Page 171 - Materials Chemistry, Second Edition
P. 171

154       Practical Design Calculations for Groundwater and Soil Remediation



           Example 5.1:   Estimate the Saturated Gasoline Vapor Concentration
           Use the information in Table 5.1 to estimate the maximum gasoline vapor
           concentration for two soil-venting projects. Both sites are impacted by acci-
           dental gasoline spills. The spill at the first site happened recently, while the
           spill at the other site occurred three years ago.


              Solution:
              The site with fresh gasoline:


               (a)  Vapor pressure of fresh gasoline is 0.34 atm at 20°C, as shown in
                   Table 5.1. The partial pressure of this gasoline in the void can be
                   found by using Equation (5.1) as:

                           P A =  P (  vap  x ) (0.34 atm)(1.0) 0.34 atm=  =
                                   )( A


                   Thus, the partial pressure of gasoline in the air is 0.34 atm
                       (=	340,000 × 10  atm), which is equivalent to 340,000 ppmV.
                                   −6
                   Use Equation (2.1) to convert the ppmV concentration into a mass
                       concentration (at 20°C), as
                   1 ppmV fresh gasoline =  [(MW of fresh gasoline)/24.05] mg/m 3
                       	  	             = (95)/24.05 = 3.95 mg/m 3
                         So, 340,000 ppmV = (340,000)(3.95)
                       	  	             = 1,343,000 mg/m  = 1,343 mg/L
                                                        3

              The site with weathered gasoline:

               (b)  Vapor pressure of the weathered gasoline is 0.049 atm, which is
                   equivalent to 49,000 ppmV.

                Use Equation (2.1) to convert the ppmV concentration into a mass
                concentration (at 20°C), as

                   1 ppmV weathered gasoline =  [(MW of weathered
                                              gasoline)/24.05] mg/m 3
                       	  	                  = (111)/24.05 = 4.62 mg/m 3
                              So, 49,000 ppmV = (49,000)(4.62)
                       	  	                  = 226,000 mg/m  = 226 mg/L
                                                           3
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