Q: Graham’s law application: Helium has four times the atomic weight of hydrogen. Relative to hydrogen, helium will diffuse at what multiple of the rate?

Introduction / Context: Graham’s law states that the rate of diffusion (or effusion) of a gas is inversely proportional to the square root of its molar mass. This principle is widely used for order-of-magnitude estimates in mass transfer and gas separation contexts. Given Data / Assumptions: Comparison is qualitative using Graham’s law: r ∝ 1/√M. Statement references atomic weight ratio He:H = 4:1, treating hydrogen on a per-atom basis in the intended textbook-style simplification. Isothermal, same pressure conditions. Concept / Approach: If helium’s mass is four times hydrogen’s reference mass, then the rate ratio is r_He / r_H = √(M_H / M_He) = √(1/4) = 1/2. Many pedagogy problems use this simplified mass ratio to reinforce the inverse square-root dependence without delving into diatomic versus monatomic complications. Step-by-Step Solution: Write Graham’s law: r_A / r_B = √(M_B / M_A).Set M_He / M_H = 4. Then r_He / r_H = √(1/4) = 1/2.Therefore helium diffuses at half the rate of hydrogen (per the problem’s stated ratio). Verification / Alternative check: Using molecular gases (He = 4, H2 = 2) would give r_He / r_H2 = √(2/4) = 1/√2. However, the stem explicitly frames a 4:1 ratio, yielding 1/2 per the simplified approach. Why Other Options Are Wrong: Options 4 and 2 invert Graham’s dependence; 1/4 corresponds to mass ratio 16:1; 1/√2 applies if comparing He to H2, which is not how the stem is framed. Common Pitfalls: Confusing atomic hydrogen with molecular hydrogen; always match the mass basis used in the question. Final Answer: 1/2

Question 1

Choosing a concentration unit for solutions: In routine chemical engineering calculations, which expression of solution concentration is generally the most convenient and broadly applicable?

Accepted Answer

Introduction / Context: Different concentration units are used for different tasks. Selecting a convenient unit simplifies material balances, equilibrium calculations, and data correlation. For vapor–liquid equilibrium and many thermodynamic models, a composition variable that is dimensionless and independent of temperature and pressure is especially valuable. Given Data / Assumptions: We seek a generally convenient unit across a range of operations and temperatures. No specific reaction stoichiometry that would favor equivalents (normality). Practical considerations such as temperature dependence (or lack thereof) matter. Concept / Approach: Mole fraction (x or y) is dimensionless, does not change with temperature or pressure (for a closed, nonreacting system), and is directly used in Raoult’s law, Dalton’s law, and activity/fugacity models. In contrast, molarity (mol/L) depends on temperature because solution volume changes. Molality (mol/kg solvent) is temperature independent and excellent for colligative properties, but it is less directly used in gas–liquid equilibrium expressions. Normality (equivalents/L) is reaction-specific and can be ambiguous unless the reaction is specified. Step-by-Step Solution: Assess each unit for general convenience across thermodynamics and separations.Note that mole fraction integrates seamlessly with Raoult’s and Dalton’s laws in equilibrium work.Conclude that “mole fraction” is the most broadly convenient choice here. Verification / Alternative check: Most VLE and EoS-based computations use mole fractions as primary composition variables; software packages likewise adopt x and y for liquid and vapor phases. Why Other Options Are Wrong: Normality: depends on reaction equivalence; not universal.Molality: useful and temperature independent, but less convenient for gas-phase calculations and phase-rule applications.Molarity: varies with temperature because volume changes, complicating wide-range calculations. Common Pitfalls: Interchanging molarity and molality; attempting to use normality without specifying the reaction; overlooking temperature effects on molarity. Final Answer: Mole fraction

Question 2

State-dependent heat capacity comparison: Which statement about the heat capacity of a substance in different physical states is incorrect in general practice?

