Question 1

A gas initially at 0 °C is cooled at constant pressure until its volume becomes one-half of the original. What is the final temperature of the gas (in °C)? Assume ideal-gas behavior.

Accepted Answer

Introduction / Context:Ideal-gas relations connect temperature and volume at constant pressure (Charles’s law). This problem checks your ability to convert Celsius to Kelvin, apply proportionality of V to T, and convert back to Celsius correctly—a common exam trap lies in mishandling the absolute temperature scale.Given Data / Assumptions:Initial temperature T1 = 0 °C = 273 K.Pressure constant; ideal-gas behavior assumed.Final volume V2 = V1/2.Concept / Approach:At constant pressure for an ideal gas: V ∝ T (in Kelvin). Therefore, V2/V1 = T2/T1. Halving volume halves absolute temperature. Always work in Kelvin for gas-law calculations, then convert to °C if needed.Step-by-Step Solution:T1 = 273 K.V2/V1 = 1/2 ⇒ T2/T1 = 1/2 ⇒ T2 = 273 * 1/2 = 136.5 K.Convert to Celsius: t2(°C) = T2 − 273 = 136.5 − 273 = −136.5 °C.Hence, the final temperature is −136.5 °C.Verification / Alternative check:If a further halving to V/4 occurred, T would be 68.25 K (−204.75 °C), showing the linear proportionality and reinforcing correct unit handling.Why Other Options Are Wrong:−136.5 K: Kelvin values cannot be negative; also the number is actually in °C.−273 °C / 0 K: correspond to zero absolute temperature, which would imply zero volume—too large a cooling step.−68.25 °C: would correspond to 204.75 K, not 136.5 K.Common Pitfalls:Using Celsius directly in Charles’s law or forgetting to convert back properly from Kelvin, leading to sign or magnitude errors.Final Answer:−136.5 °C

Question 2

Select the incorrect statement regarding heat capacities of elements and mixtures.

Accepted Answer

Introduction / Context:Thermal properties guide design of heating, cooling, and energy storage operations. Recognizing typical trends—like the Dulong–Petit law and the behavior of heat capacity with temperature and phase—prevents serious design miscalculations.Given Data / Assumptions:Dulong–Petit (room temperature) trend: Cp,m ≈ 3R for many crystalline elements.Debye model: heat capacities drop toward zero as T → 0 K.For most substances, Cp(liquid) > Cp(solid) near the melting point.Concept / Approach:Evaluate each statement against known thermodynamic behavior. Look particularly for a reversal of the typical phase trend between solid and liquid heat capacities.Step-by-Step Solution:(a) Interpretation: 6.2 kcal/kg-atom·K equals ~6.2 cal/g-atom·K, consistent with 3R ≈ 5.96 cal/mol·K → broadly correct.(b) Matches Debye behavior as T decreases.(c) Claims Cp(liquid) (d) Additivity holds for simple mixtures by mass-fraction weighting; upon solution formation, excess heat capacities appear—statement is correct.(e) Restates Dulong–Petit in molar form: acceptable.Verification / Alternative check:Data tables show Cp(H2O, liquid) > Cp(ice) near 0–25 °C; many organics also show Cp(liquid) > Cp(solid) near fusion temperatures.Why Other Options Are Wrong:Only (c) contradicts standard physical trends; others are consistent with theory or empirical data.Common Pitfalls:Confusing caloric units (cal vs kcal) per gram-atom vs kg-atom; ignoring temperature dependence and phase effects.Final Answer:Generally, compounds have lower heat capacities in the liquid state than in the solid state.

Question 3

Why is the heat of neutralisation for strong acid–strong base reactions nearly constant at about 57 kJ per kg-mole of water formed?

