Q: Gas absorption design: For an absorber operating at steady state, the operating line on an equilibrium (y–x) diagram, relative to the equilibrium curve, typically lies where?

Introduction: Graphical design of absorbers uses an equilibrium curve (y* vs. x) and an operating line (material balance). The relative position of these curves determines the driving force for mass transfer and the number of theoretical stages required. Given Data / Assumptions: We consider absorption of a solute from gas to liquid (lean solvent). Coordinates: y for gas-phase solute fraction; x for liquid-phase solute fraction. Countercurrent contact with finite solvent flow rate. Concept / Approach: For absorption, the operating line must lie above the equilibrium curve to ensure a positive driving force (y − y*) across the column. If the operating line dipped below the equilibrium curve, portions of the column would have negative or zero driving force, making absorption infeasible. For stripping (desorption), the operating line is typically below the equilibrium curve. Step-by-Step Solution: Plot equilibrium relation y* = f(x).Write the operating line from overall/segment balances: y = m x + b.Ensure y ≥ y* throughout to maintain driving force for absorption.Therefore, the operating line lies above the equilibrium curve. Verification / Alternative check: Stage-by-stage (McCabe–Thiele) stepping fails if the operating line crosses below the equilibrium curve; practical design always keeps it above with margin for approach to equilibrium. Why Other Options Are Wrong: Below or either side: Would imply loss of driving force. Velocity statement: Unrelated to y–x diagram placement. 45°-line comment: Not a general rule for absorption design. Common Pitfalls: Confusing absorption with stripping; remember the relative position reverses for desorption problems. Final Answer: line always lies above the equilibrium curve.

Q: Distillation products: What can the distillate (overhead product) from a distillation column contain in terms of number of chemical components?

Introduction: Distillation separates mixtures based on relative volatility, but each product stream is still a mixture unless the feed is a binary azeotrope-free system with perfect separation. Understanding that distillate composition can include several components helps avoid oversimplified assumptions during design and operation. Given Data / Assumptions: General multicomponent distillation. Overhead is enriched in light (more volatile) species. Thermodynamic phase (vapor or liquid) depends on condenser duty. Concept / Approach: In multicomponent systems, the top product is typically a mixture of the lighter components in proportions dictated by reflux ratio, tray efficiency, and relative volatilities. Total or partial condensers determine the phase (distillate may leave as liquid or vapor), but not the number of components present. Step-by-Step Solution: Recognize that component count is not limited to one or two unless the feed is binary or pure.Even in binary systems, each product generally contains both components unless extremely high separation is achieved.Therefore, the overhead can contain any number of components (from the light end of the mixture). Verification / Alternative check: Real plant assays for crude columns show top products containing many hydrocarbons; specialty columns for solvent recovery also produce multicomponent overheads. Why Other Options Are Wrong: Only one or two components: Too restrictive for multicomponent feeds. Only saturated liquid/vapor: Phase depends on condenser; composition can still be multicomponent. Common Pitfalls: Assuming “distillate” means pure component; purity depends on design and economics, not on a fundamental limit. Final Answer: any number of components

Q: Drying under vacuum: Why is vacuum commonly used in drying operations for certain materials?

Introduction: Vacuum drying is a key technique for heat-sensitive or oxygen-sensitive materials in pharmaceuticals, foods, and fine chemicals. Lowering the pressure reduces the boiling point of water or solvents, enabling effective drying at much lower temperatures than atmospheric drying. Given Data / Assumptions: Evaporation rate depends on temperature and pressure (vapor–liquid equilibrium). Goal: protect product quality by limiting thermal exposure and oxidation. Bound vs. unbound moisture considerations remain, but the main lever is boiling-point suppression. Concept / Approach: At reduced pressure, the saturation temperature corresponding to a given vapor pressure decreases. This allows solvent removal at temperatures well below atmospheric boiling points. While vacuum may help with some bound moisture removal indirectly, it mainly enables drying without excessive heat that could degrade sensitive products. Step-by-Step Solution: Apply phase equilibrium: lower pressure ⇒ lower boiling point.Translate to process: evaporation occurs at reduced temperature under vacuum.Conclude: the principal purpose is to reduce drying temperature to protect product quality. Verification / Alternative check: Phase diagrams for water show dramatically lower boiling temperatures at sub-atmospheric pressures (e.g., near 40–60°C at deep vacuum), matching typical vacuum dryer settings. Why Other Options Are Wrong: Very high unbound moisture: Vacuum is not required for such feeds; conventional dryers suffice. Increase temperature: Opposite of vacuum’s purpose. High bound moisture: Binding relates to material structure; vacuum alone does not specifically target bound moisture. Eliminating heat transfer: Evaporation still requires energy input. Common Pitfalls: Assuming vacuum alone guarantees fast drying; heat and mass transfer resistances still govern rates. Final Answer: reduce drying temperature.

