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## Interview MCQ Set 1

1. What is the third phase in all the two phase systems?
b) The interface
c) The container
d) The atmosphere

Answer: c [Reason:] The third phase is the container which selected as inert for separation purposes.

2. How to separate vapor-liquid-liquid system?
a) Fractional distillation
b) Fractional crystallization
c) Zone melting
d) Three phase isothermal flash

Answer: d [Reason:] A three phase isothermal flash is used to separate vapor-liquid-liquid system.

3. Why are calculations for three phase flash difficult?
a) Because of strong dependency of K values on liquid phase composition when two immiscible liquids are present
b) Because of strong dependency of K values on vapor phase composition when two immiscible liquids are present
c) Because of strong dependency of K values on liquid phase composition when two miscible liquids are present
d) Because of strong dependency of K values on liquid phase composition when three immiscible liquids are present

Answer: a [Reason:] The calculations of three phase flash are difficult due to strong dependency of K values on liquid phase composition when two immiscible liquids are present.

4. What is the enthalpy balance for three phase system?
a) hFF + Q =hvV + hLL(1) + hLL(2)
b) hFF =hvV + hLL(1) + hLLL(2)
c) hFF – Q =hvV + hL(1) + hLLL(2)
d) hFF + Q =hvV + hLL + hLLL

Answer: a [Reason:] The correct enthalpy balance is hFF + Q =hvV + hLL(1) + hLL(2).

5. The calculations of multiphase system are made by:
a) Graph
b) Hand calculations
c) Trial and error
d) Process simulators

Answer: d [Reason:] Calculations of multiphase system are made by process simulators.

6. If at a certain pressure the vapor pressure of water is 4.74pisa and the mole fraction of water is 0.923, find out the crystallizer pressure.
a) 2.5pisa
b) 4.38pisa
c) 4.56pisa
d) 4.76pisa

Answer: b [Reason:] Roult’s law is used P=psxh20. Hence P=4.74*0.923=4.38pisa.

7. If at a certain pressure the vapor pressure of water is 8pisa and the mole fraction of water is 0.9, find out the crystallizer pressure.
a) 0.64pisa
b) 7.2pisa
c) 0.66pisa
d) 0.81pisa

Answer: b [Reason:] Roult’s law is used P=psxh20. Hence P=8*0.9=7.2pisa.

8. If at a certain pressure the vapor pressure of water is 6pisa and the mole fraction of water is 0.7, find out the crystallizer pressure.
a) 3.3pisa
b) 3.5pisa
c) 3.96pisa
d) 4.2pisa

Answer: d [Reason:] Roult’s law is used P=psxh20. Hence P=6*0.7=4.2pisa.

9. If at a certain pressure the vapor pressure of water is 2pisa and the mole fraction of water is 0.9, find out the crystallizer pressure.
a) 0.45pisa
b) 1.8pisa
c) 3.6pisa
d) 0.81pisa

Answer: b [Reason:] Roult’s law is used P=psxh20. Hence P=2*0.9=1.8pisa.

10. If at a certain pressure the vapor pressure of water is 10pisa and the mole fraction of water is 0.8, find out the crystallizer pressure.
a) 5pisa
b) 6pisa
c) 7pisa
d) 8pisa

Answer: d [Reason:] Roult’s law is used P=psxh20. Hence P=10*0.8=8pisa.

## Interview MCQ Set 2

1. The Newton-Raphson method of finding roots of nonlinear equations falls under the category of which method?
a) Bracketing
b) Random
c) Open
d) Graphical

Answer: c [Reason:] It is an open method since we have to guess the square root.

2. What is the formula of finding the square root of R from equation x2- R=0?
a) Xi+1= 0.5(xi + R/xi)
b) Xi+1= 0.5(xi + R/2)
c) Xi= 0.5(xi + R/xi)
d) Xi+1= 0.5(xi + xi)

Answer: a [Reason:] The formula of finding the square root of R from equation x2– R=0 is Xi+1= 0.5(xi + R/xi).

3. If broth flow rate is 200kg/hr and solvent flow rate is 300kg/hr and POD is of diameter 1m and length 10m, what is the residence time?
a) 28s
b) 56s
c) 35s
d) 39s

Answer: a [Reason:] ttme= pi*(D2/4)L/F+S, ttime = 3.14*(1/4)10/500=56s.

