Test: Molecular Diffusion In Liquids


10 Questions MCQ Test Mass Transfer | Test: Molecular Diffusion In Liquids


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QUESTION: 1

Before collision, the distance between the two particles is mean free path.

Solution:

Explanation: The mean free path can be found after collision.

QUESTION: 2

Diffusion co-efficient in molecular diffusion is estimated by _________

Solution:

Explanation: Ficks law is the ratio of J-flux to the concentration gradient.

QUESTION: 3

Find the total flux for a particle A and B in a steady state if flux of A and B is 2.44 and 4.44 mol/sq.m sec.

Solution:

Explanation: Total flux= flux A+ flux B= 6.88.

QUESTION: 4

Find the flux of A ( xA= 0.2) if Total flux is 5 mol/sq.m sec and J flux of A is 2 mol/sq.m sec.

Solution:

Explanation: N flux of A = J flux of A + Total flux N * xA
= 2+ 0.5*2 = 3.

QUESTION: 5

 Find J flux of A ( xA = 0.5) if the total N flux is 6 mol/sq.m sec and N flux of A is 3 mol/sq.m sec.

Solution:

Explanation: N flux of A = J flux of A + Total flux N * xA
3= J flux of A + 6 * 0.5
J flux = 0.

QUESTION: 6

 Diffusivity of the liquids can be determined by

Solution:

Explanation: Wilke- Chan derived an equation for diffusivity of liquids.

QUESTION: 7

For the molecular diffusion of liquids, if the diffusing molecules having criteria of steady state diffusion of B over non diffusing A then the N flux of A is

Solution:

Explanation: For non-diffusing A then N flux of B is zero.

QUESTION: 8

 For liquid molecular diffusion of A and B, steady state equimolar counter diffusion the N flux of A is negative of N flux of B.

Solution:

Explanation: For equimolar counter diffusion, the one component flux is negative of other.

QUESTION: 9

Find the change in concentration for a steady state equimolar counter diffusion if D(AB)= 6 sq.m/sec, the change in distance is 2 m and the N flux of A is 5 mol/sq.m sec.

Solution:

Explanation: N flux of A = D(AB)/z * (Concentration difference)
Concentration difference = 5/3 =1.67.

QUESTION: 10

Find the diffusivity of AB in sq.m/sec if N flux of A 5 mol/sq.m sec, concentration difference is 2mol/cu.m and distance is 3 m.

Solution:

Explanation: N flux of A= D(AB)/z * concentration difference
D(AB) = 7.5 sq.m/sec.

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