Q: Psychrometrics terminology: The amount of heat (BTU) required to raise the temperature of 1 lb of dry gas and its accompanying vapor by 1°F is called the __________.

Introduction: In gas–vapor mixtures such as air–water systems, psychrometric properties are used to size heaters, humidifiers, and dryers. One key property is the heat required to raise the temperature of the mixture per unit mass of dry gas, accounting for both the gas and the accompanying vapor. Given Data / Assumptions: Basis: 1 lb of dry gas with its associated vapor at a specified humidity ratio. Temperature change: 1°F. Property sought: mixture heat capacity per unit mass of dry gas. Concept / Approach: “Humid heat” (c_h) is defined as the heat required to raise 1 lb of dry gas plus its accompanying vapor by 1°F. For air–water vapor, an approximate correlation is c_h ≈ c_p,air + W * c_p,vapor, where W is the humidity ratio (lb vapor/lb dry air). This distinguishes it from “specific heat,” which usually refers to a single pure substance per unit mass, and from “latent heat,” which refers to phase change energy. Step-by-Step Solution: Identify the basis: per lb of dry gas including its associated vapor.Recognize the definition that matches: humid heat.Select option “humid heat.” Verification / Alternative check: Psychrometric charts and textbooks list humid heat to compute heater duties for moist air streams: Q ≈ m_dry air * c_h * ΔT when no moisture is added or removed. Why Other Options Are Wrong: Latent heat: Energy for phase change, not sensible heating of a gas–vapor mixture. Specific heat: Generic term; here a specialized mixture property is required. Sensible heat: Refers to the heat effect, not the property name. Humid volume: A volumetric property, not a heat capacity. Common Pitfalls: Using dry-air specific heat alone and underestimating heater duty at high humidity; always account for vapor contribution with humid heat. Final Answer: humid heat

Q: Phase diagrams and partially miscible liquids (ternary systems): A plait point is the point on a binodal (solubility) curve where the tie line shrinks to a point. For a ternary liquid system that contains two pairs of partially miscible liquids, how…

Introduction / Context: Ternary liquid–liquid phase diagrams often display regions of partial miscibility. Each partially miscible binary pair produces a binodal curve terminating at a critical mixing point known as the plait point. Understanding how many plait points exist helps interpret separation behavior and solvent selection strategies. Given Data / Assumptions: Ternary liquid system (three components). Exactly two binary pairs are partially miscible; the third pair is completely miscible. Standard triangular composition diagram at constant temperature and pressure. Concept / Approach: In ternary diagrams, each binary pair that is partially miscible contributes a binodal loop along the corresponding binary edge that extends into the ternary field. Each such binodal has a critical solution point at which the tie lines coalesce to a single composition—this is the plait point. Therefore, the number of plait points equals the number of partially miscible binary pairs in the system, provided their immiscibility persists at the diagram conditions. Step-by-Step Solution: Identify partially miscible pairs: count = 2.Recognize that each partially miscible pair contributes one plait point where tie lines vanish.Therefore, total plait points in the ternary diagram = 2. Verification / Alternative check: Reference ternary maps show one plait point for every partially miscible binary edge; systems with three such pairs (e.g., water–phenol–hydrocarbon at certain T) show three plait points, confirming the one-to-one correspondence. Why Other Options Are Wrong: 0 or 1 contradict the presence of two partially miscible pairs; 3 would require all three binary pairs to be partially miscible, which is not the case here. Common Pitfalls: Confusing a plait point with a critical solution temperature; here we consider a fixed T–P diagram. Assuming a single plait point governs the whole ternary field irrespective of how many binary pairs are immiscible. Final Answer: 2

Q: Dimensionless groups in coupled transport: In simultaneous heat and mass transfer problems, which dimensionless number most directly links thermal and mass diffusivities and is widely used to correlate coupled phenomena?

