Q: Liquid–liquid extraction selectivity – temperature effect and plait point Which of the following statements about solvent selectivity in liquid–liquid extraction is/are correct?

Introduction / Context: In liquid–liquid extraction, selectivity (or separation factor) quantifies how preferentially a solute partitions between two immiscible phases. Correct statements about selectivity determine whether extraction is feasible and how sensitive performance is to temperature and phase behavior near the critical solution composition (plait point). Given Data / Assumptions: Selectivity S is the ratio of distribution coefficients or relative solubilities. Plait point is where the two liquid phases become identical in composition. Typical organic–aqueous systems show temperature-dependent partitioning. Concept / Approach: When S = 1, the solute shows no preference for either phase, making separation impossible. As temperature increases, distribution coefficients of many systems move closer together, reducing selectivity (though exceptions exist). At the plait point, the two phases are indistinguishable; selectivity becomes 1 by definition since there is no phase contrast to exploit for separation. Step-by-Step Solution: Evaluate (a): S = 1 → no driving force → correct.Evaluate (b): Many solute–solvent systems exhibit lower S at higher T → generally correct.Evaluate (c): At plait point, phases coincide → S = 1 → correct.Therefore, all three statements hold under standard interpretations.Select option (d). Verification / Alternative check: Ternary phase diagrams and tie-line data show diminishing partition differences as T approaches critical solution temperature; S tends to unity at the plait point. Why Other Options Are Wrong: Any single-statement choice omits other valid facts, while denying any of (a)–(c) contradicts core phase-equilibrium principles. Common Pitfalls: Treating statement (b) as universal; it is a frequent trend but not an absolute rule—always consult system-specific data when designing extractions. Final Answer: all (a), (b) and (c)

Q: Film theory – dependence of mass transfer coefficient on diffusivity According to the classical film theory for gas–liquid or liquid–liquid mass transfer, how does the mass transfer coefficient K depend on molecular diffusivity D?

Introduction / Context: Rate theories for mass transfer (film, penetration, surface renewal) differ in their predicted dependence of the mass transfer coefficient on diffusivity. Choosing the appropriate scaling helps compare systems, estimate rates, and understand when enhancement mechanisms dominate. Given Data / Assumptions: Film theory assumes a stagnant film of thickness δ next to the interface. Within the film, transport is by steady-state molecular diffusion. Bulk phases are well mixed beyond the film. Concept / Approach: In film theory, the mass transfer coefficient is K ≈ D/δ. If δ is treated as independent of D for a given hydrodynamic condition, the proportionality reduces to K ∝ D. In contrast, the penetration and surface renewal theories predict K ∝ D^0.5. Recognizing these different exponents allows selection of the correct model for unsteady interfacial contact or turbulent renewal conditions. Step-by-Step Solution: Start with steady-state diffusive flux: N = K * driving force.In the film: K = D/δ for a given δ.Therefore, K scales linearly with D under film-theory assumptions.Compare with other theories: penetration gives K ∝ D^0.5.Select K ∝ D as the correct film-theory dependence. Verification / Alternative check: Measured Sherwood–Reynolds–Schmidt correlations often show D^n with n between 0.33 and 0.67, but the idealized film model specifically yields n = 1 when δ is fixed by hydrodynamics. Why Other Options Are Wrong: D^0.5 corresponds to penetration/surface-renewal models, not classical steady film. D^1.5 or D^2 drastically overstate the role of diffusivity and lack theoretical basis here. Common Pitfalls: Assuming one theory fits all regimes; in highly turbulent interfaces with short contact times, D^0.5 behavior is more appropriate than the linear dependence. Final Answer: K ∝ D

Q: Distillation design – what does the Fenske equation determine? In binary or multicomponent distillation at total reflux, the Fenske equation is used to compute which design parameter?

Introduction / Context: Short-cut distillation methods separate the tasks of stage count and reflux determination. The Fenske–Underwood–Gilliland framework starts with Fenske for stage count at total reflux, Underwood for minimum reflux ratio, and Gilliland (or modern alternatives) for finite reflux design. Recognizing which equation does what avoids common exam and design errors. Given Data / Assumptions: Total reflux (no distillate or bottoms withdrawal). Constant relative volatility assumption for each key pair. Sharp split specified by key components. Concept / Approach: The Fenske equation relates the composition targets of light and heavy key components to the minimum theoretical stage number at total reflux. Because all vapor is condensed and returned, separation is maximized per stage without finite reflux penalties. Column height depends on plate efficiency and spacing, not directly from Fenske. Optimum or minimum reflux arises from Underwood and economic trade-offs, not Fenske. Step-by-Step Solution: Identify regime: total reflux.Use Fenske to link key-component splits to stage count.Conclude: result is minimum theoretical stages at total reflux.Height requires efficiency correlations; reflux comes from Underwood/Gilliland.Select option stating “minimum number of theoretical plates”. Verification / Alternative check: Design texts present Fenske for N_min, Underwood for R_min, and Gilliland for N vs R design point; this triad is standard in column synthesis. Why Other Options Are Wrong: Maximum plates: Not a defined concept; total reflux gives minimum required plates. Column height: Requires tray efficiency/packing data, not Fenske alone. Optimum reflux: Determined by economics and correlations, not Fenske. Common Pitfalls: Using Fenske at finite reflux; it strictly applies at total reflux conditions. Final Answer: minimum number of theoretical plates

Q: Distillation limits – behavior at minimum reflux ratio For a specified separation, what happens to the required number of plates (theoretical stages) as the reflux ratio approaches the minimum value?

