Question 1

Humid air enthalpy trends: For a vapour–gas (e.g., water vapour–air) mixture, the total enthalpy can be increased by which change(s)?

Accepted Answer

Introduction / Context: On a psychrometric chart and in energy balances, the enthalpy of moist air depends on both dry-bulb temperature and moisture content (humidity ratio). Understanding how each variable influences enthalpy is essential for HVAC calculations, dryers, cooling towers, and gas–vapour contacting equipment design. Given Data / Assumptions: Mixture of a non-condensable gas (air) and a condensable vapour (water). Enthalpy expressed per unit mass of dry air: h ≈ h_air(T) + ω * h_vapour(T), where ω is humidity ratio. Constant pressure near atmospheric; ideal-mixture approximations. Concept / Approach: The mixture enthalpy increases with temperature due to sensible heating of both dry air and vapour, and increases with humidity because additional vapour carries substantial latent and sensible enthalpy. Therefore, raising temperature, raising humidity, or both will increase total enthalpy. This is visible on the psychrometric chart where lines of constant enthalpy slope upward with temperature and humidity ratio. Step-by-Step Solution: h_mix = Cp_air * T + ω * (h_g(T)) approximately, per kg dry air.At constant ω, ∂h/∂T > 0 from Cp terms → h increases with T.At constant T, ∂h/∂ω = h_g(T) > 0 → h increases with humidity.If both T and ω rise, the increase is larger due to both contributions. Verification / Alternative check: Numerical examples using standard psychrometric formulas show monotonic enthalpy increase with either variable; tabulated charts confirm the same trend. Why Other Options Are Wrong: Options (a), (b), (c) are individually correct but incomplete; only “All of the above” captures every way to increase enthalpy. “None of the above” contradicts thermodynamic definitions and empirical charts. Common Pitfalls: Assuming only temperature matters; neglecting the large latent enthalpy carried by water vapour; mixing per-mass-of-dry-air and per-mass-of-mixture bases. Final Answer: All of the above

Question 2

Molecular degrees of freedom and heat capacity: For an ideal monatomic gas, the constant-volume specific heat Cv equals which multiple of the gas constant R?

Accepted Answer

Introduction / Context: Heat capacity of gases is tied to the number of accessible degrees of freedom. For monatomic ideal gases (e.g., He, Ne, Ar), only three translational degrees of freedom contribute at ordinary temperatures. Equipartition results lead directly to simple ratios between Cv, Cp, and the gas constant R. Given Data / Assumptions: Monatomic ideal gas; translational degrees of freedom = 3. Equipartition theorem: each quadratic degree contributes (1/2)R per mole to internal energy. Cp − Cv = R for ideal gases. Concept / Approach: Total molar internal energy u for a monatomic ideal gas is (3/2)R T. Therefore, Cv = (∂u/∂T)_v = (3/2)R. From Cp − Cv = R, Cp = (5/2)R and γ = Cp/Cv = 5/3 ≈ 1.667. These benchmarks are widely used in compressible-flow and thermodynamics calculations. Step-by-Step Solution: u = (3/2)R T for monatomic ideal gas.Differentiate with respect to T at constant volume: Cv = (∂u/∂T)_v = (3/2)R.Hence Cp = Cv + R = (5/2)R; confirm standard ratios.Select the correct multiple: 1.5 R. Verification / Alternative check: Experimental Cp and Cv values for He, Ne, Ar at room temperature match the (5/2)R and (3/2)R predictions within small deviations, validating the ideal monatomic model. Why Other Options Are Wrong: R: too low; would imply only two degrees or non-ideal assumptions. 2R or 3R: correspond to additional active degrees (rotations/vibrations) not present for monatomic gases at ordinary T. 0.5 R: physically impossible for monatomic ideal gases. Common Pitfalls: Applying diatomic/polyatomic heat capacities to monatomic gases; overlooking Cp − Cv = R. Final Answer: 1.5 R

Question 3

Gas solubility in water (qualitative): Among NH3, CO2, H2S, and CH4, which gas is the most soluble in liquid water under comparable conditions?