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Introduction / Context: Accurate heat capacity (Cp or Cv) values are essential in energy balances for heating/cooling, phase changes, and reactor design. While exact numbers depend on substance and temperature, there are robust trends across phases that help engineers catch mistakes and perform quick estimates. Given Data / Assumptions: Qualitative comparison among typical Cp values of solid, liquid, and gas phases at comparable temperatures. No anomalous substances considered near critical points or with unusual transitions. Concept / Approach: For most substances, heat capacity increases from solid → liquid → gas at the same temperature range. Liquids, with more accessible configurational degrees of freedom than solids, typically have larger Cp than solids. Gases (especially polyatomic) often have the highest Cp due to translational, rotational, and vibrational contributions (at higher T). Therefore any statement claiming solids have higher Cp than liquids is generally incorrect. Step-by-Step Solution: Evaluate (a): “Greater for liquid than solid” — generally true.Evaluate (b): “Lower for liquid than gas” — generally true.Evaluate (c): “Higher for solid than liquid” — contradicts the common trend; this is the incorrect statement.Evaluate (d): “Equal for solid and liquid below melting point” — not generally true either; however, the question asks for the incorrect statement in a general-practice sense, and (c) is definitively and broadly false, whereas (d) is obviously false but not the standard comparative rule; keeping one unique incorrect choice requires identifying (c) as the clearly wrong assertion in context of comparative ordering. Verification / Alternative check: Handbook data (e.g., water/ice, metals) show Cp(liquid) > Cp(solid) for typical temperatures not extremely close to phase transitions. Why Other Options Are Wrong: (a) and (b) follow common thermophysical trends.(d) “Equal” is not a standard rule; Cp varies with T and differs between phases. However, the principal egregious error is (c), which reverses the typical order. Common Pitfalls: Assuming Cp is constant across phases; ignoring temperature dependence; reversing the usual solid–liquid–gas ordering. Final Answer: Higher for solid state than for liquid state.

Question 3

Psychrometrics — percentage humidity from relative humidity: Air at 20°C and 750 mm Hg has relative humidity 80%. Given water vapor pressure at 20°C is 17.5 mm Hg, compute the percentage humidity of the air.

Accepted Answer

Introduction / Context: Humidity measures are central to drying, HVAC, and gas–liquid mass transfer. While relative humidity (RH) compares the actual partial pressure of water vapor to its saturation value at the same temperature, percentage humidity compares the humidity ratio (water per dry air) to the saturation humidity ratio at the same conditions. The two are not identical because the denominator for humidity ratio includes the dry-air term (P − p_v) rather than the total pressure P. Given Data / Assumptions: Total pressure P = 750 mm Hg. Temperature = 20°C; saturation vapor pressure p_vs = 17.5 mm Hg. Relative humidity φ = 80% → p_v = 0.80 * p_vs = 14.0 mm Hg. Ideal gas behavior; standard psychrometric definitions. Concept / Approach: Humidity ratio Y (kg H2O/kg dry air) is proportional to p_v/(P − p_v). At saturation, Y_s ∝ p_vs/(P − p_vs). The percentage humidity is defined as 100 * (Y/Y_s) = 100 * [p_v (P − p_vs)] / [p_vs (P − p_v)]. Step-by-Step Solution: Compute p_v from RH: p_v = 0.80 * 17.5 = 14.0 mm Hg.Apply formula: % humidity = 100 * [p_v (P − p_vs)] / [p_vs (P − p_v)].Substitute values: numerator = 14.0 * (750 − 17.5) = 14.0 * 732.5 = 10255.0.Denominator = 17.5 * (750 − 14.0) = 17.5 * 736 = 12880.0.Ratio = 10255.0 / 12880.0 = 0.7965 ≈ 79.65% → matches 79.62% within rounding. Verification / Alternative check: If we (incorrectly) equate percentage humidity to relative humidity, we would get 80%, which is slightly higher. The small difference arises from the (P − p_v) vs. (P − p_vs) terms in the humidity-ratio definition. Why Other Options Are Wrong: 80.38 or 78.51: result from arithmetic/rounding errors or misapplied formulas.80: equal to RH, but percentage humidity is not exactly RH except at very low pressures or special limits. Common Pitfalls: Confusing RH with percentage humidity; forgetting to use (P − p_v) and (P − p_vs) in the humidity-ratio expressions; mixing absolute/relative units. Final Answer: 79.62

Question 4

Purge calculation in isomerisation with impurity control: A butane isomerisation unit produces 70 kmol/h of pure isobutane. A continuous purge contains 85% n-butane and 15% impurity (mole%). The feed is n-butane containing 1% impurity (mole%). What …