Introduction / Context: Reflux ratio strongly influences the internal compositions in a distillation column and, consequently, the number of stages needed. Understanding the asymptotic limits (total reflux and minimum reflux) is essential for selecting an economical design point and for interpreting McCabe–Thiele constructions. Given Data / Assumptions: Binary or multicomponent distillation under constant molar overflow. Specified distillate and bottoms purities. Reflux ratio varied while separation target is fixed. Concept / Approach: At total reflux (infinite reflux ratio), the minimum number of theoretical stages is required. At the opposite extreme, the minimum reflux ratio R_min is the lowest reflux that will still achieve the desired split; however, the operating lines become tangent to the equilibrium curve, so the required number of stages grows without bound. Hence, near R_min, the column must be impractically tall, motivating operation at a higher, finite reflux where a reasonable stage count is obtained. Step-by-Step Solution: Use McCabe–Thiele: as reflux decreases toward R_min, operating lines approach the equilibrium curve.The step-off procedure yields many tiny stages → count → ∞.Therefore, at R_min the required number of theoretical stages is infinite.Contrast with total reflux → N_min finite.Select the option “number of plates tends to infinity”. Verification / Alternative check: Gilliland correlations show N increases steeply as R approaches R_min from above, consistent with the asymptotic infinity at R_min. Why Other Options Are Wrong: Zero plates or “minimum plates” at minimum reflux misstates the limit; it is the opposite of total reflux. “Most efficient” is incorrect; efficiency collapses as stages required blow up. Common Pitfalls: Confusing minimum reflux with minimum stages; they occur at opposite ends of the design spectrum. Final Answer: number of plates tends to infinity

Q: Liquid–vapor contacting hardware selection: For a given separation duty, which device typically provides the maximum interfacial contact surface area between phases?

Introduction: Mass transfer effectiveness in distillation and absorption depends on the available interfacial area and contact time between vapor and liquid phases. Different contacting devices—trays and packings—offer different specific surface areas and hydrodynamics, which in turn influence column height and energy use. Given Data / Assumptions: We compare common internals: sieve trays, bubble caps, random packing, wetted-wall columns, and spray towers. “Maximum contact surface area” refers to specific area available for mass transfer per unit column volume. Operating within typical industrial flooding and pressure drop limits. Concept / Approach: Randomly packed columns (e.g., Raschig rings, Pall rings, Berl saddles) provide very high effective surface area due to the large wetted surface of numerous packing elements and film flow over them. Tray columns (sieve or bubble cap) provide staged contacting but generally lower interfacial area per volume. Wetted-wall columns have a single liquid film on a wall, yielding relatively small area. Spray towers also deliver low area because droplet surface is limited and coalescence reduces effective contact. Step-by-Step Solution: Rank devices by typical area: packed > trays > wetted-wall/spray.Identify randomly packed columns as providing the largest specific interfacial area for many duties.Select “Randomly packed column.” Verification / Alternative check: Vendor data sheets for packings list specific surface areas (e.g., 100–300 m^2/m^3 or higher), often exceeding effective areas achievable on trays for the same volume. Why Other Options Are Wrong: Sieve and bubble cap trays: Robust staging but lower area/volume. Wetted wall: Minimal area, used mainly for accurate transfer studies. Spray tower: Low area and efficiency; large towers needed. Common Pitfalls: Assuming “more stages” means more area; stages and interfacial area are different design dimensions. Final Answer: Randomly packed column

Question 1

Liquid–liquid extraction — For systems with very low density difference between phases, which extractor type is most suitable?

Accepted Answer

Introduction: Phase disengagement in liquid–liquid extraction depends strongly on density difference and interfacial tension. When the two liquid phases are nearly the same density, gravitational separation is slow and inefficient. Equipment choice must therefore provide an enhanced driving force for separation. Given Data / Assumptions: Two immiscible (or partially miscible) liquid phases with very small density difference. Need both efficient mass transfer and reliable phase separation. Industrial-scale continuous operation desired. Concept / Approach: Centrifugal extractors (e.g., Podbielniak, disc-stack, or rotor–stator designs) impose high g-forces that dramatically accelerate coalescence and phase separation despite tiny buoyancy forces. They combine mixing and separation within a compact unit, making them ideal for low Δρ systems. Gravity-settling devices (mixer–settlers, packed/spray columns) rely on low g and thus struggle when Δρ is small; columns may also experience flooding or excessive holdup under these conditions. Step-by-Step Solution: Identify the problem: Δρ ≈ small → gravity separation is slow.Select a device that multiplies the effective g → centrifugal extractor.Note that such units integrate mixing and rapid separation.Conclude that centrifugal extractor is most suitable. Verification / Alternative check: Operational experience in pharma and specialty chemicals shows centrifugal extractors outperform gravity-based columns for closely matched densities, reducing stage volume and residence time. Why Other Options Are Wrong: A/D/E: Depend on gravity; with low Δρ, settling becomes impractically slow. C: Pulsed columns improve mass transfer but still rely on gravity for phase disengagement. Common Pitfalls: Focusing solely on mass-transfer coefficients and overlooking separation section bottlenecks caused by poor buoyancy. Final Answer: centrifugal extractor

Question 2

Selecting an absorbent — Why is a low-viscosity absorbent generally preferred?