Introduction / Context: When heat and mass transfer occur together—such as in drying, humidification, evaporative cooling, or combustion—the relative rates of thermal and species diffusion strongly influence boundary-layer behavior and overall transfer rates. A key dimensionless group captures this relationship. Given Data / Assumptions: Single-phase flows with coupled heat and mass transfer. No chemical reaction, or reaction effects considered separately. Continuum mechanics and boundary-layer concepts apply. Concept / Approach: The Lewis number is defined as Le = α / D = (k / ρcp) / D, where α is thermal diffusivity and D is mass diffusivity. Le directly compares the relative thickness and development of thermal and concentration boundary layers. Schmidt number Sc = ν / D and Prandtl number Pr = ν / α are also important, but Le explicitly couples heat and mass transfer by the ratio of diffusivities. Sherwood number Sh = k_c L / D is a mass-transfer analog of Nusselt number and is used for correlations, not for comparing α and D directly. Step-by-Step Solution: Write Le = α / D.Interpretation: Le > 1 → heat diffuses faster than mass; Le < 1 → species diffuse faster than heat.Coupled-transfer analogies (e.g., Chilton–Colburn) often invoke Le ≈ 1 to simplify correlations.Hence, the dimensionless number of interest is the Lewis number. Verification / Alternative check: In air–water systems at ambient conditions, Le ≈ 1, which supports simplified j-factor analogies linking heat and mass transfer coefficients. Why Other Options Are Wrong: Schmidt and Prandtl compare momentum with mass/heat diffusion separately; they do not directly relate mass to heat diffusion. Sherwood is a transfer coefficient correlation parameter, not a ratio of diffusivities. Common Pitfalls: Using Sc or Pr when a direct heat–mass coupling metric (Le) is required. Assuming Le = 1 universally; it varies with fluid and conditions. Final Answer: Lewis number

Q: Temperature sensitivity of liquid versus gas mass transfer: Compared with gases, how does temperature typically affect liquid-phase mass transfer coefficients?

Introduction / Context: Mass transfer coefficients depend on diffusivity and viscosity, both of which vary with temperature. The sensitivity differs markedly between liquids and gases and is essential when evaluating temperature control in extraction, absorption, and mixing operations. Given Data / Assumptions: Physical (nonreactive) mass transfer. Ordinary pressure ranges. Representative liquids and gases with typical property–temperature trends. Concept / Approach: In liquids, molecular diffusivity generally increases strongly with temperature (often following an Arrhenius-type relation) while viscosity decreases, both changes increasing mass transfer coefficients markedly. In gases, diffusivity rises with approximately T^1.5 / P and viscosity also rises with temperature, yielding a comparatively milder net effect on k. Therefore, liquid-phase mass transfer is usually more temperature-sensitive than gas-phase mass transfer. Step-by-Step Solution: Recognize k ∝ D * f(Sc, Re) and D_liquid increases sharply with T; μ_liquid decreases with T.For gases, D_gas increases with T but μ_gas also increases, partly offsetting gains in convective transport.Conclude: liquids show stronger dependence; choose option (a). Verification / Alternative check: Empirical correlations (e.g., film theory, penetration theory) show liquid-side k_L scaling strongly with D_L and ν_L; temperature exponents are larger for liquids. Why Other Options Are Wrong: (b) Opposite of observed behavior; (c) ignores clear property variations; (d) is unnecessary because a correct specific statement exists. Common Pitfalls: Assuming constant properties over wide temperature ranges in design calculations. Neglecting solubility changes with temperature which also affect driving forces. Final Answer: Affected more by temperature than that for gases

Question 1

Colligative effects: When a nonvolatile solute is dissolved in a solvent, which property (or properties) of the solution decreases compared with the pure solvent?

