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Unformatted text preview: 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 l dn l + g dn g . (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 l dn l g dn l , (4.405) dn l ( l g ) . (4.406) At this stage of the analysis, most texts, grounded in equilibrium thermodynamics, assert that l = 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 ( l g ) . (4.407) One convenient, albeit na ve, way to guarantee second law satisfaction is to let dn l dt = ( l g ) , , convenient, but na ve model (4.408) Here is some positive semi-definite scalar rate constant which dictates the time scale of approach to...
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This note was uploaded on 11/26/2011 for the course EGN 3381 taught by Professor Park-sou during the Fall '11 term at FSU.
- Fall '11