Multiple choice question for engineering
Set 1
1. Which of the following is not the equation of state?
a) Van der waals Equation
b) Charles Equation
c) Holborn Equation
d) Peng Robinson Equation
Answer
Answer: b [Reason:] Charles equation is simply P α 1/V which is not the equation of state.
2. Virial Equation is a ______
a) Arithmetic Progression
b) Geometric Progression
c) Harmonic Progression
d) None of the mentioned
Answer
Answer: b [Reason:] Virial equation is PV = RT (1 + 1/V + 1/V2 + 1/V3 + – – – – -), which is a geometric progression.
3. If the vander waal equation has only one root, what is the relation between a and b?
a) 729a = pb2
b) 243a = pb2
c) 81a = pb2
d) 9a = pb2
Answer
Answer: a [Reason:] Vander waal equation, V3 – (nb + nRT/p)V2 + (n2a/p)V – n3ab/p = 0, comparing with (V – Vc)3 = 0, => 3n(a/p)1/2 = n(ab/p)1/3, => 729a = pb2.
4. A container has O2 at 5 atm and molar volume 2 liter/mole, if a = 9.24 atm(liter/mole), b = 0.0907 liter/mole, what is the temperature of O2? (Use vander waal equation)
a) 140 K
b) 150 K
c) 170 K
d) 190 K
Answer
Answer: c [Reason:] (p + a/V2)(V – b) = RT, => T = (5 + 9.24/22)(2 – 0.0907)/0.0821 = 170 K.
5. A container has H2 at 27oC and molar volume 1 liter/mole, if a = 9.24 atm(liter/mole), b = 0.0907 liter/mole, what is the temperature of H2? (Use vander waal equation)
a) 7.5 atm
b) 12.4 atm
c) 15.6 atm
d) 17.8 atm
Answer
Answer: d [Reason:] (p + a/V2)(V – b) = RT, => p = 0.0821*300/(1 – 0.0907) – 9.24/12 = 17.8 atm.
6. What is the pressure of CO2 at 0oC, molar volume = 4 liter/mole, if B = – 0.053 liter/mole, C = 0.0026 liter2/mole2? (Use virial equation, neglect higher terms)
a) 2.4 atm
b) 4.1 atm
c) 5.5 atm
d) 6.9 atm
Answer
Answer: c [Reason:] pV = RT(1 + B/V + C/V2), => p = (0.0821*273/4)(1 – 0.053/4 + 0.0026/16) = 5.5 atm.
7. A container is filled with a gas at 10 atm with molar volume 1 liter/mole, if B = – 0.053 liter/mole, C = 0.0026 liter2/mole2, what is the temperature of gas? (Use virial equation, neglect higher terms)
a) 128.2 K
b) 156.9 K
c) 198.5 K
d) 243.6 K
Answer
Answer: a [Reason:] PV = RT(1 + B/V + C/V2), => T = 10*1/[0.0821*(1 – 0.053/1 + 0.0026/12)] = 128.2 K.
8. What is the unit of ‘b’ in Peng Robinson equation?
a) Liter/mole
b) Liter2/mole2
c) Mole/Liter
d) Mole2/Liter2
Answer
Answer: a [Reason:] Peng Robinson equation is p = RT/(V – b) – aα/[V(V + b) + b(V – b)], => unit of b = unit of molar volume = liter/mole.
9. What is the unit of λ in SRK equation?
a) Liter/mole
b) Mole/Liter
c) Atm(liter/mole)
d) It is dimensionless
Answer
Answer: d [Reason:] In SRK equation, λ is dimensionless function of reduced temperature.
10. What is the unit of ‘a’ in RK equation?
a) Liter/mole
b) Atm(liter/mole)
c) Atm(liter/mole)K0.5
d) Atm(liter/mole)K
Answer
Answer: c [Reason:] RK equation is p = RT/(V – b) – a/[T0.5V(V + b)], => unit of a = atm(liter/mole)K0.5.
Set 2
1. Which of the following is the unit of flow rate?
a) Kg/s
b) Liters/s
c) Moles/s
d) All of the mentioned
Answer
Answer: d [Reason:] Mass flow rate = m/t, => unit = kg/s, volumetric flow rate = V/t, => unit = Liters/s, molar rate = n/t, => unit = moles/s.