Accepted Answer

Introduction: Absorber performance hinges on both thermodynamics (solubility/selectivity) and hydraulics (mass transfer, pressure drop, capacity). While solvent choice often starts with solubility, the fluid's viscosity strongly affects transfer rates and column operability. This question highlights the multifaceted benefits of low viscosity. Given Data / Assumptions: Packed or tray absorber operating at near-ambient to moderate temperatures. Comparable solvent volatility and chemical stability among candidates. Mass-transfer coefficients depend on diffusivity and film thickness; hydraulics depend on liquid properties. Concept / Approach: Low-viscosity liquids exhibit thinner boundary layers and higher diffusivities, boosting liquid-side mass-transfer coefficients, which accelerates absorption. They also distribute better over packings, enhancing wetting and heat removal. Hydraulically, lower viscosity reduces friction losses (pressure drop), and delays onset of flooding (higher capacity) due to improved film flow and reduced holdup. Thus, low viscosity supports rapid absorption, improved flooding behavior, and lower pumping energy simultaneously. Step-by-Step Solution: Relate viscosity ↓ → diffusivity effective ↑ → k_L ↑.Relate viscosity ↓ → liquid film thinner, better wetting → enhanced heat transfer.Relate viscosity ↓ → friction losses ↓ and holdup ↓ → lower ΔP and higher flooding limit.Therefore, all listed advantages apply. Verification / Alternative check: Packed column design correlations (e.g., Sherwood–Lobo–Evans) and flooding charts show clear viscosity effects on k_L, pressure drop, and capacity. Why Other Options Are Wrong: A/B/C each capture only part of the benefit; the comprehensive, correct choice is all of them. E contradicts standard design practice. Common Pitfalls: Choosing a very viscous but highly soluble solvent without accounting for severe mass-transfer penalties and hydraulic limitations. Final Answer: all (a), (b) and (c)

Question 3

Solution thermodynamics — Raoult’s law applies directly to which component in a liquid solution?

Accepted Answer

Introduction: Vapor–liquid equilibrium calculations often combine Raoult’s law and Henry’s law. Knowing when each applies avoids serious errors in estimating partial pressures. This question asks which component obeys Raoult’s law in the typical dilute-solution framework. Given Data / Assumptions: Binary or multicomponent liquid solution with a volatile solvent and dilute solute. Near-ideal behavior for the solvent (activity coefficient ≈ 1). Temperature below critical and within VLE region. Concept / Approach: Raoult’s law states p_i = x_i * P_i^sat for components that behave ideally in the liquid phase—most notably the solvent at moderate dilution. The dilute solute, especially when it deviates from ideality, is better described by Henry’s law p_i = x_i * H_i. Non-volatile solutes have negligible vapor pressure and do not follow Raoult’s law in any useful sense. Step-by-Step Solution: Identify the major component (solvent) with significant vapor pressure.Apply Raoult’s law to the solvent: p_solvent = x_solvent * P_solvent^sat.For dilute solute, switch to Henry’s law if needed.Therefore, select 'solvents' as the correct component. Verification / Alternative check: Textbook VLE examples (ethanol–water near ideality) apply Raoult’s law to both at moderate composition; in dilute solutions, the solvent specifically fits Raoult, while solute tends to Henry behavior. Why Other Options Are Wrong: A is overbroad; B is contradictory; C describes a component with negligible vapor pressure; E generally uses Henry’s law constants for gases dissolved in liquids. Common Pitfalls: Applying Raoult’s law to dilute, highly non-ideal solutes; neglecting activity coefficients when deviations from ideality are significant. Final Answer: solvents

Question 4

Co-current absorber design — Which quantity does NOT need to be fixed in advance before starting the preliminary design of a co-current gas absorber?

Accepted Answer

Introduction: Absorber design begins with setting flows, compositions, and performance targets. Thermal effects can influence solvent temperature and equilibrium, but early sizing commonly assumes adiabatic operation or estimates heat effects later. This question checks which parameter need not be fixed before preliminary sizing of a co-current absorber. Given Data / Assumptions: Co-current gas and liquid flow with specified inlet compositions. Preliminary design stage: determine size (height/area) to meet outlet specs. Heat effects are typically secondary and may be approximated as adiabatic initially. Concept / Approach: Essential inputs are gas and liquid flow rates and required removal (outlet specs). These determine operating line and driving force. Heat of absorption and sensible heat changes can be included later to refine temperature profiles. Therefore, a fixed value for heat gain/loss is not mandatory at the outset, although it becomes important in detailed design, especially for exothermic absorptions (e.g., SO2 into alkaline solutions, CO2 into amines). Step-by-Step Solution: Set gas flow and composition → determines pollutant load.Choose initial solvent flow and strength → sets operating line.Specify outlet target → drives number of transfer units/height calculations.Treat heat effects as adiabatic initially; iterate later with energy balances. Verification / Alternative check: Standard design procedures (NTU/HTU or stage-based) start with mass balances and equilibrium; temperature corrections are incorporated after initial sizing, confirming that fixed heat gain/loss is not prerequisite. Why Other Options Are Wrong: B and C are essential throughputs. E (target spec) is fundamental to define the separation duty. D is incorrect because one quantity (heat gain/loss) truly need not be fixed initially. Common Pitfalls: Ignoring heat effects entirely; while not fixed initially, they should be revisited to avoid underestimating solvent rates or packing height for strongly exothermic systems. Final Answer: heat gain or loss

Question 5

Psychrometry instrumentation — Humidity of a gas–vapor mixture can be determined by measuring which property of a hygroscopic fibre sensor?

