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Multiple choice question for engineering

Set 1

1. Which of the following is defined as the ratio of partial pressure and vapor pressure?
a) Relative Humidity
b) Molal Humidity
c) Humidity
d) None of the mentioned

Answer: a [Reason:] Partial pressure divided by vapor pressure is called relative humidity.

2. What is the relative saturation of a gas with partial pressure 10 mm Hg and vapor pressure 5 mm Hg?
a) 0.5
b) 1
c) 2
d) 4

Answer: c [Reason:] Relative saturation = 10/5 = 2.

3. What is the relative humidity of a gas with partial pressure of water vapor is 5 mm Hg and vapor pressure 8 mm Hg?
a) 25%
b) 32.5%
c) 45%
d) 62.5%

Answer: d [Reason:] Relative humidity = 5/8*100 = 62.5%.

4. What is the relative saturation of a gas with partial pressure 38 mm Hg and vapor pressure 1 torr?
a) 0.05
b) 0.5
c) 2
d) 38

Answer: d [Reason:] 1 torr = 1 mm Hg, => relative saturation = 38/1 = 38.

5. What is the relative humidity of a gas with partial pressure of water vapor 360 Pa and vapor pressure 1 atm?
a) 100%
b) 225.1%
c) 355.4%
d) 405.9%

Answer: c [Reason:] Relative humidity = 360/101.3*100 = 355.4%.

6. What is called the ratio of partial pressure of vapor and partial pressure of dry gas?
a) Humidity
b) Relative humidity
c) Molal humidity
d) None of the mentioned

Answer: c [Reason:] Partial pressure of vapor divided by the partial pressure of vapor free gas is called molal humidity.

7. What is the molal humidity of a gas with pressure of gas vapor 10 mm Hg and pressure of vapor free gas 20 mm Hg?
a) 25%
b) 50%
c) 100%
d) 200%

Answer: b [Reason:] Molal humidity = 10/20*100 = 50%.

8. What is the molal saturation of a gas with 10 moles of gas vapor and 25 moles of dry gas?
a) 0.2
b) 0.4
c) 0.8
d) 1

Answer: b [Reason:] Molal saturation = 10/25 = 0.4.

9. What is the molal saturation of a gas with 10 Pa partial pressure without vapor and 0.5 atm partial pressure with vapor?
a) 0.5
b) 1
c) 5
d) 101.3

Answer: c [Reason:] Molal saturation = 0.5*101.3/10 = 5.

10. What is the molal humidity of a gas with partial pressure with vapor 1 mm Hg and partial pressure without vapor 1 atm?
a) 90%
b) 131%
c) 214%
d) 288%

Answer: b [Reason:] Molal humidity = 133.3/101.3*100 = 131%.

11. Which of the following is equal to the product of humidity and mass of dry gas?
a) Moles of water vapor
b) Mass of water vapor
c) Mass of gas vapor
d) None of the mentioned

Answer: b [Reason:] Humidity = (mass of water vapor)/(mass of dry gas).

12. What is the humidity, if the mass of dry gas is 10 Kg and mass of water vapor is 5 Kg?
a) 25%
b) 50%
c) 100%
d) 200%

Answer: b [Reason:] Humidity = 5/10*100 = 50%.

13. What is the mass of water vapor, if the mass of dry gas is 10 lb and humidity 10%?
a) 254 g
b) 453 g
c) 746 g
d) 1 Kg

Answer: b [Reason:] Mass of water vapor = 10*0.1 = 1 lb = 453 g.

14. What is the humidity if the moles of dry gas are 25 and mole of water vapor are 15?
a) 60%
b) 90%
c) 150%
d) 200%

Answer: a [Reason:] Humidity = 15/25*100 = 60%.

15. What are moles of dry gas if the moles of water vapor is 14 at 25% humidity?
a) 3.5
b) 7
c) 14
d) 56

Answer: d [Reason:] Moles of dry gas = 14/0.25 = 56.

Set 2

1. Which of the following is heat required in phase transitions?
a) Latent Heat
b) Sensible Heat
c) Depends on type of phase transition
d) None of the mentioned

Answer: a [Reason:] Heat required for phase transitions is latent heat.

2. What is the latent heat for melting?
a) Heat of vaporization
b) Heat of fusion
c) Heat of Solidification
d) None of the mentioned

Answer: b [Reason:] Latent heat for melting is called heat of vaporization.

