Q: Stoichiometry at STP: In Ca + 2 H2O → Ca(OH)2 + H2, how many cubic centimetres of H2 at STP are released when 8 g of Ca reacts with excess water? (Atomic weight of Ca = 40)

Introduction / Context: Gas-evolution stoichiometry is frequently used in laboratory calculations and industrial gas balance estimations. Converting from mass of reactant to moles of gas, then to volume at standard conditions (STP), is a core skill for chemical engineering and chemistry students. Given Data / Assumptions: Balanced reaction: Ca + 2 H2O → Ca(OH)2 + H2. Molar mass of Ca = 40 g per mol. STP molar volume: 22,400 c.c per mol of an ideal gas. Water is in excess; reaction goes to completion with respect to Ca. Concept / Approach: Use stoichiometry to relate moles of calcium to moles of hydrogen gas. The balanced equation shows a 1:1 molar ratio between Ca and H2. Convert the given mass of Ca to moles, then convert moles of H2 to volume at STP using the standard molar volume. Step-by-Step Solution: Compute moles of Ca: nCa = 8 g / 40 g mol^-1 = 0.2 mol.Use stoichiometric ratio: nH2 = nCa = 0.2 mol.Convert to volume at STP: VH2 = 0.2 mol * 22,400 c.c mol^-1 = 4,480 c.c.Select the option equal to 4,480 c.c. Verification / Alternative check: Dimensional sanity: 8 g is one-fifth of a mole of Ca; one-fifth of 22,400 c.c yields 4,480 c.c, matching the computed value. Why Other Options Are Wrong: 2,240 c.c corresponds to 0.1 mol of H2; 1,120 c.c corresponds to 0.05 mol; 0.4 is a dimensionless number, not a volume. Common Pitfalls: Using the wrong molar mass or STP molar volume; forgetting the 1:1 stoichiometric relation; mixing litres and cubic centimetres. Final Answer: 4480

Q: Thermodynamics of solutions: When the temperature of an aqueous solution is increased (no mass is added or removed), which concentration measure decreases as a result of thermal expansion?

Introduction / Context: Molarity depends on the solution volume, while mass-based concentration measures do not. Heating an aqueous solution expands the liquid, changing its volume but not the number of moles present. This question tests whether you can distinguish volume-dependent from mass-dependent concentration units under temperature change. Given Data / Assumptions: No evaporation or mass loss; only temperature increases. Water and dissolved solute are treated as incompressible over small pressure changes but expand with temperature. Solution remains single phase and well mixed. Concept / Approach: Molarity M = moles of solute / litres of solution, so it varies inversely with solution volume. Weight percent and mole fraction depend on ratios of masses or moles, which do not change with simple heating. Molality m = moles of solute per kilogram of solvent is mass based and therefore temperature independent (neglecting tiny buoyancy effects). Step-by-Step Solution: Identify which quantities depend on Vsolution: only molarity (and normality/formality) do.Upon heating: Vsolution increases due to thermal expansion.Since nsolute constant, M = n/V decreases.Mass- or mole-based measures (wt %, x, molality) remain unchanged. Verification / Alternative check: A quick calculation: if volume rises by 1%, molarity falls by ~1% for fixed moles, confirming the trend. Why Other Options Are Wrong: Weight percent and mole fraction use mass or mole ratios, unaffected by expansion. Molality uses kilograms of solvent, also mass based. Formality behaves like molarity and would also decrease, but it is not the standard unit in this context. Common Pitfalls: Confusing molality with molarity; remembering “molality = mass based, temperature independent” helps avoid this error. Final Answer: molarity

Q: Ideal-gas mixture laws: Which of the following ideal-gas relations are not directly applicable to a mixture as a mixture property (without referencing components)?

