Question 1

Dissolution energetics: On which factors does the heat of solution depend most strongly?

Accepted Answer

Introduction / Context: The heat of solution (enthalpy of solution) quantifies thermal effects when a solute dissolves in a solvent. It is a central concept for designing dissolution, crystallization, and extraction processes, as it affects temperature control, energy balances, and safety (hot or cold packs are practical examples). Given Data / Assumptions: Dissolution occurs without chemical reaction beyond solvation/ionization. Usual concentration ranges where non-ideal effects may be significant. Ambient pressure variations have negligible enthalpy effect. Concept / Approach: The heat of solution reflects competing energy changes: breaking solute–solute and solvent–solvent interactions (endothermic) versus forming solute–solvent interactions (exothermic). The balance depends on both the nature of the solute (ionic vs molecular, lattice enthalpy) and the solvent (polarity, hydrogen bonding). Concentration matters because local structure and activity coefficients change with composition, altering incremental heats of dilution and solution. Step-by-Step Discussion: Nature of solute: ions with high lattice energies can yield endothermic dissolution unless hydration compensates.Nature of solvent: polar protic solvents stabilize ions; nonpolar solvents solvate nonpolar solutes better.Concentration: at low concentration, heat of solution may approach a limiting value; at higher concentrations, heats of dilution and association can change sign/magnitude. Verification / Alternative check: Tabulated differential heats of solution show composition dependence; comparing NaCl in water vs ethanol demonstrates solvent effects; comparing NaCl vs KNO3 in water demonstrates solute effects. Why Other Options Are Wrong: Each single-factor option is incomplete; all listed factors contribute significantly. Common Pitfalls: Assuming heats of solution are constant; ignoring strong composition dependence; overlooking solvent structural effects. Final Answer: All of (a), (b), and (c).

Question 2

Gas solubility law identification: “The equilibrium mole fraction (or concentration) of a gas dissolved in a liquid is directly proportional to the partial pressure of that gas above the liquid.” This statement is ________ law.

Accepted Answer

Introduction / Context: Quantifying how gases dissolve in liquids is essential for absorption columns, carbonation of beverages, aeration in bioreactors, and environmental modeling. The classic relationship between dissolved gas concentration and its partial pressure in the contacting gas is Henry’s law, valid at sufficiently low solubility and moderate pressures for many systems. Given Data / Assumptions: Low to moderate gas solubility; dilute solution in the liquid phase. Constant temperature; Henry’s constant depends strongly on temperature. No chemical reaction between gas and solvent (otherwise, modified forms apply). Concept / Approach: Henry’s law: x_gas,liq = H_p^−1 * p_gas (pressure-based form) or C = H_c * p depending on convention. It states linear proportionality between dissolved concentration (or mole fraction) and gas partial pressure at a given temperature. Raoult’s law covers vapor–liquid equilibrium of volatile solvents (liquid-phase activity proportional to mole fraction). Amagat’s law applies to gas mixture volumes, not dissolution. Step-by-Step Solution: Match the wording “directly proportional to partial pressure” to Henry’s law.Exclude Raoult’s (relates solvent partial pressure to liquid mole fraction) and Amagat’s (additivity of gas volumes).Conclude the statement is Henry’s law. Verification / Alternative check: Plotting dissolved oxygen vs oxygen partial pressure at fixed temperature produces a straight line through the origin under Henry’s regime. Why Other Options Are Wrong: Raoult’s concerns volatile solvents, not a sparingly soluble gas in a liquid. Amagat’s concerns gas volumes, not solubility. “None of these” is incorrect because Henry’s law fits exactly. Common Pitfalls: Confusing Henry’s with Raoult’s; mixing units and definitions of Henry’s constant (several conventions exist). Final Answer: Henry’s law.

Question 3

Adiabatic flame temperature (idealized): Methane combusts with the stoichiometric amount of air: CH4 + 2 O2 → CO2 + 2 H2O (all species in the gas phase). Given ΔH°298 = −730 kJ per mole CH4 and an average Cp of 40 J·mol^−1·K^−1 for all gases, estima…

