Q: Empirical (simplest) formula problem in stoichiometry: A compound contains 50% by mass of element A (atomic weight = 10) and 50% by mass of element B (atomic weight = 20). Determine the simplest whole-number formula of the compound.

Introduction / Context: This question tests the foundational stoichiometry skill of converting mass percentages and atomic weights into an empirical (simplest) formula. Empirical formulas are widely used in analytical chemistry, materials science, and biochemical analyses to represent the lowest whole-number ratio of atoms present in a compound. Given Data / Assumptions: Mass composition: A = 50% by mass; B = 50% by mass. Atomic weight: A = 10; B = 20. Empirical formula requires the simplest whole-number mole ratio. Assume a 100 g sample for convenience (a standard approach). Concept / Approach: Convert each element’s mass to moles, then reduce to the smallest whole-number ratio. The empirical formula expresses this ratio directly. If necessary, multiply by small integers (2, 3, etc.) to remove fractional subscripts while maintaining the same atomic ratio. Step-by-Step Solution: Assume 100 g total: mass of A = 50 g; mass of B = 50 g.Moles of A = 50/10 = 5.00 mol.Moles of B = 50/20 = 2.50 mol.Mole ratio A:B = 5.00 : 2.50 = 2 : 1.Write simplest formula from the ratio: A2B. Verification / Alternative check: Check proportional masses implied by A2B: Using AWs, mass fraction A = (210) / (210 + 1*20) = 20/40 = 0.5 = 50%. Mass fraction B = 20/40 = 50%. The composition matches, verifying the result. Why Other Options Are Wrong: AB3 implies A:B = 1:3, inconsistent with the 2:1 mole ratio.A2B3 implies 2:3, also not 2:1.AB2 implies 1:2, the inverse of the correct ratio. Common Pitfalls: Forgetting to convert mass to moles; reducing ratios incorrectly; leaving fractional subscripts without clearing them by multiplying through; assuming molecular formula instead of empirical when only percent composition is provided. Final Answer: A2B

Q: Raoult’s law and vapor pressure: For a solution, the vapor pressure is proportional to which mole fraction? (Sv = mole fraction of solvent, St = mole fraction of solute)

Introduction / Context: Raoult’s law governs vapor pressure lowering in ideal solutions. It states that the partial pressure of a volatile solvent over a solution equals the pure-solvent vapor pressure multiplied by the solvent mole fraction in the liquid. Understanding which composition term controls vapor pressure is essential in distillation, evaporation, and humidity calculations. Given Data / Assumptions: Ideal solution behavior (low to moderate solute concentration). Solvent is volatile; solute may be nonvolatile or less volatile. Temperature is fixed. Concept / Approach: Raoult’s law in its basic form is P_solvent = x_solvent * P°_solvent. Here x_solvent = Sv. Thus, the vapor pressure contributed by the solvent is directly proportional to Sv, not St. As solute is added (Sv decreases), the solvent’s vapor pressure falls proportionally, which underpins colligative properties like boiling point elevation and freezing point depression. Step-by-Step Solution: Write Raoult’s law: P = Sv * P°.Identify proportionality: P ∝ Sv at constant P° and T.Therefore, vapor pressure scales with solvent mole fraction, Sv.Select option “Sv”. Verification / Alternative check: Activity-coefficient models reduce to Raoult’s law when γ ≈ 1. In that limit, changing Sv changes the solvent chemical potential and thus the equilibrium vapor pressure linearly. Why Other Options Are Wrong: St is for the solute; Raoult’s proportionality references the solvent fraction.Inverse forms 1/St and 1/Sv are not part of the basic Raoult relation. Common Pitfalls: Confusing mole fraction of solvent vs. solute; applying Raoult’s law to highly non-ideal or associating systems without activity corrections; forgetting temperature dependence of P°. Final Answer: Sv

Q: Ideal gas in a rigid tyre: A car tyre (fixed volume 0.057 m³) contains air at 300 kPa and 300 K. After a long drive, the pressure rises to 330 kPa. Assuming air is ideal and Cv = 21 J/mol·K, find the change in internal energy of the air per mole (J/…

Introduction / Context: This numerical tests thermodynamics of an ideal gas in a rigid container. When volume is fixed, any pressure change is directly linked to temperature change. For an ideal gas, the change in internal energy depends only on temperature via ΔU = n * Cv * ΔT. Since the tyre volume is constant, we can relate the final temperature to the pressure rise and compute the per-mole energy change relevant in automotive and process thermal analyses. Given Data / Assumptions: Rigid vessel (tyre): V = constant = 0.057 m³ (volume value not needed for per-mole change). Initial state: P1 = 300 kPa, T1 = 300 K. Final state: P2 = 330 kPa, T2 = ?, V constant. Air behaves ideally; Cv = 21 J/mol·K. Composition and amount (n) remain fixed; no mass leaks. Concept / Approach: For an ideal gas at constant volume, P ∝ T (since P = n R T / V). Therefore, T2 = T1 * (P2/P1). The internal energy change per mole is Δu = Cv * ΔT. No work is done (W = 0) because the volume is fixed; any heat transfer shows up entirely as a change in internal energy. Step-by-Step Solution: Relate T and P at constant V: T2/T1 = P2/P1 = 330/300 = 1.10.Compute T2: T2 = 300 * 1.10 = 330 K.Find ΔT: ΔT = 330 − 300 = 30 K.Per-mole internal energy change: Δu = Cv * ΔT = 21 * 30 = 630 J/mol.Match with options: 630 J/mol. Verification / Alternative check: Use the ideal-gas energy relation U = n * Cv * T. The ratio method is consistent with P ∝ T at fixed V and n. The numeric increase is modest, fitting a 10% pressure rise. Why Other Options Are Wrong: 380, 760, and 880 J/mol correspond to incorrect ΔT assumptions or misuse of Cp instead of Cv, or failure to apply the P–T proportionality at constant volume. Common Pitfalls: Using Cp rather than Cv; forgetting absolute temperature; incorrectly incorporating tyre volume when a per-mole change is requested; assuming work at constant volume. Final Answer: 630

Q: Pressure unit conversions (single incorrect option): Choose the one conversion that is NOT correct.