Accepted Answer

Introduction: Industrial humidity measurement uses several transduction principles: dew-point cooling, capacitance, resistance, frequency, and length change of hygroscopic materials. Fibre- or polymer-based sensors are common due to robustness and low cost. This question targets the property most often exploited with hygroscopic fibres for humidity determination. Given Data / Assumptions: A hygroscopic fibre (e.g., treated polymer or organic fibre) whose properties vary with moisture content. Measurement is performed electronically in a practical instrument. Ambient temperatures typical of HVAC/process areas. Concept / Approach: Moisture absorption changes the fibre's ionic conduction/charge carrier mobility, altering electrical resistance. Modern polymer-film or fibre sensors leverage this by forming a resistive (or capacitive) element whose impedance varies predictably with relative humidity. Electronics then convert resistance to RH via calibration curves. While mechanical length or strength may also change with humidity (e.g., hair hygrometers), resistance is the standard electrical property used in many compact sensors. Step-by-Step Solution: Expose hygroscopic fibre to gas mixture → absorbs moisture.Moisture uptake alters ionic pathways → resistance changes.Measure resistance and map to RH using calibration.Report humidity as RH or absolute humidity depending on instrument. Verification / Alternative check: Datasheets for polymer-resistive humidity sensors show monotonic R–RH characteristics; many HVAC transmitters use resistive or capacitive polymer films rather than thermal conductivity or tensile strength changes. Why Other Options Are Wrong: B: Thermal conductivity hygrometers exist but are not fibre-based sensors. C: Strength variation is used in mechanical hair hygrometers, not typical electronic fibre sensors. D/E: Humidity can indeed be sensed via resistance; optical reflectivity is not a standard sole basis in fibres. Common Pitfalls: Assuming all humidity sensors are dew-point mirrors; many field devices are polymer resistive/capacitive elements sensitive to moisture content. Final Answer: electrical resistance

Question 6

Gas absorption with chemical reaction – identify the correct practical example Which of the following absorption operations is a classic example of gas absorption accompanied by a chemical reaction in the liquid phase?

Accepted Answer

Introduction / Context: In gas–liquid operations, absorption may be purely physical (driven by solubility) or may be enhanced by a chemical reaction in the liquid phase. Correctly classifying examples helps in selecting packings, estimating heights of transfer units, and predicting rates via reaction–diffusion models. Given Data / Assumptions: Sparingly soluble acid gas SO2 can react with alkaline absorbents. Ammonia is highly soluble in water without requiring reaction. Benzene (often called benzol in coke-oven gas practice) is absorbed physically into wash oil. Concept / Approach: Absorption with reaction occurs when the absorbed species participates in a rapid acid–base, complexation, or redox reaction in the liquid, increasing driving force and sometimes changing the controlling resistance. SO2 reacts with alkali (e.g., Na2CO3/NaOH) to form bisulfite/sulfite, which lowers the dissolved SO2 activity and promotes further uptake (enhancement). Ammonia–water absorption is mostly physical (though it forms hydrogen bonds), and benzene in wash oil is purely physical solubility-driven absorption. Step-by-Step Solution: Classify each pair: NH3–H2O → physical absorption.Coke-oven benzene into wash oil → physical absorption by solubility.SO2 into alkaline liquor → chemical absorption via neutralization.Therefore, only the SO2 case exemplifies absorption with chemical reaction.Select the option that names SO2 in alkaline solution. Verification / Alternative check: Design texts list SO2–alkali and CO2–alkali systems under chemically enhanced absorption with typical Hatta number analysis and enhancement factors > 1 compared to purely physical systems. Why Other Options Are Wrong: Ammonia in water: Highly soluble but not a necessary chemical reaction system. Benzene in wash oil: Classic physical absorption into a nonpolar solvent. All of the above: Incorrect because only the alkaline SO2 system has obligatory chemical reaction. Common Pitfalls: Assuming “high solubility” implies reaction; solubility and chemical reaction are distinct mechanisms. Do not conflate absorption enhancement by reaction with mere heat of solution. Final Answer: SO2 in alkaline solution

Question 7

Physical absorption – reversibility and thermal effect Which statement best characterizes physical (non-reactive) gas absorption into a liquid under typical conditions?

Accepted Answer

Introduction / Context: Physical absorption describes the dissolution of a gas into a liquid without chemical reaction. Understanding its thermodynamic character and heat effects is crucial for absorber design, since both equilibrium and heat release influence column sizing and energy balances. Given Data / Assumptions: No chemical reaction (solubility governed by Henry’s law). Isothermal assumptions are approximate; heats of solution exist. Desorption is possible by reducing partial pressure or raising temperature. Concept / Approach: Physical absorption is fundamentally reversible: the dissolved gas can be stripped by lowering its partial pressure or increasing temperature (Le Châtelier-type response). Most gas dissolution processes are mildly exothermic due to solvent–solute interactions, so a finite heat of solution is released. While far smaller than chemisorption heats, this heat can warm the solvent and reduce solubility, affecting performance if not removed by cooling. Step-by-Step Solution: Identify the absence of chemical reaction → reversibility is expected.Recall Henry’s law behavior: y = m x; equilibrium shifts with T and partial pressure.Recognize dissolution generally releases heat (exothermic solution).Combine both truths: the best description is reversible and exothermic.Therefore select “both (b) and (c)”. Verification / Alternative check: Plant data show absorber temperature bulges; cooling reduces solvent temperature and restores capacity, consistent with exothermic absorption and reversibility via partial-pressure changes. Why Other Options Are Wrong: Irreversible: Contradicts common stripping and solvent regeneration practice. Only reversible or only exothermic: Incomplete description. Common Pitfalls: Neglecting heat effects (assuming isothermal operation) can underpredict column height since warm solvent holds less solute at the same partial pressure. Final Answer: both (b) and (c)

Question 8

Packed-tower design – relating HETP to HTU and absorption factor In a packed absorption column where both the equilibrium and operating lines are straight, how does the ratio HETP/HTU_OG vary with the absorption factor A (A = L/(mG))?