Accepted Answer

Introduction / Context: The adiabatic flame temperature (or maximum temperature rise) is a cornerstone estimate in combustion design. It assumes that all heat released by the reaction raises the sensible enthalpy of the products without losses, dissociation, or radiation. This problem requests an approximate ΔT using a constant average heat capacity for simplicity. Given Data / Assumptions: Reaction: CH4 + 2 O2 → CO2 + 2 H2O (g). ΔH°298 = −730 kJ per mol CH4 (heat released). Stoichiometric air: O2 provided by air; N2 accompanies O2 with N2/O2 ≈ 79/21. Average Cp for all gases: 40 J·mol^−1·K^−1 = 0.040 kJ·mol^−1·K^−1. Adiabatic, no dissociation, constant Cp approximation. Concept / Approach: Energy balance for an adiabatic reactor: total heat released by reaction equals sensible heat gain of products. Compute total moles of gaseous products, multiply by average Cp, and solve for ΔT = (−ΔH_reaction) / (Σ n_p Cp,avg). Temperature rise in °C equals rise in K for differences. Step-by-Step Solution: Stoichiometric air provides 2 mol O2 and accompanying N2 = 2 * (79/21) ≈ 7.52 mol.Product moles: CO2 (1) + H2O (2) + N2 (7.52) = 10.52 mol per mol CH4.Total Cp of products: 10.52 mol * 0.040 kJ·mol^−1·K^−1 = 0.4208 kJ·K^−1.ΔT ≈ 730 kJ / 0.4208 kJ·K^−1 ≈ 1735 K ≈ 1735 °C. Verification / Alternative check: The result lies near textbook adiabatic flame temperatures for methane–air when using low average Cp and ignoring dissociation; more detailed models (temperature-dependent Cp, H2O condensation at low T, dissociation) reduce the estimate. Why Other Options Are Wrong: 1225, 1335, 1525 °C underestimate ΔT given the stated −730 kJ/mol and low Cp assumption. Common Pitfalls: Forgetting to include nitrogen from air in heat capacity; mixing units (J vs kJ); using liquid-water heat of reaction instead of all-vapor basis. Final Answer: 1735

Question 4

Strong-acid concentration to pH conversion: What is the pH of an aqueous solution that contains 1 gram of hydrogen ions per litre of solution?

Accepted Answer

Introduction / Context: pH is defined as the negative base-10 logarithm of the hydronium (hydrogen ion) molar concentration: pH = −log10[H+]. This problem directly connects mass concentration of hydrogen ions to molarity and then to pH—a frequent quick check in acid–base calculations and titration problems. Given Data / Assumptions: Hydrogen ion mass in 1 L: 1 g H+ per litre. Molar mass of H+ ≈ 1 g·mol^−1. Strong-acid context with full dissociation and activity ≈ concentration. Concept / Approach: Convert grams per litre to moles per litre to find [H+]. Then apply the definition of pH. For high concentrations (≥ 1 M), activity coefficients deviate from unity; however, the textbook convention for such problems uses concentration directly unless stated otherwise. Step-by-Step Solution: Compute moles of H+: n = 1 g / (1 g·mol^−1) = 1 mol.Molarity: [H+] = 1 mol / 1 L = 1.0 M.pH = −log10(1.0) = 0. Verification / Alternative check: On the pH scale, [H+] = 10^0 M gives pH 0; [H+] = 10^−1 M gives pH 1, etc. Why Other Options Are Wrong: pH 1 corresponds to 0.1 M acid; pH 7 is neutral water; pH 10 is basic. Common Pitfalls: Confusing grams per litre with molarity; forgetting the logarithm base 10; considering mole of H2 instead of H+. Final Answer: 0

Question 5

Crystallization material balance: An aqueous solution (total 330 kg) contains 80 kg Na2SO4 (M = 142). Upon cooling, 80 kg of Na2SO4·10H2O crystals form and are removed. Determine the weight fraction of Na2SO4 remaining in the mother liquor.

Accepted Answer

Introduction / Context: Crystallization problems require careful component balances, especially when hydrates form. Here, sodium sulfate crystallizes as the decahydrate (Na2SO4·10H2O). We must track how much anhydrous salt leaves in the crystals and what remains dissolved, then compute the composition of the remaining mother liquor. Given Data / Assumptions: Initial solution mass: 330 kg, containing 80 kg Na2SO4 (anhydrous basis). Crystals removed: 80 kg Na2SO4·10H2O (molar mass 142 + 10*18 = 322 kg/kmol). No other losses; mother liquor is what remains after crystal removal. Concept / Approach: First determine how much anhydrous Na2SO4 is contained within the crystals using stoichiometric ratios from molecular weights. Subtract that from the initial Na2SO4 to find what remains dissolved. The mass of mother liquor equals initial total mass minus mass of crystals removed. Finally, compute the weight fraction of Na2SO4 in the mother liquor. Step-by-Step Solution: Moles fraction conversion: mass fraction of Na2SO4 in crystals = 142/322.Na2SO4 in 80 kg crystals = (142/322)*80 ≈ 35.25 kg.Na2SO4 remaining dissolved = 80 − 35.25 ≈ 44.75 kg.Mass of mother liquor = 330 − 80 = 250 kg.Weight fraction Na2SO4 in mother liquor = 44.75 / 250 ≈ 0.179 ≈ 0.18. Verification / Alternative check: The water tied up in crystals is (180/322)*80 ≈ 44.7 kg; plausibly removed from the initial solution’s water balance, leaving 250 kg mother liquor as computed. Why Other Options Are Wrong: 0.00 and 1.00 are extreme and contradict mass balance. 0.24 overestimates the remaining solute fraction compared with the stoichiometric removal into crystals. Common Pitfalls: Forgetting that decahydrate contains only a fraction of Na2SO4 by mass; subtracting 80 kg directly from the initial Na2SO4; ignoring that the mother liquor mass is the initial total minus crystal mass. Final Answer: 0.18

Question 6

In thermodynamics of real gases, the temperature at which a real gas most closely obeys the ideal gas law across a wide range of pressure is called the Boyle temperature. What is this term known as?