Accepted Answer

Introduction / Context: Gas solubility in water is central to scrubbing, wastewater aeration, absorption design, and environmental fate of gaseous pollutants. Chemical interactions with water, polarity, and acid–base behavior can dramatically enhance apparent solubility beyond simple physical dissolution governed by Henry’s law constants alone. Given Data / Assumptions: Gases compared: ammonia (NH3), carbon dioxide (CO2), hydrogen sulfide (H2S), methane (CH4). Comparable temperature and pressure; dilute solutions; no added electrolytes. Focus on relative solubility in pure water. Concept / Approach: NH3 is highly soluble due to its ability to hydrogen-bond and, importantly, react with water to form NH4+ and OH− (base hydrolysis), effectively consuming dissolved NH3 and increasing total ammonia species in solution. CO2 is also reactive, forming carbonic acid/bicarbonate, giving it moderate solubility. H2S exhibits weak acidity and moderate solubility. CH4 is nonpolar and exhibits very low aqueous solubility governed mostly by weak dispersion interactions, making it the least soluble of the set. Step-by-Step Solution: Consider chemical reactions: NH3 + H2O ⇌ NH4+ + OH− (favorable in water) → high apparent solubility.CO2 + H2O ⇌ H2CO3 ⇌ HCO3− + H+ → moderate enhancement.H2S ⇌ HS− + H+ → weak enhancement; limited compared to NH3.CH4: no reaction; purely physical dissolution → very low solubility. Verification / Alternative check: Henry’s law constants and published solubility data rank NH3 as highly soluble (tens of % w/w possible under pressure), CO2 and H2S as moderately soluble, and CH4 as poorly soluble at ambient conditions. Why Other Options Are Wrong: CO2, H2S: more soluble than CH4 but significantly less than NH3 due to weaker or less favorable reactions. CH4: least soluble; hydrophobic, non-reactive. N2: included as an extra distractor; also low solubility. Common Pitfalls: Comparing only physical solubility constants without accounting for acid–base reactions; assuming all gases behave inertly in water. Final Answer: NH3

Question 4

Real-gas behaviour and special temperatures: The temperature at which a real gas most closely obeys Boyle’s law over a range of pressures is called the

Accepted Answer

Introduction / Context: Real gases deviate from ideal behavior due to intermolecular forces and finite molecular size. However, at a particular temperature characteristic of each gas, second-order deviations in the pressure–volume relation vanish over a pressure range, and the gas approximately satisfies Boyle’s law (PV ≈ constant). This special temperature is known as the Boyle’s temperature. Given Data / Assumptions: Real gas described by a virial or cubic equation of state. At Boyle’s temperature, the second virial coefficient B(T) ≈ 0. We compare terminology: triple point, eutectic, inversion temperature, etc. Concept / Approach: From the virial expansion, Z = 1 + B(T)P/(RT) + … . If B(T) = 0, then Z ≈ 1 to first order in P, meaning the gas follows Boyle’s law over a finite pressure interval. This temperature, where attractive and repulsive contributions balance such that B(T) vanishes, is the Boyle’s temperature. It is distinct from inversion temperature (related to the sign change in Joule–Thomson coefficient) and from phase-equilibrium points like the triple point or eutectic point. Step-by-Step Solution: Start with virial EOS: Z = 1 + B(T)P/(RT) + …Boyle’s law requires Z ≈ 1 → set B(T) ≈ 0 at that T.The temperature satisfying B(T) = 0 is the Boyle’s temperature.Therefore, select “Boyle’s temperature”. Verification / Alternative check: Experimental P–V–T data show near-ideal compressibility at the Boyle’s temperature for many gases (e.g., N2), consistent with B(T_B) ≈ 0. Why Other Options Are Wrong: Triple point/eutectic point: phase-equilibrium points, not gas nonideality minima. Inversion temperature: pertains to Joule–Thomson expansion, not Boyle behavior. Critical temperature: where gas and liquid become indistinguishable; unrelated to B(T) = 0. Common Pitfalls: Confusing Boyle’s temperature with inversion or critical temperatures due to similar-sounding “special” temperatures. Final Answer: Boyle’s temperature

Question 5

Additivity of atomic heat (Kopp’s rule): Kopp’s rule is most useful for estimating which of the following?

Accepted Answer

Introduction / Context: Kopp’s rule provides a simple, empirical way to estimate the heat capacity of crystalline solids from their constituent elements. It relies on the approximate additivity of atomic heat capacities, and while not exact, it offers useful first estimates in materials and process calculations where precise data are unavailable. Given Data / Assumptions: Kopp’s rule applies primarily to solid inorganic compounds. Each element contributes an approximately constant atomic heat to the molar heat capacity of the solid compound. Temperature effects and structural peculiarities can cause deviations. Concept / Approach: For a solid compound A_xB_y…, the molar heat capacity at constant pressure, Cp,m, is approximated as the sum of the atomic heat capacities of the elements multiplied by their stoichiometric coefficients. This contrasts with gas-phase correlations that depend strongly on temperature and molecular degrees of freedom (translation, rotation, vibration), for which Kopp’s rule is not used. Step-by-Step Solution: Recognize that Kopp’s rule targets solids.Write Cp,m(compound) ≈ Σ ν_i * (atomic heat of element i).Select the option “Heat capacities of solids”. Verification / Alternative check: Handbooks show Kopp’s estimates often within acceptable engineering tolerance for many salts and oxides, especially at moderate temperatures, though modern data are preferred when available. Why Other Options Are Wrong: Gas heat capacities (molal or otherwise): depend on molecular degrees of freedom; Kopp’s rule is not applied. Activation energy and latent heats: unrelated physical quantities. Common Pitfalls: Using Kopp’s rule for molecular gases or polymers where it is not valid; ignoring temperature dependence. Final Answer: Heat capacities of solids