Introduction / Context: Accurate pressure conversions are indispensable in vacuum systems, gas-phase reactions, and equipment specifications. This item checks familiarity with common engineering units and prevents order-of-magnitude errors that could compromise design safety or data interpretation. Given Data / Assumptions: Standard reference values near 1 atm. Use exact or conventional rounded constants widely accepted in engineering. Concept / Approach: Memorise or quickly derive the key equivalences: 1 atm ≈ 1.01325 bar; 1 bar = 100 kPa = 10^5 Pa; 1 mm Hg ≈ 1 torr ≈ 133.3 Pa. Evaluate each option against these anchors to identify the single incorrect statement. Step-by-Step Solution: Check (a): 1 atm = 760 mm Hg = 29.92 in Hg = 14.7 psi ≈ 1.01325 bar → correct.Check (b): 1 kPa = 100 bar → incorrect by eight orders of magnitude. Correct is 1 bar = 100 kPa (so 1 kPa = 0.01 bar).Check (c): 1 mm Hg = 1 torr ≈ 133.3 Pa → correct.Check (d): 1 bar = 10^5 Pa → correct by definition. Verification / Alternative check: From SI, 1 Pa = 1 N/m^2. The bar is a convenient non-SI unit defined as 10^5 Pa, making 1 bar exactly 100 kPa. Therefore, statement (b) is clearly wrong. Why Other Options Are Wrong: Options (a), (c), and (d) match accepted engineering constants and are widely used in handbooks and instrumentation. Common Pitfalls: Inverting bar–kPa relationships; mixing absolute and gauge pressures; over-rounding constants where precision matters. Final Answer: 1 kPa = 100 bar

Q: Ideal-gas mixing in a fixed 1.0 L vessel: 500 c.c. H2 at 700 mm Hg and 500 c.c. O2 at 600 mm Hg are combined in a 1.0 L container at the same temperature. What is the final total pressure (mm Hg)?

Introduction / Context: This problem illustrates Dalton’s law of partial pressures and the ideal-gas scaling of pressure with volume at constant temperature and moles. It is common in gas handling, reaction stoichiometry before ignition, and laboratory blending operations. Given Data / Assumptions: Initial gases: 0.5 L H2 at 700 mm Hg; 0.5 L O2 at 600 mm Hg. They are mixed into a 1.0 L vessel. Temperature is the same before and after; no reaction occurs (inert mixing). Ideal-gas behavior; volumes add; pressures scale inversely with volume at constant T and n. Concept / Approach: Each gas’s partial pressure after expansion/compression to 1.0 L equals its initial pressure multiplied by the ratio Vi/Vtotal (at constant T). The final total pressure equals the sum of the new partial pressures, per Dalton’s law. Step-by-Step Solution: For H2: P_H2,final = 700 mm Hg * (0.5 L / 1.0 L) = 350 mm Hg.For O2: P_O2,final = 600 mm Hg * (0.5 L / 1.0 L) = 300 mm Hg.Total pressure: P_total = 350 + 300 = 650 mm Hg.Select 650 mm Hg. Verification / Alternative check: Using n = PV/RT separately: n_H2 = 7000.5/(R T); n_O2 = 6000.5/(R T). After mixing in 1.0 L at same T, P_total = (n_total R T)/V = [(7000.5 + 6000.5)/(R T)] * (R T)/1.0 = 650 mm Hg. Why Other Options Are Wrong: 700 or 600 mm Hg ignore one of the component contributions and the volume scaling.375 mm Hg is an arithmetic error (e.g., averaging after dividing by 2 twice). Common Pitfalls: Forgetting to scale each gas by its volume change; averaging initial pressures incorrectly; assuming reaction occurs. Final Answer: 650

Question 1

Benzene–nitrogen mixture at 900 mmHg: If the absolute humidity is 0.2 kg benzene per kg nitrogen, what is the benzene partial pressure (mmHg)?

Accepted Answer

Introduction / Context: Gas–vapor mixtures are often described by mass ratios (absolute humidity) as well as partial pressures. Converting between these requires molecular weights and total pressure. This skill is vital in drying, stripping, and environmental emissions calculations. Given Data / Assumptions: Total pressure P_total = 900 mmHg. Absolute humidity w = 0.2 kg benzene per kg nitrogen. Molecular weights: MW_benzene = 78 kg/kmol; MW_N2 = 28 kg/kmol. Concept / Approach: Absolute humidity w relates to mole fraction y via w = (y * MW_vapor)/[(1 − y) * MW_gas]. Solve for y, then compute the benzene partial pressure p_b = y * P_total. Step-by-Step Solution: Start with w = (y * 78) / [(1 − y) * 28] = 0.2.Cross-multiply: 0.2 * (1 − y) * 28 = 78 y.Compute left: 5.6 * (1 − y) = 78 y → 5.6 − 5.6y = 78y.Solve for y: 5.6 = 83.6 y → y = 5.6 / 83.6 ≈ 0.06699.Find partial pressure: p_b = 0.06699 * 900 ≈ 60.3 mmHg. Verification / Alternative check: Convert the benzene mole fraction back to mass ratio to ensure consistency: w ≈ (0.06699*78)/((0.93301)*28) ≈ 0.2 kg/kg, confirming the calculation. Why Other Options Are Wrong: 180 or 200 mmHg: would imply much higher mass ratios than 0.2 kg/kg. 720 mmHg: near total pressure; impossible given the nitrogen carrier gas. 35.0 mmHg: underestimates the required benzene content. Common Pitfalls: Using air molecular weight instead of nitrogen; forgetting that w is a mass ratio, not mole ratio; mixing up mmHg and atm units during conversions. Final Answer: 60.3

Question 2

Stoichiometric decomposition yield: From 90 kg of water, what mass of oxygen (O2) can theoretically be produced? (Use 2 H2O → 2 H2 + O2.)

