For Problems 2 and 3, use the n-hexane, n-octane data from Problem 1. Number 1 a

Question

For Problems 2 and 3, use the n-hexane, n-octane data from Problem 1. Number 1 answer is shown above on the graph. Please find problems 2 and 3!!

Problem 2 – A mixture of 100 mol containing 65% n-hexane and 35% n-octane is vaporized at 101.32 kPa abs pressure until 35 mol of vapor and 65mol of liquid are in equilibrium with each other. This occurs in a singlestage, closed system such that the liquid and vapor remain in contact untilcompleted. Calculate the composition of the liquid and the vapor.

Problem 3 –The starting mixture as Problem 3 is vaporized until 35 mol ofvapor are removed, but this is instead done under differential conditions (thevapor is removed from the system as it is produced). Calculate the averagecomposition of the distillate and the composition of the remaining mixture.

There should be no xa or ya just x and y.

ii.i-3. Boiling-Point-Diagram Calculation. The vapor-pressure data are given below for the system hexane-octane: Vapor Pressure n-Hexane n-octane T(OF) kPa mm Hg kPa mm Hg 760 155.7 68.7 101.3 16.1 121 173 136.7 79.4 175 1025 93.3 197.3 37.1 1480 200 278 107.2 2130 57.9 434 284.0 125.7 456.0 760 3420 101.3 (a) Using Raoult's law, calculate and plot the x-y data at a total pressure of 101.32 kPa. (b) Plot the boiling-point diagram

Explanation / Answer

mole fraction of n-hexanne=65%=0.65=x1

mole fraction of n-octane=0.35=x2

So composition of liquid is moles of hexane=0.65*65 moles liquid=42.25 moles

moles of octane=0.35*65=22.75 moles

Now,

Total pressure=101.32KPa=y1*p(hexane)+y2*p(octane) [Dalton's law of partial pressures, here p represents partial pressures and y1,y2 represents the molefraction of hexane and octane in vapour phase]

Also ,using Raoult's law,

p(hexane)=mole fraction* P*(hex)    [P* (hex),P*= vapour pressure of pure hexane and pure octane vapours,determined from table in problem 1]

and p(octane)=mole fraction* P*(oct)

So, p(hexane)=0.65* 197.3 KPa=128.245 kPa [T=93.3 deg C]

p(octane)=0.35*37.1 kPa=12.985 kPa

total pressure=128.245+12.985=141.23 kPa

Now,y1=p(hex)/P=128.245KPa/141.23kPa=0.91

and y2=p(oct)/P=12.985kPa/141.23kPa=0.09

hence,composition of hexane in vapour phase=0.91*35 moles =31.85 moles

and composition of octane in vapour phase=0.09*35 moles =3.15 moles

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