Question 6

Heats of neutralisation and water ionisation: Given the heat of neutralisation of strong acid–strong base (HCl + NaOH → NaCl + H2O) is −57.46 kJ per kg-mole of water formed, determine the heat of ionisation of water (H2O → H+ + OH−) in kJ per kg-mol…

Accepted Answer

Introduction / Context: The heat of neutralisation for strong acids and bases at infinite dilution is nearly constant (about −57 kJ per mole of water formed) because the net reaction effectively is H+ (aq) + OH− (aq) → H2O (l). The reverse process, water ionisation, requires the same magnitude of energy input but with opposite sign. This relationship connects calorimetry with acid–base thermodynamics. Given Data / Assumptions: Heat of neutralisation (strong acid/strong base): ΔH_neut ≈ −57.46 kJ per kg-mole H2O formed. Reaction considered for ionisation: H2O (l) → H+ (aq) + OH− (aq). Standard dilute conditions; heats are referenced per mole of water. Concept / Approach: Ionisation of water is the reverse of the neutralisation step. By Hess’s law, reversing a reaction changes the sign of its enthalpy change while preserving magnitude. Therefore, the heat of ionisation equals +57.46 kJ per kg-mole (endothermic), the energy required to dissociate water into its ions at the same reference conditions. Step-by-Step Solution: Neutralisation: H+ + OH− → H2O, ΔH = −57.46 kJ/kg-mol.Ionisation is the reverse: H2O → H+ + OH−.By Hess’s law: ΔH_ionisation = +57.46 kJ/kg-mol.Select the positive value reflecting endothermic dissociation. Verification / Alternative check: Textbook values for the heat of neutralisation converge near −57 kJ/mol at infinite dilution; reversing the reaction consistently yields +57 kJ/mol for ionisation, aligning with the concept that creating separated ions from neutral water requires input energy. Why Other Options Are Wrong: −57.46: sign error; that is neutralisation, not ionisation. 114.92 and −28.73: double or half the correct magnitude; no basis in the stoichiometry here. 0: contradicts the well-established thermochemistry. Common Pitfalls: Forgetting to reverse the sign when reversing the reaction; mixing per-mole and per-kg-mole conventions; ignoring dilution assumptions. Final Answer: 57.46

Question 7

Clausius–Clapeyron, humidity measures, and saturation: Which statement about vapor–liquid equilibrium and humidity definitions is incorrect under standard thermodynamic conventions?

Accepted Answer

Introduction / Context: Thermodynamic relations such as the Clapeyron and Clausius–Clapeyron equations describe how saturation pressure changes with temperature and how latent heat is embedded in the slope of the vapor-pressure curve. In gas–vapor mixtures (e.g., air–water), engineers also use measures like relative humidity and percentage saturation. Knowing exactly what each term means prevents design and operating errors in drying, humidification, and distillation systems. Given Data / Assumptions: Definitions follow standard chemical engineering/psychrometric conventions. Unsaturated mixture means partial pressure of vapor is less than the saturation pressure at the same temperature. Absolute humidity is mass of vapor per mass of dry carrier gas. Concept / Approach: Relative humidity is defined as p_v / p_sat(T). Percentage saturation also compares actual and saturation humidities but includes the ratio of total pressures and non-idealities; for ideal mixtures it is always less than relative humidity unless the system is at saturation. Therefore, equating the two for unsaturated mixtures is incorrect. The Clapeyron form dP/dT = λ / [T * (V_G − V_L)] and its integrated Clausius–Clapeyron approximation are standard. At the boiling point (where the carrier gas vanishes), absolute humidity (kg vapor/kg dry gas) diverges, effectively “infinite.” Step-by-Step Solution: Identify which statement conflicts with standard definitions.Use RH = p_v / p_sat and percentage saturation = (Y / Y_s) * 100, noting percentage saturation < RH for unsaturated ideal mixtures.Confirm other statements match accepted equations and limiting behavior at boiling. Verification / Alternative check: Psychrometric charts and derivations show percentage saturation equals relative humidity only at saturation; otherwise percentage saturation is lower. Why Other Options Are Wrong (or Right): (a) Correct relation between latent heat and vapor-pressure slope. (b) Correct: as dry gas fraction → 0 at boiling, absolute humidity → ∞. (c) Incorrect: the two measures are not numerically equal for unsaturated mixtures. (d) Correct general Clapeyron equation. Common Pitfalls: Confusing relative humidity with percentage saturation; mixing sign conventions in Clapeyron/Clausius–Clapeyron; forgetting limiting behavior at boiling where dry gas basis collapses. Final Answer: For an unsaturated vapor–gas mixture, percentage saturation equals relative saturation numerically.

Question 8

Combustion air requirements: Which type of fuel typically requires the greatest percentage of excess air to achieve complete combustion in practice?