3. 15 J is the latent heat for melting of ice, what is the heat of transition from 0oC ice to 15oC water if heat of transition from 0oC water to 15oC water is 10 J?
a) 10 J
b) 15 J
c) 25 J
d) 40 J

Answer: c [Reason:] H = 15 + 10 = 25 J.

4. 5 J is the latent heat for melting of ice, what is the heat of transition from 0oC ice to 15oC water if heat of transition from 0oC water to 15oC water is 12 J?
a) 5 J
b) 12 J
c) 15 J
d) 17 J

Answer: d [Reason:] H = 5 + 12 = 17 J.

5. 150 J is the latent heat for vaporization of water, what is the heat of transition from 50oC water to 100oC steam if heat of transition from 50oC water to 100oC water is 12 J?
a) 12 J
b) 152 J
c) 162 J
d) 200 J

Answer: c [Reason:] H = 150 + 12 = 162 J.

6. Heat of fusion of benzene is 10 J at 0oC, if Cp of solid benzene is 10 J/oC, and that of liquid benzene is 5 J/oC what is the enthalpy change of benzene from – 10oC to 10oC?
a) 160 J
b) 200 J
c) 240 J
d) 300 J

Answer: a [Reason:] ∆H = -10010.dT + 10 + 0105.dT = 10*10 + 10 + 5*10 = 160 J.

7. If Cp of liquid water is 10 J/oC, what is the enthalpy change of water at standard conditions from – 10oC to 0oC?
a) 10 J
b) 50 J
c) 100 J
d) Cannot be determined

Answer: d [Reason:] Since the phase of water from – 10oC to 0oC is solid ice, but Cp of liquid water is given, so it cannot be determined.

8. If Cp of liquid benzene is 15 J/oC, what is the enthalpy change of water from 5oC to 10oC?
a) 15 J
b) 50 J
c) 75 J
d) 90 J

Answer: c [Reason:] ∆H = 51015.dT = 75 J.

9. If Cp of solid benzene is 10 J/oC, what is the enthalpy change of water from 5oC to 75oC?
a) 5 J
b) 10 J
c) 75 J
d) 700 J

Answer: d [Reason:] ∆H = 57510.dT = 700 J.

10. If Cp of steam is 105 J/oC, what is the enthalpy change of steam from 106oC to 118oC?
a) 420 J
b) 640 J
c) 890 J
d) 1260 J

Answer: d [Reason:] ∆H = 106118105.dT = 1260 J.

11. If Cp of water is (23 + T) J/oC, what is the enthalpy change of water from 3oC to 5oC?
a) 54 J
b) 96 J
c) 140 J
d) 196 J

Answer: a [Reason:] ∆H = 35(23 + T).dT = 23(5 – 3) + (52 – 32)/2 = 54 J.

12. If Cp of water is 45T J/oC, what is the enthalpy change of water from 6oC to 8oC?
a) 7 J
b) 14 J
c) 28 J
d) 56 J

Answer: b [Reason:] ∆H = 68T.dT = (82 – 62)/2 = 14 J.

13. Cp of ice is 25 J/oC and that of water is 30 J/oC, what is the enthalpy change of water from – 10oC to 35oC at standard conditions, if the later heat of fusion of water is 110 J?
a) 250 J
b) 595 J
c) 960 J
d) 1410 J

Answer: d [Reason:] ∆H = -10025.dT + 110 + 03530.dT = 25*10 + 110 + 35*30 = 1410 J.

14. Cp of ice is 10 J/oC and that of water is 15 J/oC, what is the enthalpy change of water from – 15oC to 0oC at standard conditions, if the later heat of fusion of water is 110 J?
a) 150 J
b) 315 J
c) 485 J
d) 560 J

Answer: c [Reason:] ∆H = -15025.dT + 110 = 25*15 + 110 = 485 J.

15. Cp of water is 10 J/oC and that of steam is 15 J/oC, what is the enthalpy change of water from 75oC to 110oC at standard conditions, if the later heat of vaporization of water is 245 J?
a) 395 J
b) 645 J
c) 840 J
d) 999 J

Answer: b [Reason:] ∆H = 7510010.dT + 245 + 10011015.dT = 25*10 + 245 + 15*10 = 645 J.

Set 3

1. What is the change potential energy of a stone of mass 5 Kg that falls from a cliff 10 m high?
a) 5 J
b) 50 J
c) 500 J
d) 5000 J

Answer: c [Reason:] PE = 5*10*10 = 500 J.

2. What is the change in potential energy of a stone of mass 1 Kg thrown up with a velocity 10 m/s and comes down?
a) 0
b) 10 J
c) 100 J
d) 500 J

Answer: a [Reason:] Since the stone comes down to its initial position, its potential energy does not change.