Introduction / Context: Mixture behavior for ideal gases is built from component contributions. Two cornerstone mixture rules are Dalton’s law for pressures and Amagat’s law for volumes. In contrast, Boyle’s and Charles’s laws describe how a single ideal gas varies when one state variable changes, not a “mixture-as-a-single-species” without component resolution. Given Data / Assumptions: Ideal-gas behavior of each component. No interactions beyond ideal mixing. Mixture properties are sums of component partial properties. Concept / Approach: Dalton’s law: p = Σpi with each pi = yi p. Amagat’s law: V = ΣVi with Vi = yi,V V at common T, p. These are mixture-specific relations. Boyle’s and Charles’s laws apply to a single species’ p–V–T variation; applying them to a mixture without specifying composition or partial properties is not meaningful as a standalone “mixture law.” Step-by-Step Solution: Identify mixture laws: Dalton and Amagat are designed for mixtures.Recognize Boyle/Charles are single-component relations.Therefore, the items “not applicable to mixture (as mixture laws)” are Boyle’s & Charles’s laws. Verification / Alternative check: Standard thermodynamics texts present Dalton and Amagat under “ideal-gas mixtures,” but treat Boyle/Charles as single-gas limiting behaviors. Why Other Options Are Wrong: Dalton and Amagat explicitly apply to mixtures; “none of these” is false; Graham’s law concerns rates of effusion, not a mixture property law. Common Pitfalls: Assuming “ideal gas” automatically makes the mixture obey every single-gas formula without composition awareness. Final Answer: Boyle's law & Charles's law (as pure-gas relations for the mixture)

Q: Property estimation tools: The Othmer chart is primarily used for estimating which thermal property from boiling/temperature data?

Introduction / Context: The Othmer chart is a classic graphical correlation used in chemical engineering to estimate latent heats of vaporization from readily available boiling or temperature data when detailed thermophysical tables are unavailable. Recognizing what a named chart estimates helps avoid misapplication in design problems. Given Data / Assumptions: Othmer charts relate temperature/boiling data to approximate latent heats. The options listed focus on surface/solution phenomena (mixing, wetting, adsorption), which are different categories of enthalpy change. Concept / Approach: Heats of mixing, wetting, and adsorption are interfacial or solution enthalpies requiring different measurements or correlations. The Othmer approach targets vaporization enthalpy (lambda), a bulk phase-change property. Since “latent heat of vaporization” is not among options a–c, the correct choice is “none of these.” Step-by-Step Solution: Identify the property tied to Othmer charts: latent heat of vaporization.Compare with listed options: none matches vaporization heat.Therefore, select “none of these.” Verification / Alternative check: Handbooks show Othmer-type correlations linking lambda with normal boiling point or reduced temperature, confirming the intended use. Why Other Options Are Wrong: Mixing, wetting, and adsorption heats arise from molecular interactions at interfaces or in solutions and are not what Othmer charts predict. Common Pitfalls: Confusing any “heat” correlation with solution phenomena; keep phase-change enthalpies distinct from interfacial enthalpies. Final Answer: none of these

Q: Latent heat–boiling point relation: Which named relation links the latent heat of vaporization to the normal boiling point for many non-associating liquids?

Introduction / Context: Trouton’s rule states that the entropy of vaporization at the normal boiling point of many non-associating liquids is approximately constant, about 85–88 J/mol·K. This empirical regularity provides a simple link between latent heat and boiling point via ΔHvap ≈ Tb * ΔSvap. Given Data / Assumptions: Liquid is non-associating (no strong hydrogen bonding). Property evaluated at the normal boiling point. Use as an estimate, not an exact law. Concept / Approach: Because ΔSvap at Tb is roughly constant for many liquids, ΔHvap scales approximately with Tb. Antoine equation correlates vapour pressure vs temperature, not directly ΔHvap; Kopp’s rule estimates heat capacity from atomic contributions; Kistyakowsky is not the relevant common name here. Thus, the correct named relation is Trouton’s rule. Step-by-Step Solution: Recall ΔSvap ≈ constant at Tb.Therefore, ΔHvap ≈ Tb * constant.Select Trouton’s rule. Verification / Alternative check: Comparing liquids like benzene, toluene, hexane shows ΔHvap/Tb values clustering around Trouton’s constant, except for hydrogen-bonded liquids like water. Why Other Options Are Wrong: Antoine is Pvap–T; Kopp is for heat capacity; “Clapeyron integral” is a general thermodynamic relation, not the empirical rule asked. Common Pitfalls: Applying Trouton’s rule to strongly associating liquids leads to significant errors. Final Answer: Trouton's rule

