Chemical Engineering MCQs

Option A: Specific volume

Option B: Work

Option C: Pressure

Option D: Temperature

Correct Answer: Work ✔

Option A: Van’t-Hoff equation

Option B: Le-Chatelier’s principle

Option C: Arrhenius equation

Option D: None of these

Correct Answer: Van’t-Hoff equation ✔

Option A: Endothermic

Option B: Exothermic

Option C: Isothermal

Option D: Adiabatic

Correct Answer: Exothermic ✔

Option A: ∞

Option B: 0

Option C: Maximum

Option D: Minimum

Correct Answer: Minimum ✔

Option A: -94 kcal

Option B: +94 kcal

Option C: > 94 kcal

Option D: < -94 kcal

Correct Answer: +94 kcal ✔

Option A: Solubility increases as temperature increases

Option B: Solubility increases as temperature decreases

Option C: Solubility is independent of temperature

Option D: Solubility increases or decreases with temperature depending on the Gibbs free energy change

Correct Answer: Solubility increases as temperature decreases ✔

Option A: Disorder

Option B: Orderly behaviour

Option C: Temperature changes only

Option D: None of these

Correct Answer: Disorder ✔

Option A: Volume

Option B: Pressure

Option C: Temperature

Option D: All (A), B. and (C)

Correct Answer: Volume ✔

Option A: ∞

Option B: -ve

Option C: 0

Option D: +ve

Correct Answer: -ve ✔

Option A: Kinematic viscosity

Option B: Work

Option C: Temperature

Option D: None of these

Correct Answer: None of these ✔

Option A: Bomb

Option B: Separating

Option C: Bucket

Option D: Throttling

Correct Answer: Bomb ✔

Option A: Equal to its density

Option B: The reciprocal of its density

Option C: Proportional to pressure

Option D: None of these

Correct Answer: The reciprocal of its density ✔

Option A: Freezing

Option B: Triple

Option C: Boiling

Option D: Boyle

Correct Answer: Triple ✔

Option A: Steam engine

Option B: Carnot engine

Option C: Diesel engine

Option D: Otto engine

Correct Answer: Steam engine ✔

Option A: Low pressure and high temperature

Option B: Low pressure and low temperature

Option C: High pressure and low temperature

Option D: High pressure and high temperature

Correct Answer: Low pressure and low temperature ✔

Option A: Fusion

Option B: Vaporisation

Option C: Transition

Option D: None of these

Correct Answer: Transition ✔

Option A: Stirling

Option B: Brayton

Option C: Rankine

Option D: None of these

Correct Answer: Stirling ✔

Option A: Less than

Option B: More than

Option C: Equal to or higher than

Option D: Less than or equal to

Correct Answer: Less than ✔

Option A: Hess’s

Option B: Kirchoff’s

Option C: Lavoisier and Laplace

Option D: None of these

Correct Answer: A. Hess’s ✔

Option A: Adiabatic

Option B: Isometric

Option C: Isentropic

Option D: Isothermal

Correct Answer: Isometric ✔

Option A: Directly proportional to pressure

Option B: Inversely proportional to pressure

Option C: Unity at all pressures

Option D: None of these

Correct Answer: Unity at all pressures ✔

Option A: 448

Option B: 224

Option C: 22.4

Option D: Data insufficient; can’t be computed

Correct Answer: 448 ✔

Option A: F = A + PV

Option B: F = E + A

Option C: F = A – TS

Option D: F = A + TS

Correct Answer: F = A + PV ✔

Option A: Increase

Option B: Decrease

Option C: Not alter

Option D: None of these

Correct Answer: Increase ✔

Option A: (R/ΔH) (1/T1 – 1/T2)

Option B: (ΔH/R) (1/T1 – 1/T2)

Option C: (ΔH/R) (1/T2 – 1/T1)

Option D: (1/R) (1/T1 – 1/T2)

Correct Answer: B. (ΔH/R) (1/T1 – 1/T2) ✔

Option A: CO2

Option B: H2

Option C: O2

Option D: N2

Correct Answer: H2 ✔

Option A: Use of only one graph for all gases

Option B: Covering of wide range

Option C: Easier plotting

Option D: More accurate plotting

Correct Answer: Use of only one graph for all gases ✔

Option A: μ° + RT ln f

Option B: μ°+ R ln f

Option C: μ° + T ln f

Option D: μ° + R/T ln f

Correct Answer: μ° + RT ln f ✔

Option A: V/T = Constant

Option B: V ∝ 1/T

Option C: V ∝ 1/P

Option D: PV/T = Constant

Correct Answer: V/T = Constant ✔

Option A: Lewis-Randall

Option B: Margules

Option C: Van Laar

Option D: Both B. & C.

