Mechanical Engineering MCQ Questions

Q.101 Easy Thermodynamics

Which of the following statements about entropy is INCORRECT?

A Entropy is a measure of molecular disorder or randomness

B Entropy always increases for an irreversible adiabatic process

C Entropy of a reversible isothermal process equals Q/T

D Entropy of the universe decreases for spontaneous processes

Correct Answer: D. Entropy of the universe decreases for spontaneous processes

EXPLANATION

The Second Law states entropy of the universe INCREASES (ΔS_universe > 0) for spontaneous/irreversible processes, not decreases.

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Q.102 Hard Thermodynamics

A Rankine cycle operates between 8 MPa (saturation temp ≈ 295°C) and 0.01 MPa (saturation temp ≈ 45°C). The isentropic efficiency of the turbine is 85%. If inlet enthalpy to turbine is 2800 kJ/kg and exit enthalpy for isentropic expansion is 2300 kJ/kg, the actual exit enthalpy is:

A 2300 kJ/kg

B 2458 kJ/kg

C 2575 kJ/kg

D 2650 kJ/kg

Correct Answer: C. 2575 kJ/kg

EXPLANATION

ηt = (h_in - h_out_actual)/(h_in - h_out_isentropic) → 0.85 = (2800 - h_actual)/(2800 - 2300) → h_actual = 2575 kJ/kg

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Q.103 Medium Thermodynamics

The van der Waals equation accounts for real gas behavior with correction terms. Which statement is correct?

A The 'a' term corrects for intermolecular attractive forces, 'b' for molecular volume

B The 'a' term corrects for molecular volume, 'b' for intermolecular forces

C Both terms only affect pressure at very high densities

D Real gases follow ideal gas law at all pressures when molecular size is negligible

Correct Answer: A. The 'a' term corrects for intermolecular attractive forces, 'b' for molecular volume

EXPLANATION

van der Waals equation: (P + a/V²)(V - b) = RT. The 'a' term represents attractive forces; 'b' represents excluded volume.

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Q.104 Medium Thermodynamics

For a reversible process in an isolated system (adiabatic, no work), the entropy change is:

A ΔS = 0

B ΔS > 0

C ΔS < 0

D ΔS can be positive or negative

Correct Answer: A. ΔS = 0

EXPLANATION

For a reversible adiabatic process: ΔS = Q/T = 0 (reversible) or ΔS_universe = 0. Entropy of the system remains constant.

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Q.105 Medium Thermodynamics

In a Otto cycle (spark ignition engine), which process is isochoric (constant volume)?

A Intake and exhaust strokes only

B Compression and expansion (power stroke) are adiabatic

C Heat addition and heat rejection occur at constant volume

D All expansion processes

Correct Answer: C. Heat addition and heat rejection occur at constant volume

EXPLANATION

Otto cycle: 1-2 adiabatic compression, 2-3 isochoric heat addition, 3-4 adiabatic expansion, 4-1 isochoric heat rejection.

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Q.106 Easy Thermodynamics

The relationship between specific heats Cp - Cv for an ideal gas is:

A Cp - Cv = R

B Cp - Cv = 2R

C Cp = Cv (they are equal)

D Cp - Cv = R/M where M is molar mass

Correct Answer: A. Cp - Cv = R

EXPLANATION

Mayer's relation: Cp - Cv = R (per mole). This is fundamental for ideal gases.

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Q.107 Medium Thermodynamics

A throttling process (Joule-Thomson expansion) through a valve is characterized by:

A Constant enthalpy

B Constant temperature

C Constant entropy

D Zero work and zero heat transfer

Correct Answer: A. Constant enthalpy

EXPLANATION

In throttling, the process is isenthalpic (h₁ = h₂). Temperature may change depending on the Joule-Thomson coefficient.

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Q.108 Medium Thermodynamics

According to the First Law of Thermodynamics for a closed system: dU = δQ - δW. If a gas is compressed adiabatically, which is true?

A ΔU = W (work done ON the gas increases internal energy)

B ΔU = -W (internal energy decreases)

C ΔU = 0 always

D Q = W

Correct Answer: A. ΔU = W (work done ON the gas increases internal energy)

EXPLANATION

In adiabatic process, Q=0, so ΔU = -W or ΔU = W (depending on sign convention). Work done ON gas is positive, increasing internal energy.

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Q.109 Hard Thermodynamics

A reciprocating compressor compresses air from 1 bar, 300 K to 10 bar. If the process follows PVⁿ = constant with n=1.25, what is the specific work required per kg of air? (R=287 J/kg·K)

A 180 kJ/kg

B 220 kJ/kg

C 280 kJ/kg

D 320 kJ/kg

Correct Answer: B. 220 kJ/kg

EXPLANATION

W = [nRT₁/(n-1)][(P₂/P₁)^((n-1)/n) - 1] = [1.25×287×300/0.25][10^0.2 - 1] ≈ 220 kJ/kg

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Q.110 Easy Thermodynamics

The enthalpy of vaporization (latent heat) of water at 100°C is 2257 kJ/kg. This represents:

A The heat needed to convert 1 kg of liquid water to steam at constant temperature and pressure

B The heat released when steam condenses

C Both (a) and (b)

D The sensible heat of water

Correct Answer: C. Both (a) and (b)

EXPLANATION

Latent heat is absorbed during vaporization and released during condensation. Both statements are correct.

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