Accepted Answer

Introduction / Context:Calorimetry of acid–base reactions reveals a near-constant heat release when a strong acid neutralises a strong base. Understanding the ionic basis of this constancy aids in thermochemistry calculations and in interpreting departures from the ideal value for weak acids/bases.Given Data / Assumptions:Strong acids and bases are fully dissociated in aqueous solution.Observed standard heat of neutralisation ≈ −57 kJ per kg-mole of water formed at dilute conditions.Solution behaves ideally; temperature changes are modest.Concept / Approach:The net ionic reaction for strong acid–strong base neutralisation is simply H+ (aq) + OH− (aq) → H2O (l). Since the reacting species are the same regardless of the specific acid/base identities, the enthalpy change is essentially the enthalpy of water formation from these ions, giving a nearly constant value.Step-by-Step Solution:Write the full molecular reaction (e.g., HCl + NaOH → NaCl + H2O).Cancel spectator ions to obtain the net ionic equation.Note that the net process is identical across all strong acid–strong base pairs.Therefore, ΔH is approximately constant and tied to hydration states and water formation.Verification / Alternative check:For weak acids/bases, the measured heat deviates because additional enthalpy is consumed or released in ionisation steps, confirming that the constancy hinges on full dissociation and the H+/OH− reaction only.Why Other Options Are Wrong:(a) Completion is true but not the fundamental reason for identical ΔH.(b) Salt hydrolysis can be negligible, but this is not the root cause.(d) Aqueous medium is common, yet the key is which ions actually react.(e) Ionic strength varies with concentration; it does not fix ΔH at one value.Common Pitfalls:Confusing extent of reaction with identity of reacting species, or attributing constancy to specific salts rather than to the universal H+/OH− neutralisation step.Final Answer:Because in essence only H+ and OH− ions react to form water

Question 4

In fuel technology and combustion engineering, the “calorific value” of a fuel is defined as the heat of which thermochemical process?

Accepted Answer

Introduction / Context:In energy engineering, the calorific value (also called heating value) is a key fuel-quality metric used for boiler ratings, burner design, and fuel economy calculations. It expresses how much heat is released when a specified amount of fuel is completely burned and the products are cooled to a reference state (for higher heating value) or left with water as vapor (for lower heating value).Given Data / Assumptions:Standard definition from thermochemistry.Complete combustion to stable products (e.g., CO2, H2O, SO2, N2).Reference conditions are implied (usually 25°C, 1 atm).Concept / Approach:The calorific value equals the magnitude of the enthalpy change for complete combustion of the fuel. Two practical versions are used: HHV (higher heating value) assumes the water formed condenses to liquid; LHV (lower heating value) assumes water remains as vapor. Both are specific cases of heat of combustion, not heats of formation, vaporisation, or a generic “reaction.”Step-by-Step Solution:Identify the defining process for calorific value → complete combustion.Relate to enthalpy change → calorific value = absolute value of ΔH_comb (per unit mass, volume, or mole).Conclude that the correct thermochemical term is the heat of combustion.Verification / Alternative check:Fuel specification sheets list HHV and LHV; both are measured by bomb calorimetry (HHV) or computed from combustion stoichiometry and water phase assumptions—confirming they are heats of combustion.Why Other Options Are Wrong:Formation: heat of formation pertains to forming a compound from its elements, not burning a fuel.Reaction: too vague; calorific value refers specifically to combustion.Vaporisation: relates to phase change, not fuel energy content.Common Pitfalls:Confusing HHV vs LHV (water phase treatment); mixing units (kJ/kg, kJ/m^3, kJ/mol).Final Answer:combustion

Question 5

In redox chemistry, a “reduction” step is accompanied by an increase in which quantity for the species being reduced?

Accepted Answer

Introduction / Context:Redox (reduction–oxidation) reactions underpin corrosion control, electrochemistry, metallurgy, and biochemical energy transfer. Correctly identifying oxidation versus reduction is essential for balancing reactions and predicting electron flow.Given Data / Assumptions:Reduction is “gain of electrons.”Oxidation number tracks apparent electron ownership.We consider the species undergoing reduction.Concept / Approach:Reduction increases the electron count associated with a species (it accepts electrons). Consequently, its oxidation number decreases (becomes more negative or less positive). Therefore, the only quantity that increases is the number of electrons; oxidation number does not increase during reduction.Step-by-Step Solution:Define: reduction = electron gain.Effect on electron count: increases.Effect on oxidation number: decreases (opposite of option b).Choose (a) as the only correct statement.Verification / Alternative check:Example: Fe^3+ + e^- → Fe^2+. Electrons increase by 1 for the iron species; oxidation number drops from +3 to +2.Why Other Options Are Wrong:Oxidation number (b) increases only in oxidation, not reduction.Both (a) & (b) is contradictory for reduction.Neither: incorrect since electron count does change.Common Pitfalls:Mixing up “OIL RIG” mnemonic (Oxidation Is Loss, Reduction Is Gain of electrons).Final Answer:number of electrons

Question 6

Unit conversions in heat transfer rate: identify the incorrect equivalence among Watt, kcal/hr, BTU/ft^2·hr, and kcal/m^2·hr.

