90℃ toluene 53 kPa o-xylene 20 kPa 0.50 atm 25℃

90℃ toluene 53 kPa o-xylene 20 kPa 0.50 atm 25℃

 

 

At 90℃ the vapour pressure of toluene (methylbenzene) is 53 kPa

and that of o-xylene (1,2-dimethylbenzene) is 20 kPa.

What is the composition of the liquid mixture

that boils at 25℃ when the pressure is 0.50 atm?

What is the composition of the vapour produced?

 

---------------------------------------------------

 

Let, toluene = a, o-xylene = b.

 

 

 

돌턴의 분압법칙

P_tot = P_a + P_b ... (1)

( 참고 https://ywpop.tistory.com/48 )

 

 

 

라울의 법칙

P_a = X_a × P°_a ... (2)

( 참고 https://ywpop.tistory.com/2646 )

 

 

 

1 atm = 101.325 kPa 이므로,

0.50 atm × (101.325 kPa / 1 atm) = 50.66 kPa

≒ 51 kPa

---> P_tot

 

 

 

(2)식 → (1)식

P_tot = X_a × P°_a + (1 – X_a) × P°_b

( X_a + X_b = 1 )

 

51 = 53X_a + (1 – X_a)20

 

51 = 53X_a + 20 – 20X_a

 

51 – 20 = (53 – 20)X_a

 

X_a = (51 – 20) / (53 – 20) = 0.94

---> X_b = 1 – 0.94 = 0.06

---> composition of the liquid mixture

 

 

 

X_a(vap) = P_a / P_tot

= (X_a × P°_a) / P_tot

= (0.94 × 53) / 51

= 0.98

---> X_b(vap) = 1 – 0.98 = 0.02

---> composition of the vapour