Accepted Answer

Introduction / Context: Purge streams prevent the buildup of inert or contaminant species in recycle or high-conversion processes. Determining the necessary purge rate requires steady-state component balances, particularly on the impurity that must be removed from the system. Here, an isomerisation converts n-butane to isobutane (1:1 basis), and the product is specified as pure isobutane; unreacted n-butane and impurity exit mainly in the purge. Given Data / Assumptions: Product is 70 kmol/h of pure isobutane (no impurity, no n-butane). Purge composition: 85% n-butane, 15% impurity (on molar basis). Feed: n-butane with 1% impurity → feed composition: 99% n-butane, 1% impurity. Steady state; no accumulation of impurity inside the system. Isomerisation converts n-butane to isobutane one-for-one; no side reactions producing impurity. Concept / Approach: Write component balances. The impurity enters with the feed and must leave entirely in the purge (since the product is pure isobutane). Then relate the n-butane balance to the given production of isobutane, recognizing that consumed n-butane equals isobutane produced. Step-by-Step Solution: Let F = feed flow (kmol/h) and P = purge flow (kmol/h).Impurity balance: 0.01 F = 0.15 P ⇒ F = 15 P.n-Butane balance to isobutane: n-butane converted = isobutane out = 70.n-Butane in feed = 0.99 F; n-butane leaving in purge = 0.85 P.Conversion relation: 0.99 F − 0.85 P = 70.Substitute F = 15 P: 0.99(15 P) − 0.85 P = 70 ⇒ (14.85 − 0.85) P = 14.0 P = 70.Solve: P = 70 / 14 = 5 kmol/h. Verification / Alternative check: From F = 15 P, the feed is 75 kmol/h. Feed n-butane = 74.25; n-butane purged = 4.25; conversion = 74.25 − 4.25 = 70 → matches product rate, confirming consistency. Why Other Options Are Wrong: 3, 4, 6 kmol/h do not satisfy both impurity and n-butane balances simultaneously. Common Pitfalls: Forgetting to use impurity balance to relate feed and purge; assuming purge composition equals feed composition; neglecting that “pure isobutane” means zero impurity and zero n-butane in the product. Final Answer: 5 kmol/h

Question 5

Periodic table ordering principle: In the modern periodic table, elements are arranged in increasing order of their __________.

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Introduction / Context: The modern periodic table is organized to reflect recurring chemical properties (periodicity). The key variable that correctly orders the elements and explains periodic trends in valence, electronegativity, and atomic size is not mass but a more fundamental property related to nuclear charge. Given Data / Assumptions: “Modern” implies the post-Moseley arrangement that resolved anomalies in Mendeleev’s atomic-weight ordering. Chemical periodicity correlates with electron configurations, which follow the order imposed by nuclear charge. Concept / Approach: Moseley’s X-ray spectroscopy showed that the square root of the characteristic X-ray frequency varies linearly with the atomic number Z, revealing that Z (the number of protons in the nucleus) determines an element’s identity. Electron configurations, and hence chemical properties, are governed by Z, making “atomic number” the correct ordering parameter. This explains earlier anomalies (e.g., tellurium and iodine) that could not be reconciled by atomic weight alone. Step-by-Step Solution: Identify the property that uniquely specifies elements and controls electron configuration → atomic number Z.Recognize that the table’s rows and columns group elements by valence shell filling and periodic trends set by Z.Therefore, the correct completion is “atomic number.” Verification / Alternative check: Compare tellurium (Z = 52) and iodine (Z = 53): ordering by Z matches chemical properties; ordering by atomic weight would misplace them, illustrating why Z is fundamental. Why Other Options Are Wrong: Mass number varies among isotopes and is not unique to the element’s chemical behavior.Atomic weight is an average over isotopes and historically led to anomalies.Metallic characteristics are emergent trends, not a primary ordering index. Common Pitfalls: Confusing atomic weight with atomic number; overlooking isotopic variation; assuming chemical similarity derives from mass rather than electron configuration. Final Answer: atomic number

Question 6

Chemical engineering thermodynamics concept check: What is the enthalpy change called when one unit mass of a dry solid is wetted with enough liquid such that additional liquid causes no further thermal effect?

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Introduction / Context: When a porous or particulate solid contacts a liquid, heat is released or absorbed due to surface interactions and spreading of the liquid on the solid. Chemical engineers distinguish between the initial heat liberated upon first contact and the limiting amount after the solid is fully and uniformly covered. Correct terminology matters in drying, granulation, crystallization, and adsorption design. Given Data / Assumptions: Dry solid is contacted with a wetting liquid at constant initial temperature and pressure. Enough liquid is added to completely cover/penetrate accessible surface so that additional liquid no longer changes temperature. No chemical reaction between solid and liquid; effect is purely physical wetting. Concept / Approach: Two related quantities are used. The heat of wetting refers to the enthalpy change when the first portions of liquid spread on a unit mass of dry solid. The heat of complete wetting is the total enthalpy change when sufficient liquid has been added to fully wet the solid so that further addition produces no incremental thermal effect. The question clearly describes the limiting case, hence the latter term applies. Step-by-Step Solution: Interpret the stem: “sufficient liquid” and “no additional thermal effect” indicate a limiting, not incremental, condition.Map terminology: limiting condition → “complete wetting.”Therefore, select the term “heat of complete wetting.” Verification / Alternative check: In calorimetry, the integral heat effect measured up to the point of saturation (complete surface coverage) is reported as heat of complete wetting; the differential slope at low additions corresponds to heat of wetting. Why Other Options Are Wrong: Heat of mixing relates to miscible liquids/solutions; heat of adsorption concerns gas or solute binding on solid surfaces; “heat of wetting” is incremental and not necessarily the terminal value; heat of dilution involves adding solvent to a solution, not to a dry solid. Common Pitfalls: Confusing incremental versus integral definitions and overlooking the phrase indicating “no further effect.” Final Answer: heat of complete wetting

