The fugacity coefficient (φ) for an ideal gas is:

The correct answer is A: Always 1. For ideal gases, f = P (fugacity equals pressure), so φ = f/P = 1. Real gases have φ ≠ 1

The Sherwood number (Sh) in mass transfer correlations represents the dimensionless:

The correct answer is B: Ratio of convective to diffusive mass transfer. Sherwood number Sh = k_c·L/D_AB represents the ratio of convective mass transfer to diffusive mass transfer, analogous to Nusselt number in heat transfer.

A reversible process between two states A and B will have entropy change:

The correct answer is C: Equal to that of any path between same states. Entropy is a state function. ΔS is path-independent and same for all processes (reversible or irreversible) between fixed states.

In a finned tube heat exchanger, the overall surface effectiveness is 0.75. This means:

The correct answer is A: 75% of the theoretical maximum heat transfer is achieved. Overall surface effectiveness accounts for both fin and base surface contributions. A value of 0.75 indicates that 75% of the theoretical maximum heat transfer (assuming entire surface at base temperature) is actually achieved due to fin efficiency and geometric factors.

In a tubular reactor with axial dispersion, increasing the Peclet number (Pe) results in:

The correct answer is A: RTD approaching that of an ideal PFR. Peclet number Pe = uL/D. High Pe means low diffusion relative to convection, approaching ideal PFR behavior with narrow RTD.

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Chemical Engineering

Process design, thermodynamics, reactions

117 Q 5 Topics Take Mock Test

Difficulty: All Easy Medium Hard 1–10 of 117

Topics in Chemical Engineering

All Mass Transfer 100 Heat Transfer 100 Fluid Mechanics 100 Chemical Reaction Engineering 100 Thermodynamics 97

Q.1 Hard Thermodynamics

In a desalination plant using reverse osmosis, work must be applied because:

A ΔG > 0 for the desalination process

B The process is adiabatic

C Heat must be removed from the system

D Entropy of the system decreases

Correct Answer: A. ΔG > 0 for the desalination process

EXPLANATION

Desalination (salt separation) is non-spontaneous: ΔG > 0. External work must be supplied to drive the process. This applies to RO and most separation processes.

Test

Q.2 Hard Thermodynamics

For a binary ideal solution at constant T and P, if we mix 1 mole of component A and 1 mole of component B, the entropy of mixing is:

A ΔS_mix = -R(ln 0.5 + ln 0.5) = R ln 4 > 0

B ΔS_mix = 0 (ideal solution)

C ΔS_mix = -R(x_A ln x_A + x_B ln x_B) < 0

D ΔS_mix depends on temperature

Correct Answer: A. ΔS_mix = -R(ln 0.5 + ln 0.5) = R ln 4 > 0

EXPLANATION

ΔS_mix = -nR Σx_i ln x_i. For equal moles: ΔS_mix = -2R(0.5 ln 0.5 + 0.5 ln 0.5) = R ln 4 > 0, always positive.

Test

Q.3 Hard Thermodynamics

An engineer needs to liquefy natural gas (primarily methane). The gas must be cooled below the inversion temperature because:

A Below inversion temperature, Joule-Thomson coefficient is positive

B Below inversion temperature, entropy increases

C Gibbs energy becomes negative only below inversion temperature

D Molecular interactions become repulsive below inversion temperature

Correct Answer: A. Below inversion temperature, Joule-Thomson coefficient is positive

EXPLANATION

Above inversion temperature (for methane ≈ 625 K), μ_JT < 0 (heating on expansion). Below it, μ_JT > 0 (cooling on expansion), enabling liquefaction.

Test

Q.4 Hard Thermodynamics

For a polytropic process PV^n = constant, if n = γ (heat capacity ratio), the process is:

A Isothermal

B Adiabatic

C Isobaric

D Isochoric

Correct Answer: B. Adiabatic

EXPLANATION

For adiabatic process of ideal gas, PV^γ = constant where γ = Cp/Cv. This is the defining equation for adiabatic polytropic process.

Test

Q.5 Hard Thermodynamics

For a reversible process at constant T and P, the minimum work required (excluding PV work) is:

A w_net = ΔG

B w_useful = -ΔG

C w_total = ΔH

D w_useful = ΔH - TΔS

Correct Answer: B. w_useful = -ΔG

EXPLANATION

Useful work (non-PV) available = -ΔG at constant T,P. This represents maximum useful work for spontaneous process or minimum work needed for non-spontaneous process.

Test

Q.6 Hard Thermodynamics

For a spontaneous process in an isolated system, the entropy production σ satisfies:

A σ = 0 always

B σ > 0 for irreversible processes, σ = 0 for reversible

C σ < 0 for all spontaneous processes

D σ is independent of process type

Correct Answer: B. σ > 0 for irreversible processes, σ = 0 for reversible

EXPLANATION

Entropy production σ = ΔS_total ≥ 0 for isolated systems. σ > 0 for irreversible spontaneous processes; σ = 0 for reversible processes (equilibrium).

Test

Q.7 Hard Thermodynamics

The chemical potential μᵢ of a component in a mixture relates to partial molar properties by:

A μᵢ = Ḡᵢ (partial molar Gibbs energy)

B μᵢ = Ūᵢ + Pv̄ᵢ - Ts̄ᵢ

C μᵢ = (∂G/∂nᵢ)_{T,P,n_j}

D All of the above

Correct Answer: D. All of the above

EXPLANATION

All statements define chemical potential from different perspectives. μᵢ is the partial molar Gibbs energy and equals (∂G/∂nᵢ)_{T,P,n_j}.

Test

Q.8 Hard Thermodynamics

In a steam power plant, the Rankine cycle efficiency increases when:

A Boiler pressure is increased

B Condenser temperature is decreased

C Both pressure and temperature variations are optimized

D Pump work is maximized

Correct Answer: C. Both pressure and temperature variations are optimized

EXPLANATION

Rankine cycle efficiency η = 1 - T_c/T_h improves with higher boiler temperature/pressure and lower condenser temperature, following Carnot efficiency limits.

Test

Q.9 Hard Thermodynamics

In a Joule-Thomson expansion of real gas at 298 K, a positive μ_JT (inversion coefficient) means:

A Temperature increases during throttling

B Temperature decreases during throttling

C No temperature change occurs

D Process is reversible

Correct Answer: B. Temperature decreases during throttling

EXPLANATION

μ_JT = (∂T/∂P)_H > 0 means temperature decreases with pressure drop during isenthalpic expansion. For most gases at room temp (except H₂ and He), μ_JT > 0, enabling gas cooling for liquefaction.

Test

Q.10 Hard Thermodynamics

For a real gas obeying virial equation PV/nRT = 1 + B/V̄, the internal energy change with volume at constant T is:

A Zero (same as ideal gas)

B Non-zero due to intermolecular forces represented by B(T)

C (∂U/∂V)_T = T(∂B/∂T)_P/V̄²

D Both B and C

Correct Answer: D. Both B and C

EXPLANATION

For real gases, (∂U/∂V)_T ≠ 0. From thermodynamic relations: (∂U/∂V)_T = T(∂P/∂T)_V - P. Using virial equation gives (∂U/∂V)_T = T(dB/dT)/V̄². Non-ideal behavior affects internal energy.

Test

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