Q: Temperature dependence of heat capacity: As temperature rises (within ordinary ranges), how does the heat capacity of a typical substance tend to change?

Introduction / Context: Heat capacity measures how much heat is needed to raise temperature by one degree. For most solids and liquids over moderate temperature ranges, Cp increases slightly with temperature as additional vibrational or configurational modes become accessible. Many gases also show increasing Cp with temperature due to activation of vibrational degrees of freedom. Given Data / Assumptions: Moderate pressures; no phase changes. Ordinary temperature spans used in engineering data. We are asking for the general trend, not exceptions. Concept / Approach: Empirical correlations (e.g., Shomate or polynomial Cp equations) typically show positive temperature coefficients for Cp in common ranges. While there are exceptions near transitions or decomposition, the general statement “Cp increases with T” is valid and widely used for preliminary calculations. Step-by-Step Solution: Consider common data tables for water, air, metals.Cp(T) generally exhibits a gentle upward slope with T.Therefore, select “increases.” Verification / Alternative check: Plotting Cp vs T for air (200–1000 K) or for water (liquid, 0–100 °C) confirms a mild increase. Why Other Options Are Wrong: “Decreases” and “unchanged” contradict typical data; “depends” is too vague for the standard generalization taught at this level. Common Pitfalls: Confusing specific cases (e.g., near critical points) with general behavior in standard ranges. Final Answer: increases

Question 1

Counting hydrogen ions: How many H+ ions are present in 1 c.c (1 mL) of an aqueous solution with pH = 13 at 25°C?

Accepted Answer

Introduction / Context: Translating pH values into actual counts of ions sharpens intuition for how dilute acidic or basic solutions are. This is useful in biochemical buffers, environmental sampling, and micro-scale analytical chemistry where tiny volumes are involved. Given Data / Assumptions: pH = 13 at 25°C, so [H+] = 10^-13 mol L^-1. Volume considered: 1 c.c = 1 mL = 1 × 10^-3 L. Avogadro’s number NA ≈ 6.023 × 10^23 mol^-1. Concept / Approach: The number of hydrogen ions equals moles of H+ in the sample times Avogadro’s number. Compute moles using concentration times volume, then scale by NA to get the particle count. High pH means extremely low [H+], so even in a millilitre, the count is surprisingly modest compared with NA. Step-by-Step Solution: Determine [H+]: [H+] = 10^-13 mol L^-1 (from pH definition).Compute moles in 1 mL: n = 10^-13 * 10^-3 = 10^-16 mol.Convert to number of ions: N = n * NA = 10^-16 * 6.023 × 10^23.Evaluate: N = 6.023 × 10^7 ions.Match to the option: 6.023 × 10^7. Verification / Alternative check: Using pOH = 1 implies [OH-] = 0.1 mol L^-1; the product [H+][OH-] = 10^-14 holds at 25°C, reaffirming [H+] = 10^-13 and the above count. Why Other Options Are Wrong: 10^13 and 6.023 × 10^13 are many orders too high; they would correspond to concentrated acidic conditions, not pH 13.6.023 × 10^10 is still three orders too large given the tiny molar amount in 1 mL. Common Pitfalls: Confusing millilitres with litres; forgetting to multiply by Avogadro’s number; mixing up [H+] and [OH-] at high pH. Final Answer: 6.023 × 10^7

Question 2

For estimating the molar heat capacity of a crystalline solid compound from its constituent elements, which empirical rule is used?