Correct Answer: D. Both B. & C. ✔

Option A: -2 RT ln 0.5

Option B: -RT ln 0.5

Option C: 0.5 RT

Option D: 2 RT

Correct Answer: -RT ln 0.5 ✔

Option A: 72

Option B: 92

Option C: 142

Option D: 192

Correct Answer: 92 ✔

Option A: Ice at the base contains impurities which lowers its melting point

Option B: Due to the high pressure at the base, its melting point reduces

Option C: The iceberg remains in a warmer condition at the base

Option D: All A, B. and C

Correct Answer: Due to the high pressure at the base, its melting point reduces ✔

Option A: Vapor pressure

Option B: Partial pressure

Option C: Chemical potential

Option D: None of these

Correct Answer: Partial pressure ✔

Option A: Volume, mass and number of moles

Option B: Free energy, entropy and enthalpy

Option C: Both A. and B

Option D: None of these

Correct Answer: C. Both A. and B ✔

Option A: Binary solutions

Option B: Ternary solutions

Option C: Azeotropic mixture only

Option D: None of these

Correct Answer: Binary solutions ✔

Option A: Molar heat capacity

Option B: Internal energy

Option C: Viscosity

Option D: None of these

Correct Answer: Internal energy ✔

Option A: An open system of constant composition

Option B: A closed system of constant composition

Option C: An open system with changes in composition

Option D: A closed system with changes in composition

Correct Answer: A closed system with changes in composition ✔

Option A: Enthalpy

Option B: Internal energy

Option C: Either A. or B

Option D: Neither A. nor B

Correct Answer: Internal energy ✔

Option A: Low pressure & high temperature

Option B: High pressure & low temperature

Option C: Low pressure & low temperature

Option D: None of these

Correct Answer: Low pressure & high temperature ✔

Option A: ΔF = ΔH + T [∂(ΔF)/∂T]P

Option B: ΔF = ΔH – TΔT

Option C: d(E – TS) T, V < 0

Option D: dP/dT = ΔHvap/T.ΔVvap

Correct Answer: ΔF = ΔH + T [∂(ΔF)/∂T]P ✔

Option A: Concentration of the constituents only

Option B: Quantities of the constituents only

Option C: Temperature only

Option D: All A, B. and C

Correct Answer: D. All A, B. and C ✔

Option A: Isothermal compression

Option B: Isothermal expansion

Option C: Adiabatic expansion

Option D: Adiabatic compression

Correct Answer: Adiabatic compression ✔

Option A: Supersaturated

Option B: Superheated

Option C: Both A. and B

Option D: Neither A. nor B

Correct Answer: Superheated ✔

Option A: Heat pump

Option B: Heat engine

Option C: Carnot engine

Option D: None of these

Correct Answer: Heat pump ✔

Option A: More in vapour phase

Option B: More in liquid phase

Option C: Same in both the phases

Option D: Replaced by chemical potential which is more in vapour phase

Correct Answer: Same in both the phases ✔

Option A: Increases

Option B: Decreases

Option C: Remain constant

Option D: Increases linearly

Correct Answer: Increases ✔

Option A: Reversible isothermal

Option B: Irreversible isothermal

Option C: Reversible adiabatic

Option D: None of these

Correct Answer: Reversible adiabatic ✔

Option A: No heat and mass transfer

Option B: No mass transfer but heat transfer

Option C: Mass and energy transfer

Option D: None of these

Correct Answer: No mass transfer but heat transfer ✔

Option A: (∂E/∂ni)S, v, nj

Option B: (∂G/∂ni)T, P, nj = (∂A/∂ni) T, v, nj

Option C: (∂H/∂ni)S, P, nj

Option D: All (A), B. and (C)

Correct Answer: D. All (A), B. and (C) ✔

Option A: Chemical potentials of a given component should be equal in all phases

Option B: Chemical potentials of all components should be same in a particular phase

Option C: Sum of the chemical potentials of any given component in all the phases should be the same

Option D: None of these

Correct Answer: Chemical potentials of a given component should be equal in all phases ✔

