Page 178 - Materials Chemistry, Second Edition
P. 178

Vadose Zone Soil Remediation                                     161



              •  Subsurface temperature = 25°C
              •  Dry bulk density of soil = 1.6 g/cm 3
              •  Total bulk density of soil = 1.8 g/cm 3

           Estimate the extracted soil vapor concentration at the start of the soil-venting
           project.

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

                            Molecular weight = 78.1
                                          H = 5.55 atm/M
                                        P vap  = 95.2 mm-Hg
                                     log  K  = 2.13
                                          ow

               (b)  Use Table  2.4 to convert Henry’s constant to a dimensionless
                   value:

                   H  = H/RT = (5.55)/[(0.082)(273 +	25)] = 0.23 (dimensionless)
                     *

                   Use Equation (2.28) to find K ,
                                             oc
                       K  = 0.63K  = 0.63 (10 2.13 ) = (0.63)(135) = 85
                                ow
                        oc
                   Use Equation (2.26) to find K ,
                                             p
                       K  = f K  = (0.03) (85) = 2.6 L/kg
                              oc
                        p
                           oc
               (c)  Use Equation (2.40) to estimate the soil vapor concentration of
                   benzene in equilibrium with this benzene concentration in soil:

                                 ( w )  +  ( b )  K p  + ()  
                                     ρ
                                 φ
                                             a φ
                              
                                                ×
                          X =    H  H          G
                                      t ρ     
                                               
                              
                                (0.35)(0.45)  +  (1.6)(2.6)  + (0.35)(1 45%) 
                                                      −
                              
                                                             
                         500 =    0.23  0.23                ×  G
                                            1.8             
                                                            
                   So, G = 47.5 mg/L = 47,500 mg/m 3
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