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Give some typical values of Molecular Diffusivity in gases, liquids and solids A

ID: 473603 • Letter: G

Question

Give some typical values of Molecular Diffusivity in gases, liquids and solids A vapour mixture of hydrocarbons A and B is being rectified by contact with a liquid mixture of the same components. Component A is transferred from the liquid to the vapour phase, while component B is transferred in the opposite direction. Both components are diffusing through a gas film 0.1 mm thick. The temperature is 95 degree C and the pressure is one atmosphere. At this temperature the latent heats of vaporisation may be assumed to be equal. The mole fraction of component A is 0.8 on one side of the film and 0.2 on the other side of the film. Calculate the rate of diffusion of A and B in kg hr^-1 through 1 m^2 of area. DATA: Diffusivities at 95 degree C and 1 am = 0.1986 cm^2 s^-1. 1 kmole of ideal gas occupies 22.4 m^3 at 273 K. and 1 atm. The molecular weight of A is 46 kg kmol^-1 The molecular weight of B is 60 kg kmol^-1

Explanation / Answer

The order of diffusivity for gases is 10-5 m2/s, for liquid it is 10-9 m2/s and for solids it is 10-12 m2/s

We know that NA= XA*(NA+NB)+JA

NA= Convective molar flux and JA= flux dues to molecular diffusion= -DAB*dCA/dZ

Where NA= -NB for equimolar counter diffusion

NA= JA= -DAB*dCA/dZ

Or NA*(Z2-Z1) =DAB*(CA2-CA1)

NA = DAB*(CA2-CA1)/(Z2-Z1)

Z2-Z1= thickness

NA = DAB*(PA2-PA1)/RT(Z2-Z1)

= {(DABP/RT*(Z2-Z1)} ( mole fraction of A at 2- mole fractin of A at 1)

= 0.1986cm2/s*1*(0.8-0.2)/(82.06cm3.atm/mole.K*(95+273)*0.1*1 cm= 3.94*10-5 moles/cm2.s=3.94*10-5*10-4 moles/m2.s =3.94*10-9 moles/m2.s

For 1m2 area, rate of diffusion of A = 3.94*10-9 moles/s

Rate of diffusion of A = 3.94*10-9* 46g/gmole= 1.815*10-7 gm/s = 1.815*10-10 kg/s

Since NA=-NB, Rate of diffusion of B = -3.94*10-9 moles/s, in terms of mass it is = -3.94*10-9*60 *10-3 kg/s=2.37*10-10 kg/s

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