Thermodynamics

where is the solubility. Expressing the pressure in bars and noting that theunit of molar concentration is kmol of species i per m 3 , the unit of solubilityis kmol/m 3 · bar. Solubility data for selected gas–solid combinations aregiven in Table 16–3. The product of solubility of a gas and the diffusioncoefficient of the gas in a solid is referred to as the permeability, which is ameasure of the ability of the gas to penetrate a solid. Permeability isinversely proportional to thickness and has the unit kmol/s · m · bar.Finally, if a process involves the sublimation of a pure solid such as ice orthe evaporation of a pure liquid such as water in a different medium such asair, the mole (or mass) fraction of the substance in the liquid or solid phaseis simply taken to be 1.0, and the partial pressure and thus the mole fractionof the substance in the gas phase can readily be determined from the saturationdata of the substance at the specified temperature. Also, the assumptionof thermodynamic equilibrium at the interface is very reasonable for puresolids, pure liquids, and solutions except when chemical reactions areoccurring at the interface.TABLE 16–3Chapter 16 | 813Solubility of selected gases andsolids (from Barrer)(for gas i, r – i,solid side /P i,gas side)Gas Solid T K kmol/m 3 · barO 2 Rubber 298 0.00312N 2 Rubber 298 0.00156CO 2 Rubber 298 0.04015He SiO 2 298 0.00045H 2 Ni 358 0.00901EXAMPLE 16–8Mole Fraction of Water Vapor Just over a LakeDetermine the mole fraction of the water vapor at the surface of a lake whosetemperature is 15°C, and compare it to the mole fraction of water in the lake(Fig. 16–24). Take the atmospheric pressure at lake level to be 92 kPa.Solution The mole fraction of water vapor at the surface of a lake is to bedetermined and to be compared to the mole fraction of water in the lake.Assumptions 1 Both the air and water vapor are ideal gases. 2 The amountof air dissolved in water is negligible.Properties The saturation pressure of water at 15°C is 1.7057 kPa (Table A–4).Analysis There exists phase equilibrium at the free surface of the lake,and thus the air at the lake surface is always saturated at the interfacetemperature.The air at the water surface is saturated. Therefore, the partial pressure ofwater vapor in the air at the lake surface will simply be the saturation pressureof water at 15°C,The mole fraction of water vapor in the air at the surface of the lake is determinedfrom Eq. 16–22 to bey v P vPP v P sat @ 15°C 1.7057 kPa1.7057 kPa92 kPa 0.0185 or 1.85 percentWater contains some dissolved air, but the amount is negligible. Therefore,we can assume the entire lake to be liquid water. Then its mole fractionbecomesy water,liquid side 1.0 or 100 percentLake15°CAir92 kPaSaturated airy H2 O,air side = 0.0185y H2 O,liquid side ≅ 1.0FIGURE 16–24Schematic for Example 16–8.Discussion Note that the concentration of water on a molar basis is100 percent just beneath the air–water interface and less than 2 percentjust above it even though the air is assumed to be saturated (so this is thehighest value at 15°C). Therefore, large discontinuities can occur in the concentrationsof a species across phase boundaries.