4. If broth flow rate is 600kg/hr and solvent flow rate is 400kg/hr and POD is of diameter 1m and length 5m, what is the residence time?
a) 28s
b) 14s
c) 30s
d) 35s

Answer: b [Reason:] ttme= pi*(D2/4)L/F+S, ttime = 3.14*(1/4)5/1000=14s.

5. If broth flow rate is 200kg/hr and solvent flow rate is 300kg/hr and POD is of diameter 1m and length 15m, what is the residence time?
a) 72s
b) 76s
c) 82s
d) 85s

Answer: a [Reason:] ttme= pi*(D2/4)L/F+S, ttime = 3.14*(1/4)15/500=85s.

6. If broth flow rate is 200kg/hr and solvent flow rate is 300kg/hr and POD is of diameter 2m and length 5m, what is the residence time?
a) 98s
b) 111ss
c) 113s
d) 119s

Answer: a [Reason:] ttme= pi*(D2/4)L/F+S, ttime = 3.14*(4/4)5/500=113s.

7. If broth flow rate is 200kg/hr and solvent flow rate is 300kg/hr and POD is of diameter 1m and length 6m, what is the residence time?
a) 28s
b) 20s
c) 30s
d) 34s

Answer: d [Reason:] ttme= pi*(D2/4)L/F+S, ttime = 3.14*(1/4)6/500=34s.

8. If broth flow rate is 200kg/hr and solvent flow rate is 300kg/hr and POD is of diameter 1m and length 7m, what is the residence time?
a) 28s
b) 20s
c) 40s
d) 35s

Answer: c [Reason:] ttme= pi*(D2/4)L/F+S, ttime = 3.14*(1/4)7/500=40s.

9. If broth flow rate is 200kg/hr and solvent flow rate is 300kg/hr and POD is of diameter 3m and length 5m, what is the residence time?
a) 200s
b) 250s
c) 230s
d) 254s

Answer: a [Reason:] ttme= pi*(D2/4)L/F+S, ttime = 3.14*(9/4)5/500=254s.

10. If broth flow rate is 200kg/hr and solvent flow rate is 500kg/hr and POD is of diameter 1m and length 5m, what is the residence time?
a) 28s
b) 20s
c) 30s
d) 35s

Answer: b [Reason:] ttme= pi*(D2/4)L/F+S, ttime = 3.14*(1/4)5/700=20s.

## Interview MCQ Set 3

1. What is the p-v-T equation of state?
a) Relationship between volume and temperature
b) Relationship between pressure, molar volume and temperature
c) Relationship between pressure and temperature
d) Relationship between pressure and molar volume

Answer: b [Reason:] The Relationship between pressure, molar volume and temperature is the p-v-T equation of state.

2. What does the ideal gas equation neglects?
a) Temperature of molecules
b) Friction between molecules
c) Volume occupied by the molecules
d) Pressure occupied by the molecules

Answer: c [Reason:] The ideal gas equation neglects the volume occupied by the molecules hence it is only applicable at low pressures.

3. What is the van der waal’s equation of non-ideal gas?
a) [P + a(n/v)2](v/n –b) =RT
b) [P + a(n/v)2] =RT
c) [P + (n/v)2](v/n –b) =RT
d) [P + a(n/v)2](v/n ) =RT

Answer: a [Reason:] [P + a(n/v)2](v/n –b) =RT is the correct equation for real gases.

4. What is the formula for acentric factor?
a) Z = (-log(Ps/Pc)Tr=o.7) – 1000
b) z = (-log(Ps/Pc)Tr=o.7) – 1.000
c) ω = (-log(Ps/)Tr=o.7) – 1.000
d) ω = (-log(Ps/Pc)Tr=o.7) – 1.000

Answer: d. [Reason:] ω = (-log(Ps/Pc)Tr=o.7) – 1.000 is the correct equation for determining acentric factor based on the law of corresponding states.

5. What is the value of ω for symmetric molecules?
a) 1
b) 0.264
c) 0
d) 0.490

Answer: c [Reason:] The value of ω is 0 for symmetric molecules.

