Ethyl acetate (EA) has a vapor pressure of 118.3 mm Hg at 29.5 degree C, and its

Question

Ethyl acetate (EA) has a vapor pressure of 118.3 mm Hg at 29.5 degree C, and its normal boiling point is 70.0 degree C. A gas stream at 70.0 degree C and 2.10 atm absolute contains 35.0 mole % ethyl acetate and the balance nitrogen. The stream flows into a condenser, where its temperature and pressure are reduced to 29.5 degree C and 1.17 atm absolute, respectively, and some of the ethyl acetate is condensed. The pressures of the gas stream and ethyl acetate condensate flowing out of the condenser are both at 1.17 atm absolute at 29.5 degree C. The flow rate of the ethyl acetate condensate is 100.0 mol/h. What is the mole fraction, y_1, of ethyl acetate in the gas stream leaving the condenser?

Explanation / Answer

Writing overall balance,

Moles of mixturee at inlet= moles of mixture leaving = 100 mol/hr + moles of gas leaving

Writing nitogen balance, moles of nitrogen in the feed = 0.65*n0 = moles of nitrogen leaving

Leaving stream

At 29.5 deg,c , vapor pressure of ethyl acetate = 118.3 mm Hg=0.155 atm = partial pressure of ethyl acetate

partial pressure of ethyl acetate/ partial pressure of nitrogen = moalr flow rate of ethyl acetate/ molar flow rate of nitrogen

0.155/(1.17-0.155)= molar flow rate of ethyl acetate/ molar flow rate of nitrogen

Molar flow rate of ethy l acetate = molar flow rate of nitrogen* 0.153

Inlet

Entering stream is at 70 deg.c, vapor pressure = 760 mm Hg =1atm ( Normal boiling point)

Molar flow rate of ethyl acetate/ molar flow rate of nitrogen    = partial pessure of ethyl actate/ partial presure of nitrogen

Molar flow rate of ethyl acetate= Molar flow rate of nitrogen *1/(2.1-1)= 0.91* molar flow rate of nitrogen

Balance

Moles of ethyl acetate removed = moles of ethyl acetate entering/hr – moles of ethyl acetate leaving /hr= molar flow rate of nitrogen*(0.91-0.153)= 100

Molar flow rate of nitrogen =100/(0.91-0.153) =132 moles/hr

Nitroge does not under go any condensatino and is the tie substance.

At inlet molar flow rate of ethyl acetate = 0.91*132=120.12 moles/hr

Moles of ethyl acetate removed = 100 moles/hr

So the outlet gas contains 120.12- 100 = 20.12 moles/ hr of ethyl acetate

Molar flow rate of nitrogen at inlet = molar flow rate of nitrogen at outlet = 132 mole/hr

Composition of ethyl acetate in the outlet gas = 20.12/(132+20.12)=0.1322

Oulet gas temperature , T =25 deg.c =25+273=298K, P=1.17 atm, V= nRT/P, R =0.0821 L.atm/mole.K

=152.12*0.0821*298/1.17 L/hr =3181 L/hr= 3.18 m3/hr

Inlet gas molar flow rate = 132+120.12= 252.12, T = 70+273=343K, P= 2.1 atm, V= 252.12*0.0821*343/2.1=3381 L/hr= 3.381 m3/hr

Mole fraction ethyl acetate at inlet = 120.12/(132+120.12)= 0.48

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