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For a single component system at vapor liquid equilibrium, we know that: dG¹ = -s¹dT + VLdpsat : 0 dGV = -sVdT + VV dpsat dGV = dGL 0 If we now re-arrange Eq. 1, we can obtain Eq. 4: = SL = dpsat dT Similarly, from Eq. 2, we can obtain Eq. 5: dpsat dT Now by combining Eq. 4 and Eq. 5, we obtain: SL sv VL VV (Eq. 6) Is the above thermodynamic derivation correct or not? Explain your answer. VL = SV VV (Eq. 1) (Eq. 2) (Eq. 3) (Eq. 4) (Eq. 5)

For a single component system at vapor liquid equilibrium, we know that: dG¹ = -s¹dT + VLdpsat : 0 dGV = -sVdT + VV dpsat dGV = dGL 0 If we now re-arrange Eq. 1, we can obtain Eq. 4: = SL = dpsat dT Similarly, from Eq. 2, we can obtain Eq. 5: dpsat dT Now by combining Eq. 4 and Eq. 5, we obtain: SL sv VL VV (Eq. 6) Is the above thermodynamic derivation correct or not? Explain your answer. VL = SV VV (Eq. 1) (Eq. 2) (Eq. 3) (Eq. 4) (Eq. 5)

Introduction to Chemical Engineering Thermodynamics
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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For a single component system at vapor liquid equilibrium, we know that: dG¹ = −s¹dT + VLdpsat dGV = -sVdT + VV dpsat dGV = dG¹ we now re-arrange Eq. 1, we can obtain Eq. 4: = = SL = S dpsat dT Similarly, from Eq. 2, we can obtain Eq. 5: dpsat dT Now by combining Eq. 4 and Eq. 5, we obtain: SL sv (Eq. 6) VL VV Is the above thermodynamic derivation correct or not? Explain your answer. Pos B = 0 sv - 0 (Eq. 1) (Eq. 2) (Eq. 3) (Eq. 4) (Eq. 5)
Transcribed Image Text:For a single component system at vapor liquid equilibrium, we know that: dG¹ = −s¹dT + VLdpsat dGV = -sVdT + VV dpsat dGV = dG¹ we now re-arrange Eq. 1, we can obtain Eq. 4: = = SL = S dpsat dT Similarly, from Eq. 2, we can obtain Eq. 5: dpsat dT Now by combining Eq. 4 and Eq. 5, we obtain: SL sv (Eq. 6) VL VV Is the above thermodynamic derivation correct or not? Explain your answer. Pos B = 0 sv - 0 (Eq. 1) (Eq. 2) (Eq. 3) (Eq. 4) (Eq. 5)
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