Page 172 - Materials Chemistry, Second Edition
P. 172

Vadose Zone Soil Remediation                                     155



              Discussion:
                1.  The saturated vapor concentration of the weathered gasoline can
                   be a few times smaller than that of the fresh gasoline. In this
                   example, it is more than five times smaller.
                2.  The calculated vapor concentrations are essentially the same as
                   those listed in Table 5.1.


           Example 5.2:   Estimate Saturated Vapor Concentrations
                        of a Binary Mixture
           A site is impacted by an industrial solvent. The solvent consists of 50% tol-
           uene and 50% xylenes by weight. Soil venting is being considered for site
           remediation. Estimate the maximum vapor concentration of the extracted
           vapor. The subsurface temperature of the site is 20°C.

              Solution:
               (a)  From Table 2.5, the following physicochemical properties were
                   obtained:

                            Molecular weight of toluene = 92.1
                           Molecular weight of xylenes = 106.2
                                        P vap  of toluene = 22 mm-Hg
                                        P vap  of xylenes = 10 mm-Hg

               (b)  The mole fractions of toluene and xylenes in the solvent can be
                   found as:

                   Basis: 1,000 g solvent
                              Moles of toluene =  mass/MW = [(50%)(1,000)]
                                               ÷ (92.1) = 5.43 moles
                              Moles of xylenes =  mass/MW = [(50%)(1,000)]
                                               ÷ (106.2) = 4.71 moles
                         Mole fraction of toluene = (5.43)/(5.43 + 4.71) = 0.536
                         Mole fraction of xylenes = 1 − 0.536 = 0.464
               (c)  The saturated vapor concentration can be found by using
                   Equation (5.1) as:

                                         x ) (22mm-Hg)(0.536)=
                            P toluene =  P (  vap )( A
                                  = 11.79 mm-Hg = 0.0155 atm
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