Accepted Answer

Introduction / Context: For packed towers, designers often connect the concept of a theoretical plate (HETP, height equivalent to a theoretical plate) to transfer-unit methods (HTU–NTU). When the equilibrium and operating lines are linear, simple relationships exist between HETP, HTU, and the absorption factor A that guide performance estimates and packing selection. Given Data / Assumptions: Linear equilibrium: y* = m x. Straight operating line with constant L and G. Absorption factor A = L/(mG). Concept / Approach: For straight lines, the number of transfer units NTUOG reduces to a logarithmic function that mirrors the theoretical-stage expression. Under these conditions, an analytical link shows that HETP/HTUOG is a function of A, becoming unity at A = 1 and increasing as A departs upward from 1 (absorption becomes easier with greater solvent rate relative to equilibrium slope). Thus statements about monotonic increase with A and equality to 1 at A = 1 are both true. Step-by-Step Solution: Define A = L/(mG).For linear lines, derive NTUOG with logarithmic integration.Relate theoretical stages to NTU to obtain HETP/HTUOG as a function of A.Evaluate at A = 1 → ratio = 1.Observe that increasing A increases HETP/HTUOG per the derived function. Verification / Alternative check: Standard packed-column correlations and derivations in mass transfer texts show the equality at A = 1 and increasing ratio with A for systems with linear lines. Why Other Options Are Wrong: “Neither” contradicts known analytical results for linear systems. Common Pitfalls: Applying the relationship outside its assumptions (strong curvature in equilibrium, variable L or G due to heat effects) can lead to errors; always check line linearity and constancy of flows. Final Answer: both (a) and (b)

Question 9

Cooling tower make-up water – sources of loss to be replaced In industrial cooling towers, make-up water must be added to compensate for which combined losses from the circulating water system?

Accepted Answer

Introduction / Context: Cooling towers reject heat by evaporating a small portion of circulating water into the atmosphere. Water balance and chemistry control require adding make-up water to offset specific losses and maintain cycles of concentration within limits that prevent scaling, corrosion, and biofouling. Given Data / Assumptions: Open recirculating cooling system with tower. Water quality management via blowdown is practiced. Minor leaks and drift are present within typical design allowances. Concept / Approach: Three principal loss mechanisms exist: (1) evaporation (latent heat removal), (2) drift (entrained droplets carried out with exhaust air), and (3) blowdown (intentional purge to control dissolved solids), plus incidental leakage. Make-up water must replace the sum of these to sustain circulation rate and chemistry. Evaporation is the largest component; drift is minimized by eliminators; blowdown is set by cycles of concentration. Step-by-Step Solution: Write water balance: Make-up = Evaporation + Drift + Blowdown + Leakage.Recognize evaporation is proportional to heat load and approach.Account for drift via eliminator performance (typically <0.02% of flow).Set blowdown from cycles of concentration and evaporation rate.Conclude all listed losses require make-up replacement. Verification / Alternative check: Cooling tower manuals and CTI guidelines present the same mass balance and define drift and blowdown targets to size make-up supply lines. Why Other Options Are Wrong: Choosing only one or two loss types understates true make-up requirements and leads to rising TDS, scaling, or pump cavitation due to low basin levels. Common Pitfalls: Ignoring blowdown in calculations; although smaller than evaporation, it is vital for controlling concentration and preventing operational problems. Final Answer: all (a), (b) and (c)

Question 10

Liquid–liquid extraction selectivity – temperature effect and plait point Which of the following statements about solvent selectivity in liquid–liquid extraction is/are correct?

Accepted Answer

Introduction / Context: In liquid–liquid extraction, selectivity (or separation factor) quantifies how preferentially a solute partitions between two immiscible phases. Correct statements about selectivity determine whether extraction is feasible and how sensitive performance is to temperature and phase behavior near the critical solution composition (plait point). Given Data / Assumptions: Selectivity S is the ratio of distribution coefficients or relative solubilities. Plait point is where the two liquid phases become identical in composition. Typical organic–aqueous systems show temperature-dependent partitioning. Concept / Approach: When S = 1, the solute shows no preference for either phase, making separation impossible. As temperature increases, distribution coefficients of many systems move closer together, reducing selectivity (though exceptions exist). At the plait point, the two phases are indistinguishable; selectivity becomes 1 by definition since there is no phase contrast to exploit for separation. Step-by-Step Solution: Evaluate (a): S = 1 → no driving force → correct.Evaluate (b): Many solute–solvent systems exhibit lower S at higher T → generally correct.Evaluate (c): At plait point, phases coincide → S = 1 → correct.Therefore, all three statements hold under standard interpretations.Select option (d). Verification / Alternative check: Ternary phase diagrams and tie-line data show diminishing partition differences as T approaches critical solution temperature; S tends to unity at the plait point. Why Other Options Are Wrong: Any single-statement choice omits other valid facts, while denying any of (a)–(c) contradicts core phase-equilibrium principles. Common Pitfalls: Treating statement (b) as universal; it is a frequent trend but not an absolute rule—always consult system-specific data when designing extractions. Final Answer: all (a), (b) and (c)

Question 11

Film theory – dependence of mass transfer coefficient on diffusivity According to the classical film theory for gas–liquid or liquid–liquid mass transfer, how does the mass transfer coefficient K depend on molecular diffusivity D?

Accepted Answer

Introduction / Context: Rate theories for mass transfer (film, penetration, surface renewal) differ in their predicted dependence of the mass transfer coefficient on diffusivity. Choosing the appropriate scaling helps compare systems, estimate rates, and understand when enhancement mechanisms dominate. Given Data / Assumptions: Film theory assumes a stagnant film of thickness δ next to the interface. Within the film, transport is by steady-state molecular diffusion. Bulk phases are well mixed beyond the film. Concept / Approach: In film theory, the mass transfer coefficient is K ≈ D/δ. If δ is treated as independent of D for a given hydrodynamic condition, the proportionality reduces to K ∝ D. In contrast, the penetration and surface renewal theories predict K ∝ D^0.5. Recognizing these different exponents allows selection of the correct model for unsteady interfacial contact or turbulent renewal conditions. Step-by-Step Solution: Start with steady-state diffusive flux: N = K * driving force.In the film: K = D/δ for a given δ.Therefore, K scales linearly with D under film-theory assumptions.Compare with other theories: penetration gives K ∝ D^0.5.Select K ∝ D as the correct film-theory dependence. Verification / Alternative check: Measured Sherwood–Reynolds–Schmidt correlations often show D^n with n between 0.33 and 0.67, but the idealized film model specifically yields n = 1 when δ is fixed by hydrodynamics. Why Other Options Are Wrong: D^0.5 corresponds to penetration/surface-renewal models, not classical steady film. D^1.5 or D^2 drastically overstate the role of diffusivity and lack theoretical basis here. Common Pitfalls: Assuming one theory fits all regimes; in highly turbulent interfaces with short contact times, D^0.5 behavior is more appropriate than the linear dependence. Final Answer: K ∝ D

Question 12

Distillation design – what does the Fenske equation determine? In binary or multicomponent distillation at total reflux, the Fenske equation is used to compute which design parameter?

