Ethanol vapour is being absorbed from a mixture of alcohol vapour and water vapo

Question

Ethanol vapour is being absorbed from a mixture of alcohol vapour and water vapour by means of a non-volatile solvent in which alcohol is soluble but water is not. The temperature is 97 °C, and the total pressure is 760 mm Hg. The alcohol vapour can be considered to be diffusing through a film of alcohol-water vapour mixture 0.1 mm thick. The mole percent of the alcohol in the vapour at the outside of the film is 80 percent, and that on the inside, next to the solvent, is 10 percent. The diffusivity of alcohol-water vapour mixture at 25 °C and 1 atm is 0.15 cm'/s. Calculate the rate of diffusion of alcohol vapour in kilograms per hour if the area of the film is 10 m*

Explanation / Answer

the problem is the case of diffusion of component ( ethanol) throuh non-diffusing component (Water)

NA( molar flux)= CDAB*ln{(1-y2)/(1-y1}/(Z2-Z1)

C= total concentration = P/RT, P= pressure in atm=1 atm, R= 0.0821 L.atm/mole= 0.0821*10000cc.atm/mole=82.1 cc.atm/mole, T= 25 deg,c= 25+273= 298K, C= 1/(82.1* 298)= 4.1*10-5 moles/cc, Z2-Z1=0.1mm =0.01cm

y2= mole fraction of ethanol at lower end= 0.1, y1= mole fraction of ethanol at higher end= 0.8

DAB= 0.15 cm2/s, NA = 4.1*10-5* 0.15* ln{(1-0.1)/(1-0.8)}/0.01 =0.000922 moles/cm2.s

given area =10m2, 1 m= 100cm, 10m2= 10*100*100= 100000 cm2

Hence rate of transfer = 0.000922*105 moles/s=92.2 moles/sec

in 1 hr rate of transfer = 92.2*3600 moles/hr =331974.3 moles/hr

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