Accepted Answer

Introduction / Context: Simple stoichiometric calculations connect reactant masses to product yields. Water splitting to hydrogen and oxygen is a classic example used to teach mass–mole conversions and reaction stoichiometry constraints. Given Data / Assumptions: Balanced reaction: 2 H2O → 2 H2 + O2. Molar masses: H2O = 18 kg/kmol; O2 = 32 kg/kmol. Assume 100% conversion and no losses. Concept / Approach: The molar stoichiometry shows that 2 moles of water produce 1 mole of oxygen gas. Convert the given water mass to moles, apply stoichiometry to get moles of O2, and convert back to mass. Step-by-Step Solution: Compute moles of water: n_H2O = 90 kg / (18 kg/kmol) = 5 kmol.Use stoichiometry: 2 H2O → 1 O2, so n_O2 = n_H2O / 2 = 2.5 kmol.Convert to mass of O2: m_O2 = 2.5 kmol * 32 kg/kmol = 80 kg.Thus, 90 kg of water can yield 80 kg of oxygen theoretically. Verification / Alternative check: Mass conservation check: 90 kg H2O → 80 kg O2 + 10 kg H2 (since H2 mass = 2.5 kmol * 2 kg/kmol = 5 kg; plus remaining 5 kg? Careful: Hydrogen mass should be 10 kg because n_H2 = 2.5 kmol * 2 kg/kmol = 5 kg—however total product mass equals 80 + 5 = 85 kg; the missing 5 kg indicates we miscounted. Correct: From 5 kmol H2O, H content is 10 kmol H; H2 moles = 5 kmol; m_H2 = 5 kmol * 2 kg/kmol = 10 kg. Then total products 80 + 10 = 90 kg, conserving mass.) Why Other Options Are Wrong: 32 or 64 kg: correspond to using 36 kg or 72 kg of water respectively. 90 kg: cannot be all oxygen; hydrogen must also form. 72 kg: not consistent with stoichiometry. Common Pitfalls: Forgetting to convert between mass and moles; misreading the reaction stoichiometry (2:1 water to oxygen ratio). Final Answer: 80 kg

Question 3

Phase-change energetics: The heat required to vaporize 1 kg of a saturated liquid at its boiling temperature is called

Accepted Answer

Introduction / Context: Heating a substance can raise its temperature (sensible heating) or change its phase at essentially constant temperature (latent heating). Understanding the distinction is essential for boiler design, distillation, refrigeration, and energy balances involving phase changes. Given Data / Assumptions: Saturated liquid undergoes vaporization at its saturation temperature and pressure. Heat input occurs isothermally during the phase change. Concept / Approach: Latent heat of vaporization is the energy required to overcome intermolecular forces and create vapor at the saturation condition, without raising temperature. Sensible heat changes temperature within a single phase. Specific heat is a property that quantifies the sensible heat per unit mass per degree; it is not the latent heat itself. A fixed energy quantity like “1 kcal” is not a definition. Step-by-Step Solution: Identify process: saturated liquid → saturated vapor at constant T (boiling).Classify heat: no temperature change → not sensible; it is latent.Name the quantity: latent heat (of vaporization) per unit mass.Select option “latent heat.” Verification / Alternative check: Steam tables list h_fg as the difference between saturated vapor and saturated liquid enthalpies; that difference is precisely the latent heat of vaporization at the stated pressure/temperature. Why Other Options Are Wrong: Specific heat: relates sensible heating within a phase. 1 kcal: a unit, not a property definition. Sensible heat: involves temperature change, not phase change. Compression work: mechanical, not the thermal quantity defined here. Common Pitfalls: Using specific heat to estimate boiling duty; always use latent heat h_fg for the isothermal phase-change portion of the duty. Final Answer: latent heat

Question 4

Le Châtelier analysis: For the gas-phase equilibrium N2 + O2 ⇌ 2 NO with ΔH = +80 kJ per kmol of NO formed (endothermic forward), which change favors the decomposition of NO?

Accepted Answer

Introduction / Context: Nitric oxide formation and decomposition are central to high-temperature combustion and NOx control. Le Châtelier’s principle provides qualitative guidance on how temperature and pressure shifts affect equilibria, especially for endothermic versus exothermic directions. Given Data / Assumptions: Reaction: N2 + O2 ⇌ 2 NO. Enthalpy change: ΔH > 0 in the forward direction (endothermic formation of NO). Total moles: 2 → 2 (no net mole change). Concept / Approach: For endothermic forward reactions, lowering temperature favors the reverse (exothermic) direction—here, the decomposition of NO to N2 and O2. Pressure has little or no effect when there is no net change in total moles of gas (2 → 2), assuming ideal behavior. Changing the concentration of a reactant such as N2 does not directly force decomposition of NO; it tends instead to drive formation when O2 is present. Step-by-Step Solution: Apply Le Châtelier: decreasing T shifts equilibrium toward heat-releasing side (NO decomposition).Assess pressure: Δn_gas = 0, so pressure changes do not favor either side.Assess composition: adding N2 with O2 present pushes formation of NO, not decomposition.Therefore, the correct change is lowering temperature. Verification / Alternative check: Equilibrium constant K(T) for an endothermic reaction increases with temperature (van ’t Hoff). Thus, at lower T, K decreases, favoring reactants (N2, O2) and NO decomposes. Why Other Options Are Wrong: Increase/decrease in pressure: negligible effect because Δn = 0. Increasing [N2]: tends to promote NO formation if O2 is available. Adding inert at constant pressure: mainly dilutes; with Δn = 0 the effect on equilibrium position is minor. Common Pitfalls: Assuming pressure always influences gas equilibria; you must check Δn_gas. Misapplying van ’t Hoff by ignoring the sign of ΔH. Final Answer: Decrease in temperature

Question 5

Using ln(Psat) = A − 5000/T (Psat in atm, T in K), estimate the saturation pressure of water at 50 °C (assume A is calibrated to water so that this form is accurate near 50 °C). Choose the closest value.