Accepted Answer

Introduction / Context: The ideal gas law PV = nRT is an approximation that becomes exact only when intermolecular forces are negligible. Real gases deviate from ideality due to attractions and repulsions. There exists a special temperature at which the first departures from ideality cancel out over a range of pressures, giving behavior that closely tracks the ideal gas law. This temperature is known as the Boyle temperature and is a standard concept in physical chemistry and chemical engineering thermodynamics. Given Data / Assumptions: Single-component real gas considered. Behavior evaluated over a moderate pressure range near low pressure. Use of virial equation to quantify deviation from ideality. Concept / Approach: Real-gas behavior can be represented by the virial equation of state: Z = PV/(nRT) = 1 + B(T)P/RT + C(T)(P/RT)^2 + … . The second virial coefficient B(T) captures the dominant effects of intermolecular forces at low to moderate pressures. The Boyle temperature TB is defined by B(TB) = 0. At T = TB, the compressibility factor Z ≈ 1 over a finite pressure range because the first correction term vanishes, causing the gas to mimic ideal behavior better than at other temperatures. Step-by-Step Solution: Write virial expansion: Z = 1 + B(T)P/RT + higher terms.Define Boyle temperature TB such that B(TB) = 0.At T = TB, Z ≈ 1 for small to moderate P because the leading correction is zero.Therefore, the temperature sought is the Boyle temperature. Verification / Alternative check: From pair potential models (e.g., Lennard-Jones), B(T) changes sign with T; a specific T exists where attractions and repulsions balance so that the second virial term vanishes. Experimental compressibility plots Z vs. P at TB pass nearly through Z = 1 with minimal slope, confirming the definition. Why Other Options Are Wrong: Reduced: refers to dimensionless variables (Pr, Tr). Critical: associated with phase transition at Pc, Tc. Inversion: relates to Joule–Thomson coefficient changing sign. Pseudo-critical: used for gas mixtures, not this concept. Common Pitfalls: Confusing Boyle temperature with inversion temperature; the former concerns virial behavior near low pressure, the latter concerns throttling temperature effects. Final Answer: Boyle

Question 7

Gas mixture at 25 °C: Equal masses of oxygen (O2) and methane (CH4) are placed in an empty vessel. What fraction of the total pressure is due to oxygen?

Accepted Answer

Introduction / Context: When different gases occupy the same container at the same temperature, Dalton’s law of partial pressures states that each gas exerts a pressure proportional to its mole fraction. Therefore, converting given mass information into moles is the key step to finding the pressure fraction of a component in a gas mixture. Given Data / Assumptions: Equal masses of O2 and CH4. Ideal-gas behavior at 25 °C. Molar masses: O2 = 32 g/mol, CH4 = 16 g/mol. Concept / Approach: Partial pressure fraction equals mole fraction for ideal gases at a common T and V. With equal masses, species with lower molar mass contributes more moles. Compute the mole ratio n_O2 : n_CH4 from the mass ratio and molar masses, then find the oxygen mole fraction x_O2 and hence the fraction of total pressure exerted by oxygen. Step-by-Step Solution: Let the equal mass be m for each component.n_O2 = m/32, n_CH4 = m/16 = 2m/32.Total moles n_T = m/32 + 2m/32 = 3m/32.Mole fraction x_O2 = (m/32)/(3m/32) = 1/3.By Dalton’s law, p_O2/P_total = x_O2 = 1/3. Verification / Alternative check: Assume m = 32 g. Then n_O2 = 1 mol, n_CH4 = 2 mol, total = 3 mol. Oxygen’s fraction is 1/3 regardless of absolute pressure or temperature (as long as both gases share T and V and behave ideally). Why Other Options Are Wrong: 2/3 corresponds to methane’s mole fraction in this setup, not oxygen’s. 1/2 would require equal moles, which is not true for equal masses here. 1/4 or 3/5 do not match the stoichiometric mole split derived from the molar masses. Common Pitfalls: Using mass fractions instead of mole fractions in gas-mixture partial pressure calculations; forgetting that lighter gases contribute more moles for the same mass. Final Answer: 1/3

Question 8

Clausius–Clapeyron relation: For which types of phase-change processes is the Clausius–Clapeyron equation applicable?

