Thermodynamics filled in class notes_Part_67

Thermodynamics filled in class notes_Part_67 - 4.11...

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Unformatted text preview: 4.11. EQUILIBRIUM IN A TWO-COMPONENT SYSTEM 141 10 5 10 6 10 7 10 8 10 9 10 3 10 4 10 5 P (Pa) P vap (Pa) Figure 4.1: Pressure of water vapor as a function of total pressure for example problem. Now g f = g g , so one gets v f ( P − P sat ) = RT ln P vap P sat . (4.423) Solving for P vap , one gets P vap = P sat exp parenleftbigg v f ( P − P sat ) RT parenrightbigg , (4.424) = (2 . 339 kPa )exp parenleftBig . 001002 m 3 kg parenrightBig (100 kPa − 2 . 339 kPa ) parenleftBig . 4615 kJ kg K parenrightBig (293 . 15 K ) , (4.425) = 2 . 3407 kPa. (4.426) The pressure is very near the saturation pressure. This justifies assumptions that for such mixtures, one can take the pressure of the water vapor to be that at saturation if the mixture is in equilibrium. If the pressure is higher, the pressure of the vapor becomes higher as well. Figure 4.1 shows how the pressure of the equilibrium vapor pressure varies with total pressure. Clearly, a very high total pressure, on the order of 1...
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Thermodynamics filled in class notes_Part_67 - 4.11...

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