Page 175 - Materials Chemistry, Second Edition
P. 175

158       Practical Design Calculations for Groundwater and Soil Remediation



           Example 5.3:   Determine if the Free-Product Phase Is Present in
                        Subsurface Using the Saturated Vapor Concentration
           A subsurface is impacted by a spill of 1,1-dichloroethane (1,1-DCA). The 1,1-
           DCA concentrations of the soil samples from the impacted zone were between
           6,000 to 9,000 mg/kg. The subsurface has the following characteristics:

              •  Porosity = 0.4
              •  Organic content in soil = 0.02
              •  Degree of water saturation = 30%
              •  Subsurface temperature = 20°C
              •  Dry bulk density of soil = 1.6 g/cm 3
              •  Total bulk density of soil = 1.8 g/cm 3

           Determine if the free-product phase of 1,1-DCA is present in the subsurface.
           What could be the maximum 1,1-DCA concentration in soil if the free-prod-
           uct phase of 1,1-DCA is absent?

              Solution:
               (a)  From Table 2.5, the following physicochemical properties of 1,1-
                   DCA were obtained:

                               Molecular weight = 99.0
                                             H = 4.26 atm/M
                                           P vap  = 180 mm-Hg
                                        log  K  = 1.80
                                             ow
               (b)  Use  Equation  (5.1)  to  determine  the  saturated  1,1-DCA  vapor
                   concentration:

                         P vap  = 180 mm-Hg = 0.237 atm = 237,000 ppmV

               (c)  Convert the saturated vapor concentration into a mass concen-
                   tration by using Equation (2.1):

                         1 ppmV of 1,1-DCA = (99.0)/24.05 = 4.12 mg/m 3
                                   So,  G   = 237,000 ppmV = (237,000)(4.12)
                                        sat
                                          = 976,000 mg/m  = 976 mg/L
                                                        3
               (d)  Use Table 2.4 to convert the Henry’s constant to a dimensionless
                   value:

                    H  = H/RT = (4.26)/[(0.082)(273+20)] = 0.177 (dimensionless)
                     *
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