a) A vapor mixture of 0.200 moles of methanol(1) and 0.300 moles of ethanol(2) i

Question

a) A vapor mixture of 0.200 moles of methanol(1) and 0.300 moles of ethanol(2) is compressed at 20 oC. Calculate the pressure at which the first liquid forms and the mole fraction of methanol in the first liquid (x1).

b) A liquid mixture of 0.200 moles of methanol and 0.300 moles of ethanol is decompressed at 20 oC. Calculate the pressure at which the first vapor forms and the mole fraction of methanol in the first vapor (y1).

c) A flask contains 0.200 moles of methanol and 0.300 moles of ethanol at 20 oC and 0.0800 bar total pressure. Calculate the mole fraction of methanol in the liquid (x1) and in the vapor (y1). Also calculate the number of moles of methanol in the liquid phase (n1liquid).

Data: The vapor pressures of pure methanol(1) and pure ethanol(2) at 20 oC are p1* = 0.1183 bar and p2* = 0.0593 bar.

Explanation / Answer

a) According to Raoult's law, at at a given temperature, the vapour pressure of a volatile component in an ideal mixture is equal to the mole fraction of the component in the mixture multiplied by vapour pressure of pure component.

Number of moles of Methanol= 2

Number of moles of ethanol= 3

Total number of moles= 5

According to Raoult's law, partial pressure of methanol

PMeOH = XMeOH. P0MeOH

XMeOH= mole fraction of methanol in the mixture and P0MeOH is the vapour presure of pure methanol at 20 degrees

XMeOH= Mole fraction of methanol= 2/5

Partial pressure of MeOH= (2/5) X 44.48 Torr (where 0.0593 Bar= 44.48 torr approx) = 17.792 Torr

Similarly, XEtOH= 3/5

PEtOH= (3/5) X 88.73 (where 0.1183 bar= 88.73 torr approx) = 53.24 Torr

Ptotal= PMeOH + PEtOH = 71.03 torr

XMeOH = 17.792/71.03 = 0.25

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