Option A: Bucket

Option B: Throttling

Option C: Separating

Option D: A combination of separating & throttling

Correct Answer: A combination of separating & throttling ✔

Option A: Cold reservoir approaches zero

Option B: Hot reservoir approaches infinity

Option C: Either A. or B

Option D: Neither A. nor B

Correct Answer: C. Either A. or B ✔

Option A: Zero

Option B: Unity

Option C: Infinity

Option D: None of these

Correct Answer: Zero ✔

Option A: Decrease on addition of Cl2

Option B: Increase on addition of an inert gas at constant pressure

Option C: Decrease on increasing the pressure of the system

Option D: None of these

Correct Answer: None of these ✔

Option A: Compressibility

Option B: Work done under adiabatic condition

Option C: Work done under isothermal condition

Option D: Co-efficient of thermal expansion

Correct Answer: Work done under isothermal condition ✔

Option A: Are more or less constant (vary from 0.2 to 0.3)

Option B: Vary as square of the absolute temperature

Option C: Vary as square of the absolute pressure

Option D: None of these

Correct Answer: Are more or less constant (vary from 0.2 to 0.3) ✔

Option A: Pressure

Option B: Temperature

Option C: Both A. & B

Option D: Neither A. nor B

Correct Answer: C. Both A. & B ✔

Option A: Zero

Option B: +ve

Option C: -ve

Option D: Dependent on the path

Correct Answer: Zero ✔

Option A: Increase the partial pressure of I2

Option B: Decrease the partial pressure of HI

Option C: Diminish the degree of dissociation of HI

Option D: None of these

Correct Answer: Diminish the degree of dissociation of HI ✔

Option A: Amount of energy transferred

Option B: Direction of energy transfer

Option C: Irreversible processes only

Option D: Non-cyclic processes only

Correct Answer: Direction of energy transfer ✔

Option A: First law

Option B: Zeroth law

Option C: Third law

Option D: Second law

Correct Answer: Zeroth law ✔

Option A: Becomes zero

Option B: Becomes infinity

Option C: Equals 1 kcal/kmol °K

Option D: Equals 0.24 kcal/kmol °K

Correct Answer: Becomes infinity ✔

Option A: Not have a sub-atmospheric vapour pressure at the temperature in the refrigerator coils

Option B: Not have unduly high vapour pressure at the condenser temperature

Option C: Both A. and B

Option D: Have low specific heat

Correct Answer: C. Both A. and B ✔

Option A: dP/dT = ΔH/TΔV

Option B: ln P = – (ΔH/RT) + constant

Option C: ΔF = ΔH + T [∂(ΔF)/∂T]P

Option D: None of these

Correct Answer: B. ln P = – (ΔH/RT) + constant ✔

Option A: Kelvin’s

Option B: Antoine’s

Option C: Kirchoff’s

Option D: None of these

Correct Answer: Kirchoff’s ✔

Option A: Isothermal

Option B: Irreversible

Option C: Adiabatic

Option D: Reversible

Correct Answer: Irreversible ✔

Option A: Pressure

Option B: Temperature

Option C: Volume

Option D: Molar concentration

Correct Answer: Pressure ✔

Option A: Gibbs-Duhem

Option B: Maxwell’s

Option C: Clapeyron

Option D: None of these

Correct Answer: Clapeyron ✔

Option A: Is the analog of linear frictionless motion in machines

Option B: Is an idealised visualisation of behaviour of a system

Option C: Yields the maximum amount of work

Option D: Yields an amount of work less than that of a reversible process

Correct Answer: Yields an amount of work less than that of a reversible process ✔

Option A: < 0

Option B: > 0

Option C: = 0

Option D: None of these

Correct Answer: < 0 ✔

Option A: Same as Carnot cycle

Option B: Same as reverse Carnot cycle

Option C: Dependent on the refrigerant’s properties

Option D: The least efficient of all refrigeration processes

Correct Answer: Same as reverse Carnot cycle ✔

Option A: 0

Option B: < 0

Option C: < 1

Option D: > 1

Correct Answer: > 1 ✔

Option A: Pressure

Option B: Temperature

Option C: Both A. & B

Option D: Neither A. nor B

Correct Answer: Temperature ✔

Option A: Does not depend upon temperature

Option B: Is independent of pressure only

Option C: Is independent of volume only

Option D: Is independent of both pressure and volume

Correct Answer: Is independent of both pressure and volume ✔

Option A: Reversible and isothermal

Option B: Irreversible and constant enthalpy

Option C: Reversible and constant entropy

Option D: Reversible and constant enthalpy

Correct Answer: Irreversible and constant enthalpy ✔

Option A: Specific heat at constant pressure (Cp)

Option B: Specific heat at constant volume (Cv)