Accepted Answer

Introduction / Context: “Excess air” is the percentage of air supplied above the stoichiometric requirement. Real burners and furnaces never mix fuel and oxidizer perfectly; some extra air ensures complete oxidation of combustibles and low CO/soot emissions. Understanding which fuels need more excess air guides burner design, stack O2 targets, and efficiency estimates. Given Data / Assumptions: Comparison of typical industrial fuels operated in air. Focus on practical plant conditions, not ideal laboratory mixing. Complete combustion target with reasonable safety margin. Concept / Approach: Mixing quality and reaction homogeneity largely dictate excess air. Gases mix at the molecular level, so they require the least excess air (often 0–10%). Liquids atomize into droplets; mixing is good but less perfect than gases (10–20% typical). Solid fuels (e.g., coal, biomass) introduce heterogeneity: variable particle size, ash, moisture, and diffusion limits around particles. As a result, solids need substantially higher excess air (20–50% or more) to ensure burnout and avoid CO generation. Step-by-Step Solution: Compare mixing behavior: gas > liquid > solid.Relate mixing quality to required excess air for complete oxidation.Conclude solids demand the highest percentage of excess air in practice. Verification / Alternative check: Combustion handbooks and boiler guidelines target higher stack O2 for coal stokers/grates than for natural gas burners, confirming higher excess-air needs for solids. Why Other Options Are Wrong: Liquids and gases achieve better mixing and atomization/dispersion, requiring less excess air. “Nuclear” does not involve oxidation of chemical fuel with air. Biomass slurries behave as solids in terms of heterogeneity but are a subset; “solid fuels” is the broad, correct category. Common Pitfalls: Assuming stoichiometric air is sufficient in real furnaces; ignoring ash, moisture, and particle burnout times for solids. Final Answer: Solid fuels.

Question 9

Stoichiometry concept check: In any chemical reaction, the “limiting reactant” determines which key outcome?

Accepted Answer

Introduction / Context: In stoichiometry, the limiting reactant is the reagent that is consumed first and thereby caps how much product can form. This concept applies across batch and flow reactors and is fundamental to yield, feed preparation, cost estimation, and waste minimization. Given Data / Assumptions: Balanced chemical equation available. Feeds may be off-stoichiometric (excess of one or more reagents). No side reactions considered for the basic definition. Concept / Approach: The equilibrium constant is a thermodynamic property determined by temperature, independent of feed ratios. The intrinsic rate constant depends on mechanism and temperature (Arrhenius), not on feed stoichiometry. In contrast, the limiting reactant defines the theoretical maximum conversion to products: once it is fully consumed, the reaction cannot proceed further regardless of other species remaining in excess. Step-by-Step Solution: Write the balanced reaction and compute molar requirements per stoichiometric coefficients.Compare actual feed molar amounts to stoichiometric needs.The reactant with the smallest “available moles / stoichiometric coefficient” ratio is limiting and sets the maximum extent. Verification / Alternative check: Material balance on the limiting species directly yields the maximum possible moles of product via stoichiometric ratios. Why Other Options Are Wrong: (a) Equilibrium constant is not controlled by feed amounts. (c) Rate constant is kinetic, not stoichiometric. (d) Selectivity may be influenced by feeds but is not the defining role of “limiting reactant.” (e) Residence time is a design variable, not defined by which reactant is limiting. Common Pitfalls: Confusing “limiting reactant” with “rate-limiting step” in mechanisms; mixing up thermodynamic (K) and kinetic (k) parameters. Final Answer: The maximum possible conversion (extent of reaction).

Question 10

Compressibility factor Z for an ideal gas: Select the correct statement about Z for an ideal gas over all temperatures and pressures.

Accepted Answer

Introduction / Context: The compressibility factor Z = PV/(nRT) measures deviation from ideal-gas behavior. For real gases, Z departs from unity, especially near condensation conditions or at high pressures/low temperatures. Understanding the exact behavior for an ideal gas avoids confusion in state estimation and property calculations. Given Data / Assumptions: Ideal-gas equation of state, PV = nRT. No interactions or excluded volume. All temperatures and pressures are mathematically admissible under the model. Concept / Approach: By definition, an ideal gas is one whose P–V–T relationship follows PV = nRT. Substituting into Z = PV/(nRT) yields Z = 1 identically. Boyle’s temperature is a real-gas concept where first-order deviations vanish (B2 ≈ 0), making Z ≈ 1 over a range, but that pertains to real gases only—not the ideal-gas postulate. Step-by-Step Solution: Start with the ideal-gas equation PV = nRT.Compute Z = PV/(nRT) = 1.Conclude Z is exactly 1 for all T and P in the ideal model. Verification / Alternative check: Virial EOS for real gases: Z = 1 + B2P/RT + …; setting all virial coefficients to zero recovers the ideal gas where Z = 1 identically. Why Other Options Are Wrong: (a), (d), (e) contradict the identity Z ≡ 1. (c) confuses real-gas Boyle’s temperature with the ideal-gas definition. Common Pitfalls: Applying real-gas intuition (e.g., Z trends) to the ideal gas; misusing Boyle’s temperature as a universal condition. Final Answer: Z equals exactly 1 at all temperatures and pressures.

