Answer:
a) boiling pressure:
P = 0.759 atm
b) mole fractions in the vapor phase:
Xethanol(v) = 0.2192
Xwater(v) = 0.7808
Explanation:
Raoult's law:
- Pa = (P*a)×(Xa,(l))
- Xa(l) + Xb(l) = 1
∴ a: water
∴ b: ethanol
∴ Xb(l) = 0.1238
⇒ Xa(l) = 1 - 0.1238 = 0.8762
vapor pressure of the pure components at T = 85.3°C:
∴ P*a(85.3°C) ≅ 400 torr (0.53 atm).....from literature
∴ P*b(85.3°C) ≅ 800 torr (1.053 atm).....from literature
total pressure of the mix:
⇒ Pt = ((0.53 atm)(0.8762)) + ((1.053 atm)(0.1238))
⇒ Pt = 0.5947 atm
boiling point:
- Xb(l) = Xa(l) = 0.5
- P = (P*b - Pt)(0.5) + P*a
⇒ P = ((1.053 atm - 0.5947 atm)(0.5)) + 0.53 atm
⇒ P = 0.759 atm
assuming that the gas system is ideal:
- (Xb(v))(Pt) = (P*b)(Xb(l))
fraction in the vapor fase, ethanol:
⇒ Xb(v) = (P*b)(Xb(l)) / Pt
⇒ Xb(v) = ((1.053 atm)(0.1238)) / (0.5947 atm)
⇒ Xb(v) = 0.2192
∴ Xa(v) = 1 - Xb(v)
⇒ Xa(v) = 1 - 0.2192 = 0.7808