Question 7

Psychrometrics terminology: The percentage ratio of a vapor’s partial pressure to the vapor pressure of the pure liquid at the same temperature is called what?

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Introduction / Context: In gas–vapor systems (e.g., humid air), several saturation measures are used. Distinguishing relative saturation from percentage saturation and other humidity metrics is essential for drying calculations, cooling towers, and environmental control. Given Data / Assumptions: Definition refers to partial pressure of vapor (p_v) and saturation vapor pressure at the same temperature (p_v,sat). Expressed as a percentage. Concept / Approach: relative saturation (%) = 100 * (p_v / p_v,sat) This measure compares how “full” the gas is relative to the maximum vapor pressure at that temperature. Percentage saturation is a different quantity that adjusts for the presence of carrier gas (mole or mass basis) and is typically slightly lower than relative saturation for the same state. Step-by-Step Solution: Identify the exact definition in the stem: ratio of partial pressure to saturation pressure at the same temperature.Match with standard term: relative saturation (also commonly called relative humidity for water vapor in air). Verification / Alternative check: Psychrometric charts and tables list RH (%) precisely as p_v/p_v,sat × 100 under ideal-gas approximations, which aligns with the definition of relative saturation used in engineering practice. Why Other Options Are Wrong: “Percentage saturation” is a distinct measure using humidity ratios; options claiming independence from mixture composition or vapor nature are incorrect—real-gas nonidealities and species identity affect relations; “relative humidity by mass” is nonstandard wording. Common Pitfalls: Interchanging relative saturation with percentage saturation; mixing pressure- and humidity-ratio-based definitions. Final Answer: relative saturation

Question 8

Free (unrestrained) expansion of an ideal gas: Why does it not cool during Joule expansion into a vacuum?

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Introduction / Context: In a Joule (free) expansion, an ideal gas expands into a vacuum without doing work and without heat exchange. Many learners expect cooling, but an ideal gas does not cool in this process. Understanding why sharpens the distinction between ideal and real gases in throttling/expansion devices. Given Data / Assumptions: Ideal-gas behavior: intermolecular potential energy is negligible. Process is adiabatic (Q = 0) and does no external work (W = 0). Internal energy U of an ideal gas depends only on temperature. Concept / Approach: From the first law of thermodynamics, ΔU = Q − W. For free expansion of an ideal gas, Q = 0 and W = 0, so ΔU = 0. Since U = U(T) for an ideal gas, ΔU = 0 implies ΔT = 0, hence no cooling. The microscopic reason is the absence of intermolecular attractions that would convert potential energy to kinetic energy changes during expansion. Step-by-Step Solution: Apply first law: ΔU = 0 → ΔT = 0 (ideal gas).Link to molecular model: no attractive forces → U independent of volume.Therefore, temperature remains constant; no cooling occurs. Verification / Alternative check: Real gases may warm or cool depending on the Joule effect, but for the idealized case, caloric equation of state shows U = f(T) only, confirming constant T when ΔU = 0. Why Other Options Are Wrong: Collisions do not dissipate energy in an ideal gas (elastic), but that alone is not the defining reason. Inversion temperature refers to throttling (Joule–Thomson), not free expansion. No external work is done in free expansion, so option (d) is inaccurate. Heat leakage (e) is outside the stated idealized setup. Common Pitfalls: Confusing Joule free expansion with Joule–Thomson throttling through a valve. Final Answer: molecules do not exert attractive forces (U independent of volume)

Question 9

Stoichiometry of neutralization: When 56 g of CaO are mixed with 63 g of HNO3, how many grams of Ca(NO3)2 are formed?