3. A fluid in supplied to a cliff 10 m high, what is the change in specific potential energy of fluid?
a) 10 J/Kg
b) 100 J/Kg
c) 200 J/Kg
d) 500 J/Kg

Answer: b [Reason:] Change in specific potential energy = 10*10 = 100 J/Kg.

4. A fluid is supplied to a cliff 20 m high and comes down by 5 m, what is the change in specific potential energy of fluid?
a) 100 J/Kg
b) 150 J/Kg
c) 200 J/Kg
d) 250 J/Kg

Answer: b [Reason:] Change in specific potential energy = 10*(20 – 5) = 150 J/Kg.

5. A ball is thrown up and someone catches it on the cliff, what is the change in potential energy?
a) Positive
b) Negative
c) Zero
d) Cannot say

Answer: a [Reason:] Change in potential energy will be positive as the change in kinetic energy will be negative.

6. What is the kinetic energy of an object of mass 10 Kg moving with velocity 2 m/s?
a) 10 J
b) 20 J
c) 40 J
d) 50 J

Answer: b [Reason:] KE = ½*10*22 = 20 J.

7. What is the kinetic energy of an object of mass 2 Kg and velocity 5 m/s?
a) 20 J
b) 25 J
c) 35 J
d) 50 J

Answer: b [Reason:] KE = ½*2*52 = 25 J.

8. What is the specific kinetic energy of a fluid flowing at rate 0.1 m3/s and 1 m diameter pipe?
a) 0.008 J/Kg
b) 0.016 J/Kg
c) 0.028 J/Kg
d) 0.064 J/Kg

Answer: a [Reason:] v = 0.1/∏(0.5)2, => specific KE = v2/2 = 0.008 J/Kg.

9. A stone is thrown up with a velocity and comes down, what is the change in kinetic energy of stone?
a) Positive
b) Negative
c) Zero
d) Cannot say

Answer: c [Reason:] According to conservation of energy, during a process change in total energy will always be zero, if the stone comes to the same level the change in potential energy becomes zero so the change in kinetic energy also becomes zero.

10. A stone is release from a cliff and comes down, what is the change in kinetic energy of stone?
a) Positive
b) Negative
c) Zero
d) Cannot say

Answer: a [Reason:] Since its velocity increases as it comes down, so change in kinetic energy is positive.

Set 4

1. Which of the following is not a unit of pressure?
a) Bar
b) N/m2
c) Kg/m2
d) Torr

Answer: c [Reason:] Pressure is force per unit area so Kg/m2 cannot be a unit of pressure.

2. What is the value of 760 Torr in Pascal?
a) 1.01325 X 105
b) 6.025 X 105
c) 12.05 X 104
d) 3.0125 X 105

Answer: a [Reason:] 760 Torr = 1 atm = 1.01325 X 105.

3. How many Pascal does 1 bar have?
a) 104
b) 105
c) 106
d) 107

Answer: b [Reason:] 1 bar = 100 kPa = 105 Pa.

4. Which of the following equals 760 mm of Hg?
a) 1 atm
b) 1 Pa
c) 1 bar
d) 1 torr

Answer: a [Reason:] 760 mm of Hg = 101.325 kPa = 1 atm.

5. 1 psi equals how many kilopascal?
a) 5.249
b) 6.894
c) 7.489
d) 8.846

Answer: b [Reason:] 1 psi = 6894 Pa = 6.894 kPa.

6. What is the pressure of 50 kPa in bars?
a) 0.25
b) 0.5
c) 1
d) 2

Answer: b [Reason:] Pressure = 50/100 = 0.5 bar.

7. What is the pressure of 200 kPa in bars?
a) 0.25
b) 0.5
c) 1
d) 2

Answer: d [Reason:] Pressure = 200/100 = 2 bar.

8. What is the pressure of 380 torr in atm?
a) 0.5
b) 1
c) 2
d) 4

Answer: a [Reason:] Pressure = 380/760 = 0.5 atm.

9. What is the pressure of 190 torr in atm?
a) 0.25
b) 0.5
c) 0.8
d) 1

Answer: a [Reason:] Pressure = 190/760 = 0.25 atm.

10. What is the pressure of 38 mm Hg in atm?
a) 0.05
b) 0.5
c) 1
d) 5

Answer: a [Reason:] Pressure = 38/760 = 0.05 atm.