Accepted Answer

Introduction / Context: When tabulated heat capacities of complex solids are unavailable, engineers and chemists use empirical additivity rules to estimate values for preliminary design and calculations. Knowing which rule applies to solids can save significant effort in early-stage material and process evaluations. Given Data / Assumptions: We need an estimate of a crystalline solid’s molar heat capacity at constant pressure over moderate temperatures. Constituent elements and their atomic heat capacities are known or can be looked up. We accept empirical additivity with typical uncertainties. Concept / Approach: Kopp’s rule states that the molar heat capacity of a solid compound is approximately the sum of the atomic heat capacities of its constituent elements multiplied by their stoichiometric coefficients. It is useful near ambient to moderate temperatures where lattice vibrations dominate. Other named relations serve different purposes: Trouton’s rule estimates molar entropy of vaporization; the Antoine equation correlates vapor pressure with temperature; the Kistyakowsky relation appears in reaction rate or thermodynamic correlations, not in solid heat capacity additivity. Step-by-Step Solution: Identify the property: molar heat capacity of a solid compound.Select the appropriate empirical rule using atomic contributions.Kopp’s rule fits this description directly.Choose option “Kopp’s rule.” Verification / Alternative check: Comparisons with measured values for many inorganic salts show Kopp’s estimates are often within a reasonable engineering tolerance at room temperature, though deviations grow at high T or for complex structures. Why Other Options Are Wrong: Trouton’s rule deals with ΔSvap, not solid Cp.Antoine equation is for vapor pressure curves.Kistyakowsky equation is unrelated to solid heat capacity additivity. Common Pitfalls: Overextending Kopp’s rule outside its validity range; ignoring temperature dependence; mixing molar and mass-based heat capacities. Final Answer: Kopp’s rule

Question 3

Ideal gas change: if absolute temperature is tripled while pressure is simultaneously reduced to one-third, how does the gas volume change?

Accepted Answer

Introduction / Context: Quick “ratio” manipulations of the ideal gas law are common in compressor sizing, vessel relief scenarios, and reactor start-up calculations. Recognizing proportionalities avoids algebraic errors and speeds up decision-making under time pressure. Given Data / Assumptions: Ideal gas behavior applies. Initial state: P1, V1, T1; final state: P2 = P1/3, T2 = 3*T1. Amount of gas n is constant. Concept / Approach: For a fixed amount of ideal gas, V ∝ T/P. Therefore, a simultaneous change in T and P scales the volume by the factor (T2/T1) / (P2/P1). Tripling T increases V by a factor of 3; reducing P to one-third increases V by another factor of 3. Combined, the effect is multiplicative, producing a ninefold increase in volume. Step-by-Step Solution: Start from PV = nRT → V ∝ T/P (n constant).Compute ratio: (T2/T1) / (P2/P1) = 3 / (1/3) = 9.Thus V2 = 9 * V1.Pick the option indicating ninefold increase. Verification / Alternative check: Rearrange: P1V1/T1 = P2V2/T2 → V2 = V1 * (T2/T1) * (P1/P2) = V1 * 3 * 3 = 9 V1, confirming the ratio method. Why Other Options Are Wrong: Remain unaltered: ignores both effects.Increase three fold: accounts for only one of the two factors.Decrease three fold: opposite of the true combined effect. Common Pitfalls: Using Celsius instead of absolute temperature; adding factors instead of multiplying; changing n inadvertently (e.g., through leaks or phase change). Final Answer: Increase nine fold

Question 4

For gases other than monoatomic species, what is the typical molal heat capacity at constant volume (Cv, in kcal per kg-mol per K)?