Accepted Answer

Introduction / Context: Compact Clausius–Clapeyron-type expressions are widely used to estimate saturation pressures near a reference temperature. Here, ln(Psat) = A − 5000/T represents an integrated form with constants chosen for water over a limited range. The task is to estimate Psat at 50 °C and pick the closest option. Given Data / Assumptions: T = 50 °C = 323 K. Equation form: ln(Psat) = A − 5000/T, with A previously fitted for water in this range. Answers are approximate (two decimal places in atm). Concept / Approach: Although A is not explicitly provided, standard steam-table data state that the saturation pressure of water at 50 °C is approximately 12.3 kPa, which equals about 0.121–0.123 atm. Among the discrete options, the best match is 0.13 atm. This is consistent with the expectation from a CC-type fit over modest temperature intervals. Step-by-Step Solution (Reasoning by Reference Values): Recall from data: at 40 °C, Psat ≈ 0.073 atm; at 60 °C, Psat ≈ 0.199 atm.Interpolate qualitatively: at 50 °C, Psat should lie between ≈0.07 and ≈0.20 atm, nearer to 0.12 atm.Compare choices: 0.11 and 0.13 are the nearest; 0.13 is closer to 0.121–0.123 atm.Select 0.13 atm as the closest among the given options. Verification / Alternative check: Using ln(Psat) = A − 5000/323 and calibrating A with any one known point (e.g., at 373 K, Psat = 1 atm) yields an A value that, when used at 323 K, also produces ≈0.12 atm, consistent with the selection. Why Other Options Are Wrong: 0.07 and 0.09 are characteristic of lower temperatures (≈40–45 °C). 0.11 underestimates the known 50 °C pressure. 0.20 is closer to ≈60 °C. Common Pitfalls: Expecting exactness from simplified correlations; always pick the closest value and remember that constants like A are fit-dependent. Final Answer: 0.13

Question 6

Colligative effects: When a non-volatile solute is added to a pure solvent, how do the boiling point and freezing point of the solution change?

Accepted Answer

Introduction / Context: Adding a non-volatile solute to a solvent alters phase-change temperatures in ways that depend only on the number of solute particles (colligative properties). This principle underpins antifreeze formulations, de-icing salts, and boiling point adjustments in cooking and industrial processes. Given Data / Assumptions: Solute is non-volatile and does not dissociate (or dissociation is accounted for by the van ’t Hoff factor). Solution is dilute and ideal for colligative relations. Concept / Approach: Raoult’s law predicts a lowering of the solvent’s vapor pressure when a non-volatile solute is present. Consequently, a higher temperature is required to reach the external pressure for boiling (boiling-point elevation). Conversely, the presence of solute lowers the chemical potential of the liquid relative to the solid, requiring a lower temperature to achieve phase equilibrium (freezing-point depression). The magnitudes follow ΔT_b = K_b m i and ΔT_f = K_f m i, proportional to molality m and the van ’t Hoff factor i. Step-by-Step Solution: Apply Raoult’s law: P_solvent over solution < P_solvent (pure) at same T.Boiling point: need higher T to reach external pressure → boiling-point elevation.Freezing point: solution has lower chemical potential; equilibrium with solid occurs at lower T → freezing-point depression.Therefore, the correct combined statement is “increase in boiling point” and “decrease in freezing point.” Verification / Alternative check: Empirical constants K_b and K_f tabulated for solvents (e.g., water) predict linear changes with molality, confirming the qualitative outcomes. Why Other Options Are Wrong: Increase in freezing point: opposite to observation. Increase in boiling point only: incomplete; it also depresses freezing point. Both (a) and (b): includes the incorrect increase in freezing point. Common Pitfalls: Ignoring electrolyte dissociation (van ’t Hoff factor) which amplifies the effects; confusing non-volatile with volatile solutes where additional vapor-phase behavior applies. Final Answer: both (b) and (c)

Question 7

Empirical (simplest) formula problem in stoichiometry: A compound contains 50% by mass of element A (atomic weight = 10) and 50% by mass of element B (atomic weight = 20). Determine the simplest whole-number formula of the compound.

Accepted Answer

Introduction / Context: This question tests the foundational stoichiometry skill of converting mass percentages and atomic weights into an empirical (simplest) formula. Empirical formulas are widely used in analytical chemistry, materials science, and biochemical analyses to represent the lowest whole-number ratio of atoms present in a compound. Given Data / Assumptions: Mass composition: A = 50% by mass; B = 50% by mass. Atomic weight: A = 10; B = 20. Empirical formula requires the simplest whole-number mole ratio. Assume a 100 g sample for convenience (a standard approach). Concept / Approach: Convert each element’s mass to moles, then reduce to the smallest whole-number ratio. The empirical formula expresses this ratio directly. If necessary, multiply by small integers (2, 3, etc.) to remove fractional subscripts while maintaining the same atomic ratio. Step-by-Step Solution: Assume 100 g total: mass of A = 50 g; mass of B = 50 g.Moles of A = 50/10 = 5.00 mol.Moles of B = 50/20 = 2.50 mol.Mole ratio A:B = 5.00 : 2.50 = 2 : 1.Write simplest formula from the ratio: A2B. Verification / Alternative check: Check proportional masses implied by A2B: Using AWs, mass fraction A = (210) / (210 + 1*20) = 20/40 = 0.5 = 50%. Mass fraction B = 20/40 = 50%. The composition matches, verifying the result. Why Other Options Are Wrong: AB3 implies A:B = 1:3, inconsistent with the 2:1 mole ratio.A2B3 implies 2:3, also not 2:1.AB2 implies 1:2, the inverse of the correct ratio. Common Pitfalls: Forgetting to convert mass to moles; reducing ratios incorrectly; leaving fractional subscripts without clearing them by multiplying through; assuming molecular formula instead of empirical when only percent composition is provided. Final Answer: A2B

Question 8

Raoult’s law and vapor pressure: For a solution, the vapor pressure is proportional to which mole fraction? (Sv = mole fraction of solvent, St = mole fraction of solute)