2. Which of the following is not conserved for the flow of an incompressible liquid in a pipe with variable area?
a) Mass
b) Volume
c) Velocity
d) Flow Rate
Answer
Answer: c [Reason:] According to the equation of continuity, the product of area and velocity remains constant throughout the pipe, since area of the pipe is variable, so the velocity if the liquid is variable.
3. What is the mass flow rate of a liquid flowing through a pipe of which 100 grams of water was collected in 10 seconds?
a) 0.01 kg/s
b) 0.05 kg/s
c) 0.1 kg/s
d) 0.5 kg/s
Answer
Answer: a [Reason:] mass flow rate = 0.1/ 10 = 0.01 kg/s.
4. What is the volumetric flow rate if 100 kg of water is collected in 10 seconds? (Density of water = 1000 kg/m3)
a) 0.1 m3/s
b) 0.2 m3/s
c) 0.01 m3/s
d) 0.02 m3/s
Answer
Answer: c [Reason:] volume of water collected = 100/1000 = 0.1 m3, => volumetric flow rate = 0.1/10 = 0.01 m3/s.
5. A liquid is flowing through a pipe of cross section 4 m2 at a rate of 4 kg/s, at a point pipe is divided into two pipes of areas of equal areas, what will be the ratio of flow rate in both the pipes?
a) 1:3
b) 1:1
c) 1:2
d) 2:3
Answer
Answer: b [Reason:] Since areas of the pipes are equal, by continuity equation flow will divide equally in the pipes, so their ratio will be 1:1.
6. The velocity of water in a pipe of area 5 m2 is 10 m/s, what will be the flow rate of water?
a) 25 m3/s
b) 50 m3/s
c) 75 m3/s
d) 100 m3/s
Answer
Answer: b [Reason:] Flow rate = area*velocity = 5*10 = 50 m3/s.
7. A liquid of density 0.1 kg/m3 is flowing through a pipe at a rate of 15 kg/s, what will be the area of the pipe if the fluid velocity is 5 m/s?
a) 30 m2
b) 45 m2
c) 75 m2
d) 90 m2
Answer
Answer: a [Reason:] Volumetric flow rate of the liquid = 15/0.1 = 150 m3/s, => Area of pipe = 150/5 = 30 m2.
8. 1 mole of Carbon dioxide enters in a reactor at a rate of 0.25 kg/s and leaves it at the rate of 1 kg/s, what are the number of moles of carbon dioxide leaves the reactor?
a) 0.25
b) 0.5
c) 0.75
d) 1
Answer
Answer: a [Reason:] Applying reaction equilibrium equation, 1*0.25 = n*1, => n = 0.25.
9. A Composition of carbon dioxide, oxygen and nitrogen with mole ratio 1:2:5 with 4 moles of oxygen, enters a reactor with a flow rate of 100 kg/s, the ratio of rates of leaving the reactor of the three is 5:2:1, and 5 moles of nitrogen leaves the reactor, what is the total number of moles leaving the reactor?
a) 6.1
b) 6.2
c) 6.3
d) 6.4
Answer
Answer: b [Reason:] number of moles of carbon dioxide, oxygen and nitrogen are 2, 4 and 10 respectively, let the leaving flow rates of carbon dioxide, oxygen and nitrogen are 5y, 2y and y respectively, => for nitrogen, 10*100 = y*5 => y = 200 kg/s, => for oxygen, 4*100 = 400*n => n = 1, for carbon dioxide, 2*100 = 1000*n => n = 0.2, => total number of moles leaving the reactor = 5 + 1 + 0.2 = 6.2.
10. 10 moles of oxygen divides and enters two reactors flow rate 45 kg/s and leaves the reactors at the flow rates of 6 kg/s and 9 kg/s, what are number of moles of oxygen that leaves the reactor with 9 kg/s?
a) 180
b) 450
c) 60
d) 90
Answer
Answer: d [Reason:] Let number of moles entering and leaving the second reactor be n1 and n2 respectively, applying reaction equilibrium equation to whole system, => 10*45 = n*(6 + 9), => n = 30, => Applying reaction equilibrium equation to both the reactors separately, => n1*45 = n2*9 => n2 = 5n1, and (10 – n1)*45 = (30 – n2)*6, Solving both equations we get n2 = 90.