Accepted Answer

Introduction / Context: Heat capacities of gases determine energy requirements for heating, cooling, and reacting streams. An order-of-magnitude grasp helps check calculations and select appropriate correlations without constantly consulting tables. Given Data / Assumptions: Gas molar heat capacity at constant volume for a monatomic ideal gas: Cv = (3/2)R. Gas constant R ≈ 1.987 kcal per kg-mol per K. Polyatomic gases have additional degrees of freedom (rotational and vibrational) that increase Cv above monatomic values at ordinary temperatures. Concept / Approach: For monatomic gases, Cv ≈ 1.5R ≈ 2.98 kcal per kg-mol per K (often rounded to about 3). Diatomic and polyatomic gases possess rotational (and at higher T, vibrational) modes, leading to Cv values exceeding ~3 kcal per kg-mol per K at ambient to moderate temperatures. Hence, for all but monatomic gases, Cv is typically greater than 3 on this unit basis. Step-by-Step Solution: Compute monatomic benchmark: 1.5 * 1.987 ≈ 2.98 ≈ 3.Recognize added degrees of freedom for non-monoatomic gases raise Cv.Conclude: for “all except monatomic,” Cv > 3 kcal per kg-mol per K.Select option “> 3.” Verification / Alternative check: Reference data: at room temperature, N2 and O2 have Cv near 5/2 R ≈ 4.97 kcal per kg-mol per K; CO2 and other polyatomics are typically higher, consistent with “> 3.” Why Other Options Are Wrong: 3 is approximately the monatomic value; non-monoatomic commonly exceed it.“< 3” or “< 1” contradict well-established degrees of freedom arguments and data. Common Pitfalls: Mixing mass-based (per kg) with molar (per kg-mol) bases; forgetting temperature dependence of vibrational contributions at high T. Final Answer: > 3

Question 5

For a solution formed by dissolving a nonvolatile solute in a volatile solvent, how does the vapor pressure compare with that of the pure solvent at the same temperature?

Accepted Answer

Introduction / Context: Vapor pressure lowering is one of the colligative properties of solutions. It underpins practical operations like evaporation, distillation, and humidity control. Knowing the qualitative trend helps in quickly diagnosing process behavior when solutes are present. Given Data / Assumptions: Solute is nonvolatile (it does not contribute to the vapor). Solvent is volatile and obeys Raoult’s law approximately at low solute concentrations. Temperature is fixed and uniform. Concept / Approach: Raoult’s law for an ideal solution states that the partial pressure of the solvent over the solution equals its mole fraction in the liquid times the pure-solvent vapor pressure: Psolution = xsolvent * P°solvent. Since xsolvent < 1 when a solute is present, Psolution < P°solvent. This reduction is independent of the solvent’s identity to first order and depends mainly on the solute mole fraction (a colligative effect). Step-by-Step Solution: Assume a nonvolatile solute so the vapor is essentially pure solvent.Apply Raoult’s law: Psolution = xsolvent * P°solvent.Because 0 < xsolvent < 1, conclude Psolution < P°solvent.Therefore, vapor pressure is “less than” that of the pure solvent. Verification / Alternative check: Experimental data for aqueous salt solutions show decreasing water activity (and thus lower vapor pressure) with increasing solute concentration, aligning with the law. Why Other Options Are Wrong: “More than” or “equal to” would require volatile solutes raising total pressure or pure solvent, which contradicts the stated nonvolatile condition.“Depends on the solvent” ignores the colligative nature of the effect for ideal solutions. Common Pitfalls: Confusing nonvolatile with volatile solutes; forgetting that strong nonideality or electrolytes need activity coefficients but still show vapor pressure lowering. Final Answer: Less than

Question 6

Stoichiometry at STP: In Ca + 2 H2O → Ca(OH)2 + H2, how many cubic centimetres of H2 at STP are released when 8 g of Ca reacts with excess water? (Atomic weight of Ca = 40)