Accepted Answer

Introduction / Context:Thermal calculations often require converting among SI and FPS engineering units. A frequent source of error is inverting ratios or mixing area conversions. This question checks consistency across common heat-flow units.Given Data / Assumptions:1 kcal = 4184 J.1 hour = 3600 s.1 BTU = 1055.06 J; 1 ft^2 = 0.092903 m^2.Concept / Approach:Convert one base pair precisely and propagate. Since 1 kcal/hr = 4184/3600 W ≈ 1.163 W, the inverse is 1 W ≈ 0.860 kcal/hr, not 1.163 kcal/hr. For surface rates, convert energy and area consistently, confirming the tabulated factors.Step-by-Step Solution:Compute: 1 kcal/hr = 4184/3600 ≈ 1.163 W → correct.Invert: 1 W = 3600/4184 ≈ 0.860 kcal/hr → option (b) is wrong.Check: 1 BTU/ft^2·hr to kcal/m^2·hr: multiply by 10.7639 m^2/ft^2 and 0.252164 kcal/BTU → ≈ 2.712 → correct.Check: 1 kcal/m^2·hr to BTU/ft^2·hr: 3.9683/10.7639 ≈ 0.3687; and to W/m^2: 1.163 → correct.Verification / Alternative check:Cross-validate with standard tables; the only inconsistent statement remains option (b).Why Other Options Are Wrong:(a), (c), (d) match precise conversions; thus they are not “wrong.”Common Pitfalls:Forgetting to invert factors; mixing per-area conversions; rounding beyond significant figures.Final Answer:1 Watt = 1.163 kcal/hr

Question 7

A fixed-pressure gas expansion (ideal gas): A sample occupies 2 litres at N.T.P. At what temperature will it occupy 4 litres if the pressure is kept constant?

Accepted Answer

Introduction / Context:For an ideal gas at constant pressure, volume is directly proportional to absolute temperature (Charles’s law). Doubling the volume requires doubling the temperature in kelvins.Given Data / Assumptions:Initial state at N.T.P. ≈ 273 K and 1 atm.V1 = 2 L, V2 = 4 L, P constant.Ideal-gas behavior; container can expand.Concept / Approach:Use V ∝ T at constant P: V2/V1 = T2/T1. With V2/V1 = 2, T2 = 2 * T1.Step-by-Step Solution:T1 = 273 K (N.T.P.).Compute T2 = 2 * 273 = 546 K.Convert to Celsius: 546 − 273 = 273°C.Verification / Alternative check:Check proportionality: doubling V at constant P must double T in kelvins, confirming 273°C.Why Other Options Are Wrong:273 K indicates no change; 100°C and 200°C do not double the absolute temperature from 273 K.Common Pitfalls:Doubling Celsius temperature instead of absolute temperature; misreading NTP as 300 K.Final Answer:273°C

Question 8

Specific heat unit conversion: 1 BTU per pound per °F is equivalent to how many kcal per kilogram per °C?

Accepted Answer

Introduction / Context:Thermal property data appear in different unit systems. Converting specific heat capacities between FPS and SI/metric units is a routine task in energy balances.Given Data / Assumptions:1 BTU = 1055.06 J; 1 kcal = 4184 J.1 lb = 0.453592 kg.Temperature interval conversion: 1°F = 5/9 °C.Concept / Approach:Start from 1 BTU/lb·°F and convert numerator (energy), then denominator (mass and temperature interval) to kcal/kg·°C.Step-by-Step Solution:Express in SI: 1 BTU/lb·°F = 1055.06 J / (0.453592 kg * 5/9 °C).Compute denominator factor: 0.453592 * 5/9 ≈ 0.252.Value ≈ 1055.06 / 0.252 ≈ 4187 J/kg·°C.Convert to kcal/kg·°C: 4187 / 4184 ≈ 1.00.Verification / Alternative check:Common reference: 1 BTU/lb·°F ≈ 4.186 kJ/kg·°C ≈ 1 kcal/kg·°C.Why Other Options Are Wrong:Values like 2.42, 1.987, or 4.97 are inconsistent with the precise conversion and would severely misstate heat loads.Common Pitfalls:Forgetting to convert °F to °C intervals; mixing mass conversions.Final Answer:1

Question 9

An unknown alcohol (M = 74) contains 48 g carbon, 10 g hydrogen, and 16 g oxygen per mole. Determine its molecular formula.