11. 10 Kg/s of a fluid with density 10 Kg/m3 is passed into a tank of capacity 50 m3, how much time will it take to fill the tank?
a) 5 Second
b) 50 Second
c) 500 Second
d) 5000 Second
Answer
Answer: b [Reason:] Flow rate of fluid = 10/10 = 1 m3/s, => Time = 50/1 = 50 second.
12. 15 Kg/s of a fluid with density 5 Kg/m3 is passed into a tank of capacity 30 m3, how much time will it take to fill the tank?
a) 3 Second
b) 10 Second
c) 30 Second
d) 100 Second
Answer
Answer: b [Reason:] Flow rate of fluid = 15/5 = 3 m3/s, => Time = 30/3 = 10 second.
13. 20 Kg/s of a fluid with density 5 Kg/m3 is passed into a tank of capacity 60 m3, how much time will it take to fill the tank?
a) 5 Second
b) 10 Second
c) 15 Second
d) 20 Second
Answer
Answer: c [Reason:] Flow rate of fluid = 20/5 = 4 m3/s, => Time = 60/4 = 15 second.
14. 16 grams of oxygen is passed in the process in 5 minutes, what is the molar flow rate?
a) 0.1 mole/min
b) 1 mole/min
c) 10 mole/min
d) 16 mole/min
Answer
Answer: a [Reason:] Flow rate of fluid = (16/32)/5 = 0.1 mole/min.
15. 24 grams of CH4 is passed in the process in 3 minutes, what is the molar flow rate?
a) 0.1 mole/min
b) 0.3 mole/min
c) 0.5 mole/min
d) 0.7 mole/min
Answer
Answer: c [Reason:] Flow rate of fluid = (24/16)/3 = 0.5 mole/min.
Set 3
1. Which of the following is used for the pressure measurement of only liquid?
a) Manometer
b) Piezometer
c) Differential Manometer
d) None of the mentioned
Answer
Answer: b [Reason:] Piezometer is a pressure measuring device used to measure pressure of liquid only.
2. Which of the following is used to measure pressure of only gas?
a) U-tube manometer
b) Piezometer
c) Differential Manometer
d) None of the mentioned
Answer
Answer: a [Reason:] U-tube manometer is used to measure the pressure of gas only.
3. What is the difference between manometer and differential manometer?
a) Differential manometer has gas in empty space
b) Manometer has gas in empty space
c) Differential manometer has liquid in empty space
d) Manometer has liquid in empty space
Answer
Answer: a [Reason:] Differential manometer has gas in empty space while manometer has vacuum.
4. Which of the following device is used to measure velocity of flow?
a) Manometer
b) Venturi meter
c) Pitot tube
d) None of the mentioned
Answer
Answer: c [Reason:] Pitot tube is used to measure the velocity of flow.
5. A U-tube manometer is open at one end and has a gas at the other end, if the level of the mercury in both legs of manometer is equal then what is the temperature of the gas?
a) 1 Pa
b) 1 Torr
c) 1 mm Hg
d) 1 atm
Answer
Answer: d [Reason:] Since the level of mercury in both legs are equal, this means that pressure is equal on both ends, => pressure of gas = pressure of atmosphere = 1 atm.
6. A U-tube manometer is closed at both ends one end is vacuum and other end has a gas, if the difference in level of mercury is h = 10 mm, what is the pressure of gas?
a) 1.33 Pa
b) 1.45 Pa
c) 1.56 Pa
d) 1.75 Pa
Answer
Answer: a [Reason:] Since the other end of the manometer has vacuum, this means pressure of gas is equal to difference in pressure of mercury in both ends, => Pressure of gas = d*g*h = 13.6*9.8*0.01 = 1.33 Pa.
