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Unformatted text preview: Solutions to Chemical and Engineering Thermodynamics, 3e 6
6.1 (a) By Eqn. (6.23) 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. (6.23) 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.
 Spring '07
 Rethwisch

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