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INSTITUTO POLITÉCNICO NACIONAL

ESCUELA SUPERIOR DE INGENIERÍA QUÍMICA E INDUSTRIAS EXTRACTIVAS

DEPARTAMENTO DE INGENIERÍA QUÍMICA INDUSTRIAL

LABORATORIO DE TERMODINÁMICA QUÍMICA I

PRÁCTICA 4

AZEÓTROPOS

PROFESORA:
I.Q.I. ITZEL GUTIÉRREZ GONZÁLEZ

EQUIPO 3
INTEGRANTES:
GIRON NAVARRO ROCIO
GONZÁLEZ SÁNCHEZ MITZI MILENA

GRUPO:
3IM6

NOVIEMBRE DE 2010

TABLA DE DATOSEXPERIMENTALES

cloroformo | metanol | | | | | |
X1 | X2 | V1 | V2 | ηliq | T | ηvap |
0 | 1 | 0 | 80 | 1.3295 | 331.15 | 1.3295 |
0.1 | 0.9 | 14.5 | 65.5 | 1.3505 | 327.15 | 1.376 |
0.2 | 0.8 | 27 | 53 | 1.369 | 324.15 | 1.395 |
0.3 | 0.7 | 37 | 43 | 1.3845 | 322.15 | 1.409 |
0.4 | 0.6 | 46 | 34 | 1.3985 | 320.65 | 1.416 |
0.5 | 0.5 | 53 | 27 | 1.4085 | 320.05 | 1.419 |
0.6| 0.4 | 60 | 20 | 1.4185 | 319.85 | 1.422 |
0.7 | 0.3 | 66 | 14 | 1.4275 | 319.65 | 1.4255 |
0.8 | 0.2 | 71 | 9 | 1.4345 | 320.05 | 1.429 |
0.9 | 0.1 | 76 | 4 | 1.442 | 320.35 | 1.433 |
1 | 0 | 80 | 0 | 1.445 | 326.15 | 1.445 |

Constantes de antoine

| A | B | C |
cloroformo | 15.9732 | 2696.79 | -46.16 |
metanol | 18.5875 | 3626.55 | -34.29 |

Calcular las presiones desaturación con la ecuación de Antoine

lnPsat=BT+C

Psat(cloroformo) | Psatmetanol |
672.126738 | 584.721335 |
587.420558 | 494.862204 |
529.624142 | 435.334714 |
493.670211 | 399.090545 |
468.001032 | 373.603968 |
458.036033 | 363.800465 |
454.752206 | 360.58118 |
451.487172 | 357.385929 |
458.036033 | 363.800465 |
462.997175 | 368.674772 |
567.626228 | 474.305067 |Calcular y1 de grafica
y1 | y2 |
0 | 1 |
0.24 | 0.76 |
0.37 | 0.63 |
0.51 | 0.49 |
0.58 | 0.42 |
0.6 | 0.4 |
0.63 | 0.37 |
0.67 | 0.33 |
0.73 | 0.27 |
0.77 | 0.23 |
1 | 0 |

Calcular 1 y 2 a partir de la formula
i=yi*PXi*Pisat

Donde P= 585

Posteriormente calcular el logaritmo natural de y1 y y2

1 | 2 | ln1 | ln2 |
0 | 1.00047658 | 0 | 0.00047646 |
2.39011043| 0.99825769 | 0.87133957 | -0.0017438 |
2.04343026 | 1.05823745 | 0.7146299 | 0.05660475 |
2.01450275 | 1.02608294 | 0.70037239 | 0.02574858 |
1.812496 | 1.09608044 | 0.5947049 | 0.09174058 |
1.53263051 | 1.28641947 | 0.42698555 | 0.25186275 |
1.35073561 | 1.50070228 | 0.30064934 | 0.40593319 |
1.24018711 | 1.80057453 | 0.21526226 | 0.58810579 |
1.16543779 | 2.17083285 | 0.1530968 |0.7751109 |
1.08100012 | 3.6495581 | 0.07788665 | 1.29460609 |
1.03060777 | 0 | 0.03014869 | 0 |

Calcular A12 Y A21 graficando ln1 vs X1 y ln2 vs X2. Se tiene que

A21 | A12 |
1.8 | 1.24 |

Sustituir ambos valores en la ecuación de Van Laar para obtener ln1 y ln2
ln1=A12[1+A12*X1A21X2]2

ln1=A21[1+A21*X2A12X1]2
Despejar y obtener 1 y 2
VAN LAAR |
ln1 | 1 | ln2 | 2 |
1.24| 3.45561346 | 0 | 1 |
1.06993835 | 2.91519976 | 0.00685577 | 1.006879321 |
0.90240561 | 2.46552707 | 0.02988995 | 1.030341134 |
0.73913279 | 2.09411868 | 0.07358694 | 1.076362112 |
0.58231338 | 1.790175 | 0.14375206 | 1.154597799 |
0.43472992 | 1.54454585 | 0.24796767 | 1.281418497 |
0.29991938 | 1.34974998 | 0.39623647 | 1.486220727 |
0.18239121 | 1.20008358 | 0.60190294 |1.825589483 |
0.08791709 | 1.09189759 | 0.88299599 | 2.418133577 |
0.02391975 | 1.02420813 | 1.26422361 | 3.540342995 |
0 | 1 | 1.78 | 5.929856419 |
Sustituir A12, A21 en la ecuación de Margules para obtener ln1 y ln2

ln1=x22[A12+2x1A21-A12]

ln2=x12[A21+2x2A12-A21]

Sustituir valores de 1 en la formula
Yi=iXi*Pisat
Para obtener Y1 corregida
y1(van laar) | y1(Margules) |
0 | 0 |0.2927262 | 0.30019113 |
0.44642826 | 0.46212393 |
0.5301559 | 0.54800167 |
0.57285726 | 0.58757459 |
0.60466466 | 0.6139688 |
0.62954029 | 0.63332928 |
0.64833443 | 0.64818455 |
0.68393633 | 0.68224425 |
0.72954688 | 0.72849726 |
1 | 0.97030124 |

GRÁFICAS

X1 | ηliq | ηvap |
0 | 1.3295 | 1.3295 |
0.1 | 1.3505 | 1.376 |
0.2 | 1.369 | 1.395 |
0.3 | 1.3845 | 1.409 |...
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