7. A closed U-tube manometer has a gas on one end with pressure 10 Pa and other end has vacuum, what is the approximate density of the liquid if the height difference of the liquid is 10 cm?
a) 1 Kg/m3
b) 10 Kg/m3
c) 100 Kg/m3
d) 1000 Kg/m3
Answer
Answer: b [Reason:] Pressure of gas = Pressure due to height difference of liquid = d*g*h, => 10 = d*10*0.1, => d = 10 Kg/m3.
8. A closed U-tube manometer has a gas on one end with pressure 5 Pa and other end has vacuum, what is the approximate height difference of the liquid if the density of liquid is 10 Kg/m3?
a) 5 mm
b) 10 mm
c) 50 mm
d) 100 mm
Answer
Answer: c [Reason:] Pressure of gas = Pressure due to height difference of liquid = d*g*h, => 5 = 10*10*h => h = 0.05 m = 50 mm.
9. A closed manometer is filled with 3 fluids first on the left leg with density 8 kg/m3 and length 10 cm, middle one with density 10 kg/m3 and height difference 5 cm more on right leg and the third on the right leg with density 15 kg/m3, what is the length of the third liquid?
a) 1 cm
b) 2 cm
c) 3 cm
d) 4 cm
Answer
Answer: b [Reason:] Pressure on both legs should be equal, => 8*10 = 10*5 + 15*x, => x = 2 cm.
10. A barometer has liquid level 10 cm above the outside level and the gas captured inside has the pressure of 0.5 atm, what is the density of liquid?
a) 0.5 atm-s2/m2
b) 1 atm-s2/m2
c) 1.5 atm-s2/m2
d) 2 atm-s2/m2
Answer
Answer: a [Reason:] Pressure of gas + Pressure of liquid = atmospheric pressure, => 0.5 atm + d*10*0.1 = 1 atm, => d = 0.5 atm-s2/m2.
11. A barometer has liquid of density 1 kg/m3 of level 5 cm below the outside level, what is the pressure of the gas captured inside?
a) 0.5 atm + 1 Pa
b) 1 atm + 0.5 Pa
c) 1.5 atm + 1 Pa
d) 2 atm + 0.5 Pa
Answer
Answer: b [Reason:] Pressure of gas = Pressure of liquid + atmospheric pressure = 1*10*0.05 Pa + 1 atm = 0.5 Pa + 1 atm.
12. A barometer has liquid with density 1 Kg/m3, and the gas has pressure 0.5 atm, what is the height of liquid level?
a) 0.01 atm-m2-s2/kg
b) 0.05 atm-m2-s2/kg
c) 0.1 atm-m2-s2/kg
d) 0.5 atm-m2-s2/kg
Answer
Answer: b [Reason:] Pressure of gas + Pressure of Liquid = atmospheric pressure, => 0.5 atm + 1*10*h = 1 atm, => h = 0.05 atm-m2-s2/kg.
13. A barometer the liquid-1 with density 2 Kg/m3 and height 5 cm below the outside level and above that liquid-2 with density 0.5 Kg/m3 and length 10 cm, what is the gauge pressure of the gas?
a) 0.01 Pa
b) 0.05 Pa
c) 0.1 Pa
d) 0.5 Pa
Answer
Answer: d [Reason:] Gauge pressure of gas + Pressure of liquid-2 = Pressure of liquid-1, => Gauge pressure of gas = 2*10*0.05– 0.5*10*0.1 = 0.5 Pa.
14. A barometer has liquid-1 with density 5 Kg/m3 and height 10 cm below the outside level above that there is liquid-2 of length 5 cm, if the gauge pressure of the gas is 1 Pa, what is the density of liquid-2?
a) 2 Kg/m3
b) 4 Kg/m3
c) 6 Kg/m3
d) 8 Kg/m3
Answer
Answer: d [Reason:] Gauge pressure of gas + Pressure of liquid-2 = Pressure of liquid-1, => 1 + d*10*0.05 = 5*10*0.1, => d = 8 Kg/m3.
15. A barometer has liquid-1 with density 1 Kg/m3 and height 10 cm below the outside level above that a liquid-2, if the gauge pressure of gas is 0.5 Pa, what is the pressure liquid-2?
a) 0.5 Pa
b) 1 Pa
c) 1.5 Pa
d) 2 Pa
Answer
Answer: a [Reason:] Gauge pressure of gas + Pressure of liquid-2 = Pressure of liquid-1, => Pressure of liquid-2 = 1*10*0.1 – 0.5 = 0.5 Pa.
