Problem#4: Consider an NAPL mixture of PCE and fuel containing BTEX with the fol

Question

Problem#4: Consider an NAPL mixture of PCE and fuel containing BTEX with the following mole fractions: PCE-0.10, B-0.15; T= 0.15; E 0.08 and X = 0.10. The vapor pressure of the pure compounds are PCE = 0.0251 atm: B-0.126 atm; T = 0.0380 atm. E 0.0 126 atm and X = 0.0117 atm. Assume Raoults law and Henry's laws apply, estimate the aqueous concentration of PCE in water that is in equilibrium with the NAPL mixture Estimate the reduction in aqueous PCE concentration (solubility) due to the presence of BTEX. Hint: First estimate the solubility when PCE is the sole NAPL (mole fraction = 1.0)

Explanation / Answer

NAPL Raoult’s Law: C G = X t (P°/RT) X t = mole fraction of compound in NAPL [-] P° = pure compound vapor pressure [atm] R = universal gas constant [m 3-atm/mole/°K] T = temperature [°K]P° = 76 mm Hg at 20°C = 0.1 atm R = 8.205 x 10-5 m 3-atm/mol/°K T = 20°C (assumed) = 293°K

Assume 100% benzene, mole fraction X t = 1 C G = X t P°/(RT) = 4.16 mol/m 3 Molecular weight of benzene, C 6 H 6 = 78 g/mol C G = 4.16 mol/m 3 × 78 g/mol = 324 g/m 3 = 0.32 g/L C G = 0.32 g/L x 24 L/mol / (78 g/mol) x 10 6 = 99,000 ppmv One mole of ideal gas = 22.4 L at STP (1 atm, 0 C), Corrected to 20 C: 293/273*22.4 = 24.0 L/mol Gas concentration in equilibrium with pure benzene NAPL

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