Question 11

Charles’s law application at constant pressure: A gas at 0°C is cooled at constant pressure until its volume becomes half. What is the final gas temperature?

Accepted Answer

Introduction / Context: Charles’s law states that at constant pressure, the volume of a fixed mass of gas is directly proportional to its absolute temperature: V ∝ T (K). This relation is frequently used for quick estimates of temperature changes when volume changes at constant pressure in ideal-gas approximations. Given Data / Assumptions: Initial temperature T1 = 0°C = 273 K. Final volume V2 = 0.5 V1 (half the original). Ideal-gas behavior and constant pressure. Concept / Approach: From Charles’s law, V1/T1 = V2/T2. Halving the volume halves the absolute temperature. Convert temperatures to Kelvin before applying proportionality to avoid negative values and to maintain linearity of the relation. After computing T2 in Kelvin, convert back to °C for the final answer if needed. Step-by-Step Solution: Use V1/T1 = V2/T2.Set V2 = 0.5 V1 ⇒ T2 = 0.5 T1.Compute T2 = 0.5 * 273 K = 136.5 K.Convert to °C: 136.5 K − 273 = −136.5°C. Verification / Alternative check: Sanity check: Lower volume at constant pressure implies lower temperature; the result should be below 0°C but above absolute zero, which −136.5°C satisfies. Why Other Options Are Wrong: 0°C or 0 K contradict the halving relation; 0 K is unattainable and would imply zero volume. −136.5 K is invalid since Kelvin cannot be negative. −273°C would imply T = 0 K, not half of 273 K. Common Pitfalls: Using Celsius directly in proportionality; forgetting to convert back to °C after computation; proposing negative Kelvin values. Final Answer: −136.5°C.

Question 12

Concentration measures independent of T and P: In a binary liquid system, which concentration expressions are independent of temperature and pressure?

Accepted Answer

Introduction / Context: Choosing a concentration scale that is insensitive to temperature and pressure simplifies calculations when conditions vary. Scales based on mass or amount of substance (moles) rather than volume are more robust because mass and amount are invariant to ordinary T–P changes, while volume depends strongly on T and P. Given Data / Assumptions: Binary liquid solution (solute + solvent). No chemical reaction or mass loss. Comparing mass- and mole-based measures vs volume-based measures. Concept / Approach: Mass ratio (kg solute/kg solvent), mole ratio (kmol solute/kmol solvent), and molality (kmol solute per 1000 kg solvent) all use mass and/or amount of substance in numerator and denominator. Neither mass nor mole count changes with temperature or pressure (absent reaction or phase change losses). By contrast, molarity and normality use volume, which expands or contracts with T and P, making them T–P dependent. Step-by-Step Solution: Identify which definitions avoid solution volume.Confirm (a), (b), (c) use mass and/or moles only.Conclude all listed measures are T–P independent. Verification / Alternative check: Thermodynamics texts define molality as moles of solute per kilogram of solvent; it is widely used when temperature varies because it remains constant with thermal expansion. Why Other Options Are Wrong: “None of the above” is incorrect since all three listed are independent of T and P. Common Pitfalls: Confusing molality with molarity; assuming “normality” is T–P independent even though it relies on volume through equivalents per liter. Final Answer: All of the above.

Question 13

Dew point and saturation definitions: If the partial pressure of the solvent vapor equals its saturation pressure at the same temperature, the system is at its ________.

Accepted Answer

Introduction / Context: In gas–vapor systems (e.g., moist air), key temperatures define onset conditions of phase change at specified pressures or compositions. The two most common are the dew point and bubble point. Correctly identifying which condition is met by a given equality prevents misinterpretation in humidification, drying, and condenser design. Given Data / Assumptions: Carrier gas with a condensable vapor (solvent). Equality p_v = p_sat(T) at the same temperature. Total pressure fixed. Concept / Approach: For a vapor–gas mixture, when the vapor’s partial pressure equals the saturation pressure at the mixture temperature, the vapor is just saturated and will begin to condense upon any further cooling. The temperature at which this occurs, for a fixed vapor partial pressure, is the dew point. Equivalently, that temperature is a saturation temperature for the vapor at that partial pressure. Hence, saying the system is at its dew point and at a saturation temperature are both correct descriptions of the same thermodynamic condition in this context. Step-by-Step Solution: Start from the condition p_v = p_sat(T).Recognize this defines “just saturated vapor” in the mixture.Identify the temperature as the dew point (and also a saturation temperature for that vapor partial pressure). Verification / Alternative check: Psychrometric charts show the 100% relative humidity curve where p_v = p_sat(T); points along this curve are dew points for the corresponding vapor contents. Why Other Options Are Wrong: Bubble point pertains to liquid mixtures where first bubble of vapor forms. Critical temperature is unrelated to this equality for mixtures at ordinary pressures. Common Pitfalls: Confusing dew point (vapor saturation in gas) with bubble point (boiling onset in liquid); overlooking that “saturation temperature” in mixtures is context-dependent. Final Answer: Both saturation temperature and dew point.