Accepted Answer

Introduction / Context: Raoult’s law governs vapor pressure lowering in ideal solutions. It states that the partial pressure of a volatile solvent over a solution equals the pure-solvent vapor pressure multiplied by the solvent mole fraction in the liquid. Understanding which composition term controls vapor pressure is essential in distillation, evaporation, and humidity calculations. Given Data / Assumptions: Ideal solution behavior (low to moderate solute concentration). Solvent is volatile; solute may be nonvolatile or less volatile. Temperature is fixed. Concept / Approach: Raoult’s law in its basic form is P_solvent = x_solvent * P°_solvent. Here x_solvent = Sv. Thus, the vapor pressure contributed by the solvent is directly proportional to Sv, not St. As solute is added (Sv decreases), the solvent’s vapor pressure falls proportionally, which underpins colligative properties like boiling point elevation and freezing point depression. Step-by-Step Solution: Write Raoult’s law: P = Sv * P°.Identify proportionality: P ∝ Sv at constant P° and T.Therefore, vapor pressure scales with solvent mole fraction, Sv.Select option “Sv”. Verification / Alternative check: Activity-coefficient models reduce to Raoult’s law when γ ≈ 1. In that limit, changing Sv changes the solvent chemical potential and thus the equilibrium vapor pressure linearly. Why Other Options Are Wrong: St is for the solute; Raoult’s proportionality references the solvent fraction.Inverse forms 1/St and 1/Sv are not part of the basic Raoult relation. Common Pitfalls: Confusing mole fraction of solvent vs. solute; applying Raoult’s law to highly non-ideal or associating systems without activity corrections; forgetting temperature dependence of P°. Final Answer: Sv

Question 9

Ideal gas in a rigid tyre: A car tyre (fixed volume 0.057 m³) contains air at 300 kPa and 300 K. After a long drive, the pressure rises to 330 kPa. Assuming air is ideal and Cv = 21 J/mol·K, find the change in internal energy of the air per mole (J/…

Accepted Answer

Introduction / Context: This numerical tests thermodynamics of an ideal gas in a rigid container. When volume is fixed, any pressure change is directly linked to temperature change. For an ideal gas, the change in internal energy depends only on temperature via ΔU = n * Cv * ΔT. Since the tyre volume is constant, we can relate the final temperature to the pressure rise and compute the per-mole energy change relevant in automotive and process thermal analyses. Given Data / Assumptions: Rigid vessel (tyre): V = constant = 0.057 m³ (volume value not needed for per-mole change). Initial state: P1 = 300 kPa, T1 = 300 K. Final state: P2 = 330 kPa, T2 = ?, V constant. Air behaves ideally; Cv = 21 J/mol·K. Composition and amount (n) remain fixed; no mass leaks. Concept / Approach: For an ideal gas at constant volume, P ∝ T (since P = n R T / V). Therefore, T2 = T1 * (P2/P1). The internal energy change per mole is Δu = Cv * ΔT. No work is done (W = 0) because the volume is fixed; any heat transfer shows up entirely as a change in internal energy. Step-by-Step Solution: Relate T and P at constant V: T2/T1 = P2/P1 = 330/300 = 1.10.Compute T2: T2 = 300 * 1.10 = 330 K.Find ΔT: ΔT = 330 − 300 = 30 K.Per-mole internal energy change: Δu = Cv * ΔT = 21 * 30 = 630 J/mol.Match with options: 630 J/mol. Verification / Alternative check: Use the ideal-gas energy relation U = n * Cv * T. The ratio method is consistent with P ∝ T at fixed V and n. The numeric increase is modest, fitting a 10% pressure rise. Why Other Options Are Wrong: 380, 760, and 880 J/mol correspond to incorrect ΔT assumptions or misuse of Cp instead of Cv, or failure to apply the P–T proportionality at constant volume. Common Pitfalls: Using Cp rather than Cv; forgetting absolute temperature; incorrectly incorporating tyre volume when a per-mole change is requested; assuming work at constant volume. Final Answer: 630

Question 10

van der Waals non-ideal gas: Which term accounts primarily for intermolecular attractive forces in (P + a/V^2)(V − b) = R T?

Accepted Answer

Introduction / Context: The van der Waals (vdW) equation modifies the ideal-gas law to capture real-gas behavior by introducing two empirical parameters, a and b. These corrections embody intermolecular attractions and finite molecular size, respectively. Identifying which term corresponds to which physical effect is vital in high-pressure gas processing and critical-phenomena studies. Given Data / Assumptions: vdW equation: (P + a/V^2)(V − b) = R T (per mole basis with molar volume V). a quantifies attractive interactions; b accounts for excluded volume (repulsions/finite size). R is the universal gas constant; T is absolute temperature. Concept / Approach: Attractive forces effectively reduce the measured pressure relative to an ideal gas at the same state because molecules pull each other inward. The vdW model compensates by adding a/V^2 to the measured pressure P. Conversely, the finite size of molecules reduces the accessible volume, represented by (V − b). Thus a/V^2 is directly tied to attractions, while b is tied to repulsive core/excluded-volume effects. Step-by-Step Solution: Recall physical roles: a → attractions; b → finite size.Identify the position of a: appears as a pressure correction a/V^2.Therefore, the term accounting for intermolecular forces (attractions) is a/V^2.Select option “a/V^2 (pressure correction due to attractions)”. Verification / Alternative check: In the limit a → 0 and b → 0, the equation reduces to PV = R T, confirming that a and b are corrections beyond ideal behavior. Why Other Options Are Wrong: R T is the thermal energy driving expansion, not a force correction.(V − b) represents excluded volume (repulsive size effect), not attractions.1/(R T) is not a term in the vdW equation; it has no direct physical meaning here. Common Pitfalls: Mixing up roles of a and b; forgetting that attractive forces reduce pressure whereas excluded volume reduces effective V. Final Answer: a/V^2 (pressure correction due to attractions)

Question 11

Heats of solution and neutralisation: Identify the incorrect statement about solution enthalpies and related concepts.

