# Bioseparation Technology Questions and Answers – Inter-Phase and Unsteady State Mass Transfer

This set of Bioseparation Technology Multiple Choice Questions & Answers (MCQs) focuses on “Inter-Phase and Unsteady State Mass Transfer”.

1. The transfer of solute from one medium to another medium across an interface is called _____________
a) Inter-phase mass transfer
d) Convective mass transfer

Explanation: Inter-phase or interfacial mass transfer is the transport of solute from one medium to another medium across an interface. The transport of mass is within a single phase and it is directly dependent on the concentration gradient of the species being transported in that specific phase.

2. Which separation process of bioseparation involves the inter-phase mass transfer of solutes?
a) Liquid-liquid extraction, membrane separation
b) Liquid-liquid extraction, membrane separation, leaching, chromatography
c) Leaching, chromatography
d) Membrane separation, leaching, chromatography

Explanation: Liquid-liquid extraction, membrane separation, leaching, chromatography are the separation processes of bioseparation which involves inter-phase mass transfer of solutes. The way of describing the inter-phase mass transfer is to use mass-transfer coefficients based on the difference between the bulk solutions in one phase and the equilibrium constant in the other phase.

3. The absorption of sulphur dioxide from air by water is not an example of inter-phase mass transfer.
a) False
b) True

Explanation: The absorption of sulphur dioxide from air is an example of inter-phase mass transfer. In this SO2 diffuses and passes through an inter-phase which is between the water and air and the gas diffuses to the immiscible liquid phase i.e. water. Mass transfer occurs in every stage due to the equilibrium state of concentration gradient which is available between the liquid and gaseous phase.

4. What will be the equation of solute flux across the interface of liquid medium?
a) NA = -(D + E) $$\frac{dC_A}{dx}$$
b) NA = kA ΔcA
c) NA = D1 $$\frac{(C_1 – C_{i1})}{x_1}$$
d) NA = -DA $$\frac{dC_A}{dZ}$$

Explanation: The solute flux across the interface of liquid medium is NA = D1 $$\frac{(C_1 – C_{i1})}{x_1}$$ where, D1 is the diffusivity in liquid, x1 is the length of flow of solute, C1 is the solute concentration at x1 distance from interface in liquid, Ci1 is the interfacial solute concentration in liquid. NA = -(D + E) $$\frac{dC_A}{dx}$$, NA = kA ΔcA is flux for convective mass transfer and NA = -DA $$\frac{dC_A}{dZ}$$ is the molar flux when solutes flow in direction of z.

5. Estimate the solute flux across the interface of liquid medium when the solute concentration is 45000 kg-moles/m3, interfacial solute concentration in liquid is 40000 kg-moles/m3, diffusivity is 2.5 × 10-11 m2/s and the length of the tube is 5m3.
a) 2.5 × 10-11 kg-moles/ m2
b) 2.5 × 10-12 kg-moles/ m2.s
c) 2.5 × 10-10 kg-moles/ m2.s
d) 2.5 × 10-8 kg-moles/ m2.s

Explanation: NA = D1 $$\frac{(C_1 – C_{i1})}{x_1}$$ where, D1 is the diffusivity, C1 is the solute concentration at x1 distance from interface in liquid, x1 is the length of flow of solute, Ci1 is the interfacial solute concentration in liquid. Therefore, NA = 2.5 × 10-11 $$\frac{(45000-40000)}{5}$$ = 2.5 × 10-8 kg-moles/ m2.

6. Estimate the interfacial solute concentration in the two liquids when the solute concentration is 45000 kg-moles/m3, interfacial solute concentration in liquid is 40000 kg-moles/m3, diffusivity is 2.5 × 10-11 m2/s and the length of the tube is 5m3 and the partition coefficient is 0.25.
a) 10000 kg-moles/m3
b) 20000 kg-moles/m3
c) 30000 kg-moles/m3
d) 40000 kg-moles/m3

Explanation: Using Ci2 = K Ci1 where, Ci2 is the interfacial solute concentration in the two liquids, K is the partition coefficient and Ci1 is the interfacial solute concentration in liquid. Therefore, the interfacial solute concentration in the two liquids is Ci2 = 0.25 × 40000 = 10000 kg-moles/m3.