11. What is the pressure of 1900 mm Hg in atm?
a) 0.5
b) 1
c) 1.5
d) 2.5

Answer: d [Reason:] Pressure = 1900/760 = 2.5 atm.

12. What is the pressure of 19 mm Hg in pascal?
a) 1012
b) 2533
c) 3654
d) 6412

Answer: b [Reason:] Pressure = 19/760*101325 = 2533 Pa.

13. What is the pressure of 38 mm Hg in Torr?
a) 0.05
b) 0.5
c) 1
d) 38

Answer: d [Reason:] Pressure = 38 Torr.

14. What is the pressure of 380 Torr in bar?
a) 0.13
b) 0.49
c) 0.5
d) 1

Answer: b [Reason:] Pressure = 380/760*100000/101325 = 0.49 bar.

15. What is the pressure of 1900 Torr in bar?
a) 2.1
b) 2.3
c) 2.4
d) 2.5

Answer: c [Reason:] Pressure = 1900/760*100000/101325 = 2.4 bar.

Set 5

1. Solution-1 at the rate F combines with solution-2 at the rate D, their combination gives the product at the rate P, what is the equation of material balance?
a) F + D = P
b) F – P = D
c) D – F = P
d) None of the mentioned

Answer: a [Reason:] The general equation of material balance is F + D = P.

2. Solution-1 containing 30% sulfuric acid flowing at the rate 10 Kg/min combines with Solution-2 containing 20% sulfuric acid flowing at the rate 5 Kg/min, if their product is out at the rate 20 Kg/min what is the percentage of sulfuric acid in the product?
a) 10%
b) 20%
c) 30%
d) 40%

Answer: b [Reason:] Equation of material balance, 0.3(10) + 0.2(5) = x(20), => x = 0.2, => Percentage of sulfuric acid in the product = 20%.

3. Solution-1 containing 10% nitric acid flowing at the rate 10 Kg/min combines with Solution-2 containing 40% nitric acid flowing at the rate 5 Kg/min, if their product contains 30% nitric acid, what is the flow rate of product?
a) 10 Kg/min
b) 20 Kg/min
c) 30 Kg/min
d) 40 Kg/min

Answer: a [Reason:] Equation of material balance, 0.1(10) + 0.4(5) = 0.3(x), => x = 10 Kg/min.

4. Solution-1 containing 10% nitric acid and 20% acetic acid at the rate F combines with Solution-2 containing 20% nitric acid and 10% acetic acid at the rate D, the product formed contains 30% nitric acid and 20% acetic acid at the rate P, how many independent equations containing F, D, and P are possible?
a) 1
b) 2
c) 3
d) 4

Answer: c [Reason:] The equations of material balance are: 0.1F + 0.2D = 0.3P, 0.2F + 0.1D = 0.2P, and F + D = P.

5. Solution-1 containing 10% hydrochloric acid and 20% acetic acid at the rate F combines with Solution-2 containing 20% hydrochloric acid and 30% acetic acid at the rate D, the product formed contains 20% hydrochloric acid and 30% acetic acid at the rate P, how many independent equations containing F, D, and P are possible?
a) 1
b) 2
c) 3
d) 4

Answer: b [Reason:] The equations of material balance are: 0.1F + 0.2D = 0.2P, 0.2F + 0.3D = 0.3P and F + D = P, but if we subtract first two of them then we get the third, so number of independent equations are 2.

6. Solution-1 containing 40% nitric acid and 20% carbon dioxide at the rate F combines with Solution-2 containing 30% nitric acid and 40% carbon dioxide at the rate D, the product formed contains 30% nitric acid and 20% carbon dioxide at the rate P, If P = xF what is x?
a) 1
b) 2
c) 3
d) 4

Answer: b [Reason:] The equations of material balance are: 0.4F + 0.2D = 0.3P, 0.2F + 0.4D = 0.3P, and F + D = P, Solving these we get D = F = P/2, => x = 2.

7. Solution-1 containing 40% acetic acid and 20% carbon dioxide at the rate F combines with Solution-2 containing 20% acetic acid and 40% carbon dioxide at the rate D, the product formed contains 30% acetic acid at the rate P, what is the percentage of carbon dioxide in product?
a) 10%
b) 20%
c) 30%
d) 40%

Answer: c [Reason:] The equations for material balance are: 0.4F + 0.2D = 0.3P, F + D = P, and 0.2F + 0.4D = yP, solving these we get y = 0.3, => percentage of carbon dioxide = 30%.

