Introductory Chemical Engineering Thermodynamics

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Unformatted text preview: Solutions to Chemical and Engineering Thermodynamics, 3e 6 6.1 (a) By Eqn. (6.2-3) FG H IJ HN K i = Gi ; P , S , N j i but Gi = Hi - TSi . Thus FG H IJ HN K i = Hi - TSi P , S , N j i (b) Since U = U S ,V , N f F U IJ dU = G H S K a FG U IJ dV + FG U IJ H V K H N K F U IJ dN = TdS - PdV + G H N K dS + V,N S ,N i i i i i S ,V , N ji dN i S ,V , N ji (1) However, we also have U = H - PV ; dU = dH - PdV - VdP , and, by Eqn. (6.2-3) dU = VdP + TdS + Gi dNi - PdV -VdP = TdS - PdV + Gi dNi Equating (1) and (2) shows that Gi = A = A( T ,V , N ) dA = or dA = - SdT - PdV + i (2) FG U IJ HN K i . Next we start from S ,V , N j i FG A IJ H T K dT + V,N FG A IJ H V K T, N dV + i FG A IJ H N K i dN i T ,V , N j i FG A IJ HN K i dN i T ,V , N ji (3) However, we also have that A = U - TS ; Solutions to Chemical and Engineering Thermodynamics, 3e Chapter 6 dA = dU - TdS - SdT = TdS - PdV + Gi dNi - TdS - SdT or dA = -SdT - PdV + Gi dNi Comparing (3) and (4) yields Gi = (4) FG A IJ HN K i T ,V , N ji ...
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This homework help was uploaded on 02/06/2008 for the course CBE 103 taught by Professor Rethwisch during the Spring '07 term at University of Iowa.

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