Accepted Answer

Introduction: Adding a nonvolatile solute to a solvent changes several properties known as colligative properties. Understanding the direction of change for boiling point, freezing point, and vapor pressure is important in designing evaporation and crystallization operations and in interpreting laboratory measurements. Given Data / Assumptions: Solute is nonvolatile and does not dissociate unusually. Ideal or near-ideal solution behavior is assumed for trends. Comparison is with the pure solvent at the same pressure. Concept / Approach: Adding solute lowers the solvent’s escaping tendency, thereby reducing the solution’s vapor pressure (Raoult’s law). Lower vapor pressure causes boiling point elevation (so boiling point increases, not decreases) and freezing point depression (freezing point decreases). Thus, among the listed properties, freezing point and vapor pressure decrease upon solute addition. Step-by-Step Solution: Vapor pressure: decreases due to reduced mole fraction of solvent.Boiling point: increases because a higher temperature is needed for the solution’s vapor pressure to reach ambient pressure.Freezing point: decreases due to disruption of solid–liquid equilibrium.Select properties that decrease: freezing point and vapor pressure. Verification / Alternative check: Standard equations: ΔTb = Kb * m > 0 (elevation), ΔTf = Kf * m > 0 (depression magnitude), and p_solution = x_solvent * p^0_solvent < p^0_solvent. Why Other Options Are Wrong: Boiling point: It increases. Density: May change either way depending on solute; not a colligative certainty. Common Pitfalls: Confusing the directions of change—remember: vapor pressure ↓, boiling point ↑, freezing point ↓ for nonvolatile solutes. Final Answer: both (b) and (c)

Question 2

Gas absorption design: For an absorber operating at steady state, the operating line on an equilibrium (y–x) diagram, relative to the equilibrium curve, typically lies where?

Accepted Answer

Introduction: Graphical design of absorbers uses an equilibrium curve (y* vs. x) and an operating line (material balance). The relative position of these curves determines the driving force for mass transfer and the number of theoretical stages required. Given Data / Assumptions: We consider absorption of a solute from gas to liquid (lean solvent). Coordinates: y for gas-phase solute fraction; x for liquid-phase solute fraction. Countercurrent contact with finite solvent flow rate. Concept / Approach: For absorption, the operating line must lie above the equilibrium curve to ensure a positive driving force (y − y*) across the column. If the operating line dipped below the equilibrium curve, portions of the column would have negative or zero driving force, making absorption infeasible. For stripping (desorption), the operating line is typically below the equilibrium curve. Step-by-Step Solution: Plot equilibrium relation y* = f(x).Write the operating line from overall/segment balances: y = m x + b.Ensure y ≥ y* throughout to maintain driving force for absorption.Therefore, the operating line lies above the equilibrium curve. Verification / Alternative check: Stage-by-stage (McCabe–Thiele) stepping fails if the operating line crosses below the equilibrium curve; practical design always keeps it above with margin for approach to equilibrium. Why Other Options Are Wrong: Below or either side: Would imply loss of driving force. Velocity statement: Unrelated to y–x diagram placement. 45°-line comment: Not a general rule for absorption design. Common Pitfalls: Confusing absorption with stripping; remember the relative position reverses for desorption problems. Final Answer: line always lies above the equilibrium curve.

Question 3

Distillation products: What can the distillate (overhead product) from a distillation column contain in terms of number of chemical components?

