speed etc
Loss of VOC to volume is
𝒅𝑪
𝒅𝒕
Ash = r As
After Integrating the above equation:
C = CS + (C0 –CS) exp – (
−𝒌𝒕
𝒉
)
C0 = VOC concentration in air a time zero
C = VOC concentration in air any time t

4. Example
Determine the time required for benzene (C6H6) and DDT (C14H9Cl5) to
vaporize to half their original concentrations from a wastewater treatment
plant holding tank of depth 2 m. The mass transfer coefficient for benzene
is 0.144 m/h and for DDT is 9.34 x 10-3 m/h. Values for transfer
coefficients are given in techobanoglous and Schroeber (1987).
Solution:
t1/2 = 0.69 h/k
For benzene,k= 0.144 m/h
DDT, k = 9.34 x 10-3 m/h
Therefore for benzene , t1/2 = 0.69 x
2
0.144
= 9.5 h
Therefore for DDT ,t1/2 = 0.69 x
2
9.34 x 10−3
= 147 h ( 6 days)