Question 14

Saturated vapor, sensible heat, and phase-diagram endpoints: Which statement is correct regarding saturation and phase-equilibrium curves?

Accepted Answer

Introduction / Context: Understanding terminology around saturation, sensible/latent heat, and phase diagrams is fundamental in thermodynamics and process design. Misusing these terms leads to incorrect assessments of condensation, boiling, and energy duties in heat exchangers and separation equipment. Given Data / Assumptions: Pure-substance phase diagram behavior. Standard definitions of sensible and latent heat. Saturated states defined by equality of partial and saturation pressures at a given temperature. Concept / Approach: The vapor-pressure curve (liquid–vapor equilibrium line) of a pure substance spans from the triple point (where solid, liquid, and vapor coexist) to the critical point (where liquid and vapor become indistinguishable). A saturated vapor exists on this line below the critical temperature; above the critical temperature, no saturation state exists—only supercritical fluid/gas. Sensible heat refers to heat causing temperature change; latent heat refers to heat added or removed at constant temperature during a phase change at equilibrium. Step-by-Step Solution: Evaluate (a): Above T_c, no saturated states exist → false.Evaluate (b): Saturation requires p_v = p_sat(T), not p_v less than p_sat → false.Evaluate (c): Heat at constant temperature during phase change is latent heat, not sensible → false.Evaluate (d): Correct description of the vapor-pressure curve endpoints. Verification / Alternative check: Consult any standard phase diagram; the liquid–vapor line indeed ends at the triple and critical points. Why Other Options Are Wrong: (a) Mislabels supercritical or superheated states as “saturated.” (b) Misstates the saturation condition; equality, not “exceeds,” defines saturation. (c) Confuses sensible with latent heat. Common Pitfalls: Interchanging dew/bubble/saturation conditions; calling any constant-temperature heat exchange “sensible.” Final Answer: The vapor-pressure vs. temperature curve of a pure substance terminates at the triple point and the critical point.

Question 15

Definition of isotopes: Isotopes are atoms that have the same ________ but differ in mass number because of differing neutron counts.

Accepted Answer

Introduction / Context: Isotopes play a key role in nuclear chemistry, radiometric dating, and analytical methods. They are variants of the same chemical element and therefore share chemical behavior while differing in nuclear properties that depend on neutron count. Given Data / Assumptions: Isotopes share the same element identity. Chemical identity is defined by proton number (atomic number Z). Mass number A = Z + N, where N is number of neutrons. Concept / Approach: All atoms of the same element must have the same atomic number (Z). Isotopes differ in neutron number N, which changes the mass number A but not the element’s chemical identity. Average atomic masses on periodic tables are weighted by natural isotopic abundances and are not integers. Electron configurations can vary by ionization/excitation but ground-state configurations are element-specific and not a defining criterion for isotopes. Step-by-Step Solution: Define isotopes: same Z, different N → different A.Show example: ^12C, ^13C, ^14C all have Z = 6, different N.Conclude the defining “same” quantity is atomic number (protons). Verification / Alternative check: Nuclear notation A_ZX explicitly encodes Z as the element identity; changing Z changes the element, so isotopes cannot share mass number or neutron number necessarily, only Z. Why Other Options Are Wrong: Mass number: by definition different among isotopes. Number of neutrons: differs among isotopes. Average atomic mass: a tabulated, weighted value differs per sample. Electron configuration: not the defining feature for isotope identity. Common Pitfalls: Confusing “isotopes” (same Z) with “isobars” (same A) or “isotones” (same N). Final Answer: Atomic number (number of protons).

Question 16

Temperature dependence of heat of reaction: Which named relation provides the effect of temperature on the heat (enthalpy) of reaction?