6. How can the compressibility factor be determined form generalized equation and Redich-kwong equation?
a) Z3 – Z2 + (A-B-B2)Z – AB =0
b) Z3 – Z2 + (A-B-B2) – AB =0
c) Z3 – Z2 + (A-B-B2)Z – A =0
d) Z3 – Z2 + (A-B)Z – AB =0

Answer: a. [Reason:] The generalized equation PV= ZRT and the redich kwong equation are combined to give Z3 – Z2 + (A-B-B2)Z – AB =0 where A and B are constants which can be determined.

7. If A =0.2724 and B= 0.05326 find the value of compressibility factor Z.
a) 0.7867
b) 0.7314
c) 0.8656
d) 10.435

Answer: b [Reason:] Z3 – Z2 + (A-B-B2)Z – AB =0 is the equation used. Hence by solving Z3 – Z2 + (0.2724-0.05326-0.053262)Z – 0.2724*0.05326 =0 we get Z= 0.7314.

8. What is the correct representation of Soave-Redich-Kwong equation?
a) P = [RT/v-b] – [a/v2+bv+a].
b) P = [RT/v-b] – [a/v2+bv].
c) P = [RT/v-b] – [a/v2+bv+b].
d) P = [RT/v-b] – [a/v2].

Answer: b [Reason:] P = [RT/v-b] – [a/v2+bv] is the Soave-Redich-Kwong equation.

9. What is the value of b if the temperature is 477.6K and the saturation pressure is 2829kPa?
a) 836.7 units
b) 2334.666 units
c) 0.0626 units
d) 0.4356 units

Answer: c [Reason:] b = 0.08664RT/Pc is the equation is used to determine b. Hence b= 0.08664 *8.314*477.6/2829000= 2334.666 units

10. If the critical temperature of propane is 369.8K and temperature of mixture is 477.6K and the saturation pressure is 4250kPa , what is the value of a?
a) 234.5 units
b) 678.9 units
c) 765.6 units
d) 836.7 units

Answer: d [Reason:] a= 0.42748R2Tc2.5/PcT0.5 is the equation used to calculate a. Hence a= 0.427(8.314)2(369.8)2.5/4250000*(477.6)0.5.

## Interview MCQ Set 4

1. Calculate the initial weight of moisture if one kg block of borax laundry soap with initial moisture content of 20.2% is dried and brought to 20%.
a) 0.168
b) 0.234
c) 0.200
d) 0.300

Answer: a [Reason:] W=100X/100+X and hence W= 0.168(1)=0.168kg.

2. Calculate the initial weight of moisture if one kg block of borax laundry soap with initial moisture content of 15 % is dried and brought to 14.8%.
a) 0.168
b) 0.130
c) 0.200
d) 0.300

Answer: b [Reason:] W=100X/100+X and hence W= 0.130(1)=0.130kg.

3. Calculate the initial weight of moisture if one kg block of borax laundry soap with initial moisture content of 20.2% is dried and brought to 20%.
a) 0.168
b) 0.234
c) 0.156
d) 0.115

Answer: d [Reason:] W=100X/100+X and hence W= 0.115(1)=0.15kg.

4. Calculate the initial weight of moisture if one kg block of borax laundry soap with initial moisture content of 18% is dried and brought to 17.8%.
a) 0.168
b) 0.146
c) 0.152
d) 0.234

Answer: c [Reason:] W=100X/100+X and hence W= 0.152(1)=0.152kg.

5. Calculate the initial weight of moisture if one kg block of borax laundry soap with initial moisture content of 25% is dried and brought to 24.8%.
a) 0.16
b) 0.27
c) 0.200
d) 0.300

Answer: c [Reason:] W=100X/100+X and hence W= 0.2(1)=0.2kg.

6. Calculate the initial weight of moisture if one kg block of borax laundry soap with initial moisture content of 20.2% is dried and brought to 20%.
a) 0.168
b) 0.234
c) 0.200
d) 0.237

Answer: d [Reason:] W=100X/100+X and hence W= 0.237(1)=0.237kg.

7. Calculate the initial weight of moisture if one kg block of borax laundry soap with initial moisture content of 27% is dried and brought to 26.5%.
a) 0.324
b) 0.234
c) 0.225
d) 0.212

Answer: d [Reason:] W=100X/100+X and hence W= 0.212(1)=0.212kg.

8. Calculate the initial weight of moisture if one kg block of borax laundry soap with initial moisture content of 20.2% is dried and brought to 20%.
a) 0.168
b) 0.259
c) 0.254
d) 0.257

Answer: b [Reason:] W=100X/100+X and hence W= 0.259(1)=0.259kg.