8. Solution-1 containing 20% w1 and w2 at the rate 10 Kg/min combines with Solution-2 containing 10% w1, 20% w2 and w3 at the rate 20 Kg/min, the product formed contains and 10% w1, 30% w2 and 20% w3, what is the percentage of w2 in solution-1?
a) 5%
b) 10%
c) 25%
d) 50%

Answer: d [Reason:] Rate of product, P = 10 + 20 = 30 Kg/min, equation of material balance for w2: x(10) + 0.2(20) = 0.3(30), => w2 = 0.5, => percentage of w2 = 50%.

9. CH4 at the rate 10 mole/min is reacted with O2 at the rate 10 mole/min, what is the rate of formation of CO2 in the product?
a) 5 mole/min
b) 10 mole/min
c) 20 mole/min
d) 40 mole/min

Answer: a [Reason:] The balanced chemical reaction is: CH4 + 2O2 -> CO2 + 2H2O, this means O2 is the limiting reagent, => rate of formation of CO2 = 10/2 = 5 mole/min.

10. CH4 at the rate 10 mole/min is reacted with O2 at the rate 30 mole/min, what is the rate of formation of CO2 in the product?
a) 5 mole/min
b) 10 mole/min
c) 20 mole/min
d) 40 mole/min

Answer: b [Reason:] The balanced chemical reaction is: CH4 + 2O2 -> CO2 + 2H2O, this means CH4 is the limiting reagent, => rate of formation of CO2 = 10 mole/min.

11. Solution-1 containing 30% acetic acid and 20% nitrogen at the rate F combines with Solution-2 containing 20% acetic acid and 10% nitrogen at the rate D and Solution-3 containing 10% acetic acid and 30% nitrogen at the rate N the product formed contains 20% acetic acid at the rate P, what is the percentage of nitrogen in product?
a) 10%
b) 20%
c) 30%
d) Cannot be determined

Answer: d [Reason:] The equations for material balance are: 0.3F + 0.2D + 0.1N = 0.2P, F + D + N = P, and 0.2F + 0.1D + 0.3N = yP, since there are 5 variables and 3 equation, the equations cannot be solved.

12. Solution-1 containing 30% acetic acid and 20% nitrogen at the rate F combines with Solution-2 containing 20% acetic acid and 10% nitrogen at the rate D and Solution-3 containing 10% acetic acid and 30% nitrogen at the rate N the product formed contains 20% acetic acid at the rate P, what is the percentage of nitrogen in product?
a) 10%
b) 20%
c) 30%
d) Cannot be determined

Answer: d [Reason:] The equations for material balance are: 0.3F + 0.2D + 0.1N = 0.2P, F + D + N = P, and 0.2F + 0.1D + 0.3N = yP, since there are 5 variables and 3 equation, the equations cannot be solved.

13. Solution-1 containing 30% acetic acid and 20% nitrogen at the rate F combines with Solution-2 containing 20% acetic acid and 10% nitrogen at the rate D and Solution-3 containing 10% acetic acid and 30% nitrogen at the rate N the product formed contains 20% acetic acid at the rate 10 Kg/min, what is the percentage of nitrogen in product?
a) 10%
b) 20%
c) 30%
d) Cannot be determined

Answer: d [Reason:] The equations for material balance are: 0.3F + 0.2D + 0.1N = 0.2(10), F + D + N = 10, and 0.2F + 0.1D + 0.3N = y(10), since there are 4 variables and 3 equation, the equations cannot be solved.

14. Solution-1 containing 30% acetic acid and 100% nitrogen at the rate 10 Kg/min combines with Solution-2 containing 20% acetic acid and 10% nitrogen at the rate 20 Kg/min and Solution-3 containing 10% acetic acid and 30% nitrogen at the rate 30 Kg/min the product formed contains 20% acetic acid, what is the percentage of nitrogen in product?
a) 10%
b) 20%
c) 30%
d) Cannot be determined

Answer: a [Reason:] The equations for material balance are: P = 10 + 20 + 30 = 60 Kg/min, and 0.1(10) + 0.1(20) + 0.3(30) = y(60), => y = 0.1, => Percentage of nitrogen in the product = 10%.

15. Solution-1 containing 30% acetic acid and 20% nitrogen at the rate 10 Kg/min combines with Solution-2 containing 20% acetic acid and 10% nitrogen at the rate 20 Kg/min and Solution-3 containing 10% acetic acid and 30% nitrogen at the rate N the product formed contains 20% acetic acid at the rate P, what is the percentage of nitrogen in product?
a) 10%
b) 17.5%
c) 25.5%
d) 32.5%