Accepted Answer

Introduction / Context: Engineering calculations often require adjusting heats of reaction from a reference temperature to a process temperature. The appropriate tool is Kirchhoff’s equation, which connects the temperature dependence of reaction enthalpy to the difference in heat capacities of products and reactants. Given Data / Assumptions: Standard reference temperature T_ref with known ΔH_rxn(T_ref). Heat capacities (Cp) of species as functions of temperature. Ideal-mixture assumption for enthalpy calculations. Concept / Approach: Kirchhoff’s equation: ΔH_rxn(T) = ΔH_rxn(T_ref) + ∫_{T_ref}^{T} ΔCp dT, where ΔCp = Σ ν_p Cp_p − Σ ν_r Cp_r. This relation states that the change in reaction enthalpy with temperature equals the integral of the net heat-capacity change over that temperature range. Other named equations serve different purposes: Maxwell’s relations are thermodynamic identities; the Antoine equation correlates vapor pressure with temperature; Kistiakowsky refers to certain empirical kinetics/thermochemistry correlations, not the general ΔH(T) dependence. Step-by-Step Solution: Start from ΔH_rxn(T_ref) as a known baseline.Compute ΔCp(T) from species Cp polynomials and stoichiometric coefficients.Integrate ΔCp from T_ref to process T to obtain ΔH_rxn(T). Verification / Alternative check: For small temperature intervals and nearly constant ΔCp, use ΔH_rxn(T) ≈ ΔH_rxn(T_ref) + ΔCp * (T − T_ref) as a reliable approximation. Why Other Options Are Wrong: Maxwell’s relations connect derivatives of thermodynamic potentials; they do not give ΔH(T). Antoine equation correlates vapor pressure; unrelated to reaction enthalpy. Kistiakowsky correlations are not the canonical ΔH(T) relation for general reactions. Common Pitfalls: Confusing standard heats (at 298 K) with process conditions; neglecting temperature-dependent Cp, especially for wide temperature ranges. Final Answer: Kirchhoff’s equation.

Question 17

For a real gas that follows the van der Waals equation of state, how does the difference in specific heats (Cp − Cv) compare with the ideal-gas value? Explain on the basis of thermodynamic identities and stability criteria.

Accepted Answer

Introduction / Context: The difference between the constant-pressure and constant-volume specific heats, Cp − Cv, is a fundamental quantity that links caloric and thermal equations of state. For an ideal gas, Cp − Cv equals the universal gas constant R on a molar basis. For real gases described by the van der Waals model, non-ideality changes thermal expansion and compressibility, which in turn modifies Cp − Cv. This question checks your understanding of the general identity relating Cp − Cv to measurable volumetric properties and why real-gas behavior typically yields a value larger than R under stable conditions. Given Data / Assumptions: Real gas modeled by the van der Waals equation. Stable single-phase states (no crossing into spinodal region). Calorically simple behavior (Cp and Cv defined in the usual way). Concept / Approach: The general identity is Cp − Cv = T * V * α^2 / κ_T (molar basis: replace V with molar volume). Here α is the volumetric thermal expansion coefficient and κ_T is the isothermal compressibility. For an ideal gas, α = 1/T and κ_T = 1/P, giving Cp − Cv = R. For a van der Waals gas at the same T and P, κ_T is usually smaller (the fluid is less compressible) and α does not decrease proportionally, so the ratio α^2/κ_T is larger, making Cp − Cv greater than R in stable regions. Step-by-Step Solution: Use identity: Cp − Cv = T V α^2 / κ_T.Insert ideal-gas limits: α_id = 1/T, κ_T,id = 1/P → Cp − Cv = R.For van der Waals fluids, κ_T < κ_T,id at comparable T, P (less compressible), while α often remains of similar order.Therefore Cp − Cv (real) > Cp − Cv (ideal) = R, for stable states. Verification / Alternative check: Another exact form is Cp − Cv = −T * (∂P/∂T)_V^2 / (∂P/∂V)_T. For stable fluids, (∂P/∂V)_T < 0, making the fraction positive. Real-gas deviations generally make the magnitude larger than the ideal case at the same conditions. Why Other Options Are Wrong: R: true only for ideal gases. < R or 0.5 R: contradict typical real-gas trends in stable regions. Negative: impossible for a mechanically stable single phase because (∂P/∂V)_T is negative. Common Pitfalls: Confusing per-mole vs per-mass bases; forgetting that α and κ_T govern Cp − Cv in all fluids; applying the ideal-gas result blindly to real fluids at high pressures. Final Answer: > R

Question 18

In aqueous chemistry, which process describes ions from dissolved salts reacting with water to generate acidity or alkalinity of the solution?