7. The concentrations of the solute at various locations within the transfer zone do not change with time in unsteady state mass transfer.
a) True
b) False

Explanation: The concentrations of the solute at various locations within the transfer zone changes with time in unsteady state mass transfer while the concentration does not change in the case of steady state mass transfer.

8. The equation for unsteady state molecular diffusion in 3-dimension for the Cartesian co-ordinate system is __________
a) DAB ($$\frac{\delta^2 C_A}{\delta x^2} + \frac{\delta^2 C_A}{\delta y^2} + \frac{\delta^2 C_A}{\delta z^2}$$) = –$$\frac{\sigma C_A}{\delta t}$$
b) JA = –$$\frac{\varepsilon D(C_{A1} – C_{A2})}{\tau(x_1 – x_2)}$$
c) JA = $$\frac{\varepsilon D(C_{A1} – C_{A2})}{\tau(x_1 – x_2)}$$
d) DAB ($$\frac{\delta^2 C_A}{\delta x^2} + \frac{\delta^2 C_A}{\delta y^2} + \frac{\delta^2 C_A}{\delta z^2}$$) = $$\frac{\sigma C_A}{\delta t}$$

Explanation: The equation for unsteady state molecular diffusion in 3-dimension for the Cartesian co-ordinate system is DAB ($$\frac{\delta^2 C_A}{\delta x^2} + \frac{\delta^2 C_A}{\delta y^2} + \frac{\delta^2 C_A}{\delta z^2}$$) = $$\frac{\sigma C_A}{\delta t}$$. The equation depends on the geometry of the system and accordingly the partial differential equation is modified and used for the study of unsteady mass transfer.

9. What will be the flux of solute across the interface at any time t?
a) NA (0, t) = √$$\frac{D_{AB}}{\pi t}$$ (CA1 – CA0)
b) NA = -(D + E) $$\frac{dC_A}{dx}$$
c) NA = D1 $$\frac{(C_1 – C_{i1})}{x_1}$$
d) NA = -DA ($$\frac{dC_A}{dZ}$$)

Explanation: The flux of solute across the interface at any time t can be obtained using equation NA (0, t) = √$$\frac{D_{AB}}{\pi t}$$ (CA1 – CA0), this equation was obtained using $$\frac{C_{A1} – C_A}{C_{A1} – C_{A0}}$$ = erf ⁡($$\frac{x}{2\sqrt{D_{AB}t}}$$) in which the initial conditions are at t = 0, CA = CA0 and the boundary conditions are at x = 0, CA = CA1.

10. Determine the flux of the antibiotic across the gel-water interface when a thick gel slab of agar is exposed on one side to a well-mixed aqueous solution of an antibiotic, the concentration of the antibiotic solution is 0.01 kg-moles/m3. Assume that the concentration in the solution does not change with the time and the change time is 5 minutes, diffusivity in the gel is 8 x 10-11 m2/s.
a) 3.9 × 10-11 kg-moles/m2s
b) 2.91 × 10-11 kg-moles/m2s
c) 4.9 × 10-11 kg-moles/m2s
d) 2.91 × 10-10 kg-moles/m2s

Explanation: The flux of the antibiotic across the interface is NA (0, t) = √$$\frac{D_{AB}}{\pi t}$$ (CA1 – CA0) therefore, the flux is NA (0, 600) = $$\sqrt{\frac{8 × 10^{-11}}{3.142 × 300}}$$ (0.001 – 0) = 2.91 × 10-10 kg-moles/m2.

Sanfoundry Global Education & Learning Series – Bioseparation Technology.

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