Accepted Answer

Introduction / Context:Finding a molecular formula from elemental composition and molecular weight is a staple of chemical stoichiometry and fuels characterization.Given Data / Assumptions:Molecular mass = 74 g/mol.Per mole: C = 48 g, H = 10 g, O = 16 g.Atomic weights: C = 12, H = 1, O = 16.Concept / Approach:Convert each element’s mass to moles, then assemble the atomic ratios to get the molecular formula that matches the given molecular mass.Step-by-Step Solution:Moles of C = 48/12 = 4.Moles of H = 10/1 = 10.Moles of O = 16/16 = 1.Empirical ratio → C4H10O; this already sums to 412 + 101 + 16 = 74 g/mol.Write as an alcohol: C4H9OH (equivalent to C4H10O).Verification / Alternative check:Known C4 alcohols (1-butanol, isobutanol, sec-butanol, tert-butanol) all have formula C4H10O, consistent with the calculation.Why Other Options Are Wrong:C3H21OH, (C2H4)2H2·OH, and C2H33OH have impossible or non-integer hydrogen counts and do not yield M = 74.Common Pitfalls:Forgetting that an alcohol’s “OH” adds O and H to the carbon–hydrogen skeleton; mis-summing atomic masses.Final Answer:C4H9OH

Question 10

For an ideal gas, the average translational kinetic energy per molecule depends only on which parameter?

Accepted Answer

Introduction / Context:Kinetic theory links macroscopic temperature to microscopic molecular motion. In the ideal gas model, translational kinetic energy is solely temperature dependent.Given Data / Assumptions:Ideal gas behavior.Equipartition theorem applies.Focus on translational degrees of freedom.Concept / Approach:The average translational kinetic energy per molecule is (3/2) k_B T, where k_B is Boltzmann’s constant and T is absolute temperature. It is independent of the gas species or molecular size in the ideal approximation.Step-by-Step Solution:Write relation: E_trans,avg = (3/2) k_B T.Note species independence: no molecular parameter other than T appears.Hence, the correct determinant is absolute temperature.Verification / Alternative check:Root-mean-square speed v_rms ∝ sqrt(T/M) varies with molar mass, but energy ~ (1/2) m v^2 averages to (3/2) k_B T, removing species dependence.Why Other Options Are Wrong:Nature or size of the gas affects collision dynamics and real-gas deviations, not the ideal-gas energy–temperature relation.Common Pitfalls:Confusing speed dependence on molar mass with energy dependence on temperature.Final Answer:absolute temperature

Question 11

Osmotic pressure control: which action will increase the osmotic pressure (π) of a solution, all else being equal?

Accepted Answer

Introduction / Context:Osmotic pressure is critical in membrane separations, food preservation, and biological cells. It scales with solute particle concentration and absolute temperature.Given Data / Assumptions:Dilute solution obeying van’t Hoff relation: π = i * C * R * T.i is the van’t Hoff factor; C is molar concentration; T is absolute temperature.Closed system with a semipermeable membrane conceptually present.Concept / Approach:From π = i C R T, increasing π requires increasing either i (more dissociation), C (more moles of solute per unit volume), or T (higher temperature). The listed actions either decrease T, decrease C, or effectively dilute the solution—none of which raise π.Step-by-Step Solution:Option (a): lowering temperature reduces T → π decreases.Option (b): increasing volume at fixed solute moles reduces C → π decreases.Option (c): dilution reduces C → π decreases.Thus, none of the listed actions increase π.Verification / Alternative check:To increase π in practice: add solute (increase C), choose a solute with higher dissociation i, or raise the solution temperature.Why Other Options Are Wrong:Each proposed action moves π in the opposite direction predicted by the van’t Hoff equation.Common Pitfalls:Assuming vessel expansion increases pressure (it lowers concentration instead); confusing osmotic pressure with hydrostatic pressure.Final Answer:none of these

Question 12

Which concentration unit remains independent of temperature changes: molarity, normality, or molality?