Accepted Answer

Introduction / Context: The Clausius–Clapeyron equation connects the slope of a phase boundary in the pressure–temperature plane to the latent heat and volume change of the transition. It is fundamental for predicting how saturation pressure varies with temperature and applies to any two-phase equilibrium involving a first-order phase transition. Given Data / Assumptions: Two phases in equilibrium (e.g., solid–vapor, solid–liquid, liquid–vapor). Latent heat and molar volume change defined at the coexistence condition. Reversible, equilibrium phase change. Concept / Approach: The general form is dP/dT = ΔH_trans / (T ΔV_trans), where ΔH_trans is the molar latent heat for the specific transition and ΔV_trans is the change in molar volume between phases. This derivation makes no restriction on which pair of phases is considered, so it holds for sublimation (solid–vapor), melting/fusion (solid–liquid), and vaporisation (liquid–vapor). The “Clausius” simplification often refers to liquid–vapor with ΔV ≈ V_vapor and ideal vapor behavior, but the base Clapeyron equation itself is universal to first-order transitions. Step-by-Step Solution: Recognize that Clapeyron comes from equality of Gibbs free energy of two phases and its temperature/pressure derivatives.Identify that ΔH and ΔV are pathway-independent state changes at coexistence.Conclude the relation applies to any two-phase equilibrium line (solid–liquid, solid–vapor, liquid–vapor).Therefore the correct choice is “all (a), (b) & (c).” Verification / Alternative check: Textbook P–T phase diagrams show slope predictions (e.g., positive for most melting curves, negative for water’s solid–liquid line near 0 °C), all derived from the Clapeyron equation. Why Other Options Are Wrong: Choosing only one phase change ignores the general derivation that does not depend on which two phases are in equilibrium. None of these is contradicted by the standard proof. Common Pitfalls: Confusing the general Clapeyron equation with the “Clausius” ideal-vapor approximation, which is a special case for vaporisation only. Final Answer: all (a), (b) & (c)

Question 9

According to Kirchhoff’s equation in thermochemistry, the heat of reaction depends primarily on which variable (through heat capacities)?

Accepted Answer

Introduction / Context: Standard heats of reaction are tabulated at reference temperatures (often 298 K), but process conditions may differ. Kirchhoff’s equation provides a way to adjust the heat of reaction from one temperature to another using heat capacities, enabling accurate energy balances at operating conditions. Given Data / Assumptions: Heats of formation and reaction are state functions. Heat capacities Cp(T) of reactants and products are known or approximated. Pressure effects on heats are usually negligible for condensed phases and small for gases over moderate ranges. Concept / Approach: Kirchhoff’s equation states that d(ΔH_rxn)/dT = ΔCp, where ΔCp is the sum of heat capacities of products minus reactants. Integrating between T1 and T2 gives ΔH_rxn(T2) = ΔH_rxn(T1) + ∫_(T1→T2) ΔCp dT. Thus, temperature is the controlling variable for adjusting heats of reaction via heat capacities. Neither reaction order nor molecularity appears in the thermodynamic relation, and pressure effects are secondary unless very high pressures are involved for gases. Step-by-Step Solution: Write differential form: d(ΔH_rxn)/dT = ΔCp.Integrate over temperature interval to obtain ΔH_rxn(T2).Conclude that temperature changes alter the heat of reaction through ΔCp.Hence, the correct controlling variable is temperature. Verification / Alternative check: Using tabulated Cp polynomials, engineers routinely correct ΔH_rxn from 298 K to reactor temperature; these calculations match calorimetric data, validating the temperature dependence predicted by Kirchhoff’s equation. Why Other Options Are Wrong: Pressure/Volume: may affect reaction enthalpy slightly for gases but are not the basis of Kirchhoff’s relation. Molecularity/Reaction order: kinetic concepts; do not enter thermodynamic state property adjustments. Common Pitfalls: Confusing temperature correction (thermodynamics) with rate changes (kinetics); neglecting ΔCp variation with temperature when large ranges are involved. Final Answer: temperature

Question 10

At 25 °C, a solution has equal concentrations of hydrogen ions [H+] and hydroxide ions [OH−]. What will be the pH of this solution?

Accepted Answer

Introduction / Context: The pH scale quantifies acidity via hydrogen ion concentration, while pOH relates to hydroxide ions. At 25 °C, the ion product of water Kw equals 1.0 × 10^-14, leading to a convenient relationship between pH and pOH. Understanding neutrality and the meaning of equal [H+] and [OH−] is foundational in aqueous chemistry, biochemistry, and environmental engineering. Given Data / Assumptions: Temperature is 25 °C (298 K). Neutral water satisfies [H+][OH−] = Kw = 1.0 × 10^-14. The solution has [H+] = [OH−]. Concept / Approach: If [H+] = [OH−] and their product equals Kw, then each must be 1.0 × 10^-7 M. pH is defined as −log10[H+]. Therefore, pH = 7. This is the conventional neutrality point at 25 °C. Note that neutrality shifts slightly with temperature since Kw depends on T; however, for the stated 25 °C, pH 7 is exactly neutral. Step-by-Step Solution: Write Kw = [H+][OH−] = 1.0 × 10^-14 at 25 °C.Given [H+] = [OH−] ⇒ [H+]^2 = 1.0 × 10^-14.Solve: [H+] = 1.0 × 10^-7 M.Compute pH = −log10(1.0 × 10^-7) = 7. Verification / Alternative check: pOH = −log10[OH−] = 7; pH + pOH = 14 at 25 °C, matching the standard relation derived from Kw. Why Other Options Are Wrong: 0 or 14 are extreme acid/base values, not neutral. 10 or 5 indicate basic or acidic conditions, respectively, which contradict equal [H+] and [OH−]. Common Pitfalls: Forgetting that neutrality is temperature-dependent; at temperatures other than 25 °C, neutral pH is not exactly 7, but here 7 is correct by definition. Final Answer: 7

Question 11

Avogadro’s law application: If 1 Nm³ of O2 contains N molecules at given conditions, how many molecules are present in 2 Nm³ of SO2 at the same T and P?

