Sistema binario tetracloruro de carbono y metanol

Sistema Tetracloruro de Carbono-Metanol

Datos Azeotrópicos:
P= 760 mm Hg
T= 55.7 °C
X1= 0.445
CCl4= X1

Constantes de Antoine:
| Tetracloruro de Carbono | Metanol |
A | 6.87926 | 8.08097 |
B | 1212.021 | 1582.271 |
C | 226.409 | 239.726 |

Cálculos generales:

log10P1sat=6.87926-1212.02155.7+226.409=2.58297

P1sat=102.58297=382.80 mm Hglog10P2sat=8.08097-1582.27155.7+239.726=2.72508

P2sat=102.72508=530.98 mm Hg

En Composición Azeotrópica:

γ1=PP1sat=760 mm Hg382.80 mm Hg=1.98537

γ2=PP2sat=760 mm Hg530.98 mm Hg=1.43132

Constantes de van Laar

A=log10γ11+X2log10γ2X1log10γ12 B=log10γ21+X1log10γ1X2log10γ22

A=Log101.985371+(0.555)(log101.43132)0.445(log101.98537)2=0.8129699

B=log10γ21+0.445(log101.98537)0.555(log101.41322=0.9995547Cálculos de grafico P Vs X1, Y1: (2 ejemplos para X1 y Y1)
P1sat= 382.80 mm Hg
P2sat= 530.98 mm Hg
A= 0.8129699
B= 0.9995547
X1= 0.2

log10γ1=AX22ABX1+X22=0.8129699(0.8)20.81296990.99955470.2+0.82=0.561439286

γ1=100.561439286=3.6428

log10γ2=BX12BAX2+X12=0.9995547(0.2)20.99955470.81296990.8+0.22=0.028539786

γ2=100.028539786=1.0679

Calcular la presión:

P=X1γ1P1sat+X2γ2P2satP=0.2382.800.561439286+0.8530.980.028539786

P=732.5196 mm Hg

Calcular Y1:

Y1=X1γ1P1satP=0.23.6428382.80732.5196

Y1=0.38073

Para X1= 0.6

log10γ1=AX22ABX1+X22=0.81296990.420.81296990.99955470.6+0.42=0.164956432

γ1=100.164956432=1.4620

log10γ2=BX12BAX2+X12=0.9995547(0.6)20.99955470.81296990.4+0.62=0.301869804

γ2=100.301869804=2.00387

Calcular la presión:P=X1γ1P1sat+X2γ2P2sat

P=0.6382.801.4620+0.4530.982.00387

P=761.40 mm Hg

Cálculos de grafico T Vs X1, Y1: (2 ejemplos para X1 y Y1)
P= 760 mmHg
A= 0.8129699
B= 0.9995547
Para X1= 0.2

T (°C) | P1sat | P2sat | γ1 | γ2 | P’ | P-P’ |
56.62 | 395.311 | 551.698 | 3.6428 | 1.0679 | 759.34 | 0.655598 |
56.64 | 395.5867 | 552.1554 | 3.6428 | 1.0679 | 759.93 | 0.073921 |
56.66 | 395.8623 |552.61366 | 3.6428 | 1.0679 | 760.52 | -0.51834 |

Para T= 56.62 °C
log10P1sat=6.87926-1212.02156.62+226.409=2.596939107

P1sat=102.596939107=395.311 mm Hg

log10P2sat=8.08097-1582.27156.62+239.726=2.741701037

P2sat=102.741701037=551.698 mm Hg

log10γ1=AX22ABX1+X22=0.81296990.820.81296990.99955470.2+0.82=0.561439286

γ1=100.561439286=3.6428log10γ2=BX12BAX2+X12=0.9995547(0.2)20.99955470.81296990.8+0.22=0.028539786

γ2=100.028539786=1.0679

P'=X1γ1P1sat+X2γ2P2sat

P'=0.2395.3113.6428+0.8551.6981.0679

P'=759.34 mm Hg

Para T= 56.64 °C
log10P1sat=6.87926-1212.02156.64+226.409=2.597241692

P1sat=102.597241692=395.5867 mm Hg

log10P2sat=8.08097-1582.27156.64+239.726=2.742061353

P2sat=102.742061353=552.1554 mm Hg

P'=X1γ1P1sat+X2γ2P2satP'=0.2395.58673.6428+0.8552.15541.0679

P'=759.93 mm Hg

Para T= 56.66 °C
log10P1sat=6.87926-1212.02156.66+226.409=2.597544235

P1sat=102.597544235=395.8623 mm Hg

log10P2sat=8.08097-1582.27156.66+239.726=2.742421621

P2sat=102.742421621=552.61366 mm Hg

P'=X1γ1P1sat+X2γ2P2sat

P'=0.2395.86233.6428+0.8552.613661.0679

P'=760.52 mm Hg

Calcular Y1:

Y1=X1γ1P1satP=0.23.6428395.5867760

Y1=0.37922P= 760 mmHg
A= 0.8129699
B= 0.9995547
Para X1= 0.6

T (°C) | P1sat | P2sat | γ1 | γ2 | P’ | P-P’ |
55.64 | 381.9969 | 529.6461 | 1.4620 | 2.0039 | 759.63 | 0.3692 |
55.66 | 382.2650 | 530.0886 | 1.4620 | 2.0039 | 760.22 | -0.2207 |
55.68 | 382.5333 | 530.5315 | 1.4620 | 2.0039 | 760.81 | -0.81 |

Para T= 55.64 °C
log10P1sat=6.87926-1212.02155.64+226.409=2.582059868P1sat=102.582059868=381.9969 mm Hg

log10P2sat=8.08097-1582.27155.64+239.726=2.723985784

P2sat=102.723985784=529.6461 mm Hg

log10γ1=AX22ABX1+X22=0.81296990.420.81296990.99955470.6+0.42=0.164956432

γ1=100.164956432=1.4620

log10γ2=BX12BAX2+X12=0.9995547(0.6)20.99955470.81296990.4+0.62=0.301869804

γ2=100.301869804=2.0039

P'=X1γ1P1sat+X2γ2P2sat

P'=0.6381.99691.4620+0.4529.64612.0039...