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Thermodynamics filled in class notes_Part_66

Thermodynamics filled in class notes_Part_66 - 139 4.11...

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4.11. EQUILIBRIUM IN A TWO-COMPONENT SYSTEM 139 Because T is fixed at 100 C and the material is a two phase mixture, the pressure is also fixed at a constant. Here there are two phases at saturation; g for gas and l for liquid. Equation (4.399) reduces to μ l dn l + μ g dn g 0 . (4.402) Now for the pure H 2 O if a loss of moles from one phase must be compensated by the addition to another. So one must have dn l + dn g = 0 . (4.403) Hence dn g = dn l . (4.404) So Eq. (4.402), using Eq. (4.404) becomes μ l dn l μ g dn l 0 , (4.405) dn l ( μ l μ g ) 0 . (4.406) At this stage of the analysis, most texts, grounded in equilibrium thermodynamics, assert that μ l = μ g , ignoring the fact that they could be different but dn l could be zero. That approach will not be taken here. Instead divide Eq. (4.406) by a positive time increment, dt 0 to write the second law as dn l dt ( μ l μ g ) 0 . (4.407) One convenient, albeit na¨ ıve, way to guarantee second law satisfaction is to let dn l dt = κ ( μ l μ g ) , κ 0 , convenient, but na¨ ıve model (4.408) Here κ is some positive semi-definite scalar rate constant which dictates the time scale of approach to equilibrium. Note that Eq. (4.408) is just a hypothesized model. It has
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