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Cambio de volumen molar a volumen total:

V  x1 V1  x2 V2  x1 x2 (45 x1  25 x2 ) n V  V t  n1 V1  n2 V2 



45 n12 n2 25 n1 n22  n2 n2

n1 V1 n2 V2 45 n12 n2 25 n1 n22    n n n3 n3

cm3 mol

cm3

Obtención de la expresión para el volumen molar parcial de (1):

  nV    n1

 

2  2 n 2 n1  2 n12 n  2  n  2 n1 n   25 n 2    n4 n4    



 V 1  110  0  45 n2  n2

 110  90 x1 x2  90 x12 x2  25 x22  50 x1 x22  110  90 x1 x2 (1  x1 )  x22 (25  50 x1 )  110  90 x1 x22  x22 (25  50 x1 )  110  x22 (90 x1  25  50 x1 )  110  x22 ( 40 x1  25) Obtención de la expresión para el volumen molar parcial de (2):

45 n12 n2 25 n1 n22  cm3 2 2 n n 2  n  2 n2 n   2 n 2 n2  2n22 n   V 2  0  90  45 n12   25 n 1   n4 n4    

n V  V t  n1 V1  n2 V2    nV    n2 



 n1

 90  45 x12  90 x12 x2  50 x1 x2  50 x1 x22  90  x12 ( 45  90 x2 )  50 x1 x2 (1  x2 )  90  x12 (45  90 x2 )  50 x12 x2  90  x12 ( 45  90 x2  50 x2 )  90  x12 (45  40 x2 ) Para

x1  0.4

x2  1  x1

x2  0.6 3

V  110 x1  90 x2  x1 x2 ( 45 x1  25 x2) 2

Vmp1  110  x2  ( 40 x1  25) 2

Vmp2  90  x1  ( 45  40 x2)

3

Vmp1  124.76

cm

mol 3

Vmp2  93.36

Comprobando por la regla de adicionabilidad: x1 Vmp1  x2 Vmp2  105.92

V  105.92

cm

mol

cm

mol

3

V1  110

n1 

n1V1 V1

n  n1  n2

cm

mol

3

cm

V2  90

moles de 1

n1  6.8182

n2 

moles de solución

n  23.4848

3

n1V1  750

mol

n2V2  1500

cm

n2V2

n2  16.6667

V2

x1 

n1

x1  0.2903

n

Volumen molar de solución preparada: 3

V  110 x1  90 x2  x1 x2 ( 45 x1  25 x2)

Volumen total de solución:

Vt  n  V

cm

V  102.1537

mol 3

Vt  2399.0635

cm

a) V 1  110  x22 ( 40 x1  25) V 2  90  x12 (45  40 x 2 ) b)

Regla de adicionabilidad:



V  x1 Vmp1  x2 Vmp2





?  x1 110  x22 (40 x1  25)  x 2 90  x12 (45  40 x2 )



?  110 x1  40 x12 x22  25 x1 x22  90 x2  45 x12 x 2  40 x12 x 2 22 V  110 x1  90 x2  x1 x2 (45 x1  25 x2

3

cm

moles de 2 n2 x2  x2  0.7097 n

c)

Ecuación de Gibbs-Duhen

x

i

x

d Mi 0

i

x1 d V 1  x2 d V 2  0

x1

d Mi 0 dV 1 dV 2  x2 0 dx1 dx1

dV 1  0  x22 (40)  (40 x1  25) 2 x2  40 x22  80 x1x2  50 x2 dx1

dV 2  0  x12 (40)  (45  40 x2 ) 2 x1  40 x12  90 x1  80 x1 x2 dx1









x1 40 x22  80 x1 x2  50 x2  x 2 40 x12  90 x1  80 x1 x2  ? 40 x1 x22  80 x12 x2  50 x1 x2  40 x12 x2  90 x1 x2  80 x1 x 22  ? 40 x1 x2  40 x 12 x2  40 x1 x 22  ? 40 x1 x2 (1  x1  x 2)  ? 40 x1 x2 (0)  ? 00

d)

dV 1 dx1

0

0

x1  0

x1  i

i1



x2  1  x1 i

10

V  110 x1  90 x2  x1  x2  45 x1  25 x2 i

i

i

i

 i 2 40x1i  25

Vmp1  110  x2 i

2

40 x1  90 x1  80 x1 x2  0

x2  1

x1 0

i  1  11

i

2

40 x2  80 x1 x2  50 x2  0

x2  0

x11

dV 2 dx1 e)

x1  1

i

i



Vid  110 x1  90 x2

i

i

 i 2 45  40x2i

Vmp2  90  x1 i

i

i

3

cm

mol

TABLA DE VALORES i

x1 

V 

i

1 2 3 4 5 6 7 8 9 10 11

Vid 

i

i

Vmp1  i

Vmp2  i

0

90

90

135

90

0.1

94.43

92

133.49

90.09

0.2

98.64

94

131.12

90.52

0.3

102.51

96

128.13

91.53

0.4

105.92

98

124.76

93.36

0.5

108.75

100

121.25

96.25

0.6

110.88

102

117.84

100.44

0.7

112.19

104

114.77

106.17

0.8

112.56

106

112.28

113.68

0.9

111.87

108

110.61

123.21

1

110

110

110

135

GRAFICA 140 135 130 Vi Vidi

125 120

Vmp1i115 Vmp2i110 105 100 95 90

0

0.1

0.2

0.3

0.4

0.5 x1i

0.6

0.7

0.8

0.9

1