Accepted Answer

Introduction / Context: Short-cut distillation methods separate the tasks of stage count and reflux determination. The Fenske–Underwood–Gilliland framework starts with Fenske for stage count at total reflux, Underwood for minimum reflux ratio, and Gilliland (or modern alternatives) for finite reflux design. Recognizing which equation does what avoids common exam and design errors. Given Data / Assumptions: Total reflux (no distillate or bottoms withdrawal). Constant relative volatility assumption for each key pair. Sharp split specified by key components. Concept / Approach: The Fenske equation relates the composition targets of light and heavy key components to the minimum theoretical stage number at total reflux. Because all vapor is condensed and returned, separation is maximized per stage without finite reflux penalties. Column height depends on plate efficiency and spacing, not directly from Fenske. Optimum or minimum reflux arises from Underwood and economic trade-offs, not Fenske. Step-by-Step Solution: Identify regime: total reflux.Use Fenske to link key-component splits to stage count.Conclude: result is minimum theoretical stages at total reflux.Height requires efficiency correlations; reflux comes from Underwood/Gilliland.Select option stating “minimum number of theoretical plates”. Verification / Alternative check: Design texts present Fenske for N_min, Underwood for R_min, and Gilliland for N vs R design point; this triad is standard in column synthesis. Why Other Options Are Wrong: Maximum plates: Not a defined concept; total reflux gives minimum required plates. Column height: Requires tray efficiency/packing data, not Fenske alone. Optimum reflux: Determined by economics and correlations, not Fenske. Common Pitfalls: Using Fenske at finite reflux; it strictly applies at total reflux conditions. Final Answer: minimum number of theoretical plates

Question 13

Stefan’s law (Stefan diffusion) in gas-phase mass transfer Stefan’s law primarily describes mass transfer under which mechanism/condition?

Accepted Answer

Introduction / Context: When a volatile species evaporates into a stagnant or slowly moving gas, the observed flux includes the effect of diffusion and, in some cases, a compensating bulk flow. Stefan’s law gives a convenient closed-form expression for steady diffusion of a species through a stagnant gas film, widely used for evaporation and drying calculations. Given Data / Assumptions: Binary gas mixture with species A diffusing through stagnant B. Steady state, isothermal, ideal gas behavior near the interface. No homogeneous chemical reaction in the gas phase. Concept / Approach: Stefan’s diffusion law derives from a simplified Stefan–Maxwell framework for the special case of A diffusing through non-diffusing B. It yields a logarithmic form for flux that accounts for the build-up of A in the film. Despite the presence of a compensating molar flux of A (bulk flow term appears in the derivation), the law is fundamentally a diffusion model in a stagnant film and is classed under diffusion-controlled mass transfer correlations. Step-by-Step Solution: Set up steady diffusion of A through stagnant B across a film.Apply Stefan–Maxwell with N_B = 0 to derive Stefan’s logarithmic flux.Interpret the result as diffusion with an effective driving force.Recognize the bulk term is a consequence of species continuity but the rate law is diffusion-based.Select “diffusion”. Verification / Alternative check: Drying and evaporation models use Stefan’s law for gas-side control, validating its classification as a diffusion relation in a stagnant film. Why Other Options Are Wrong: Bulk flow alone ignores the essential concentration-gradient-driven diffusion. “Both equally” is imprecise; the formulation is a diffusion law including the stagnant-gas effect. “Neither” contradicts the derivation and application. Common Pitfalls: Confusing Stefan’s law with equimolar counterdiffusion; its hallmark is diffusion of one species through a non-diffusing second species, not equal counterflows. Final Answer: diffusion

Question 14

Distillation limits – behavior at minimum reflux ratio For a specified separation, what happens to the required number of plates (theoretical stages) as the reflux ratio approaches the minimum value?

Accepted Answer

Introduction / Context: Reflux ratio strongly influences the internal compositions in a distillation column and, consequently, the number of stages needed. Understanding the asymptotic limits (total reflux and minimum reflux) is essential for selecting an economical design point and for interpreting McCabe–Thiele constructions. Given Data / Assumptions: Binary or multicomponent distillation under constant molar overflow. Specified distillate and bottoms purities. Reflux ratio varied while separation target is fixed. Concept / Approach: At total reflux (infinite reflux ratio), the minimum number of theoretical stages is required. At the opposite extreme, the minimum reflux ratio R_min is the lowest reflux that will still achieve the desired split; however, the operating lines become tangent to the equilibrium curve, so the required number of stages grows without bound. Hence, near R_min, the column must be impractically tall, motivating operation at a higher, finite reflux where a reasonable stage count is obtained. Step-by-Step Solution: Use McCabe–Thiele: as reflux decreases toward R_min, operating lines approach the equilibrium curve.The step-off procedure yields many tiny stages → count → ∞.Therefore, at R_min the required number of theoretical stages is infinite.Contrast with total reflux → N_min finite.Select the option “number of plates tends to infinity”. Verification / Alternative check: Gilliland correlations show N increases steeply as R approaches R_min from above, consistent with the asymptotic infinity at R_min. Why Other Options Are Wrong: Zero plates or “minimum plates” at minimum reflux misstates the limit; it is the opposite of total reflux. “Most efficient” is incorrect; efficiency collapses as stages required blow up. Common Pitfalls: Confusing minimum reflux with minimum stages; they occur at opposite ends of the design spectrum. Final Answer: number of plates tends to infinity

Question 15

Psychrometrics – precise definition of relative humidity Relative humidity (RH) is defined as which ratio under standard psychrometric convention?