Accepted Answer

Introduction / Context: Solutions of salts can be acidic, basic, or neutral. The phenomenon responsible for producing acidity or alkalinity when salt ions interact with water is crucial in water treatment, buffer design, and environmental chemistry. Distinguishing hydrolysis from other terms like hydration, electrolysis, and dialysis avoids conceptual errors. Given Data / Assumptions: Water is the solvent, salts are the solutes. We consider acid–base behavior arising from ion–water reactions at ambient conditions. Concept / Approach: Hydrolysis is the reaction of an ion (often the conjugate base of a weak acid or the conjugate acid of a weak base) with water to produce hydroxide or hydronium ions, shifting pH away from neutrality. For example, acetate ion reacts as CH3COO− + H2O ⇌ CH3COOH + OH− (basic solution), whereas ammonium reacts as NH4+ + H2O ⇌ NH3 + H3O+ (acidic solution). Hydration merely denotes water molecules coordinating around ions without changing pH significantly; electrolysis involves electrical decomposition; dialysis is membrane separation and not a chemical pH-generating reaction. Step-by-Step Solution: Identify reaction type that changes [H3O+] or [OH−]: hydrolysis.Contrast with hydration: no chemical transformation, just solvation shells.Discard electrolysis: requires electric current to drive redox reactions.Discard dialysis: physical separation through a semi-permeable membrane. Verification / Alternative check: pH predictions from K_a, K_b, and K_w relations for salt solutions are all derived from hydrolysis equilibria of the ions involved. Why Other Options Are Wrong: Hydration: solvation only; minimal pH effect. Electrolysis: external current; not salt–water acid/base chemistry. Dialysis: separation, not reaction. Peptization: dispersion of particles, unrelated to pH from salts. Common Pitfalls: Calling any ion–water interaction “hydration”; reserve “hydrolysis” for reactions that alter pH via H3O+ or OH− generation. Final Answer: hydrolysis

Question 19

Unit conversion check: 1 kcal per kg per °C is equivalent to how many BTU per lb per °F for specific heat capacity?

Accepted Answer

Introduction / Context: Specific heat capacity is frequently cited in both SI and English units. Converting correctly between 1 kcal/(kg·°C) and BTU/(lb·°F) prevents significant energy-balance errors in heat-transfer and process calculations. Given Data / Assumptions: 1 kcal = 3.96832 BTU (thermochemical). 1 kg = 2.20462 lb. Temperature interval: 1 °C = 1.8 °F. Concept / Approach: Convert the compound unit stepwise: multiply by the energy conversion, divide by the mass conversion, and divide by the temperature-interval conversion. Algebraically, value in BTU/(lb·°F) = 1 * (3.96832 BTU/kcal) * (1 kg / 2.20462 lb) * (1 °C / 1.8 °F). The mass and temperature conversions together cancel the numerical advantage of the energy conversion, yielding very nearly 1. Step-by-Step Solution: Start: 1 kcal/(kg·°C).Apply energy conversion: 1 × 3.96832 → 3.96832 BTU/(kg·°C).Convert mass: 3.96832 / 2.20462 → ≈ 1.800 BTU/(lb·°C).Convert temperature interval: 1.800 / 1.8 → ≈ 1.00 BTU/(lb·°F). Verification / Alternative check: Recognize that the “specific heat of water” near room temperature is ≈ 1 kcal/(kg·°C) and ≈ 1 BTU/(lb·°F); the two definitions were historically aligned to make conversions convenient. Why Other Options Are Wrong: 2.42 and 4.97 come from partial conversions (energy only) without accounting for mass and temperature units. 0.24 is the conversion between BTU/lb·°F and cal/g·°C, not the value asked here. None of these is incorrect because 1 is exact to within rounding. Common Pitfalls: Forgetting that degrees Celsius and Fahrenheit use different interval sizes; always include the factor of 1.8 when converting per-degree quantities. Final Answer: 1

Question 20

Enriching air with pure oxygen: What is the required mole ratio of air to added O2 to obtain an oxygen-enriched stream containing 50% O2 by volume (assume dry air contains 21% O2)?

Accepted Answer

Introduction / Context: Air enrichment by blending pure oxygen with air is used in combustion intensification and medical or environmental applications. Determining the correct blend ratio is a straightforward material-balance problem using mole fractions (equal to volume fractions for gases at common T and P). Given Data / Assumptions: Dry air contains 21% O2 by volume (mole basis). Target mixture contains 50% O2 by volume. Pure O2 is blended with air; inert nitrogen comes only from air. Concept / Approach: Let x = moles of air, y = moles of pure O2 added. Oxygen in the blend = 0.21x + y; total moles = x + y. The design requirement is (0.21x + y)/(x + y) = 0.50. Solve for y/x to get the mixing ratio, then invert to obtain air:oxygen. Step-by-Step Solution: Set (0.21x + y)/(x + y) = 0.50.Cross-multiply: 0.21x + y = 0.50x + 0.50y.Group y terms: y − 0.50y = 0.50x − 0.21x → 0.50y = 0.29x.Solve: y/x = 0.58 → air:O2 = x:y = 1 : 0.58 = 1.724 ≈ 1.72. Verification / Alternative check: Assume x = 1 mol air, y = 0.58 mol O2. O2 fraction = (0.21 + 0.58)/(1 + 0.58) = 0.79/1.58 ≈ 0.50, confirming the calculation. Why Other Options Are Wrong: 0.58 is the inverse ratio (O2 per air), not air per O2. 0.50 and 0.20 are unrelated to the derived ratio. 2.50 gives an O2 fraction far from 50%. Common Pitfalls: Using mass instead of mole/volume fractions for gas blending; forgetting that the inerts (chiefly N2) come only from air and dilute the final O2 fraction. Final Answer: 1.72

Stoichiometry Questions