Accepted Answer

Introduction / Context: Avogadro’s law states that equal volumes of any gases at the same temperature and pressure contain equal numbers of molecules, regardless of chemical identity. This principle allows direct proportionality between gas volume and number of molecules, forming the basis for molar volume and normal-cubic-meter calculations in industry. Given Data / Assumptions: Both gases compared at the same temperature and pressure. Volumes are “normal” cubic meters (Nm³), i.e., referenced to standard conditions. Ideal-gas approximation holds sufficiently well at these conditions. Concept / Approach: Number of molecules is proportional to volume at fixed T and P: n ∝ V. If 1 Nm³ contains N molecules, then any gas occupying 2 Nm³ at the same T and P contains 2N molecules, independent of its chemical composition (O2 vs. SO2). Step-by-Step Solution: Use Avogadro’s law: n1/n2 = V1/V2 at the same T, P.Given 1 Nm³ O2 → N molecules.For 2 Nm³ SO2: molecule count = 2 × N = 2N.Chemical identity does not affect the count at fixed T, P, and V. Verification / Alternative check: At standard conditions, one kilomole of any ideal gas occupies ~22.414 Nm³. Thus, molecules per Nm³ are fixed by Avogadro’s number, independent of gas species; doubling volume doubles molecules. Why Other Options Are Wrong: N implies equal volumes, but here volume is doubled. N/2 or N/4 contradict the direct proportionality to volume. 4N would require quadrupling the volume, not doubling. Common Pitfalls: Thinking heavier gases contain fewer molecules at the same volume; mass differs, but molecule count per volume at fixed T and P is the same for all gases under ideal behavior. Final Answer: 2N

Question 12

Humidity terminology: The ratio of the existing moles of vapor per mole of vapor-free gas to the moles that would be present if the mixture were saturated at the same T and P is called

Accepted Answer

Introduction / Context: In gas–vapor systems (such as air–water vapor), several saturation measures are used in mass-transfer and drying calculations. Distinguishing relative humidity, relative saturation, and percentage saturation avoids confusion and ensures correct graphical and computational use of psychrometric charts. Given Data / Assumptions: Gas–vapor mixture at specified temperature and pressure. Vapor is the condensable species; the carrier gas is non-condensable. Definitions based on molar ratios (vapor per mole of dry gas). Concept / Approach: Relative humidity (on a mole ratio basis) is defined as RH = (y / y_s), where y is the actual moles of vapor per mole of vapor-free gas and y_s is the saturation moles per mole of vapor-free gas at the same temperature and pressure. By contrast, relative saturation and percentage saturation involve ratios based on partial pressures and include corrections for total pressure and non-idealities; these often differ numerically from simple RH, especially at higher humidities or pressures. Step-by-Step Solution: Identify the definition in the prompt: existing vapor moles per mole of dry gas divided by saturated vapor moles per mole of dry gas.Map to standard terminology: this equals relative humidity by mole ratio.Therefore, choose “relative humidity.”Note: percentage saturation = (y / y_s) × (1 − y_s) / (1 − y) × 100%, which differs from simple RH. Verification / Alternative check: Psychrometric relations show RH equals the ratio of actual to saturation partial pressures under ideal gas assumptions; since y ∝ p_v/(P − p_v), using the same T, P definition recovers the mole-ratio form given above. Why Other Options Are Wrong: Relative saturation and percentage saturation use different denominators or correction factors; they are not the simple mole-ratio definition cited. None of these is invalid because the described ratio matches the standard RH definition. Humid volume is a property (mixture volume per mass of dry gas), not a saturation ratio. Common Pitfalls: Interchanging RH with percentage saturation; always check whether the ratio is defined on a mole basis and whether corrections for total pressure are included. Final Answer: relative humidity

Question 13

At the boiling point, a liquid is in equilibrium with its vapor. On average, which energy measure is equal for molecules in the two phases at that temperature?

Accepted Answer

Introduction / Context: Phase equilibrium at the boiling point means liquid and vapor coexist at the same temperature and pressure with no net mass transfer on average. Molecular interpretations help explain macroscopic equilibrium: temperature reflects the average translational kinetic energy of molecules, while potential energy reflects interactions (which differ across phases). Given Data / Assumptions: Liquid–vapor equilibrium at the boiling temperature. System is at steady state with no net heat or mass accumulation. Equipartition of translational energy linking temperature and average kinetic energy. Concept / Approach: Temperature is directly proportional to the average translational kinetic energy of molecules. At equilibrium, both phases have the same temperature; therefore, the molecules in liquid and in vapor have the same average kinetic energy. However, potential energy differs: vapor molecules are farther apart with higher potential energy on average due to weaker cohesive forces, which is why latent heat is required to vaporize the liquid without changing temperature. Step-by-Step Solution: Recognize: T is common to both phases at equilibrium.Average kinetic energy ∝ T; hence equal in both phases.Potential energy differs because intermolecular spacing and interactions differ between liquid and vapor.Thus, choose “kinetic energy.” Verification / Alternative check: Thermodynamically, enthalpy differs by the latent heat h_fg at the same T, which is largely due to a change in potential energy and configuration rather than kinetic energy differences; this supports the conclusion that kinetic energies are equal while total energies are not. Why Other Options Are Wrong: Potential energy and intermolecular forces are not equal across phases; they differ and account for latent heat. Total energy includes potential contributions and thus is not equal. Vibrational energy only is not the defining equal measure across phases at the same T; the key equality concerns average translational kinetic energy. Common Pitfalls: Assuming “same temperature” implies all forms of energy are identical; only the average translational kinetic energy correlates uniquely with temperature in this context. Final Answer: kinetic energy

