3. (30 points total) Air in a 600-ft\' container is at ambient condition (25 C &

Question

3. (30 points total) Air in a 600-ft' container is at ambient condition (25 C &1 atm) with the initial mole fraction of water 0.06. In that container, 4.5 lbm dry silica gel are placed. The equilibrium curve for the Air-Silica gel system is Y-0.45X. Silica gel adsorbs water, therefore water is the only transferring component. Y is the ratio of the mass of water to the mass of dry air in the air phase, while X is the ratio of the mass of water to the mass of dry silica gel (dry solid) in the solid phase. Calculate the initial and final partial pressure of water in the air and the final moisture content of the solid phase in terms of weight fraction

Explanation / Answer

mole fraction of water=0.06, partial pressure of water= mole fraction* total pressure= 0.06*1=0.06 atm

volume of air = 600 ft3, 1ft3= 28.32 L, 600ft3= 600*28.32 L=16992 L

T =25deg.c= 25+273= 298K, P= 1atm ,R =0.0821 L.atm/mole.K

n= no of moles = PV/RT =1*16992/(0.0821*298) moles =694.5 moles

moles of air = mole fraction of air* total moles = (1-0.06)*694.5 =652.85 moles

mass of dry air = moles* molar mass =652.85*29 =18993 gm

moles of water= 0.06*694.5*18 gm =750.1gm

mass of silicsa =4.5 lbm, 1 lbm=0.4535 kg and 4.5 lbm= 4.5*0.4535kg*1000gm/kg = 2041 gm

let x= mass of water transferred to silica phase

Equilibrium relation is Y= 0.45X

(750.1-x)/18993 =0.45*x/2041

750.1/18993- x/18993 = 0.45x/2041

0.0394-5.3*10-5x= 0.00022048x

x*(0.00022048+5.3*10-5)= 0.0394

x= 144 gm

moles of water in the air phase = (mass of total water in air- mass of water transferred to silica)/18= 33.67 moles

moles of dry air = 652.85

moisture content ( weight fraction) = mass of water now/ total mass = 144/(18993+144)= 0.0075

mole fraction of water vapor now= 33.67/(652.85+33.67)= 0.049

partial pressure of water now ( i.e after adsorption in silica)= 0.049*1=0.049 atm

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