Accepted Answer

Introduction / Context: Relative humidity is a fundamental measure in drying, HVAC, and environmental control. Precision matters: RH must reference saturation properties at the actual gas (air) temperature, not an arbitrary or “room” temperature, to be thermodynamically meaningful and comparable across conditions. Given Data / Assumptions: Ideal-gas behavior for water vapor in air is assumed for typical calculations. Saturation vapor pressure depends strongly on temperature. Humidity ratio and partial pressure definitions are standard. Concept / Approach: Relative humidity RH is defined as RH = p_v / p_vs(T_air), where p_v is the water vapor partial pressure and p_vs(T_air) is the saturation vapor pressure evaluated at the same gas temperature. An equivalent definition uses humidity ratios: RH = (w / w_s) at the gas temperature. Any definition referencing a different temperature than the gas temperature is incorrect because saturation properties change markedly with temperature. Step-by-Step Solution: State the pressure-based definition: RH = p_v / p_vs at T_air.Note the humidity-ratio equivalence at the same T_air.Emphasize that using a fixed “room temperature” or unrelated temperature gives wrong RH.Therefore, choose the option specifying saturation at the gas temperature.Reject options that fix a temperature irrespective of actual gas conditions. Verification / Alternative check: Psychrometric charts are constructed at a specified total pressure and evaluate saturation curves at the current air temperature; RH is read as a ratio at that temperature, confirming the definition. Why Other Options Are Wrong: Room-temperature reference: Incorrect when air is not at that exact temperature. Fixed 25°C ratio: Arbitrary and wrong for general conditions. None of these: Incorrect because a correct definition exists in option (b). Common Pitfalls: Confusing RH with absolute humidity or humidity ratio; RH is temperature dependent, so the same moisture content can yield very different RH at different temperatures. Final Answer: partial pressure of the vapor to the saturation vapor pressure at the gas (air) temperature

Question 16

Separation methods for liquid solutions: Which of the following unit operations cannot be used to separate the components of an existing liquid–liquid or liquid solution mixture?\r (Select the operation that is not a solution-separation method.)

Accepted Answer

Introduction: This question tests your understanding of which mass-transfer or phase-change operations are suitable for separating components already present in a liquid solution. While several unit operations are routinely used to split solutes or solvent from a liquid mixture, others are designed for gas–liquid systems and therefore are not appropriate for separating a liquid solution into its constituents. Given Data / Assumptions: The feed is a liquid solution or a liquid–liquid mixture whose components need to be separated. We compare common unit operations used in chemical engineering. We select the operation that does not achieve separation for a liquid solution. Concept / Approach: Separation of liquid solutions typically uses differences in volatility (distillation or evaporation), solubility at different temperatures (fractional crystallisation), or distribution of a solute between two immiscible liquids (liquid–liquid extraction). Gas absorption, by contrast, removes a gaseous solute from a gas stream into a liquid; it is not a method for splitting the components of an existing liquid solution. Step-by-Step Solution: List solution-separation methods: distillation/evaporation, crystallisation, liquid extraction.Identify absorption as a gas–liquid contact process, not a liquid-solution separation.Select the non-applicable method: absorption. Verification / Alternative check: Process design textbooks classify absorption under gas cleaning operations, whereas crystallisation, extraction (L–L), and evaporation belong to liquid-solution separations. Why Other Options Are Wrong: Fractional crystallisation: Separates based on solubility differences. Liquid extraction: Transfers a solute between immiscible liquids. Evaporation: Removes solvent to concentrate or precipitate solute. Fractional distillation: Splits volatile components by relative volatility. Common Pitfalls: Confusing absorption (gas into liquid) with extraction (solute between two liquids). They are distinct operations with different applications. Final Answer: absorption

Question 17

Liquid–vapor contacting hardware selection: For a given separation duty, which device typically provides the maximum interfacial contact surface area between phases?

Accepted Answer

Introduction: Mass transfer effectiveness in distillation and absorption depends on the available interfacial area and contact time between vapor and liquid phases. Different contacting devices—trays and packings—offer different specific surface areas and hydrodynamics, which in turn influence column height and energy use. Given Data / Assumptions: We compare common internals: sieve trays, bubble caps, random packing, wetted-wall columns, and spray towers. “Maximum contact surface area” refers to specific area available for mass transfer per unit column volume. Operating within typical industrial flooding and pressure drop limits. Concept / Approach: Randomly packed columns (e.g., Raschig rings, Pall rings, Berl saddles) provide very high effective surface area due to the large wetted surface of numerous packing elements and film flow over them. Tray columns (sieve or bubble cap) provide staged contacting but generally lower interfacial area per volume. Wetted-wall columns have a single liquid film on a wall, yielding relatively small area. Spray towers also deliver low area because droplet surface is limited and coalescence reduces effective contact. Step-by-Step Solution: Rank devices by typical area: packed > trays > wetted-wall/spray.Identify randomly packed columns as providing the largest specific interfacial area for many duties.Select “Randomly packed column.” Verification / Alternative check: Vendor data sheets for packings list specific surface areas (e.g., 100–300 m^2/m^3 or higher), often exceeding effective areas achievable on trays for the same volume. Why Other Options Are Wrong: Sieve and bubble cap trays: Robust staging but lower area/volume. Wetted wall: Minimal area, used mainly for accurate transfer studies. Spray tower: Low area and efficiency; large towers needed. Common Pitfalls: Assuming “more stages” means more area; stages and interfacial area are different design dimensions. Final Answer: Randomly packed column

Question 18

Concentration units (molality): A solution contains 0.30 kmol of a solute dissolved in 600 kg of solvent.\r What is the molality of the solution (in mol per kg of solvent)?