Set 4
1. What is the degree of freedom of a system with 2 phases and 1 component?
a) 1
b) 2
c) 3
d) 4
Answer
Answer: a [Reason:] F = 2 – 2 + 1 = 1.
2. What is the number of phases in a system with 2 degrees of freedom and 2 components?
a) 1
b) 2
c) 3
d) 4
Answer
Answer: b [Reason:] P = 2 – 2 + 2 = 2.
3. What is the number of components in a system with 3 degrees of freedom and 2 phases?
a) 1
b) 2
c) 3
d) 4
Answer
Answer: c [Reason:] C = 2 – 2 + 3 = 3.
4. What is the degree of freedom of a system with 3 phases and 5 components?
a) 1
b) 2
c) 3
d) 4
Answer
Answer: d [Reason:] F = 2 – 3 + 5 = 6.
5. What is the degrees of freedom of a system with 5 phases and 5 components?
a) 1
b) 2
c) 3
d) 4
Answer
Answer: b [Reason:] F = 2 – 5 + 5 = 2.
6. What is the degrees of freedom of a system with pure water?
a) 2
b) 4
c) 5
d) 7
Answer
Answer: a [Reason:] F = 2 – 1 + 1 = 2.
7. What is the degrees of freedom of a system with mixture of water and ice?
a) 1
b) 2
c) 3
d) 4
Answer
Answer: a [Reason:] F = 2 – 2 + 1 = 1.
8. What is the degrees of freedom of a system with mixture of water, liquid benzene and oxygen?
a) 1
b) 2
c) 3
d) 4
Answer
Answer: c [Reason:] F = 2 – 2 + 3 = 3.
9. What is the degrees of freedom of a system with mixture of ice, water and neon?
a) 1
b) 2
c) 3
d) 4
Answer
Answer: c [Reason:] F = 2 – 3 + 2 = 3.
10. What is the degrees of freedom of a system with mixture of copper pyrites, liquid benzene and helium?
a) 1
b) 2
c) 3
d) 4
Answer
Answer: b [Reason:] F = 2 – 3 + 3 = 2.
11. What is the degree of freedom of a gas mixture composed of CO2, SO2, SO3, CH4, H2CO3?
a) 2
b) 3
c) 5
d) 7
Answer
Answer: c [Reason:] F = 2 – 1 + 4 = 5.
12. What is the degree of freedom of a gas mixture composed of O2, H2O and H2O2?
a) 1
b) 2
c) 3
d) 4
Answer
Answer: c [Reason:] F = 2 – 1 + 2 = 3.
13. What is the degree of freedom of a mixture composed Liquid H2O, solid Br2 and gaseous CO2, NO2?
a) 1
b) 2
c) 3
d) 4
Answer
Answer: c [Reason:] F = 2 – 3 + 4 = 3.
14. What is the degree of freedom of a mixture composed of liquid O2, gaseous H2 and ice?
a) 1
b) 2
c) 3
d) 4
Answer
Answer: a [Reason:] F = 2 – 3 + 2 = 1.
15. What is the degree of freedom of a gaseous mixture composed of CH4, CO2, H2CO3, H2O and H2O2?
a) 1
b) 2
c) 3
d) 4
Answer
Answer: d [Reason:] F = 2 – 1 + 3 = 4.
Set 5
1. Water having CP = 10 J/oC, is heated from 10oC to 50oC, what is the enthalpy change?
a) 100 J
b) 250 J
c) 400 J
d) 500 J
Answer
Answer: c [Reason:] ∆H = 10∫5010.dT = 10*40 = 400 J.
2. Liquid benzene having CP = 29 J/oC, is heated from 5oC to 10oC, what is the enthalpy change?
a) 29 J
b) 145 J
c) 195 J
d) 290 J
Answer
Answer: b [Reason:] ∆H = 5∫1029.dT = 29*5 = 145 J.
3. Ice having CP = 15 J/oC, is heated from -10oC to -4oC, what is the enthalpy change?
a) 30 J
b) 60 J
c) 90 J
d) 120 J
Answer
Answer: c [Reason:] ∆H = -10∫-415.dT = 15*6 = 90 J.
