Page 173 - Materials Chemistry, Second Edition

156 Practical Design Calculations for Groundwater and Soil Remediation

Thus, partial pressure of toluene = 0.0155 atm = 15,500 ppmV.

x ) (10 mm-Hg)(0.464)=
P xylenes = P ( vap )( A
= 4.64 mm-Hg = 0.0061 atm

Thus, partial pressure of xylenes = 0.0061 atm = 6,100 ppmV.
The volumetric (or molar) composition of the extracted vapor
= (15,500)/[15,500 + 6,100] = 71.8% (toluene)

(d) The mass concentration can be found by using Equation (2.1) as:

1 ppmV toluene = (92.1)/24.05 = 3.83 mg/m 3
So, 15,500 ppmV = (15,500)(3.83) = 59,400 mg/m = 59.4 mg/L
3
1 ppmV xylenes = (106.2)/24.05 = 4.42 mg/m 3
So, 15,500 ppmV = (6,100)(4.42) = 27,000 mg/m = 27.0 mg/L
3
The weight composition of the extracted vapor
= (59.4)/[59.4 +27.0] = 68.8% (toluene)

Discussion:
1. The toluene concentration in the extracted vapor is 68.8% by
weight and 71.8% by volume. Both are higher than its concentra-
tion in the liquid solvent, 50% by weight. The higher percentage
of toluene in the vapor is mainly due to its higher vapor pressure.
2. This saturated vapor concentration would be higher than the
actual concentration of the extracted vapor due to the fact that
(1) not all the air flows through the impacted zone and (2) limita-
tions on mass transfer exist.

As mentioned, the presence or absence of a free-product phase greatly
affects the extracted vapor concentration. Equation (2.40) in Chapter 2 can be
used as a starting point for discussion.

 [( w + ρ K p + φ H 
φ
( )]
)( )
X =  b a  × C
 t ρ 
  ( w ) + ( b ) K p + ()   (2.40)
ρ
φ
a φ

 ×
=   H H   G
 t ρ 
 
Let soil saturation concentration (X ) correspond to the COC concentra-
sat
tion in soil at which the adsorptive limits of soil grains, the solubility of soil