Accepted Answer

Introduction / Context:Engineers select concentration units carefully, because temperature changes alter solution volumes and therefore change certain measures of concentration. Picking a temperature-invariant unit simplifies calculations across varying thermal conditions.Given Data / Assumptions:Molarity (mol/L) and normality (equivalents/L) use solution volume in the denominator, which expands or contracts with temperature.Molality (mol solute per kg solvent) uses mass, which is independent of temperature.Concept / Approach:Because volume is temperature dependent while mass is not, any concentration defined per unit volume changes with temperature. Molality is defined per unit mass of solvent and therefore remains constant when the solution warms or cools (assuming no evaporation or mass loss).Step-by-Step Solution:Identify definitions: M = mol/L; N = eq/L; m = mol/kg solvent.Recognize thermal expansion affects L but not kg.Thus only molality is temperature independent.Verification / Alternative check:Laboratory practice: standard solutions specified by molarity are prepared at a reference temperature; molal solutions avoid this dependency.Why Other Options Are Wrong:Molarity and normality vary with volume changes due to temperature, so they are not invariant.“None of these” is incorrect because molality does meet the criterion.Common Pitfalls:Confusing molarity (volume based) with molality (mass based); ignoring solvent evaporation effects in open systems.Final Answer:molality

Question 13

Pressure-head equivalence: 1 kg per square metre corresponds to how many millimetres of water column?

Accepted Answer

Introduction / Context:Engineering pressure units are often interconverted between force/area (or weight/area) and a fluid head such as millimetres of water column. Being fluent with this conversion helps in calibrating manometers, specifying blower performance, and reading differential pressure indicators.Given Data / Assumptions:Quantity given: 1 kg/m^2, interpreted as weight per unit area.Standard gravity: g ≈ 9.80665 m/s^2.Density of water near room temperature: rho ≈ 1000 kg/m^3.Definition: 1 mm H2O is the hydrostatic pressure from a 1 mm water column.Concept / Approach:Convert weight per area into pascals, then compare with the pressure represented by 1 mm of water head. Hydrostatic relation: P = rho * g * h. Also, 1 kg/m^2 corresponds to P = g N/m^2 because 1 kg of mass under gravity exerts g newtons of weight.Step-by-Step Solution:Compute pressure from 1 kg/m^2: P = 1 * g ≈ 9.80665 Pa.Pressure from 1 mm H2O: P_mm = rho * g * h = 1000 * 9.80665 * 0.001 = 9.80665 Pa.Therefore, 1 kg/m^2 equals exactly the pressure of 1 mm of water column (at standard conditions).Verification / Alternative check:Dimensional check confirms both are pressures in pascals and numerically identical using standard gravity and water density.Why Other Options Are Wrong:10, 100, and 1000 mm H2O would be 10×, 100×, and 1000× larger pressures; 0.1 is 10× too small.Common Pitfalls:Confusing kilogram-force with kilogram of mass; forgetting to include g; using centimetres of water instead of millimetres.Final Answer:1

Question 14

Metric micro-scale: one micron (micrometre) equals how many centimetres?

Accepted Answer

Introduction / Context:Unit conversions at micro-scale are essential in filtration, particle technology, microfabrication, and microscopy. Knowing how micrometres relate to centimetres avoids costly specification errors in screens and membranes.Given Data / Assumptions:1 micrometre (1 micron) = 1 µm = 10^-6 m.1 metre = 100 centimetres = 10^2 cm.Concept / Approach:Convert micrometres to metres, then metres to centimetres by multiplying by 100. Use exponent rules to simplify the power of 10.Step-by-Step Solution:Start: 1 µm = 10^-6 m.Convert to cm: 10^-6 m * 100 cm/m = 10^-6 * 10^2 cm.Combine exponents: 10^-4 cm.Verification / Alternative check:Since 1 cm = 10,000 µm, the reciprocal check gives 1 µm = 1/10,000 cm = 10^-4 cm.Why Other Options Are Wrong:10^-2 is too large (that is 0.01 cm = 100 µm); 10^-6 is metres, not centimetres; 10^-8 and 10^-5 are inconsistent with the two-step conversion.Common Pitfalls:Skipping the metre-to-centimetre step or mixing up powers of ten.Final Answer:10^-4

Question 15

Thermodynamic terminology: a vapor existing above its critical temperature is properly called a what?