Accepted Answer

Introduction: Distillation separates mixtures based on relative volatility, but each product stream is still a mixture unless the feed is a binary azeotrope-free system with perfect separation. Understanding that distillate composition can include several components helps avoid oversimplified assumptions during design and operation. Given Data / Assumptions: General multicomponent distillation. Overhead is enriched in light (more volatile) species. Thermodynamic phase (vapor or liquid) depends on condenser duty. Concept / Approach: In multicomponent systems, the top product is typically a mixture of the lighter components in proportions dictated by reflux ratio, tray efficiency, and relative volatilities. Total or partial condensers determine the phase (distillate may leave as liquid or vapor), but not the number of components present. Step-by-Step Solution: Recognize that component count is not limited to one or two unless the feed is binary or pure.Even in binary systems, each product generally contains both components unless extremely high separation is achieved.Therefore, the overhead can contain any number of components (from the light end of the mixture). Verification / Alternative check: Real plant assays for crude columns show top products containing many hydrocarbons; specialty columns for solvent recovery also produce multicomponent overheads. Why Other Options Are Wrong: Only one or two components: Too restrictive for multicomponent feeds. Only saturated liquid/vapor: Phase depends on condenser; composition can still be multicomponent. Common Pitfalls: Assuming “distillate” means pure component; purity depends on design and economics, not on a fundamental limit. Final Answer: any number of components

Question 4

Drying under vacuum: Why is vacuum commonly used in drying operations for certain materials?

Accepted Answer

Introduction: Vacuum drying is a key technique for heat-sensitive or oxygen-sensitive materials in pharmaceuticals, foods, and fine chemicals. Lowering the pressure reduces the boiling point of water or solvents, enabling effective drying at much lower temperatures than atmospheric drying. Given Data / Assumptions: Evaporation rate depends on temperature and pressure (vapor–liquid equilibrium). Goal: protect product quality by limiting thermal exposure and oxidation. Bound vs. unbound moisture considerations remain, but the main lever is boiling-point suppression. Concept / Approach: At reduced pressure, the saturation temperature corresponding to a given vapor pressure decreases. This allows solvent removal at temperatures well below atmospheric boiling points. While vacuum may help with some bound moisture removal indirectly, it mainly enables drying without excessive heat that could degrade sensitive products. Step-by-Step Solution: Apply phase equilibrium: lower pressure ⇒ lower boiling point.Translate to process: evaporation occurs at reduced temperature under vacuum.Conclude: the principal purpose is to reduce drying temperature to protect product quality. Verification / Alternative check: Phase diagrams for water show dramatically lower boiling temperatures at sub-atmospheric pressures (e.g., near 40–60°C at deep vacuum), matching typical vacuum dryer settings. Why Other Options Are Wrong: Very high unbound moisture: Vacuum is not required for such feeds; conventional dryers suffice. Increase temperature: Opposite of vacuum’s purpose. High bound moisture: Binding relates to material structure; vacuum alone does not specifically target bound moisture. Eliminating heat transfer: Evaporation still requires energy input. Common Pitfalls: Assuming vacuum alone guarantees fast drying; heat and mass transfer resistances still govern rates. Final Answer: reduce drying temperature.

Question 5

Psychrometrics terminology: The amount of heat (BTU) required to raise the temperature of 1 lb of dry gas and its accompanying vapor by 1°F is called the __________.

Accepted Answer

Introduction: In gas–vapor mixtures such as air–water systems, psychrometric properties are used to size heaters, humidifiers, and dryers. One key property is the heat required to raise the temperature of the mixture per unit mass of dry gas, accounting for both the gas and the accompanying vapor. Given Data / Assumptions: Basis: 1 lb of dry gas with its associated vapor at a specified humidity ratio. Temperature change: 1°F. Property sought: mixture heat capacity per unit mass of dry gas. Concept / Approach: “Humid heat” (c_h) is defined as the heat required to raise 1 lb of dry gas plus its accompanying vapor by 1°F. For air–water vapor, an approximate correlation is c_h ≈ c_p,air + W * c_p,vapor, where W is the humidity ratio (lb vapor/lb dry air). This distinguishes it from “specific heat,” which usually refers to a single pure substance per unit mass, and from “latent heat,” which refers to phase change energy. Step-by-Step Solution: Identify the basis: per lb of dry gas including its associated vapor.Recognize the definition that matches: humid heat.Select option “humid heat.” Verification / Alternative check: Psychrometric charts and textbooks list humid heat to compute heater duties for moist air streams: Q ≈ m_dry air * c_h * ΔT when no moisture is added or removed. Why Other Options Are Wrong: Latent heat: Energy for phase change, not sensible heating of a gas–vapor mixture. Specific heat: Generic term; here a specialized mixture property is required. Sensible heat: Refers to the heat effect, not the property name. Humid volume: A volumetric property, not a heat capacity. Common Pitfalls: Using dry-air specific heat alone and underestimating heater duty at high humidity; always account for vapor contribution with humid heat. Final Answer: humid heat