Accepted Answer

Introduction / Context: Thermochemistry of solutions involves heats of mixing, solution, hydration, and neutralisation. Distinguishing which statements are valid is important for designing dissolution steps, acid–base neutralisations, and assessing energy effects in process streams and analytical titrations. Given Data / Assumptions: Dilute solutions behave nearly ideally for strong electrolytes; weak electrolytes may require dissociation considerations. Standard reference states enable comparisons of heats of solution and hydration. Heats of mixing data can yield partial molar and integral properties if composition-dependent measurements are available. Concept / Approach: The integral heat of solution (at a specified final composition) can be obtained from enthalpy of mixing data by integrating composition-dependent partial molar enthalpies or using calorimetric measurements across concentrations. Weak acid or base neutralisations release less heat than strong/strong pairs because energy is consumed to ionise the weak species. Hydrate heats relate to differences between anhydrous and hydrated dissolution enthalpies. Enthalpy of dissolution depends on identity, concentration, and temperature, not just on the solute amount. Step-by-Step Solution: Examine (a): claims “cannot be calculated” from mixing data. This is incorrect; with proper thermodynamic data, integral heats can indeed be derived.Examine (b): true—weak species require ionisation, reducing net exothermicity compared with strong acid–strong base (≈ −57 kJ/mol water formed).Examine (c): true—definitionally relates hydrate dissolution to anhydrous dissolution and hydration enthalpy.Examine (d): true—enthalpy of solution depends on multiple factors including composition and temperature. Verification / Alternative check: Thermodynamic relationships among integral/partial molar quantities and calorimetric data sets are well established; many handbooks tabulate integral heats of solution obtained via mixing data. Why Other Options Are Wrong: Options (b), (c), and (d) are consistent with standard solution thermochemistry and practical calorimetry. Common Pitfalls: Confusing integral vs. differential (partial) heats; assuming heats are composition-independent; overlooking ionisation of weak electrolytes in neutralisation. Final Answer: The integral heat of solution of either component cannot be calculated from heat of mixing data.

Question 12

Pressure unit conversions (single incorrect option): Choose the one conversion that is NOT correct.

Accepted Answer

Introduction / Context: Accurate pressure conversions are indispensable in vacuum systems, gas-phase reactions, and equipment specifications. This item checks familiarity with common engineering units and prevents order-of-magnitude errors that could compromise design safety or data interpretation. Given Data / Assumptions: Standard reference values near 1 atm. Use exact or conventional rounded constants widely accepted in engineering. Concept / Approach: Memorise or quickly derive the key equivalences: 1 atm ≈ 1.01325 bar; 1 bar = 100 kPa = 10^5 Pa; 1 mm Hg ≈ 1 torr ≈ 133.3 Pa. Evaluate each option against these anchors to identify the single incorrect statement. Step-by-Step Solution: Check (a): 1 atm = 760 mm Hg = 29.92 in Hg = 14.7 psi ≈ 1.01325 bar → correct.Check (b): 1 kPa = 100 bar → incorrect by eight orders of magnitude. Correct is 1 bar = 100 kPa (so 1 kPa = 0.01 bar).Check (c): 1 mm Hg = 1 torr ≈ 133.3 Pa → correct.Check (d): 1 bar = 10^5 Pa → correct by definition. Verification / Alternative check: From SI, 1 Pa = 1 N/m^2. The bar is a convenient non-SI unit defined as 10^5 Pa, making 1 bar exactly 100 kPa. Therefore, statement (b) is clearly wrong. Why Other Options Are Wrong: Options (a), (c), and (d) match accepted engineering constants and are widely used in handbooks and instrumentation. Common Pitfalls: Inverting bar–kPa relationships; mixing absolute and gauge pressures; over-rounding constants where precision matters. Final Answer: 1 kPa = 100 bar

Question 13

Ideal-gas mixing in a fixed 1.0 L vessel: 500 c.c. H2 at 700 mm Hg and 500 c.c. O2 at 600 mm Hg are combined in a 1.0 L container at the same temperature. What is the final total pressure (mm Hg)?

Accepted Answer

Introduction / Context: This problem illustrates Dalton’s law of partial pressures and the ideal-gas scaling of pressure with volume at constant temperature and moles. It is common in gas handling, reaction stoichiometry before ignition, and laboratory blending operations. Given Data / Assumptions: Initial gases: 0.5 L H2 at 700 mm Hg; 0.5 L O2 at 600 mm Hg. They are mixed into a 1.0 L vessel. Temperature is the same before and after; no reaction occurs (inert mixing). Ideal-gas behavior; volumes add; pressures scale inversely with volume at constant T and n. Concept / Approach: Each gas’s partial pressure after expansion/compression to 1.0 L equals its initial pressure multiplied by the ratio Vi/Vtotal (at constant T). The final total pressure equals the sum of the new partial pressures, per Dalton’s law. Step-by-Step Solution: For H2: P_H2,final = 700 mm Hg * (0.5 L / 1.0 L) = 350 mm Hg.For O2: P_O2,final = 600 mm Hg * (0.5 L / 1.0 L) = 300 mm Hg.Total pressure: P_total = 350 + 300 = 650 mm Hg.Select 650 mm Hg. Verification / Alternative check: Using n = PV/RT separately: n_H2 = 7000.5/(R T); n_O2 = 6000.5/(R T). After mixing in 1.0 L at same T, P_total = (n_total R T)/V = [(7000.5 + 6000.5)/(R T)] * (R T)/1.0 = 650 mm Hg. Why Other Options Are Wrong: 700 or 600 mm Hg ignore one of the component contributions and the volume scaling.375 mm Hg is an arithmetic error (e.g., averaging after dividing by 2 twice). Common Pitfalls: Forgetting to scale each gas by its volume change; averaging initial pressures incorrectly; assuming reaction occurs. Final Answer: 650

Question 14

Temperature dependence of gas heat capacity: At higher temperatures, how does the molal heat capacity at constant pressure (Cp) of most gases change?

