A stream containing 5 % mol acetone (1) and 95 % mol dichloromethane (2) is fed

Question

A stream containing 5 % mol acetone (1) and 95 % mol dichloromethane (2) is fed to an agitated tank at a flow rate of 5000 cm3/min. It is diluted with a stream of pure acetone to a final concentration of 45 % mol of acetone. What are The volumetric flow rates of The acetone and product streams? What is The partial molar volume of acetone in The outlet stream? The pure component volumes are 73.4 cm3/mol and 64.3 cm3/mol for acetone and dichloromethane respectively. The following expression applies for The volume change of mixing: deltaV mix = (0.5118 - 0.0678X1 + 0.1964X!2) xxx2 cm3/mol

Explanation / Answer

basis =1 mole of mixtrue entering the tank

moel of acetone =0.05 moles of dichloromethane

volume of 1 mole of mixture =0.05*73.4 +0.95*64.3=64.755 cm3

volumetric flow rate entering the tank=5000cm3/min

moles of mixture entering = 5000/64.755=77.21 moles/min ,moles of acetone = 77.21*5/100=3.86 moles/min. mole of dichloromethane entering =77.21*0.95= 77.21-3.86=73.35 moles/min

let x be the mole of acetone/min added and y the mixed products stream/min

there are two stream entering ( mixture of acetone and dichloroomethane ), pure acetone and one stream leaving ( 45% mol acetone and 55% mo dichloromethane

mass balance gives input =out put ( steady state condition)

overalll balance gives

77.21+x=y (1)

x=y-77.21 (2)

making dichloromethane balance gives

73.35 =x*0 ( no dichloromethane in pure acetone stream)+0.45y

y= 73.35/0.45=163 mole/min

moles of products =163mol/min

from Equation (2)

x= 163-77.21=85.79 mole/min,volumetricc flow rate = 85.79 mol/mi * 73.4 cm3/mol=6297 cm3/min

volume of 1 mol of 45% mol acetone mixture = 0.45mol*73,4 cm3/mol +0.55*64.3 mol cm3/mol=68.4 cm3

moles of 45% mol mixture leaving = 163.72 mol/min * 68.4 cm3/mol=11198.45 cm3/min

delVmix= (0.5118-0.0678x1+0.1964x12)x1x2

when x1=0.45

delvmix=0.1289 = x1V1- +x2V2- - {x1V1+x2V2)

V1- and V2 -represent partial molar volume and V1 and V2 represent pure component volumes

0.1290 = 0.45*V1- +0.55*V2- (0.45*73.4 +0.55*64.3)=0.45V1- +0.55V2- -68.3

0.45V1- +0.55V2- =0.1290+68.3 =68.43 (2)

(0.45/0.55)+0.55/0.55V2= 68.43/0.55 =0.82V1- +V2- =152.1 (3)

for x=0.55(assumed value)

delvmix= 0.132145 = 0.55*V1- +0.45*V2- -(0.55*73.4+0.45*64.3)= 0.55V1- + 0.45V2- 69.3

0.55V1- +0.45V2- =69.3 (4)

(0.55/0.45)V1-+(0.45/0.45)V2-= 69.3/0.45=154 , 1.22V1-+V2-= 154 (5)

subtracting Eq. (5) and Eq.4 gives

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