Accepted Answer

Introduction / Context:Correct phase terminology is crucial in phase-equilibrium calculations and equipment selection. Above the critical temperature, a substance cannot be liquefied by pressure alone, which changes how we refer to the phase.Given Data / Assumptions:Temperature is above the critical temperature of the substance.Standard definitions from thermodynamics apply.Concept / Approach:For T > T_c, there is no distinct liquid–vapor phase transition upon compression; the substance behaves as a gas. The term “vapor” is commonly reserved for T below T_c where condensation by pressure is possible. “Supercritical fluid” refers to conditions above both the critical temperature and critical pressure and is not one of the provided canonical options.Step-by-Step Solution:Recognize that at T > T_c, liquefaction by pressure is impossible.Therefore, the appropriate term is “gas” (gaseous state), not “vapor.”Select the option labeled “gaseous.”Verification / Alternative check:Phase diagrams show the vapor–liquid dome ending at the critical point; to the right (higher T) the region is simply gas unless at high pressure where the term supercritical fluid applies.Why Other Options Are Wrong:“Saturated”/“sub-cooled” describe near-boiling or below-saturation conditions; “unsaturated” is nonstandard here; “supercritical vapor” is imprecise terminology.Common Pitfalls:Using “vapor” and “gas” interchangeably without regard to T_c; calling everything above T_c “supercritical” regardless of pressure.Final Answer:gaseous

Question 16

Identify the correct 1.987-value for the universal gas constant R by unit system.

Accepted Answer

Introduction / Context:The universal gas constant R has the same physical value expressed in different unit systems. Remembering the common numeric representations avoids unit mistakes in thermodynamic equations of state and energy balances.Given Data / Assumptions:The numeric value 1.987 is associated with R in certain units.1 kcal = 1000 cal; 1 kg-mole (kmol) = 1000 moles; 1 Btu is the British thermal unit.Concept / Approach:The most familiar is R = 1.987 cal/mol·K. Multiplying numerator and denominator by 1000 gives R = 1.987 kcal/kg-mole·K. In English units, R ≈ 1.986–1.987 Btu/lb-mole·°R. Thus, the 1.987 figure corresponds to both kcal/kg-mole·K and Btu/lb-mole·°R.Step-by-Step Solution:From cal/mol·K to kcal/kg-mole·K: (1.987 cal/mol·K) × (1 kcal/1000 cal) × (1000 mol/kg-mole) = 1.987 kcal/kg-mole·K.Standard English-units value: R ≈ 1.9869 Btu/lb-mole·°R.Therefore, both (b) and (c) are valid for the 1.987 numeral.Verification / Alternative check:Other common forms: R = 8.314 kJ/kmol·K and R = 10.731 psia·ft^3/(lb-mole·°R) corroborate consistency after unit conversions.Why Other Options Are Wrong:(a) uses °C instead of K; while Δ°C equals ΔK, R is conventionally tabulated per K. (e) with kJ has numeric 8.314, not 1.987.Common Pitfalls:Forgetting that kg-mole is kmol; mixing absolute units (K, °R) with relative scales (°C, °F).Final Answer:both (b) & (c)

Question 17

Purpose of recycling streams in a chemical process: what principal benefits does recycle provide?

Accepted Answer

Introduction / Context:Recycle streams are ubiquitous in process flowsheets. They can improve economics, stabilize operations, and reduce utility usage by returning unreacted materials, concentrating valuable species, and conserving sensible/latent heat.Given Data / Assumptions:Generic chemical process with a recycle loop.Steady-state operation is implied.Concept / Approach:Recycling unreacted feed raises overall conversion (and thus yield) without needing extreme single-pass conversions. Recycle from separation steps can enrich the reactor feed/product composition. Thermal recycle (e.g., hot vapor/liquid returns, reflux) conserves heat and reduces external utilities.Step-by-Step Solution:Identify recycle effects: material reuse elevates overall yield.Separation-recycle pairs can tighten composition near desired ranges, enriching product streams.Energy carried back with recycle reduces heating/cooling demand, aiding heat conservation.Verification / Alternative check:Typical examples include ammonia synthesis loops (H2/N2 recycle), distillation reflux (product enrichment and heat integration), and purge-controlled polymerization loops.Why Other Options Are Wrong:Choosing any single benefit ignores the multi-faceted advantages; a claim of reduced conversion per pass is not a benefit and is not generally true.Common Pitfalls:Confusing per-pass conversion with overall conversion; neglecting the need for purge to control inerts when recycling.Final Answer:all (a), (b) & (c)

Question 18

At the melting point, how does the vapor pressure of a substance in the solid state compare to that in the liquid state?