Question 14

Solubility versus temperature: For most salts, solubility increases with temperature. Which of the following has solubility that is nearly independent of temperature?

Accepted Answer

Introduction / Context: Understanding how solubility varies with temperature is crucial in crystallization, evaporation, and precipitation operations. While many salts show significant solubility increases with temperature (endothermic dissolution), some exhibit weak dependence, and a few show retrograde behavior (solubility decreases at higher T). Recognizing patterns helps in selecting operating temperatures to control crystal size and yield. Given Data / Assumptions: Aqueous solutions under atmospheric pressure. Solubilities compared over common process temperatures (e.g., 0–100 °C). Qualitative trend recognition rather than exact numeric values. Concept / Approach: Sodium chloride (NaCl) is a classic example of a salt whose solubility in water changes only slightly with temperature—its dissolution enthalpy is small, so temperature has little effect. In contrast, potassium nitrate (KNO3) shows a strong positive temperature dependence (used in hot crystallization). Sodium thiosulphate (hypo) and sodium carbonate hydrates exhibit more pronounced changes. Anhydrous sodium sulfate displays complex hydration equilibria (Glauber’s salt formation) that introduce marked temperature effects. Step-by-Step Solution: Identify the salt with minimal temperature sensitivity: NaCl.Contrast: KNO3 has strongly increasing solubility with T.Hydrated salts (e.g., Na2CO3·H2O) and thiosulfate show larger T effects due to hydration equilibria.Therefore, the correct selection is sodium chloride. Verification / Alternative check: Reference solubility curves show NaCl rising from roughly 35–39 g/100 g water between 0 and 100 °C—a modest change compared with KNO3 (about 13 → 240 g/100 g water) or sodium thiosulfate with large increases and hydrates changing stability with T. Why Other Options Are Wrong: Sodium carbonate monohydrate: hydration/dehydration affects solubility markedly with T. Anhydrous sodium sulphate: exhibits complex hydrated phases; solubility varies strongly. Hypo (sodium thiosulfate): shows a strong positive T dependence. Potassium nitrate: textbook case of strong solubility increase with T. Common Pitfalls: Assuming all salts follow the same trend; always check specific solubility curves and hydrate formation behavior. Final Answer: sodium chloride

Question 15

Surface phenomena: How does the surface tension of water change as temperature increases (within the liquid range)?

Accepted Answer

Introduction / Context: Surface tension reflects the energetic cost of creating new surface area and is governed by intermolecular cohesive forces. In applications such as distillation, bubble formation, atomization, coating, and detergency, knowledge of how temperature affects surface tension is essential for predicting droplet size, foaming, and wetting behavior. Given Data / Assumptions: Pure water over its liquid temperature range well below the critical point. Atmospheric pressure; no surfactants or impurities. Concept / Approach: As temperature rises, thermal agitation weakens the net cohesive forces at the liquid surface. The free energy difference between molecules at the surface and in the bulk diminishes, reducing the surface tension. Empirically, water’s surface tension decreases monotonically with temperature and approaches zero at the critical point where the distinction between liquid and vapor disappears. Step-by-Step Solution: Identify the trend: increasing T disrupts hydrogen bonding coherence at the interface.Consequent effect: less energy required to create surface area → lower surface tension.Therefore, as temperature increases, surface tension decreases.This qualitative rule holds for most liquids, including water. Verification / Alternative check: Data show ~72 mN/m at 20 °C falling to ~59 mN/m near 100 °C for water, confirming the strong negative slope with temperature. Why Other Options Are Wrong: Increases or increases linearly: contradict empirical data and molecular reasoning. Remains constant: not supported; the change is significant across typical process temperatures. Oscillates: no such periodic behavior exists for pure water. Common Pitfalls: Ignoring contamination or surfactants, which can drastically lower surface tension even at fixed T; the question concerns pure water only. Final Answer: decreases

Question 16

Thermodynamics vocabulary — vapor–liquid equilibrium: The pressure at which a liquid and its vapor can coexist in equilibrium at a given temperature is called the __________ vapor pressure.

