Question

Using Raoult’s law, estimate the boiling pressure and mole fractions in the vapor phase that is in equilibrium with a liquid having 0.1238 mole fraction of ethanol, in a mixture ethanol/water at 85.3o C.

Answer 1

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

• Pt = Pa + Pb

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