Accepted Answer

Introduction / Context:Understanding vapor pressures of solid and liquid phases is essential for sublimation, drying, and melting operations. Around the melting point, the liquid generally has a higher tendency to evaporate than the solid.Given Data / Assumptions:Substance at its melting point (equilibrium between solid and liquid).Standard behavior (no unusual polymorphic anomalies considered).Concept / Approach:Vapor pressure reflects the escaping tendency of molecules. Liquids typically have higher surface molecular mobility than solids; consequently, at the same temperature (near the melting point), the equilibrium vapor pressure over the liquid exceeds that over the solid (the sublimation pressure).Step-by-Step Solution:Compare phase tendencies: liquid molecules are less constrained.Therefore, P_vap,liquid > P_vap,solid at the same T.Select “less” for the solid relative to the liquid.Verification / Alternative check:Clausius–Clapeyron considerations show different slopes; empirical data (e.g., ice vs liquid water near 0°C) support lower sublimation pressure than liquid vapor pressure at comparable temperatures.Why Other Options Are Wrong:“More” contradicts observed behavior; “same” would only occur under special rare conditions; “depends” is unnecessarily vague here.Common Pitfalls:Confusing triple-point relationships with melting-point comparisons; mixing sublimation and vaporization concepts.Final Answer:less

Question 19

For a neutral aqueous solution with pH = 7 at 25°C, what is the value of [H+][OH−]?

Accepted Answer

Introduction / Context:The ionic product of water, Kw, underpins acid–base calculations. At 25°C, neutral water has equal hydrogen and hydroxide ion concentrations, and their product is a small constant value central to pH and pOH relationships.Given Data / Assumptions:Neutral solution at 25°C: pH = 7 implies [H+] = 10^-7 mol/L.In neutrality, [H+] = [OH−].Concept / Approach:Kw = [H+][OH−] = 10^-14 at 25°C. Because both [H+] and [OH−] are 10^-7 mol/L, their product is 10^-14, which is far less than 1 in magnitude.Step-by-Step Solution:At pH 7: [H+] = 10^-7 mol/L.Neutrality: [OH−] = 10^-7 mol/L.Product: [H+][OH−] = 10^-14, which is numerically < 1.Verification / Alternative check:Also, pKw = 14 at 25°C → Kw = 10^-14 confirms the result.Why Other Options Are Wrong:0 is impossible in water; 1 is off by 14 orders of magnitude; > 1 is physically inconsistent at 25°C; “exactly 10^-7” confuses concentration with the product.Common Pitfalls:Confusing pH with [H+]; forgetting temperature dependence of Kw (though the question specifies 25°C).Final Answer:< 1

Question 20

Trouton’s ratio and Kistyakowsky correlation: for which liquids is the Kistyakowsky equation used to estimate λb/Tb?

Accepted Answer

Introduction / Context:Trouton’s rule states that many liquids have λ_b/T_b ≈ 21 cal/mol·K at their normal boiling points. Deviations occur, especially for associating or polar liquids. The Kistyakowsky equation refines the estimate for certain classes of liquids.Given Data / Assumptions:Trouton’s ratio λ_b/T_b around 21 holds best for simple, non-associating liquids.Kistyakowsky correlation targets such systems.Concept / Approach:Non-polar, relatively symmetric molecules tend to follow Trouton’s rule closely, and the Kistyakowsky equation provides a semi-empirical correction for improved accuracy. Polar and hydrogen-bonded liquids (e.g., water, alcohols) deviate significantly and are not well described by this simple relation.Step-by-Step Solution:Identify the class of liquids with near-constant λ_b/T_b values: non-polar.Note that Kistyakowsky's correlation is intended for those relatively ideal, non-associating liquids.Therefore, select “non-polar.”Verification / Alternative check:Data tables show hydrocarbons and noble-element compounds conform better to Trouton/Kistyakowsky than strongly polar or hydrogen-bonding liquids.Why Other Options Are Wrong:Polar and associating liquids diverge from the ~21 value; “both” or “neither” contradict the intended scope of the equation.Common Pitfalls:Overgeneralizing Trouton’s rule to all liquids, including hydrogen-bonded ones.Final Answer:non-polar

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