Question 6

Phase diagrams and partially miscible liquids (ternary systems): A plait point is the point on a binodal (solubility) curve where the tie line shrinks to a point. For a ternary liquid system that contains two pairs of partially miscible liquids, how…

Accepted Answer

Introduction / Context: Ternary liquid–liquid phase diagrams often display regions of partial miscibility. Each partially miscible binary pair produces a binodal curve terminating at a critical mixing point known as the plait point. Understanding how many plait points exist helps interpret separation behavior and solvent selection strategies. Given Data / Assumptions: Ternary liquid system (three components). Exactly two binary pairs are partially miscible; the third pair is completely miscible. Standard triangular composition diagram at constant temperature and pressure. Concept / Approach: In ternary diagrams, each binary pair that is partially miscible contributes a binodal loop along the corresponding binary edge that extends into the ternary field. Each such binodal has a critical solution point at which the tie lines coalesce to a single composition—this is the plait point. Therefore, the number of plait points equals the number of partially miscible binary pairs in the system, provided their immiscibility persists at the diagram conditions. Step-by-Step Solution: Identify partially miscible pairs: count = 2.Recognize that each partially miscible pair contributes one plait point where tie lines vanish.Therefore, total plait points in the ternary diagram = 2. Verification / Alternative check: Reference ternary maps show one plait point for every partially miscible binary edge; systems with three such pairs (e.g., water–phenol–hydrocarbon at certain T) show three plait points, confirming the one-to-one correspondence. Why Other Options Are Wrong: 0 or 1 contradict the presence of two partially miscible pairs; 3 would require all three binary pairs to be partially miscible, which is not the case here. Common Pitfalls: Confusing a plait point with a critical solution temperature; here we consider a fixed T–P diagram. Assuming a single plait point governs the whole ternary field irrespective of how many binary pairs are immiscible. Final Answer: 2

Question 7

Dimensionless groups in coupled transport: In simultaneous heat and mass transfer problems, which dimensionless number most directly links thermal and mass diffusivities and is widely used to correlate coupled phenomena?

Accepted Answer

Introduction / Context: When heat and mass transfer occur together—such as in drying, humidification, evaporative cooling, or combustion—the relative rates of thermal and species diffusion strongly influence boundary-layer behavior and overall transfer rates. A key dimensionless group captures this relationship. Given Data / Assumptions: Single-phase flows with coupled heat and mass transfer. No chemical reaction, or reaction effects considered separately. Continuum mechanics and boundary-layer concepts apply. Concept / Approach: The Lewis number is defined as Le = α / D = (k / ρcp) / D, where α is thermal diffusivity and D is mass diffusivity. Le directly compares the relative thickness and development of thermal and concentration boundary layers. Schmidt number Sc = ν / D and Prandtl number Pr = ν / α are also important, but Le explicitly couples heat and mass transfer by the ratio of diffusivities. Sherwood number Sh = k_c L / D is a mass-transfer analog of Nusselt number and is used for correlations, not for comparing α and D directly. Step-by-Step Solution: Write Le = α / D.Interpretation: Le > 1 → heat diffuses faster than mass; Le < 1 → species diffuse faster than heat.Coupled-transfer analogies (e.g., Chilton–Colburn) often invoke Le ≈ 1 to simplify correlations.Hence, the dimensionless number of interest is the Lewis number. Verification / Alternative check: In air–water systems at ambient conditions, Le ≈ 1, which supports simplified j-factor analogies linking heat and mass transfer coefficients. Why Other Options Are Wrong: Schmidt and Prandtl compare momentum with mass/heat diffusion separately; they do not directly relate mass to heat diffusion. Sherwood is a transfer coefficient correlation parameter, not a ratio of diffusivities. Common Pitfalls: Using Sc or Pr when a direct heat–mass coupling metric (Le) is required. Assuming Le = 1 universally; it varies with fluid and conditions. Final Answer: Lewis number

Question 8

Solid–liquid extraction (leaching) rate control: In practical leaching of a solid particle by a solvent, which step most commonly limits the overall rate of extraction?