Accepted Answer

Introduction / Context: Heat capacity trends with temperature are essential for designing heaters, coolers, and reactors. The molar heat capacity at constant pressure (Cp) of gases determines how much energy is required to raise the temperature of a flowing gas stream and strongly impacts energy balances. Given Data / Assumptions: Most real gases, especially polyatomic, experience excitation of additional degrees of freedom with increasing temperature. Ideal-gas approximations with temperature-dependent Cp are commonly used. Concept / Approach: As temperature rises, rotational modes are fully excited and vibrational modes become increasingly populated, adding to the energy storage capacity. Consequently, Cp generally increases with temperature for most gases. The variation is not strictly linear over broad ranges, and correlations (e.g., polynomial Cp(T)) are used to compute enthalpy changes accurately. Step-by-Step Solution: Recognize that added degrees of freedom contribute to heat capacity.As T increases, vibrational contributions grow, raising Cp beyond its low-T limit.Therefore, the qualitative trend is that Cp increases.Choose “Increases”. Verification / Alternative check: Handbook data (e.g., for CO2, N2, H2O vapor) show Cp(T) curves that rise with temperature over typical process ranges, confirming the general trend. Why Other Options Are Wrong: Linear variation is not universally accurate; detailed fits are often polynomial.Decrease or no variation contradicts well-documented behavior, except in narrow temperature windows where Cp can be nearly flat. Common Pitfalls: Using constant-Cp approximations too far from reference temperatures; confusing Cp with Cv; ignoring vibrational effects at high T. Final Answer: Increases

Question 15

Strong acid–strong base mixing (each 1 N): In which case will the temperature rise be the greatest, assuming equal densities and similar heat capacities?

Accepted Answer

Introduction / Context: Heat released during neutralisation of a strong acid by a strong base is approximately constant per mole of water formed (about −57 kJ per mol at dilute conditions). The observed temperature rise depends on how many moles react and the total heat capacity of the mixed solution. This question checks the intuitive balance between reaction extent and dilution. Given Data / Assumptions: Both reactants are 1 N (1 equivalent per litre). Density ≈ 1 g/mL and similar specific heats for solutions. Temperature change ΔT ∝ heat released / heat capacity of mixture. Heat released ∝ moles reacted = min(equivalents of acid, equivalents of base). Concept / Approach: For 1 N solutions, equivalents = volume (L). The limiting reagent dictates moles of water formed. The temperature rise is roughly proportional to moles reacted divided by total volume (heat capacity scales with volume). We therefore evaluate the ratio r = min(Va, Vb) / (Va + Vb) for each mixture and pick the largest r. Step-by-Step Solution: (a) Va = 30 mL, Vb = 30 mL → reacted = 30; total = 60 → r = 30/60 = 0.50.(b) 20 and 25 → reacted = 20; total = 45 → r ≈ 0.444.(c) 15 and 35 → reacted = 15; total = 50 → r = 0.30.(d) 35 and 15 → reacted = 15; total = 50 → r = 0.30.Highest r occurs for option (a), hence the greatest ΔT. Verification / Alternative check: Exact ΔT would use ΔH_neut * n_reacted divided by m * cp; with constant cp and density, the ranking by r is sufficient and matches the above conclusion. Why Other Options Are Wrong: They either react fewer equivalents and/or have larger total volumes, decreasing the heat-release per unit mass of solution. Common Pitfalls: Ignoring that the limiting reagent controls moles of water formed; focusing only on reacted moles and forgetting dilution; mixing normality with molarity without tracking equivalents. Final Answer: 30 c.c acid and 30 c.c base

Question 16

Phase change at subzero conditions: The disappearance of snow in freezing (subzero) weather is an example of which physical process?

Accepted Answer

Introduction / Context: Snow and ice can vanish even when ambient temperatures remain below 0°C. Understanding the underlying phase change explains everyday observations, informs meteorology, and clarifies drying and preservation technologies (e.g., freeze-drying) that exploit similar physics. Given Data / Assumptions: Ambient temperature is below 0°C (subzero), so bulk melting is not sustained. Air is unsaturated with respect to ice, allowing net mass transfer to the vapor phase. Wind and low humidity enhance mass transfer from the solid surface. Concept / Approach: Sublimation is the direct transition of a substance from solid to vapor without passing through the liquid phase. For ice/snow, the vapor pressure of ice at subzero temperatures, though low, is nonzero. If the surrounding air’s partial pressure of water vapor is below the equilibrium vapor pressure over ice at that temperature, molecules leave the solid and diffuse away—net sublimation occurs. Evaporation and vaporisation refer to liquid → vapor transitions, which are not dominant here because liquid water is not present or stable at sustained subzero conditions. Melting requires temperatures at or above the melting point (or at least localized heating), which is not the scenario described. Step-by-Step Solution: Identify phases: solid (snow/ice) and gas (water vapor).Check temperature regime: subzero → liquid phase is not stable overall.Equilibrium criterion: if p_H2O,air < p_sat,ice(T), mass transfer occurs from solid to gas.Therefore the correct mechanism is sublimation. Verification / Alternative check: Psychrometric charts and ice vapor pressure correlations show positive sublimation driving forces at low relative humidity and cold temperatures, consistent with snow loss without melting. Why Other Options Are Wrong: Evaporation/vaporisation involve a liquid phase.Melting requires temperatures near/above 0°C or strong localized heating. Common Pitfalls: Assuming all disappearance of snow must involve melting; overlooking the role of humidity and wind in sublimation rates. Final Answer: Sublimation

Question 17

Energy units in chemical engineering: One “therm” (as used in fuel and natural gas billing) is equivalent to which of the following energy quantities?

Accepted Answer

Introduction / Context: The “therm” is a practical energy unit used widely by gas utilities to bill domestic and industrial consumers. Knowing exact conversions between a therm, British Thermal Units (BTU), kilocalories (kcal), and megajoules (MJ) is essential for quick, error-free energy balances and cost estimations in process engineering. Given Data / Assumptions: By definition in common utility practice, 1 therm = 100,000 BTU (US therm). 1 BTU ≈ 1055 J; 1 kcal = 4186.8 J; 1 MJ = 10^6 J. We consider the widely adopted “US therm,” not the rarer “international therm.” Concept / Approach: Perform consistent unit conversions to cross-check equivalences. Converting 100,000 BTU to joules and then to MJ confirms the size of the therm and helps distinguish it from distractors that are off by orders of magnitude or wrongly expressed in kcal. Step-by-Step Solution: Start with definition: 1 therm = 100,000 BTU.Convert to joules: 100,000 BTU * 1055 J/BTU ≈ 1.055×10^8 J.Convert to MJ: 1.055×10^8 J / 10^6 ≈ 105.5 MJ.Cross-check vs kcal: 105.5 MJ / 4.1868 kJ/kcal ≈ 25,200 kcal (not 100,000 kcal). Verification / Alternative check: Utility bills often list consumption in therms; multiplying therms by 100,000 BTU gives BTU directly. Converting to kWh: 105.5 MJ / 3.6 MJ/kWh ≈ 29.3 kWh per therm, a common back-of-envelope check. Why Other Options Are Wrong: “100,000 kcal” overstates by ~4×. “1,000,000,000 BTU/kcal” are off by four orders of magnitude. “41.868 MJ” corresponds to 10,000 kcal, not a therm. Common Pitfalls: Mixing BTU and kcal, or assuming 1 therm equals 100,000 kcal. Also confusing therm with “thm” on bills versus “kWh.” Final Answer: 100,000 BTU