4. Steam having CP = 105 J/oC, is heated from 100oC to 104oC, what is the enthalpy change?
a) 210 J
b) 420 J
c) 630 J
d) 840 J
Answer
Answer: b [Reason:] ∆H = 100∫104105.dT = 105*4 = 420 J.
5. Cyclohexane having CP = 48 J/oC, is heated from 50oC to 55oC, what is the enthalpy change?
a) 240 J
b) 480 J
c) 720 J
d) 960 J
Answer
Answer: a [Reason:] ∆H = 50∫5548.dT = 48*5 = 240 J.
6. What is the value of CP – CV for ideal gases?
a) 0
b) R/2
c) R
d) 2R
Answer
Answer: c [Reason:] For ideal gases CP = CV + R.
7. A mixture contains 20% O2 with CP = 15 J/oC and 80% CO2 with CP = 20 J/oC, what is the average CP of mixture?
a) 3 J/oC
b) 9 J/oC
c) 16 J/oC
d) 19 J/oC
Answer
Answer: d [Reason:] Average CP = 0.2*15 + 0.8*20 = 19 J/oC.
8. A mixture contains 30% Ar with CP = 30 J/oC, 30% N2 with CP = 20 J/oC and 40% CH4 with CP = 10 J/oC, what is the average CP of mixture?
a) 7 J/oC
b) 12 J/oC
c) 19 J/oC
d) 26 J/oC
Answer
Answer: c [Reason:] Average CP = 0.3*30 + 0.3*20 + 0.4*10 = 19 J/oC.
9. A mixture contains 20% O2 with CP = 10 J/oC, 30% N2 with CP = 20 J/oC and 50% CO2 with CP= 40 J/oC, what is the average CP of mixture?
a) 20 J/oC
b) 28 J/oC
c) 34 J/oC
d) 42 J/oC
Answer
Answer: b [Reason:] Average CP = 0.2*10 + 0.3*20 + 0.5*40 = 2 + 6 + 20 = 28 J/oC.
10. A mixture contains 20% SO2 with CP= 35 J/oC, 30% NO3 with CP = 20 J/oC and 50% CO2 with CP = 40 J/oC, what is the average CP of mixture?
a) 12 J/oC
b) 17 J/oC
c) 25 J/oC
d) 33 J/oC
Answer
Answer: d [Reason:] Average CP = 0.2*35 + 0.3*20 + 0.5*40 = 7 + 6 + 20 = 33 J/oC.
11. Heat Capacity of CO2 is given by 12 + 3T + T2, what is the enthalpy change from 0oC to 6oC?
a) 169 J
b) 198 J
c) 244 J
d) 360 J
Answer
Answer: b [Reason:] ∆H = 0∫6 (12 + 3T + T2).dT = 12*6 + 3*36/2 + 63/3 = 198 J.
12. Heat Capacity of water is given by 4 + T2, what is the enthalpy change from 0oC to 3oC?
a) 7 J
b) 14 J
c) 21 J
d) 32 J
Answer
Answer: c [Reason:] ∆H = 0∫3 (4 + T2).dT = 4*3 + 33/3 = 21 J.
13. CP-T curve covers 10 J from 2oC to 10oC, what is the enthalpy change?
a) 10 J
b) 20 J
c) 80 J
d) 100 J
Answer
Answer: a [Reason:] Enthalpy change is the area covered under CP-T curve.
14 CP-T curve of CO2 forms a rectangle of height 10 J/oC from 5oC to 10oC?
a) 10 J
b) 50 J
c) 100 J
d) 250 J
Answer
Answer: b [Reason:] Enthalpy change is the area covered under CP-T curve = Area of rectangle = 10*(10 – 5) = 50 J.
15. CP-T curve of H2 forms a rectangle of height 20 J/oC from 4oC to 5oC?
a) 20 J
b) 80 J
c) 100 J
d) 140 J
Answer
Answer: c [Reason:] Enthalpy change is the area covered under CP-T curve = Area of rectangle = 20*(5 – 4) = 20 J.