9. Calculate the initial weight of moisture if one kg block of borax laundry soap with initial moisture content of 30% is dried and brought to 28%.
a) 0.168
b) 0.289
c) 0.276
d) 0.248

Answer: d [Reason:] W=100X/100+X and hence W= 0.248(1)=0.248kg.

10. Calculate the initial weight of moisture if one kg block of borax laundry soap with initial moisture content of 24% is dried and brought to 20%.
a) 0.194
b) 0.236
c) 0.876
d) 0.239

Answer: a [Reason:] W=100X/100+X and hence W= 0.194(1)=0.194kg.

## Interview MCQ Set 5

1. If YN+1= 0.0005
XN= 0.001
Y2= 0.01174
YL=XL= 0.05
L/V= 0.2295
Calculate the number of ideal countercurrent washing stages required.
a) 2.5
b) 2.6
c) 2.8
d) 2.95

Answer: d [Reason:] N= log(XN-YN+1/YL-Y2)/log(L/V), hence N= 2.95.

2. If YN+1= 0.0005
XN= 0.001
Y2= 0.01174
YL=XL= 0.05
L/V= 0.35
Calculate the number of ideal countercurrent washing stages required.
a) 2.5
b) 4.1
c) 2.2
d) 6.14

Answer: b [Reason:] N= log(XN-YN+1/YL-Y2)/log(L/V), hence N= 4.1.

3. If YN+1= 0.0005
XN= 0.001
Y2= 0.01174
YL=XL= 0.05
L/V= 0.1165
Calculate the number of ideal countercurrent washing stages required.
a) 2.0
b) 2.6
c) 2.8
d) 2.95

Answer: a [Reason:] N= log(XN-YN+1/YL-Y2)/log(L/V), hence N= 2.0.

4. If YN+1= 0.0005
XN= 0.001
Y2= 0.01174
YL=XL= 0.05
L/V= 0.55
Calculate the number of ideal countercurrent washing stages required.
a) 4.6
b) 6.4
c) 7.2
d) 2.95

Answer: c [Reason:] N= log(XN-YN+1/YL-Y2)/log(L/V), hence N= 7.2.

5. If YN+1= 0.0005
XN= 0.001
Y2= 0.01174
YL=XL= 0.05
L/V= 0.256
Calculate the number of ideal countercurrent washing stages required.
a) 3.18
b) 2.6
c) a)8
d) 3.67

Answer: d [Reason:] N= log(XN-YN+1/YL-Y2)/log(L/V), hence N= 3.18.

6. If YN+1= 0.0005
XN= 0.001
Y2= 0.01174
YL=XL= 0.05
L/V= 0.318
Calculate the number of ideal countercurrent washing stages required.
a) 2.5
b) 2.6
c) 3.8
d) 3.18

Answer: c [Reason:] N= log(XN-YN+1/YL-Y2)/log(L/V), hence N= 3.8.

7. If YN+1= 0.0005
XN= 0.001
Y2= 0.01174
YL=XL= 0.05
L/V= 0.108
Calculate the number of ideal countercurrent washing stages required.
a) 6.13
b) 2.45
c) 1.95
d) 2.95

Answer: c [Reason:] N= log(XN-YN+1/YL-Y2)/log(L/V), hence N= 1.94.

8. If YN+1= 0.0005
XN= 0.001
Y2= 0.01174
YL=XL= 0.05
L/V= 0.44
Calculate the number of ideal countercurrent washing stages required.
a) 2.5
b) 5.3
c) 5.4
d) 5.5

Answer: b [Reason:] N= log(XN-YN+1/YL-Y2)/log(L/V), hence N= 5.3.

9. If YN+1= 0.0005
XN= 0.001
Y2= 0.01174
YL=XL= 0.05
L/V= 0.24
Calculate the number of ideal countercurrent washing stages required.
a) 3.0
b) 3.7
c) 2.8
d) 3.95

Answer: a [Reason:] N= log(XN-YN+1/YL-Y2)/log(L/V), hence N= 3.0.

10. If YN+1= 0.0005
XN= 0.001
Y2= 0.01174
YL=XL= 0.05
L/V= 0.29
Calculate the number of ideal countercurrent washing stages required.
a) 3.5
b) 3.6
c) 3.8
d) 3.95