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Introduction / Context: Accurate stoichiometric calculations underpin reactor sizing and yield estimation. This problem involves the neutralization of a basic oxide with nitric acid to form a salt and water, emphasizing limiting-reagent identification and mass–mole conversions. Given Data / Assumptions: Chemical reaction: CaO + 2 HNO3 → Ca(NO3)2 + H2O. Molar masses: CaO = 56 g/mol; HNO3 = 63 g/mol; Ca(NO3)2 = 164 g/mol. Masses provided: 56 g CaO and 63 g HNO3. Concept / Approach: Convert masses to moles, identify the limiting reactant using stoichiometric coefficients, then compute product moles from the limiting reagent. Finally, convert moles of product to mass using its molar mass. Step-by-Step Solution: n_CaO = 56 g / 56 g/mol = 1.0 mol.n_HNO3 = 63 g / 63 g/mol = 1.0 mol.Stoichiometry requires 2 mol HNO3 per 1 mol CaO; available HNO3 (1.0 mol) is limiting.From 2 HNO3 → 1 Ca(NO3)2, so 1.0 mol HNO3 yields 0.5 mol Ca(NO3)2.Mass of product = 0.5 mol × 164 g/mol = 82 g. Verification / Alternative check: If HNO3 were doubled (126 g), CaO would then be limiting, and 1 mol CaO would give 1 mol Ca(NO3)2 = 164 g, confirming the proportionality of our 82 g result. Why Other Options Are Wrong: 164 g assumes excess acid; 41 g and 8.2 g are fractional values inconsistent with stoichiometry; 123 g lacks any stoichiometric basis. Common Pitfalls: Forgetting the 2:1 acid-to-oxide stoichiometry; using atomic instead of formula molar masses. Final Answer: 82 g

Question 10

Trend in heat capacity by phase: For most substances, in which phase is the heat capacity typically the greatest?

Accepted Answer

Introduction / Context: Heat capacity indicates how much energy is required to raise temperature. Recognizing broad trends across phases helps in estimating energy requirements in heating/cooling operations when detailed data are unavailable. Given Data / Assumptions: Comparison is qualitative and typical; exceptions exist. Consider molar or mass heat capacities at common conditions. Concept / Approach: Liquids generally exhibit higher heat capacities than solids due to additional configurational degrees of freedom (molecular rearrangements) and stronger interactions than gases, which at comparable conditions possess lower heat storage per unit mass or per mole for many common substances. Water is a prime example with an unusually high heat capacity among liquids. Step-by-Step Solution: Compare typical values: water C_p ≈ 4.18 kJ/kg·K (liquid); ice ≈ 2.1 kJ/kg·K (solid); water vapor near 1.9–2.1 kJ/kg·K at 1 atm.Most organic liquids also exceed their solid counterparts in C_p.Hence, “liquid phase” is the best general choice. Verification / Alternative check: Handbooks commonly list C_p(liquid) > C_p(solid) for the same substance near ambient; gases depend on composition and temperature but are often lower on a per-mass basis for common materials. Why Other Options Are Wrong: “Solid” typically lower; “gas” can be higher on a molar basis for some species but not as a broad rule; “no consistent trend” ignores widely observed patterns; “plasma” is beyond the intended scope. Common Pitfalls: Confusing per mole versus per mass comparisons; cherry-picking exceptions. Final Answer: liquid phase

Question 11

Viscosity benchmark for water: At room temperature, the viscosity of liquid water is closest to which unit value?

Accepted Answer

Introduction / Context: Viscosity benchmarks enable quick checks of rheological calculations and pump sizing. Engineers often memorize a few anchor values, and water at room temperature is the most important of these. Given Data / Assumptions: Room temperature ≈ 20–25°C. Dynamic viscosity μ units: Pa·s, poise (P), centipoise (cP). Kinematic viscosity ν uses stokes (St) and is ν = μ/ρ. Concept / Approach: Water at ≈ 20°C has μ ≈ 1.0 mPa·s = 0.001 Pa·s. Since 1 P = 0.1 Pa·s and 1 cP = 0.001 Pa·s, water’s viscosity ≈ 1 cP. A poise (1 P) is 100× larger; a stoke (1 St) is a unit of kinematic viscosity, not dynamic. Step-by-Step Solution: Convert: 0.001 Pa·s → 1 cP (because 1 cP = 10^-3 Pa·s).Therefore, the closest standard value is 1 cP. Verification / Alternative check: At 20°C, tabulated μ_water ≈ 1.002 cP; at 25°C ≈ 0.89 cP. Both are of order unity in cP, confirming the benchmark. Why Other Options Are Wrong: 1 P is too large; 1 St is not a dynamic viscosity; “both” mixes incompatible units; 10 cP corresponds to far more viscous liquids like light oils. Common Pitfalls: Confusing dynamic with kinematic viscosity and mixing poise with stoke without density conversion. Final Answer: 1 centipoise

Question 12

Molecular kinetic energy comparison: At the same temperature and pressure, which gas sample has the highest kinetic energy per mole?

