Page 71 - Materials Chemistry, Second Edition

54 Practical Design Calculations for Groundwater and Soil Remediation

2. Most technical articles do not talk about the units of K . Actually, K
p
p
has a unit of “(volume of solvent)/(mass of adsorbent),” and it is equal
to mL/g or L/kg in most, if not all, of the correlation equations.

2.4.4 Solid–Liquid–Vapor Equilibrium
As mentioned in the beginning of this section, an NAPL may end up in
four different phases as it enters a vadose zone. We have just discussed the
equilibrium systems of liquid–vapor and soil–liquid. Now we move one step
further to discuss the system including liquid, vapor, and solid (and free
product in some of the applications).
The soil moisture in the vadose zone is in contact with both soil grains and
air in the void, and the COC in each phase can travel to the other phases. The
dissolved concentration in the liquid, for example, is affected by the concen-
trations in the other phases (i.e., soil, vapor, and free product). If the entire
system is in equilibrium, these concentrations are related by the equilibrium
equations mentioned previously. In other words, if the entire system is in
equilibrium and the COC concentration of one phase is known, the concen-
trations at other phases can be estimated using the equilibrium relationships.
Although, in real applications, the equilibrium condition does not always
exist, the estimate from such a condition serves as a good starting point or as
the upper or the lower limit of the real values.

Example 2.32: Solid–Liquid–Vapor-Free-Product
Equilibrium Concentrations

Free-product phase of 1,1,1-trichloroethane (1,1,1-TCA) was found in the sub-
surface at a site. The soil is silty, with an organic content of 2%. The subsur-
face temperature is 20°C. Estimate the maximum concentrations of TCA in
the air void, in the soil moisture, and on the soil grains.

Solution:
(a) Since the free product is present, the maximum vapor con-
centration will be the vapor pressure of the TCA liquid at that
temperature.
From Table 2.5, the vapor pressure of TCA is 100 mm-Hg at 20°C.

100 mm-Hg = (100 mm-Hg) ÷ (760 mm-Hg/atm) = 0.132 atm
G = 0.132 atm = 132,000 ppmV

Use Equation (2.1) to convert ppmV to mg/m (MW = 133.4 from
3
Table 2.5)
132,000 ppmV = (132,000)(133.4/24.05) mg/m 3
G = 732,200 mg/m = 732.2 mg/L
3