Test: Concentration And Velocity


15 Questions MCQ Test Mass Transfer | Test: Concentration And Velocity


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This mock test of Test: Concentration And Velocity for Chemical Engineering helps you for every Chemical Engineering entrance exam. This contains 15 Multiple Choice Questions for Chemical Engineering Test: Concentration And Velocity (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Concentration And Velocity quiz give you a good mix of easy questions and tough questions. Chemical Engineering students definitely take this Test: Concentration And Velocity exercise for a better result in the exam. You can find other Test: Concentration And Velocity extra questions, long questions & short questions for Chemical Engineering on EduRev as well by searching above.
QUESTION: 1

 In which of the following conditions mass transfer will occur spontaneously? C and z is concentration and distance respectively.

Solution:

Explanation: Mass transfer occurs in the direction of decreasing concentration.

QUESTION: 2

Which among the following is always true for mass transfer to occur?

Solution:

Explanation: Mass transfer occurs to attain an equilibrium position or to minimize the energy of the system. Mass occur can occur even if there is no difference in concentration, pressure and temperature. Example: – A ball rolls down a slope to minimize its potential energy.

QUESTION: 3

A solution contains 0.3 moles of solute A, 0.2 moles of B and 0.5 moles of C. What will be the mole fraction of A in the mixture?

Solution:

Explanation: Mole Fraction of A= Moles of A/Total moles

QUESTION: 4

A gas mixture containing 21% O2 and 79% N2 flows through a pipe of circular cross-section of diameter 2cm at atmospheric pressure. The velocity of O2 is 0.04m/s and N2 is 0.035m/s.

Q. What is the value of molar average velocity (m/s) of the mixture?

Solution:

Explanation: Molar average velocity= (C1u1+C2u2+……….)/C
=y1u1+y2u2…..
= 0.21*0.04+0.79*0.035
=0.036

QUESTION: 5

A gas mixture containing 21% O2 and 79% N2 flows through a pipe of circular cross-section of diameter 2cm at atmospheric pressure. The velocity of O2 is 0.04m/s and N2 is 0.035m/s.

Q. What will be the value of mass average velocity (m/s)?

Solution:

Explanation: Mass average velocity= (ρ1u12u2)/ ρ
ρ1= (p1*M1)/RT
ρ= (p*M)/RT.
Dividing these two equations
ρ1/ ρ = (p1*M1)/pM =(y1*M1)/M
M=y1*M1+y2*M2=0.79*28+0.21*32=28.84
Mass average velocity= (y1*M1*u1+y2*M2*u2)/M
= (0.21*32*0.04+0.79*0.035)/28.84
= 0.981

QUESTION: 6

A gas mixture containing 21% O2 and 79% N2 flows through a pipe of circular cross-section of diameter 2cm at atmospheric pressure. The velocity of O2 is 0.04m/s and N2 is 0.035m/s.

Q. What is the value of Volume average velocity?

Solution:

Explanation: Volume average velocity= Molar average velocity.

QUESTION: 7

In a gas mixture if molecular weights of all the species are equal than which one will be more, Mass average velocity or Molar average velocity?

Solution:

Explanation: Average molar mass of the mixture=M(x1+x2+………+ xn)=M
(x1+x2+………+ xn=1)
Mass average velocity=Σuiρi/ρ=uiPui/P
Molar average velocity=Σciui/c=ρuiPi/P (P=cRT).

QUESTION: 8

The velocity of various components of mixture(21% A, 78% B and 1% C) are:
Velocity is in the form: v⃗=xi+yj+zk

Velocity of component A= (0.1, 0, 0)
Velocity of component B= (0, 0.1, 0)
Velocity of component B= (0, 0, 0.1)
Calculate the value of average molar velocity?

Solution:

Explanation: Molar average velocity = (c1u1+c2u2+c3u3)/C.

QUESTION: 9

Calculate the value of mass average velocity in the previous question?

Solution:

Explanation: (y1*M1*u1+y2*M2*u2)/M.

QUESTION: 10

For the calculation of Mass average velocity, which velocity of the molecule is used?

Solution:

Explanation: None.

QUESTION: 11

Which among the following can be the unit of Flux?

Solution:

Explanation: Dimension analysis.
Flux= rate at which a species passes through a unit area, in unit time.

QUESTION: 12

A molecule in a mixture moves with a velocity of 0.03m/s in (+x) direction. The concentration of the particle is 0.5 mol/liter. A observer is running in (-x) direction with a velocity of 2m/s. calculate the value of molar flux (mol/m2.s) relative to the observer.

Solution:

Explanation: Molar flux with respect to observer= C(u-U)
(C=500mol/m3, u=0.03m/s, U= (-) 2m/s).

QUESTION: 13

In the previous question what will be the value of flux (mol/m2.s) with respect to a stationary observer?

Solution:

Explanation: Here U=0.

QUESTION: 14

 Under which of the following conditions mass average velocity and molar average velocity are equal?

Solution:

Explanation: None.

QUESTION: 15

The mass flow rate of a component in a mixture in X-direction is 2kg/s. The area vector at a particular cross-section is (1.1, 3.2, 5.4) cm2. Calculate the value of flux (kg/cm2.s) through the cross-section. 

Solution:

Explanation: For flux calculation area normal to the flow direction should be taken into account

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