Accepted Answer

Introduction / Context: In kinetic theory, the average translational kinetic energy per molecule depends only on absolute temperature, not on the molecular identity. Expressed per mole, this leads to a universal value proportional to temperature, which is critical for understanding equipartition and temperature scales. Given Data / Assumptions: Ideal-gas assumption. Same temperature for all gases being compared. Per mole basis. Concept / Approach: The mean translational kinetic energy per molecule is (3/2) k_B T. On a per mole basis, multiply by Avogadro’s number to obtain (3/2) R T, which is identical for any ideal gas at the same temperature. Pressure does not alter this equality for ideal gases; it affects density and mean free path, not the per-mole kinetic energy at a given T. Step-by-Step Solution: KE_molar = (3/2) R T for an ideal gas.At fixed T, KE_molar is the same for Cl2, N2, C2H6, He, etc.Therefore, all gases have the same kinetic energy per mole. Verification / Alternative check: Although heavier molecules move more slowly (lower rms speed), their higher mass exactly compensates to give the same average translational energy at the same temperature. Why Other Options Are Wrong: Picking a specific gas ignores the temperature-only dependence of KE in the ideal-gas model. Common Pitfalls: Confusing speed with kinetic energy; rms speed varies with molecular mass, while KE per mole at fixed T does not. Final Answer: all gases have the same kinetic energy per mole

Question 13

Mass velocity (mass flux): What is the correct SI-consistent unit for mass velocity?

Accepted Answer

Introduction / Context: Mass velocity, also called mass flux, is a key transport parameter in momentum, heat, and mass transfer correlations. Using the correct units is essential for dimensional consistency when applying film, boundary-layer, or packed-bed equations. Given Data / Assumptions: Mass velocity G = ṁ / A, where ṁ is mass flow rate and A is cross-sectional area. Engineering time base often uses hour in plant calculations; SI base would use seconds. Concept / Approach: From the definition, the units are mass per area per time. Therefore, a consistent engineering unit is kg/(m^2·hr). Pure kg/hr lacks area normalization; kg/m·hr is dimensionally incorrect; kg/m^2 without time omits the dynamic aspect. Step-by-Step Solution: Start with G = ṁ / A.Units: (kg/hr) / (m^2) = kg/(m^2·hr).Hence, the correct choice is kg/m^2·hr. Verification / Alternative check: In SI base, kg/(m^2·s) is equally valid; converting hours to seconds simply changes the numerical value, not the dimensional form. Why Other Options Are Wrong: Options lacking area or time components are dimensionally incomplete or incorrect; mixed units like kg/m·hr mis-handle the area dimension. Common Pitfalls: Confusing superficial velocity (length/time) with mass velocity (mass/area/time). Final Answer: kg/m^2 · hr

Question 14

Effect of solute on aqueous specific heat: As the concentration of a solute increases in an aqueous solution, how does the specific heat usually change?

Accepted Answer

Introduction / Context: Specific heat capacity directly affects energy required for heating and cooling process streams. Water has a very high specific heat, and dissolving solutes typically lowers the mixture’s heat capacity—knowledge that is crucial for evaporators, crystallizers, and heat exchangers. Given Data / Assumptions: Aqueous solutions at moderate concentrations. No phase change; temperature range where properties vary smoothly. Concept / Approach: Most inorganic salts and many organics reduce the ability of water to store thermal energy per unit mass, leading to a decrease in specific heat with increasing solute concentration. The strong hydrogen-bond network of water contributes to its large C_p; solutes disrupt this structure and replace water with lower-C_p material. Step-by-Step Solution: Consider a mass-weighted mixture: C_p,mixture ≈ Σ (mass fraction_i * C_p,i).Since C_p,solute < C_p,water for many common solutes, increasing solute fraction lowers the mixture’s C_p.Therefore, the general trend is a decrease. Verification / Alternative check: Handbook data for salt solutions (NaCl, CaCl2, KNO3) show monotonic decreases of C_p with concentration at constant temperature. Why Other Options Are Wrong: “Increases” contradicts common data; “unchanged” ignores strong composition dependence; while some unusual systems may deviate slightly, the engineering rule-of-thumb is a decrease. Common Pitfalls: Assuming linearity across all ranges; at very high concentrations or near phase changes, behavior can be nonlinear. Final Answer: decreases

Question 15

Graham’s law application: Helium has four times the atomic weight of hydrogen. Relative to hydrogen, helium will diffuse at what multiple of the rate?

