Termodinamica

CALCULOS

El punto azeotropico se obtiene de la grafica de T vs x1y1, de la cual se obtienen los siguientes datos:
T= 45.5 ºC----------------------- 318.65 K
x1=y1=0.63



Cloroformo
LnPsat/ mmHg = 15.9732 – [(2696.79/318.65)-46.16]= 6.0763

Psat/ mmHg= 435.41 mmHg


Metanol
Ln Psat/ mmHg = 18.5875 – [(3626.79/318.65)-34.29]= 5.834
Psat/ mmHg= 341.765 mmHg


Para 1
1= (x1P )/(x1 Psat-cloroformo)-  = ((0.63)(585) )/((0.63)(435.41)) = 1.3434

Para 2
2 = (x1P )/(x1 Psat-metanol)  = ((0.63)(585) )/((0.63)(341.765)) = 1.7117A manera de comprobación se suman los valores de y1 con el de y2 los cuales deben de sumar 1.
Para x1=0.1 ; x2=0.9

y1= ([(x1)(Psat)(y1)])/P  = ([(0.1)(559.858)(2.959)])/585 = 0.2832y1 y2
0 1
0.2832 0.7168
0.4378 0.5622
0.5236 0.4764
0.5725 0.4275
0.6009 0.3991
0.6187 0.3813
0.6343 0.3657
0.6607 0.3393
0.7303 0.2697
0.999 0.001
y2 = 1- 0.2832 = 0.7168Calcular In y1 ; In y2 con ecuaciones de Van Laar
A1-2 = 1.2
A2-1= 1.9
Ln y1 =(A 1-2)/((1+〖((A 1-2)(x1))/((A2-1)(x2)))〗^2 )  1.2/((1+〖((1.2)(0))/((1.9)(1)))〗^2 )= 1.2

Ln y1 =(A 2-1)/((1+〖((A 2-1)(x2))/((A1-2)(x1)))〗^2 )  1.9/((1+〖((1.9)(1))/((1.2)(0)))〗^2 ) = 0

Tabla de resultados de la ecuación de Van Laar
Ln y1 Ln y2
1.2 01.0478 0.0082
0.8950 0.0353
0.7432 0.0862
0.5942 0.1668
0.4507 0.2847
0.3164 0.4496
0.1961 0.6743
0.0965 0.9751
0.0268 1.3740
0 1.9


Calculo de In y1; y2 con la ecuación de MargulesA1-2 = 1.3
A2-1= 2.2

Para y1
Ln y1 = x22 [ A1-2 + 2x1(A2-1 – A1-2)]
Ln y1 = 12 [ 1.2 + 2(0)( 1.9 – 1.2)]= 1.2

Para y2
Ln y2 = x12 [ A2-1 + 2x2(A1-2 – A2-1)]
Ln y2 = 02 [ 1.9 + 2(1)(1.2 –1.9)] = 0
Tabla de resultados de la ecuación de Van Laar
Ln y1 Ln y2
1.2 0
1.0854 0.0064
0.9472 0.0312
0.7938 0.0828
0.6336 0.1696
0.475 0.3
0.3264 0.4824
0.1962 0.7252
0.0928 1.0368
0.0246...