(a)方法1 45-15x2 H1=H+x2,=H+(1-x1) dh dx F1=150-45x-5x+(-x)-45-15x2) H,=105-15x2+10x3J/mol H h=H X1 H2=150-45x1-5x3-x1(-45-15x H=150+10x,J/mol (D)
( ) 1 1 1 1 2 1 dx dH H x dx dH H = H + x = + − 1 2 1 dx dH H = H − x 2 1 1 45 15x dx dH = − − ( )( ) 2 1 1 3 H1 =150− 45x1 −5x1 + 1− x − 45−15x H x x J / mol (C ) 3 1 2 1 =105−15 1 +10 ( ) 2 1 1 3 H2 =150− 45x1 −5x1 − x − 45−15x H x J / mol ( D ) 3 2 =150+10 1 (a) 方法1
(a)方法2 H=150-45x,-5x3 nH=150n-45x,n-5x nH=150n-451-5-2 a(nH) an 150 45-5×3 3-2On H 5 T P an an H=n1+ O
( ) 1 3 3 2 1 2 1 1 1 1 2 5 1 150 45 5 3 2 n n n n n n n n n nH H T ,P,n − − − − = = (a) 方法2 3 1 1 H =150 − 45x −5x 2 3 3 1 1 150 45 5 n n nH = n − x n − x 2 3 1 150 45 1 5 n n nH = n − n − n = n1 + n2 1 1 1 2 = = n n n n
H,=105-5×3x2+10x3 nH=150n-45n,-5 2 an 2 an H 150 5n n' an T P H=150+10x3
( ) 2 3 3 1 2 2 2 2 150 5 1 n n n n n n n nH H T ,P,n − − = = 2 3 1 150 45 1 5 n n nH = n − n − 3 2 150 10 1 H = + x 3 1 2 H1 =105−53x1 +10x
(b)H=150-45x-5x3J/mol B H,=150-45×1-5×1=100J/mol H=150-45×0-5×03=150J/mol (c) Hi=lim H,=105//mol 0 H2=lim H2=lim H2=150+10=160//mol x2->0
(b) H x x J / mol (B) 3 1 1 =150− 45 −5 H 150 45 1 5 1 100 J / mol 3 1 = − − = (c) H lim H J / mol x 1 105 0 1 1 = = → H H H J mol x x lim lim 150 10 160 / 2 1 2 0 2 2 1 = = = + = → → H 150 45 0 5 0 150 J / mol 3 2 = − − =
4.2.3 Gibbs- Duhem方程 nM= ∑ d(nM)=∑M)+∑(Mh)(4-18) nM=fT, P, n, n2 a(nM) dT+ anM dP ++ ∑(Mah) aT IP n aP Tn oM OM d(nm=n dT+ dp t aT Px aP ∑(Ma)(4-19) T
4.2.3 Gibbs-Duhem 方程 nM =ni Mi ( ) =( )+( ) (4−18) d nM ni dMi Mi dni nM f (T,P,n ,n , ,n , ) = 1 2 i ( ) ( ) ( ) +( ) + = i i P,n T ,n dP M dn P nM dT T nM d nM( ) + ( ) (4 −19) + = i i P,x T ,x dP M dn P M dT n T M d nM n