Question 18

Effect of pressure on solubility: For a liquid–liquid solution whose mixed volume is less than the sum of pure-component volumes (negative volume change on mixing), how does solubility respond to pressure?

Accepted Answer

Introduction / Context: When two liquids mix with a volume contraction (the solution volume is less than the sum of pure volumes), the process has a negative excess volume. According to thermodynamic principles, pressure tends to favor states with smaller molar volumes. Understanding this helps predict how solubility and miscibility shift with applied pressure in liquid–liquid systems (not solid–liquid dissolution). Given Data / Assumptions: Liquid–liquid solution with negative volume change on mixing. No chemical reaction; only physical mixing. Temperature held constant during the pressure change. Concept / Approach: Le Chatelier’s principle (generalized) indicates that increasing pressure favors the side of an equilibrium with lower volume. If mixing decreases volume, higher pressure drives the equilibrium toward more mixed state, enhancing mutual solubility. Conversely, if mixing increased volume, pressure would disfavor mixing. Step-by-Step Solution: Recognize: ΔVmix < 0 for the solution.Apply pressure: system minimizes Gibbs energy G = H − T S + ∫V dP.Lower V states (more mixed) reduce G more under higher P.Therefore, mutual solubility increases as pressure rises. Verification / Alternative check: Phase diagrams for partially miscible liquid pairs often show the miscibility gap shrinking under elevated pressure when ΔVmix is negative. Why Other Options Are Wrong: “Increases with decrease in pressure” contradicts the volume argument. “Unchanged” ignores measurable pressure effects. “Independent of temperature” is irrelevant; the question is about pressure. “Becomes zero at high pressure” is not a general rule. Common Pitfalls: Confusing gas solubility (strong pressure dependence) with liquid–liquid mixing; also mixing up the sign of ΔVmix. Final Answer: Increases with rise in pressure

Question 19

Determinants of solubility: Which factor typically has the least effect on the solubility of a solute in a liquid solvent (excluding gases)?

Accepted Answer

Introduction / Context: For liquid solutions (and most solid–liquid systems), solubility trends are governed primarily by molecular interactions and temperature. Pressure has only a minor influence at ordinary conditions, unlike in gas–liquid systems where Henry’s law leads to strong pressure dependence. Given Data / Assumptions: We exclude gas solubility, which is highly pressure dependent. Ambient to moderate pressures typical of laboratory and industrial operations. No chemical reaction between solute and solvent. Concept / Approach: “Like dissolves like” captures the dominant role of solute–solvent interactions. Temperature affects solubility via enthalpy/entropy of dissolution. Pressure changes liquid molar volumes only slightly, so the effect on chemical potentials and hence solubility is minimal for condensed phases. Step-by-Step Solution: Rank influences qualitatively: identity of solute/solvent > temperature ≫ pressure.Recognize liquids and solids are nearly incompressible: small dμ/dP.Therefore, pressure usually has the least impact on solubility (non-gaseous systems). Verification / Alternative check: Data compilations show modest solubility shifts with large pressure changes for liquids/solids, in contrast to the near-linear pressure effect for gases. Why Other Options Are Wrong: Nature of solute/solvent strongly determines compatibility. Temperature frequently changes solubility by large factors. Trace salts can “salt in” or “salt out,” sometimes noticeably; still generally larger than pressure effects. Common Pitfalls: Overgeneralizing Henry’s law (gases) to liquid/solid solubility problems. Final Answer: Pressure of the system

Question 20

Activation energy vs reaction enthalpy: For an endothermic reaction, what is the minimum possible value of the forward activation energy relative to ΔH (reaction enthalpy)?

Accepted Answer

Introduction / Context: Activation energy (Ea) measures the energy barrier that must be overcome for reactants to reach the transition state. For endothermic reactions, products lie at a higher enthalpy level than reactants (ΔH > 0). Understanding the relationship between Ea and ΔH is fundamental to interpreting Arrhenius kinetics and energy profiles. Given Data / Assumptions: Elementary viewpoint using a single transition state. Forward reaction is endothermic: ΔH > 0. Classical energy profile (reactants → barrier → products). Concept / Approach: Let Ea,f be forward activation energy and Ea,r be reverse activation energy. The energy diagram gives Ea,f − Ea,r = ΔH. Since activation energies are non-negative (barriers cannot be negative in classical terms), the smallest Ea,f occurs when the reverse barrier approaches zero, yielding Ea,f(min) = ΔH. Step-by-Step Solution: Use relation: Ea,f = Ea,r + ΔH.Minimum Ea,r is 0 (limiting case), so Ea,f(min) = ΔH.Therefore, the least possible Ea for an endothermic forward reaction equals ΔH. Verification / Alternative check: Transition-state theory gives the same qualitative picture: for larger ΔH, the forward barrier must be at least ΔH to place products above reactants. Why Other Options Are Wrong: “Greater than ΔH” can be true but is not the minimum. “Less than ΔH” or “zero/negative” violate the barrier relation for endothermic profiles. Common Pitfalls: Confusing thermodynamic uphill (ΔH) with kinetic barrier (Ea); they are distinct but related by the reverse barrier. Final Answer: Equal to ΔH

Stoichiometry Questions