Mass Transfer Questions and Answers – Diffusion in a Binary Gas Mixture

This set of Mass Transfer Multiple Choice Questions & Answers (MCQs) focuses on “Diffusion in a Binary Gas Mixture”.

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.
a) 2.713
b) 1.713
c) 0.1713
d) 0.01713
View Answer

Answer: b
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.

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.
a) 4.242*10-4
b) 4.242*10-5
c) 4.242*10-3
d) 4.242*10-2
View Answer

Answer: a
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.

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
a) NA=2NB
b) 2NA=NB
c) NA=4NB
d) 4NA=NB
View Answer

Answer: a
Explanation: Latent heat of vaporization of A is provided from condensation of B. NA∆HA=NB∆HB.
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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.
a) 4.73h
b) 14.6h
c) 5.87h
d) 16.2h
View Answer

Answer: a
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.

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?
a) 3.64h
b) 4.73h
c) 5.94h
d) 2.36h
View Answer

Answer: a
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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6. The molar average velocity of the components in a binary mixture in which they are in equimolar counter diffusion is
a) Equal to mass average velocity
b) Zero
c) Always negative
d) Always positive
View Answer

Answer:b
Explanation: NA=-NB.

7. A cylinder of sulfur is kept in a large volume of stagnant air for sublimation. How much time will be required for its complete sublimation?
a) One fourth the time required by sphere of same mass
b) Half the time required by a cube of same mass
c) One fourth the time required by a cube of same mass
d) Infinite
View Answer

Answer: d
Explanation: No or negligible mass transfer in stagnant air.
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8. For a gas mixture (A+B) equimolar counterdiffusion takes place. Which among the following statements are correct for molar flux?
a) If temperature is doubled, the flux will get halved
b) If diffusion length is halved, the flux will get halved
c) If temperature is halved, the flux will get halved
d) If temperature is doubled and length is halved, the flux will become four times
View Answer

Answer: a
Explanation: NA=(D*(P1-P2))/RTL……….(for equimolar counter diffusion).

Sanfoundry Global Education & Learning Series – Mass Transfer.

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To practice all areas of Mass Transfer, here is complete set of 1000+ Multiple Choice Questions and Answers.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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