Accepted Answer

Introduction / Context: Leaching or solid–liquid extraction removes a soluble constituent from a solid matrix into a contacting solvent. Understanding the controlling resistance—internal diffusion within the particle or external film diffusion—guides particle-size selection, agitation, and equipment design. Given Data / Assumptions: Porous solid containing a soluble solute. Isothermal operation; solvent does not chemically react with the matrix. Well-agitated slurry such that external films can be minimized. Concept / Approach: Overall extraction rate often depends on the slowest step. In many practical systems with adequate agitation, dissolution at the surface is rapid and external mass-transfer resistance is relatively small, leaving internal diffusion through solvent-filled pores as the controlling mechanism. This is particularly true for larger particles or tortuous pore networks. Step-by-Step Solution: Identify candidate resistances: interfacial dissolution, internal diffusion, external diffusion.Assess typical magnitudes: interfacial dissolution is fast; agitation reduces external film thickness.Internal diffusion depends on pore tortuosity and path length and is often rate-limiting.Therefore, select internal diffusion (option b) as the common controlling step. Verification / Alternative check: Shrinking-core/ash-layer models predict linear-to-parabolic kinetics when diffusion through the product/porous layer controls, aligning with many experimental leaching curves. Why Other Options Are Wrong: (a) Interfacial phase change/dissolution is seldom controlling for soluble solids in good solvents. (c) External diffusion can dominate only at low agitation or very viscous slurries; it is not the general case. (d) 'All' overstates the contribution of interfacial phase change in typical leaching. Common Pitfalls: Ignoring the impact of particle size: reducing size shortens diffusion paths and accelerates extraction. Overlooking clogging or swelling that can alter internal diffusivity during operation. Final Answer: Diffusion of solute through solvent-filled pores to the outside of the particle (internal diffusion)

Question 9

Temperature sensitivity of liquid versus gas mass transfer: Compared with gases, how does temperature typically affect liquid-phase mass transfer coefficients?

Accepted Answer

Introduction / Context: Mass transfer coefficients depend on diffusivity and viscosity, both of which vary with temperature. The sensitivity differs markedly between liquids and gases and is essential when evaluating temperature control in extraction, absorption, and mixing operations. Given Data / Assumptions: Physical (nonreactive) mass transfer. Ordinary pressure ranges. Representative liquids and gases with typical property–temperature trends. Concept / Approach: In liquids, molecular diffusivity generally increases strongly with temperature (often following an Arrhenius-type relation) while viscosity decreases, both changes increasing mass transfer coefficients markedly. In gases, diffusivity rises with approximately T^1.5 / P and viscosity also rises with temperature, yielding a comparatively milder net effect on k. Therefore, liquid-phase mass transfer is usually more temperature-sensitive than gas-phase mass transfer. Step-by-Step Solution: Recognize k ∝ D * f(Sc, Re) and D_liquid increases sharply with T; μ_liquid decreases with T.For gases, D_gas increases with T but μ_gas also increases, partly offsetting gains in convective transport.Conclude: liquids show stronger dependence; choose option (a). Verification / Alternative check: Empirical correlations (e.g., film theory, penetration theory) show liquid-side k_L scaling strongly with D_L and ν_L; temperature exponents are larger for liquids. Why Other Options Are Wrong: (b) Opposite of observed behavior; (c) ignores clear property variations; (d) is unnecessary because a correct specific statement exists. Common Pitfalls: Assuming constant properties over wide temperature ranges in design calculations. Neglecting solubility changes with temperature which also affect driving forces. Final Answer: Affected more by temperature than that for gases

Mass Transfer Questions