Test: Diffusion In A Binary Gas Mixture


5 Questions MCQ Test Mass Transfer | Test: Diffusion In A Binary Gas Mixture


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QUESTION: 1

Oxygen diffuses through a stagnant layer of air, 1mm thick, ambient temperature 28°C and 1atm total pressure. The partial pressure of oxygen on two sides of layer is P1=0.9atm and P2=0.1atm respectively. Calculate the value of Molar flux (in mol/m2.s) with respect to an observer moving with molar average velocity. Calculate this value at the end of the path where P2=0.1atm. Use R=8.205*10-5m3atmK-1mol-1

Solution:

Explanation: This is a case of diffusion of A through non-diffusing B.
Molar flux of oxygen, NA= (DAB*P ln⁡[(P-P2)/(P-P1)])/RTl
Putting all the required values in the above equation.
NA=1.904 mol/m2.s
Molar average velocity, U= (NA+NB)/C
U=NA/C (NB=0)
U=NART/P
U=4.702*10-2m/s
Now, UA=UC/CA
UA=U/yA=0.4702 m/s (yA is mole fraction of A at the end)
JA=CA(uA-U)
=PA*( uA-U)/RT
=1.713mol/m2.s.

QUESTION: 2

In order to prepare Lithium nitride, air is passed gently over a reactor. Suppose that Nitrogen diffuse to a length 5cm before getting reacted with Lithium. Nitrogen reacts with lithium quickly so that partial pressure of Nitrogen at that point is 0. Reactor is a cylinder of length 10cm and diameter 2cm. Assume air to be a mixture of Nitrogen and Oxygen only. The temperature is 298K and total pressure 1atm. Diffusivity of N2 in O2 is 2*10-5m2/s. Calculate the Molar flux (mol/m2.s) of N2 at a distance of 2.5cm from the top with respect to an observer moving with twice the mass average velocity a direction towards the liquid surface. (R=8.205*10-5m3atmK-1mol-1). Assume z=0 at the top of the reactor. 

Solution:

Explanation:
Case: Diffusion of A through non-diffusing B
N2 in air= 79%
O2 in air= 21%
Molar flux of N2, NA=0.025mol/m2.s
Pressure of N2 at the point z=2.5cm=
( DAB*P ln⁡[(P-P5)/(P-P0)])/RTl=(DAB*P ln⁡[(P-P5)/(P-P2.5)])/RTl
P2.5=0.542atm
Velocity of N2= NART/P2.5=112.78*10-5m/s
Molar flux of O2-=0
Velocity of O2=0
Average molar mass=0.542*28+0.458*32=29.832
Mass average velocity= (0.542*28*112.78*10-5)/29.832=57.37*10-5m/s
Required molar flux at the distance of z=5, is
CA(uA-2V)= NA-2PAV/RT=4.242*10-4mol/m2.s.

QUESTION: 3

A binary liquid mixture is separated by distillation process. The more volatile A vaporizes while less volatile B gets transported in opposite direction. Latent heat of vaporization of A and B is 0.5unit and 1unit respectively. What will be the relation between molar flux of A and B 

Solution:

Explanation: Latent heat of vaporization of A is provided from condensation of B. NA∆HA=NB∆HB.

QUESTION: 4

 In an accident a container containing ethanol(s.g 0.789) falls down in a room of dimension 1*1*2(all in m). Ethanol forms a layer of 2mm. Ethanol vaporizes and diffuses through a stagnant film of air of thickness 2.5mm. What will be the time required for the water layer to disappear completely. The diffusivity of ethanol in air is 2.567*10-5 m2/s. the total pressure is 1.013 bar and the pressure of ethanol in bulk air is 0.02244 bar and at the ethanol-air interface is 0.02718bar. The temperature is 25.2°C.

Solution:

Explanation: This is the case of diffusion of through non diffusing B.
NA=DP/RTZ*ln⁡[(P-P1)/(P-P2)] NA=(2.567*10-5*1.013)/(0.08317*298.2*2.5*10-3)*ln⁡[(1.013-0.02244)/(1.013-0.02718)] = 2.01*10-6 kmol/m2.s=9.259*10-5kg/m2.s
Amount of ethanol = 2*10-3*1=0.002m3=1.578kg
Time for evaporation= 1.578/9.259*10-5=17042.87s=4.73h.

QUESTION: 5

 In the previous question if the floor has micro pores and water penetrates the floor at a constant rate of 0.1kg/m2h, what will be the time required for the water layer to disappear? 

Solution:

Explanation: Combined rate of disappearance= 0.1+9.259*10-5*3600=0.433kg/m2h
Time for disappearance= 1.578/0.433=3.64h.

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