Accepted Answer

Introduction / Context: Graham’s law states that the rate of diffusion (or effusion) of a gas is inversely proportional to the square root of its molar mass. This principle is widely used for order-of-magnitude estimates in mass transfer and gas separation contexts. Given Data / Assumptions: Comparison is qualitative using Graham’s law: r ∝ 1/√M. Statement references atomic weight ratio He:H = 4:1, treating hydrogen on a per-atom basis in the intended textbook-style simplification. Isothermal, same pressure conditions. Concept / Approach: If helium’s mass is four times hydrogen’s reference mass, then the rate ratio is r_He / r_H = √(M_H / M_He) = √(1/4) = 1/2. Many pedagogy problems use this simplified mass ratio to reinforce the inverse square-root dependence without delving into diatomic versus monatomic complications. Step-by-Step Solution: Write Graham’s law: r_A / r_B = √(M_B / M_A).Set M_He / M_H = 4. Then r_He / r_H = √(1/4) = 1/2.Therefore helium diffuses at half the rate of hydrogen (per the problem’s stated ratio). Verification / Alternative check: Using molecular gases (He = 4, H2 = 2) would give r_He / r_H2 = √(2/4) = 1/√2. However, the stem explicitly frames a 4:1 ratio, yielding 1/2 per the simplified approach. Why Other Options Are Wrong: Options 4 and 2 invert Graham’s dependence; 1/4 corresponds to mass ratio 16:1; 1/√2 applies if comparing He to H2, which is not how the stem is framed. Common Pitfalls: Confusing atomic hydrogen with molecular hydrogen; always match the mass basis used in the question. Final Answer: 1/2

Question 16

Gas densities and molecular weights: If gas X has a density exactly twice that of gas Y at the same temperature and pressure, and the molecular weight of gas Y is M, what is the molecular weight of gas X?

Accepted Answer

Introduction / Context: In ideal-gas behavior, density is directly linked to molecular weight at fixed temperature and pressure. This question tests the proportional relationship between density and molecular weight—useful in identifying gases and checking gas-mixture calculations. Given Data / Assumptions: Both gases are compared at the same temperature and pressure. Ideal-gas approximation applies. Density of X is twice the density of Y. Molecular weight of Y is M. Concept / Approach: For an ideal gas, ρ = P * MW / (R * T). At the same P and T, density ρ is proportional to molecular weight MW. Hence, if ρX = 2 * ρY, then MWX = 2 * MWY. Step-by-Step Solution: Use ρ ∝ MW at fixed T, P.Given ρX = 2 * ρY ⇒ MWX = 2 * MWY.MWY = M ⇒ MWX = 2M. Verification / Alternative check: Compare molar volumes: At the same T and P, equal moles occupy the same volume; doubling density must arise from doubling mass per mole, that is, doubling molecular weight. Why Other Options Are Wrong: M/2 or M/4: These would reduce density, contradicting the statement. M: Would keep densities equal, not doubled. Common Pitfalls: Confusing density proportionality when T or P change; here they are explicitly the same for both gases. Final Answer: 2M

Question 17

Osmosis calculation at 0°C: One gram-mole of a non-electrolyte is dissolved to make 22.4 litres of solution at 0°C. What is the osmotic pressure (in atm) of this solution?

Accepted Answer

Introduction / Context: Osmotic pressure is central to membrane separations, food science, and bioprocessing. For dilute solutions of non-electrolytes, it mirrors the ideal-gas law, enabling quick head calculations for reverse osmosis or dialysis design. Given Data / Assumptions: Moles of solute n = 1 mol. Solution volume V = 22.4 L. Temperature T = 0°C = 273 K. R = 0.082057 L·atm·mol^-1·K^-1; ideal dilute solution. Concept / Approach: For dilute, non-electrolyte solutions, osmotic pressure π follows πV = nRT, analogous to PV = nRT for gases. Step-by-Step Solution: π = n * R * T / V.Substitute: π = 1 * 0.082057 * 273 / 22.4.Compute numerator: 0.082057 * 273 ≈ 22.4.Thus π ≈ 22.4 / 22.4 = 1.0 atm. Verification / Alternative check: This setup intentionally mirrors one mole of ideal gas at STP occupying about 22.4 L and exerting ~1 atm, hence the neat numerical result for π. Why Other Options Are Wrong: 0.5, 1.5, 2: Do not match the direct calculation from π = nRT/V. Common Pitfalls: For electrolytes, include van’t Hoff factor i. Also ensure volume is solution volume and temperature is in Kelvin for consistency. Final Answer: 1

Question 18

Saturated gas mixtures: For a gas saturated with a condensable vapour under ideal-gas assumptions, the volumetric (mole) fraction of vapour depends on vapour saturation pressure and total pressure. It is independent of which factor?

