Para um dos sistemas binários listados na tabela, com base na Eq. (10.5) e na equação de Wilson, prepare um diagrama t - x - y para P = 101,33 k P a .

Passo 1

O arquivo WILSON.mcd reproduz a tabela de parâmetros de Wilson na página 474 e inclui os coeficientes Antoine necessários.

Coeficientes Antoine:

1 – propanol A 1 : = 16.1154         B 1 : = 3483.67 ⋅ K         C l : = 205.807 ⋅ K  

Água A 2 : = 16.3872         B 2 : = 3885.70 ⋅ K         C 2 : = 230.170 ⋅ K  

P s a t 1 ⁡ ( T ) : = e x p ⁡ A 1 - B 1 ( T - 273.15 ⋅ K ) + C 1 ⋅ k P a

P s a t 2 ⁡ ( T ) : = e x p ⁡ A 2 - B 2 ( T - 273.15 ⋅ K ) + C 2 ⋅ k P a

Parâmetros da equação de Wilson

V 1 : = 75.14 ⋅ c m 3 m o l V 2 : = 18.07 ⋅ c m 3 m o l

a 12 : = 775.48 ⋅ c a l m o l a 21 : = 1351.90 ⋅ c a l m o l

Λ 12 ( T ) : = V 2 V 1 ⋅ e x p ⁡ - a 12 R ⋅ T Λ 21 ( T ) : = V 1 V 2 ⋅ e x p ⁡ - a 21 R ⋅ T

γ 1 ( x 1 , x 2 , T ) : = e x p e x p   x 2 ⋅ Λ 12 ( T ) x 1 + x 2 ⋅ Λ 12 ( T ) - Λ 21 ( T ) x 2 + x 1 ⋅ Λ 21 ( T )   ( x 1 + x 2 ) ⋅ Λ 12 ( T )

γ 2 ( x 1 , x 2 , T ) : = e x p e x p   - x 1 ⋅ Λ 12 ( T ) x 1 + x 2 ⋅ Λ 12 ( T ) - Λ 21 ( T ) x 2 + x 1 ⋅ Λ 21 ( T )   ( x 2 + x 1 ) ⋅ Λ 21 ( T )

Passo 2

T-x,y diagrama P : = 101.33 ⋅ k P a

Achado: T : = ( 90 + 273.15 ) ⋅ K

Dados P = x 1 ⋅ γ l ( x 1,1 - x 1 , T ) ⋅ P s a t 1 ( T ) …

+ ( 1 - × 1 ) ⋅ γ 2 ( x 1,1 - x 1 , T ) ⋅ P s a t 2 ⁡ ( T )

T e q ⁡ ( x 1 ) : =   F i n d   ( T )

y e q ⁡ ( x 1 ) : = x 1 ⋅ γ 1 ( x 1,1 - x 1 , T e q ⁡ ( x 1 ) ) ⋅ P s a t 1 ⁡ ( T e q ⁡ ( x 1 ) ) P

x : = 0,0.05 … 1.0

Resposta

T ,   x , y   D i a g r a m a   e m   P : = 101.33 ⋅ k P a

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