Accepted Answer

Introduction: Molality (m) is a temperature-independent concentration unit widely used in thermodynamics and colligative property calculations. It is defined as moles of solute per kilogram of solvent, not per kilogram of solution. Correctly identifying solute amount and solvent mass is crucial. Given Data / Assumptions: Moles of solute n_s = 0.30 kmol. Mass of solvent m_solvent = 600 kg. Molality m = moles of solute per kilogram of solvent. Concept / Approach: Use the direct definition of molality. Units can be kmol/kg or mol/kg; numerically they are consistent as long as moles and kilograms are used. Here we can compute using kmol/kg and interpret the result in mol/kg by noting that 1 kmol/kg equals 1000 mol/kg numerically, but the ratio value is unchanged when reported simply as “molality.” Step-by-Step Solution: m = n_s / m_solventSubstitute: m = 0.30 kmol / 600 kgCompute: m = 0.0005 kmol/kg = 0.5 mol/kgTherefore, molality = 0.50 Verification / Alternative check: A quick proportional check: halving the denominator from 600 to 300 would double molality from 0.5 to 1.0, which is consistent with the formula. Why Other Options Are Wrong: 0.60 or 1 or 2: Do not match the computed ratio. 0.05: Off by a factor of 10; a common slip when converting kmol to mol. Common Pitfalls: Mistakenly using mass of solution instead of mass of solvent, or converting kmol to mol and then dividing by kilograms again incorrectly. Final Answer: 0.50

Question 19

Steam distillation economics: The amount of steam required per unit quantity of distillate in steam distillation can be reduced by adjusting operating conditions.\r Which change decreases steam consumption?

Accepted Answer

Introduction: Steam distillation is used to volatilize temperature-sensitive or high-boiling organic compounds by contacting them with steam. The steam rate required per unit of organic distillate depends on vapor pressures at the operating temperature. Reducing steam demand improves energy efficiency and throughput. Given Data / Assumptions: Immiscible steam distillation behavior: total pressure ≈ p_water(T) + p_organic(T). Ratio of moles of steam to organic in the vapor ≈ p_water(T) / p_organic(T). We compare effects of raising temperature and lowering total pressure. Concept / Approach: As temperature increases, the vapor pressure of the organic generally rises faster than that of water, decreasing the ratio p_water/p_organic and thus reducing the steam required per unit organic distillate. By contrast, simply lowering total pressure forces operation at a lower temperature where both vapor pressures are smaller; this often increases the steam-to-organic ratio and does not reduce steam use for a given distillate rate. Step-by-Step Solution: Use relation: steam required per organic ≈ p_water(T) / p_organic(T).Effect of raising T: p_organic increases significantly ⇒ ratio decreases ⇒ less steam per unit distillate.Effect of lowering total pressure: operating T drops ⇒ p_organic may drop markedly ⇒ ratio increases ⇒ more steam.Conclusion: raising temperature reduces steam requirement. Verification / Alternative check: Design charts for steam distillation show reduced steam-to-feed ratios at higher boiling temperatures for many organics, consistent with the faster rise of organic vapor pressure. Why Other Options Are Wrong: Lowering total pressure: Typically raises the steam/organic ratio. Both/neither: Contradict the trend above. Increasing reflux: Not a primary lever in direct steam distillation without a rectifying section. Common Pitfalls: Confusing total pressure effects with simple distillation behavior; in steam distillation of immiscible systems, the key ratio is of vapor pressures at the operating temperature. Final Answer: raising the temperature.

Question 20

Surface phenomena: Chemisorption (chemical adsorption) is best described by which characteristic?

Accepted Answer

Introduction: Adsorption involves accumulation of molecules at surfaces and is broadly categorized into physisorption (Van der Waals) and chemisorption (chemical bonding). Distinguishing their key traits guides catalyst selection, surface treatment, and sensor design. Given Data / Assumptions: Chemisorption forms chemical bonds between adsorbate and surface sites. Physisorption arises from weak Van der Waals forces. Temperature effects and reversibility differ markedly between the two. Concept / Approach: Chemisorption typically involves higher heats of adsorption, site specificity, possible activation energy, and limited monolayer coverage. Because bonds formed are chemical in nature, desorption often requires significant energy, making the process effectively irreversible under the original conditions. Physisorption, in contrast, is generally reversible and exothermic with low heats of adsorption and multilayer potential. Step-by-Step Solution: Differentiate by bonding: chemical vs. Van der Waals.Relate bonding strength to reversibility: strong bonding implies practical irreversibility.Select the option aligning with chemisorption: an irreversible phenomenon. Verification / Alternative check: Catalysis literature shows adsorbed intermediates bound strongly to active sites; desorption often requires higher temperature or reaction, consistent with practical irreversibility. Why Other Options Are Wrong: Same as Van der Waals: Incorrect; that describes physisorption. “Adsorption of heat (endothermic)”: Adsorption is typically exothermic; wording here is misleading. Reversible: Describes physisorption more closely. No activation energy: Chemisorption often shows activation energy. Common Pitfalls: Assuming all adsorption is easily reversible; in chemisorption, strong bonding dominates behavior and kinetics. Final Answer: an irreversible phenomenon.

Mass Transfer Questions