In a distillation column operating at steady state, the mass transfer rate is governed by which of the following equations?

The correct answer is D: All of the above are valid representations. All three equations represent valid forms of mass transfer rate equations. Option A uses liquid phase coefficient, Option B is Fick's law, Option C uses gas phase coefficient, and all are interconnected through film theory.

In a recuperative heat exchanger, the effectiveness (ε) is defined as the ratio of actual heat transfer to:

The correct answer is A: Maximum possible heat transfer (C_min·ΔT_max). Effectiveness ε = Q_actual/Q_max = Q/(C_min(T_h,in - T_c,in)) for counterflow and parallel flow exchangers.

For a long horizontal pipeline with incompressible fluid, the pressure loss due to friction increases when:

The correct answer is C: Both velocity and pipe length increase. From Darcy-Weisbach: h_f = f(L/D)(V²/2g). Friction loss is proportional to length and approximately proportional to V² (in turbulent flow, f decreases slightly with V).

In the Hagen-Poiseuille equation for laminar flow through a circular pipe, the volumetric flow rate is proportional to:

The correct answer is D: Fourth power of diameter. Hagen-Poiseuille equation: Q = (πΔPd⁴)/(128μL), showing Q ∝ d⁴. This is why small diameter pipes are very sensitive to pressure drops.

The dimensionless Stanton number (St = h/(ρ·v·c_p)) in heat transfer represents:

The correct answer is A: Ratio of heat transferred to sensible heat capacity of fluid. St represents the fraction of heat that can be transferred relative to the sensible heat available in flowing fluid per unit area per unit time.

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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 Medium Thermodynamics

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

A Greater than that of an irreversible process between same states

B Less than that of an irreversible process between same states

C Equal to that of any path between same states

D Dependent on the type of process

Correct Answer: C. Equal to that of any path between same states

EXPLANATION

Entropy is a state function. ΔS is path-independent and same for all processes (reversible or irreversible) between fixed states.

Test

Q.6 Medium Thermodynamics

For a constant pressure process, the heat absorbed equals:

A Change in internal energy

B Change in enthalpy

C Change in entropy

D Work done by system

Correct Answer: B. Change in enthalpy

EXPLANATION

At constant pressure: ΔH = Q_p (definition of enthalpy). From ΔU = Q - W and W = PΔV, we get Q = ΔH.

Test

Q.7 Medium Thermodynamics

The compressibility factor Z for a real gas at high pressures is typically:

A Greater than 1 (repulsive forces dominate)

B Less than 1 (attractive forces dominate)

C Equal to 1

D Always greater than 2

Correct Answer: A. Greater than 1 (repulsive forces dominate)

EXPLANATION

At high pressures, molecular volume effect (b term) dominates, making Z > 1. At moderate pressures, Z < 1 due to intermolecular attractions.

Test

Q.8 Easy Thermodynamics

A system absorbs 500 J of heat and does 300 J of work. The change in internal energy is:

A 200 J

B 800 J

C -200 J

D 500 J

Correct Answer: A. 200 J

EXPLANATION

First law: ΔU = Q - W = 500 - 300 = 200 J (using convention W = work by system).

Test

Q.9 Medium Thermodynamics

For a binary ideal solution, Raoult's law states that:

A Partial pressure of each component = mole fraction × vapor pressure of pure component

B Total pressure = sum of mole fractions × atmospheric pressure

C Activity coefficient of each component = 1

D Vapor pressure is independent of composition

Correct Answer: A. Partial pressure of each component = mole fraction × vapor pressure of pure component

EXPLANATION

Raoult's law: P_i = x_i × P_i° for ideal solutions. Both statements A and C are equivalent for ideal solutions.

Test

Q.10 Medium Thermodynamics

The Joule-Thomson coefficient μ_JT is negative for most gases at room temperature. This means:

A Temperature increases during expansion

B Temperature decreases during expansion

C No temperature change occurs

D Entropy decreases during expansion

Correct Answer: A. Temperature increases during expansion

EXPLANATION

μ_JT = (∂T/∂P)_H. Negative coefficient means T increases with pressure drop (cooling requires very low T or high P).

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