Accepted Answer

Introduction / Context: Humidification and gas–vapour contacting rely on saturated-gas relationships. For an ideal saturated mixture, the vapour mole fraction is given by y_v = p_sat(T) / P_total. This allows quick estimation of humidity and dew-point behavior. Given Data / Assumptions: Ideal-gas behavior for both vapour and carrier gas. Gas is saturated with the vapour at temperature T. Vapour partial pressure equals its saturation pressure p_sat(T). Concept / Approach: At saturation, y_v = p_v / P = p_sat(T) / P. Therefore, y_v depends on temperature (through p_sat) and on total pressure P. It also depends on the liquid identity because different liquids have different p_sat(T). However, the identity of the noncondensable gas does not appear in y_v for ideal systems. Step-by-Step Solution: Write y_v = p_sat(T) / P.Identify dependencies: p_sat depends on T and liquid identity.Carrier gas identity does not enter, assuming ideal behavior.Hence, independence is from the nature of the gas. Verification / Alternative check: Psychrometric parallels show humidity ratio depends on p_sat(T) and pressure; whether the carrier is air, nitrogen, or helium does not change y_v ideally. Why Other Options Are Wrong: Nature of the liquid: Determines p_sat(T); crucial. Temperature: Controls p_sat; crucial. Total pressure: Explicit in y_v; crucial. Common Pitfalls: Non-ideality (solubility, association, or high pressures) can introduce dependence on gas identity; this question assumes ideal conditions. Final Answer: Nature of the noncondensable gas

Question 19

Third-law perspective: At a temperature of absolute zero (0 K), what kind of molecular energy does a classical ideal gas possess?

Accepted Answer

Introduction / Context: Absolute zero (0 K) is a limiting concept where thermal motion ceases. In classical thermodynamics and kinetic theory, the internal energy of an ideal gas is purely kinetic; at 0 K, that kinetic energy vanishes. This question checks conceptual understanding across translational, rotational, and vibrational modes. Given Data / Assumptions: Classical ideal-gas model. Temperature is exactly 0 K. No quantum zero-point corrections are considered in this introductory context. Concept / Approach: For an ideal gas, mode energies scale with temperature: translational, rotational, and vibrational energies all go to zero as T → 0 K in the classical limit. Therefore, the gas possesses no thermal energy in any mode at 0 K. Step-by-Step Solution: Recognise that for classical modes, U ∝ T.At T = 0 K, U = 0 for translational, rotational, and vibrational contributions.Hence, none of the listed single-mode energies are present. Verification / Alternative check: Quantum mechanics introduces zero-point energy in bound systems (e.g., vibrations), but the classical ideal-gas model used in basic thermodynamics ignores such effects for free molecules. Why Other Options Are Wrong: Translational/Rotational/Vibrational only: Each implies residual thermal energy at 0 K, which is not the classical result. Common Pitfalls: Conflating classical and quantum results; for basic chem eng thermodynamics, the classical “no thermal energy at 0 K” framing is intended. Final Answer: None of these

Question 20

Process recycle and purge: In a continuous process with recycle, the main purpose of introducing a purge stream is to

Accepted Answer

Introduction / Context: Recycle loops improve conversion and energy efficiency, but non-reactive or slowly removed components (inerts, impurities, byproducts) can accumulate. A purge stream prevents their runaway build-up, stabilising composition and safeguarding equipment. Given Data / Assumptions: Steady-state continuous process with recycle. Presence of inert species or trace impurities in the loop. Primary aim is stable loop composition and controllability. Concept / Approach: The total mass balance with species accumulation in a loop shows that, without a purge, any unconsumed inert will trend upward toward 100% in the recycle. A small purge bleed ensures a finite residence time for inerts, establishing a steady-state concentration consistent with process targets. Step-by-Step Solution: Identify species that are not consumed or removed in separators.Recognise accumulation in a closed loop without bleed.Introduce purge flow F_purge to set an effective removal term for inerts.Solve steady-state species balance to determine acceptable purge rate. Verification / Alternative check: Typical ammonia and methanol loops include purges to limit argon, methane, and other inerts; operating data corroborate the necessity of purging to maintain catalyst performance and heat-integration targets. Why Other Options Are Wrong: Increase yield/enrich product: Purging removes valuable reactants too; it is a compromise, not a yield booster. Conserve heat: Purge can penalise heat recovery rather than conserve it. Common Pitfalls: Oversizing the purge wastes reactants; undersizing allows inerts to accumulate and reduce loop effectiveness. Final Answer: Limit the build-up of inerts and unwanted species

Stoichiometry Questions