Accepted Answer

Introduction / Context: In chemical engineering thermodynamics and phase equilibria, precise terminology is crucial. When a pure liquid and its vapor coexist without net evaporation or condensation at a fixed temperature, the system is at vapor–liquid equilibrium (VLE). The pressure of the vapor in equilibrium with its liquid at that temperature has a standard name that appears in phase diagrams, Antoine-equation correlations, and design of separation operations such as distillation and evaporation. Given Data / Assumptions: Pure substance at a specified temperature. Liquid and vapor phases are present simultaneously, with no net mass transfer (equilibrium). No capillarity or curvature effects (bulk phases), and no noncondensable gases. Concept / Approach: For a pure component, the equilibrium pressure of the vapor above its liquid at temperature T is the saturated vapor pressure. This value increases monotonically with temperature and reaches the external pressure at the normal boiling point (when saturated vapor pressure equals 1 atm). In correlations such as the Antoine equation, log10 Psat is expressed as a function of 1/T, enabling engineers to estimate phase-change conditions and driving forces for mass transfer. Step-by-Step Solution: Identify the scenario: liquid and vapor in equilibrium at fixed T.The term used universally for the pressure of the vapor in equilibrium with its liquid is “saturated vapor pressure.”Therefore, the blank should be filled with “saturated.” Verification / Alternative check: At the normal boiling point, the saturated vapor pressure equals 1 atm; this is consistent with textbook definitions and phase diagrams where the liquid–vapor boundary (vapor pressure curve) is labeled Psat(T). Why Other Options Are Wrong: Limiting: not a standard term for VLE pressure.Real: ambiguous; all physical pressures are “real,” but the term does not denote equilibrium vapor pressure.Normal: “normal boiling point” is defined at 1 atm; “normal vapor pressure” is not a standard term. Common Pitfalls: Confusing saturated vapor pressure (equilibrium property at a given T) with partial pressure in a mixture, or with absolute/atmospheric pressure. Also, do not mix up “normal boiling point” (a temperature) with “vapor pressure” (a pressure). Final Answer: saturated

Question 17

Raoult’s law application at room conditions: A saturated aqueous solution of benzene (liquid mole fraction of benzene = xB) is in equilibrium with an air–vapor mixture at room temperature. If the pure-component vapor pressures at these conditions ar…

Accepted Answer

Introduction / Context: Determining partial pressures of volatile components above solutions is fundamental in vapor–liquid equilibrium, environmental emissions estimation, and absorber/stripper design. For ideal or near-ideal solutions at low solute concentrations, Raoult’s law provides a simple, widely used relation between the liquid composition and the partial pressure of a volatile solvent/solute in the gas phase. Given Data / Assumptions: Temperature is fixed (room temperature). Liquid phase: water containing dissolved benzene; liquid mole fraction of benzene is xB. Gas phase: air plus water vapor plus benzene vapor. Pure-component vapor pressures at the same temperature: pvB (benzene), pvw (water). Ideal solution approximation for benzene in water at low xB. Concept / Approach: Raoult’s law states: partial pressure of component i above a liquid solution is p_i = x_i · p_i^sat, where x_i is the mole fraction of i in the liquid, and p_i^sat is the pure-component saturated vapor pressure at the same temperature. The presence of other gases (air, water vapor) affects the total pressure but does not change the direct proportionality p_B = xB · pvB under the ideal assumption. Step-by-Step Solution: Identify benzene as the volatile component of interest.Apply Raoult’s law: p_B = xB * pvB.Therefore, the benzene partial pressure equals xB·pvB. Verification / Alternative check: Dalton’s law governs the gas mixture: P_total = p_air + p_w + p_B. Even though water contributes pvw (or close to it), benzene’s contribution is still xB·pvB if the liquid behaves ideally for benzene. Nonideality would introduce an activity coefficient γB (p_B = xB·γB·pvB), but γB ≈ 1 is assumed here. Why Other Options Are Wrong: pvB: ignores dilution in the liquid; only valid for pure benzene.(Patm − pvw)·xB: incorrectly scales by the “dry-air” pressure, not Raoult’s law.xB·Patm: total pressure is irrelevant to Raoult’s proportionality. Common Pitfalls: Confusing Raoult’s law (liquid-based) with Dalton’s law (gas-based), or forgetting that dissolved benzene’s partial pressure drops in proportion to its liquid mole fraction. Final Answer: xB·pvB

Question 18

Dry-basis composition after condensation: A gas has molar composition 10% H2, 10% O2, 30% CO2, and 50% H2O (by moles). If 50% of the water condenses, what is the final mole percent of H2 on a dry basis?