Option C: Joule-Thompson co-efficient

Option D: None of these

Correct Answer: Joule-Thompson co-efficient ✔

Option A: Water

Option B: Ammonia

Option C: Freon

Option D: Brine

Correct Answer: Water ✔

Option A: Conduction

Option B: Convection

Option C: Radiation

Option D: Condensation

Correct Answer: Condensation ✔

Option A: Doubling the absolute temperature as well as pressure of the gas

Option B: Reducing pressure to one fourth at constant temperature

Option C: Reducing temperature to one fourth at constant pressure

Option D: Reducing the temperature to half and doubling the pressure

Correct Answer: Reducing pressure to one fourth at constant temperature ✔

Option A: Pressure and temperature

Option B: Reduced pressure and reduced temperature

Option C: Critical pressure and critical temperature

Option D: None of these

Correct Answer: Reduced pressure and reduced temperature ✔

Option A: Δ H = 0 and ΔS = 0

Option B: Δ H ≠ 0 and ΔS = 0

Option C: Δ H ≠ 0 and ΔS ≠ 0

Option D: Δ H = 0 and ΔS ≠ 0

Correct Answer: Δ H ≠ 0 and ΔS = 0 ✔

Option A: Superheated

Option B: Desuperheated

Option C: Non-condensable

Option D: None of these

Correct Answer: Superheated ✔

Option A: Momentum

Option B: Mass

Option C: Energy

Option D: None of these

Correct Answer: Energy ✔

Option A: Heat capacity of a crystalline solid is zero at absolute zero temperature

Option B: Heat transfer from low temperature to high temperature source is not possible without external

Option C: Gases having same reduced properties behaves similarly

Option D: None of these

Correct Answer: Heat capacity of a crystalline solid is zero at absolute zero temperature ✔

Option A: (∂T/∂V)S = (∂p/∂S)V

Option B: (∂T/∂P)S = (∂V/∂S)P

Option C: (∂P/∂T)V = (∂S/∂V)T

Option D: (∂V/∂T)P = -(∂S/∂P)T

Correct Answer: (∂V/∂T)P = -(∂S/∂P)T ✔

Option A: Enhanced COP

Option B: Decreased COP

Option C: No change in the value of COP

Option D: Increased or decreased COP; depending upon the type of refrigerant

Correct Answer: Enhanced COP ✔

Option A: Third law of thermodynamics

Option B: Second law of thermodynamics

Option C: Nernst heat theorem

Option D: Maxwell’s relations

Correct Answer: Third law of thermodynamics ✔

Option A: (∂P/∂V)S = (∂P/∂V)T

Option B: (∂P/∂V)S = [(∂P/∂V)T]Y

Option C: (∂P/∂V)S = y(∂P/∂V)T

Option D: (∂P/∂V)S = 1/y(∂P/∂V)T

Correct Answer: (∂P/∂V)S = y(∂P/∂V)T ✔

Option A: -94 kcal

Option B: > -94 kcal

Option C: < – 94 kcal

Option D: Zero

Correct Answer: Zero ✔

Option A: Violates second law of thermodynamics

Option B: Involves transfer of heat from low temperature to high temperature

Option C: Both A. and B

Option D: Neither A. nor B

Correct Answer: Involves transfer of heat from low temperature to high temperature ✔

Option A: 1st

Option B: Zeroth

Option C: 3rd

Option D: None of these

Correct Answer: Zeroth ✔

Option A: Isobaric

Option B: Isothermal

Option C: Adiabatic

Option D: None of these

Correct Answer: Isothermal ✔

A. Δ S1 is always Δ SR
C. Δ S1 is always > Δ SR
D. Δ S1 is always = Δ SR

Correct Answer: Δ S1 is always > Δ SR ✔

Option A: Van Laar equation

Option B: Margules equation

Option C: Wilson’s equation

Option D: All A, B. and C

Correct Answer: D. All A, B. and C ✔

Option A: Adiabatic

Option B: Isothermal

Option C: Isometric

Option D: None of these

Correct Answer: Isothermal ✔

Option A: +ve

Option B: -ve

Option C: 0

Option D: Either of the above three; depends on the nature of refrigerant

Correct Answer: 0 ✔

Option A: A = H – TS

Option B: A = E – TS

Option C: A = H + TS

Option D: None of these

Correct Answer: B. A = E – TS ✔

Option A: Fugacity

Option B: Activity co-efficient

Option C: Free energy

Option D: None of these

Correct Answer: None of these ✔

Option A: Isothermal

Option B: Adiabatic

Option C: Isentropic

Option D: None of these

Correct Answer: Isothermal ✔