# Volatilization

Pesticide in the dissolved phase is available for volatilization. The amount of pesticide removed from the water via volatilization is:

         $$pst\_{vol,wtr}=v\_v*SA*\frac{F\_d\*pst\_{lkwtr}}{V}$$                                                     8:4.1.8

     where $$pst\_{vol,wtr}$$ is the amount of pesticide removed via volatilization (mg pst), $$v\_v$$ is the volatilization mass-transfer coefficient (m/day), $$SA$$ is the surface area of the water body (m$$^2$$), $$F\_d$$ is the fraction of total pesticide in the dissolved phase, $$pst\_{lkwtr}$$ is the amount of pesticide in the water (mg pst), and *V* is the volume of water in the water body(m$$^3$$ H$$\_2$$O).

            The volatilization mass-transfer coefficient can be calculated based on Whitman’s two-film or two-resistance theory (Whitman, 1923; Lewis and Whitman, 1924 as described in Chapra, 1997). While the main body of the gas and liquid phases are assumed to be well-mixed and homogenous, the two-film theory assumes that a substance moving between the two phases encounters maximum resistance in two laminar boundary layers where transfer is a function of molecular diffusion. In this type of system the transfer coefficient or velocity is:

            $$v\_v=K\_l\*\frac{H\_e}{H\_e+R*T\_K*(K\_l/K\_g)}$$                                                         8:4.1.9

where $$v\_v$$ is the volatilization mass-transfer coefficient (m/day), $$K\_l$$ is the mass-transfer velocity in the liquid laminar layer (m/day), $$K\_g$$ is the mass-transfer velocity in the gaseous laminar layer (m/day), $$H\_{e}$$ is Henry’s constant (atm m$$^3$$ mole$$^{-1}$$), $$R$$ is the universal gas constant (8.206 $$\*$$ 10$$^{-5}$$ atm m$$^3$$ (K mole)$$^{-1}$$), and $$T\_K$$ is the temperature     ($$K$$).

For lakes, the transfer coefficients are estimated using a stagnant film approach:

            $$K\_l=\frac{D\_l}{z\_l}$$                                      $$K\_g=\frac{D\_g}{z\_g}$$                              8:4.1.10

where $$K\_l$$ is the mass-transfer velocity in the liquid laminar layer (m/day), $$K\_g$$ is the mass-transfer velocity in the gaseous laminar layer (m/day), $$D\_l$$ is the liquid molecular diffusion coefficient (m$$^2$$/day), $$D\_g$$ is the gas molecular diffusion coefficient (m$$^2$$/day), $$z\_l$$ is the thickness of the liquid film (m), and $$z\_g$$ is the thickness of the gas film (m).

            Alternatively, the transfer coefficients can be estimated with the equations:

             $$K\_l=K\_{l,O\_2}\*(\frac{32}{MW})^{0.25}$$                                                        8:4.1.11

             $$K\_g =168\*\mu\_w\*(\frac{18}{MW})^{0.25}$$                                                 8:4.1.12

where $$K\_l$$ is the mass-transfer velocity in the liquid laminar layer (m/day), $$K\_g$$ is the mass-transfer velocity in the gaseous laminar layer (m/day), $$K\_{l,O\_2}$$ is the oxygen transfer coefficient (m/day), $$MW$$ is the molecular weight of the compound, and $$\mu\_w$$ is the wind speed (m/s). Chapra (1997) lists several different equations that can be used to calculate $$K\_{l,O\_2}$$.        


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