1. For what kind of mixtures DAB=DBA holds?
a) Ideal
b) Real
c) For both real and ideal
d) This relation is never true.
Explanation: Ideal
2. In the expression for molar flux NA=(NA+NB)CA/C – DABdCA/dz, the terms representing bulk flow and molecular diffusion are, respectively
a) (NA+NB)CA/C, DABdCA/dz
b) (NA+NB)CA/C, DAdCA/dz
c) (NA+NB)CA/C, (NA+NB)CA/C
d) (DABdCA/dz, DABdCA/dz)
Explanation: (NA+NB)CA/C, DABdCA/dz
3. Consider loss of ethanol vapor by diffusion from a half-filled open test tube. At what point in the diffusion path will the contribution of the bulk flow term to the molar flux be maximum?
a) At the liquid-gas interface
b) In the bulk liquid
c) In the bulk gas
d) None of the mentioned
Explanation: At the liquid-gas interface concentration will be maximum. Bulk liquid is not a part of the diffusion path.
4. A reaction 3A+2B+C=1.5D+E occurs in a reactor at steady state. What will be the value of the flux ratio NA/ND
a) -2
b) -0.5
c) 2
d) -0.5
Explanation: NA/ND= (rate of disappearance of A)/(rate of appearance of D)
= -3/1.5
= -2.
5. For a component A of a mixture, concentration as a function of x is given:
CA=5e-10x (x is in cm and CA in mol/liter)
Calculate the value of diffusion velocity (m/s)of the component A at the point x=0, if diffusivity of A in the mixture is 2.567*10-5m2/s.
a) 0.2567
b) 2.567
c) 0.0025
d) 3.541
Explanation: By Fick’s law, JA=(-)DAB*dC/dx
CA(uA-U)= (-)DABdC/dx……1
Diffusion velocity= (uA-U)=(-DABdC/dx)/CA
dC/dx = -50e-10x
CA= 5 mol/lit
Put all the values in equation 1.
6. A sheet of Fe 1.0 mm thick is exposed to an oxidizing gas on one side and a deoxidizing gas on the other at 725°C. After reaching steady state, the Fe membrane is exposed to room temperature, and the C concentrations at each side of the membrane are 0.012 and 0.075 wt%. Calculate the diffusion coefficient (m2/sec) if the diffusion flux is 1.4×10-8kg/m2-sec.
a) 9.87*10-12
b) 9.87*10-13
c) 9.87*10-11
d) 9.87*10-10
Explanation: Convert from wt% to kg/m3 .
Concentration in kg/m3=[C1/(C1/ρ1+C2/ρ2)]*1000
0.012% and 0.075% in kg/m3 are 0.270 and 1.688 respectively.
Putting all the values in the equation JA=(-)DABdC/dx.
7. At which condition molar flux with respect to a stationary observer and with respect to an observer moving with molar average velocity?
a) In a very dilute solution
b) In a highly concentrated solution
c) At moderate concentration
d) Never
Explanation: In a very dilute solution, the contribution of the bulk flow term becomes negligible.
NA= (NA+NB)CA/C – DABdCA/dz (general equation)
(NA+NB)CA/C =0 (in dilute solution)
NA= -DABdCA/dz = JA.
8. Which among the following is the statement of the ‘Fick’s Law’?
a) The molar flux of species relative to an observer moving with the molar average velocity is proportional to the concentration gradient of the species.
b) The mass flux of species relative to an observer moving with the molar average velocity is proportional to the concentration gradient of the species.
c) The molar flux of species relative to an observer moving with the mass average velocity is proportional to the concentration gradient of the species.
d) The molar flux of species relative to a stationary observer is proportional to the concentration gradient of the species.
Explanation: The molar flux of species relative to an observer moving with the molar average velocity is proportional to the concentration gradient of the species.
9. The gas A diffuses through non-diffusing B from point 1 to point 2. The total pressure is 2atm and yA1=0.1 and yA2=0.Then the ratio (dPA/dz)1/(dPA/dz)2 is
a) 10
b) 1.11
c) 0.9
d) 2.3
Explanation: 0.9
10. In industries titanium is hardened through diffusion of carbon. The concentration of carbon at 1mm into the surface of the titanium slab is 0.25kg/m3 and at 3mm the concentration is 0.68kg/m3. The rate at which the carbon is entering into its surface is 1.27*10-19kg/m2.s. calculate the value of diffusion coefficient of carbon.
a) 5.91*10-12
b) 5.91*10-10
c) 5.91*10-11
d) 5.91*10-13
Explanation: J=-D(∂C/∂x)
∂C/∂x= Δc/Δx=(0.25-0.68)/(0.003-0.001)=-215kg/m4
1.27*10-19kg/m2.s= -D*(-215kg/m4)
D=5.91*10-12m2/s.