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Introduction / Context: Process gas compositions are frequently reported on a “dry basis,” excluding water vapor. When moisture is partially removed by condensation, engineers must recalculate the composition of the remaining stream, which directly affects combustion calculations, emissions reporting, and downstream separations. Given Data / Assumptions: Initial composition (mole %): H2 = 10, O2 = 10, CO2 = 30, H2O = 50. Fifty percent of the water vapor condenses and leaves; noncondensable components do not change in amount. Ideal-gas behavior; temperature and pressure changes do not create additional reactions. Dry basis means excluding water from the denominator after condensation. Concept / Approach: Select a convenient basis (e.g., 100 mol of feed gas). Remove 50% of the water by condensation. Then compute the new “dry gas” composition by dividing each noncondensable component’s moles by the total moles of the noncondensable gases (i.e., after excluding remaining water). Step-by-Step Solution: Choose a 100 mol basis: H2 = 10 mol; O2 = 10 mol; CO2 = 30 mol; H2O = 50 mol.Condense 50% of H2O: removed = 25 mol H2O; remaining H2O = 25 mol.Noncondensables remain: H2 = 10; O2 = 10; CO2 = 30.Dry basis total (exclude water): 10 + 10 + 30 = 50 mol.H2 mole fraction on dry basis = 10 / 50 = 0.20 → 20%. Verification / Alternative check: If one mistakenly divides by the wet total (including the remaining 25 mol H2O), the percentage would be 10/75 = 13.33%, which is not “dry basis.” The correct method excludes water entirely from the denominator when reporting dry-basis composition. Why Other Options Are Wrong: 10%: this is the original value (wet basis), not the dry-basis value after condensation.5% and 18.18%: arise from incorrect denominators (e.g., partially excluding components) or arithmetic mistakes. Common Pitfalls: Confusing wet-basis and dry-basis definitions; forgetting to remove the condensed water from the total; using percent of original instead of recomputed normalized percentages. Final Answer: 20%

Question 19

Phase rule reasoning for a binary system (non-reacting): For which case does the number of degrees of freedom equal the number of components in a two-component system?

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Introduction / Context: Gibbs’ phase rule, F = C − P + 2 for non-reacting systems, relates the degrees of freedom (independent intensive variables), the number of components C, and the number of phases P. Understanding when F equals C is useful for interpreting phase diagrams and designing separation processes involving partially miscible liquids. Given Data / Assumptions: Non-reacting system. Binary (C = 2) unless otherwise stated. Pressure and temperature are the standard intensive variables considered. Concept / Approach: Apply the phase rule F = C − P + 2. For a binary system (C = 2): If P = 1 (single phase), F = 2 − 1 + 2 = 3, which does not equal C.If P = 2 (two phases), F = 2 − 2 + 2 = 2, which equals C.Therefore, for a binary system, F equals the number of components when two phases coexist (e.g., two partially miscible liquids in equilibrium), giving two degrees of freedom (commonly T and overall composition at fixed pressure, or T and P with composition constrained). Step-by-Step Solution: Identify that we want F = C with C = 2.Solve F = 2 − P + 2 = 2 ⇒ −P + 4 = 2 ⇒ P = 2.Recognize that a partially miscible binary liquid system often exhibits two liquid phases at certain T.Hence, the case “partially miscible two-liquid system with two phases present” satisfies the condition. Verification / Alternative check: At a fixed pressure (e.g., 1 atm), the condensed-phase rule (F = C − P + 1) might be used; for P = 2 and C = 2, F = 1 (equal to C − 1). The original question assumes the full form, consistent with including pressure as a variable. Why Other Options Are Wrong: Only soluble liquid components: P = 1 → F = 3 ≠ 2.Two liquids plus an extra solute: this is C = 3, not a binary system; it changes the equality sought.None of these: incorrect because the two-liquid, two-phase case satisfies F = C for C = 2. Common Pitfalls: Forgetting which version of the phase rule is applied; ignoring whether the system is binary; confusing “components” with “species present in each phase.” Final Answer: Partially miscible two-liquid system with two phases present.

Question 20

Charts related to vapor-pressure correlations: Which chart is commonly associated with the Antoine (Antonie) equation representation of vapor pressure?

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Introduction / Context: Engineers often linearize empirical correlations to ease data plotting and interpolation. The Antoine equation expresses vapor pressure as a function of temperature, typically in the logarithmic form. A specific chart, widely used in older design practice and hand-calculations, presents vapor pressure–temperature data on axes that render the relation near-linear, simplifying estimation and extrapolation. Given Data / Assumptions: Antoine equation relates log10 Psat to 1/T (or to T in certain forms) with substance-specific constants. Charts are tools to visualize or linearize this relationship. Concept / Approach: The Cox chart is a semi-empirical plot that linearizes vapor-pressure data for many substances by choosing appropriate temperature and pressure scales. When Antoine (or similar) relations hold, substances appear approximately as straight lines on Cox charts, enabling quick interpolation between tabulated points without solving equations. Step-by-Step Solution: Recognize that Antoine-type correlations are customarily plotted on Cox charts.Match the chart name to the equation usage: “Cox” is the correct association. Verification / Alternative check: Process handbooks and classic design texts provide Cox charts for families of compounds along with Antoine constants; these charts were historically used before ubiquitous digital tools. Why Other Options Are Wrong: Ostwald: commonly associated with viscosity or dilution laws, not vapor pressure charts.Mollier’s (h–s) chart is for steam/thermodynamics, unrelated to Antoine equation plotting.Enthalpy–concentration: used in humidification or solutions, not specific to Antoine vapor-pressure plotting. Common Pitfalls: Confusing different thermodynamic charts; assuming any enthalpy-related chart is tied to vapor-pressure data; overlooking that Antoine is logarithmic in pressure. Final Answer: Cox

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