Thermodynamics filled in class notes_Part_66

Thermodynamics filled in class notes_Part_66 - 4.11....

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

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.

Page1 / 2

Thermodynamics filled in class notes_Part_66 - 4.11....

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online