Accepted Answer

Introduction / Context: Gas-evolution stoichiometry is frequently used in laboratory calculations and industrial gas balance estimations. Converting from mass of reactant to moles of gas, then to volume at standard conditions (STP), is a core skill for chemical engineering and chemistry students. Given Data / Assumptions: Balanced reaction: Ca + 2 H2O → Ca(OH)2 + H2. Molar mass of Ca = 40 g per mol. STP molar volume: 22,400 c.c per mol of an ideal gas. Water is in excess; reaction goes to completion with respect to Ca. Concept / Approach: Use stoichiometry to relate moles of calcium to moles of hydrogen gas. The balanced equation shows a 1:1 molar ratio between Ca and H2. Convert the given mass of Ca to moles, then convert moles of H2 to volume at STP using the standard molar volume. Step-by-Step Solution: Compute moles of Ca: nCa = 8 g / 40 g mol^-1 = 0.2 mol.Use stoichiometric ratio: nH2 = nCa = 0.2 mol.Convert to volume at STP: VH2 = 0.2 mol * 22,400 c.c mol^-1 = 4,480 c.c.Select the option equal to 4,480 c.c. Verification / Alternative check: Dimensional sanity: 8 g is one-fifth of a mole of Ca; one-fifth of 22,400 c.c yields 4,480 c.c, matching the computed value. Why Other Options Are Wrong: 2,240 c.c corresponds to 0.1 mol of H2; 1,120 c.c corresponds to 0.05 mol; 0.4 is a dimensionless number, not a volume. Common Pitfalls: Using the wrong molar mass or STP molar volume; forgetting the 1:1 stoichiometric relation; mixing litres and cubic centimetres. Final Answer: 4480

Question 7

Thermodynamics of solutions: When the temperature of an aqueous solution is increased (no mass is added or removed), which concentration measure decreases as a result of thermal expansion?

Accepted Answer

Introduction / Context: Molarity depends on the solution volume, while mass-based concentration measures do not. Heating an aqueous solution expands the liquid, changing its volume but not the number of moles present. This question tests whether you can distinguish volume-dependent from mass-dependent concentration units under temperature change. Given Data / Assumptions: No evaporation or mass loss; only temperature increases. Water and dissolved solute are treated as incompressible over small pressure changes but expand with temperature. Solution remains single phase and well mixed. Concept / Approach: Molarity M = moles of solute / litres of solution, so it varies inversely with solution volume. Weight percent and mole fraction depend on ratios of masses or moles, which do not change with simple heating. Molality m = moles of solute per kilogram of solvent is mass based and therefore temperature independent (neglecting tiny buoyancy effects). Step-by-Step Solution: Identify which quantities depend on Vsolution: only molarity (and normality/formality) do.Upon heating: Vsolution increases due to thermal expansion.Since nsolute constant, M = n/V decreases.Mass- or mole-based measures (wt %, x, molality) remain unchanged. Verification / Alternative check: A quick calculation: if volume rises by 1%, molarity falls by ~1% for fixed moles, confirming the trend. Why Other Options Are Wrong: Weight percent and mole fraction use mass or mole ratios, unaffected by expansion. Molality uses kilograms of solvent, also mass based. Formality behaves like molarity and would also decrease, but it is not the standard unit in this context. Common Pitfalls: Confusing molality with molarity; remembering “molality = mass based, temperature independent” helps avoid this error. Final Answer: molarity

Question 8

Ideal-gas mixture laws: Which of the following ideal-gas relations are not directly applicable to a mixture as a mixture property (without referencing components)?

Accepted Answer

Introduction / Context: Mixture behavior for ideal gases is built from component contributions. Two cornerstone mixture rules are Dalton’s law for pressures and Amagat’s law for volumes. In contrast, Boyle’s and Charles’s laws describe how a single ideal gas varies when one state variable changes, not a “mixture-as-a-single-species” without component resolution. Given Data / Assumptions: Ideal-gas behavior of each component. No interactions beyond ideal mixing. Mixture properties are sums of component partial properties. Concept / Approach: Dalton’s law: p = Σpi with each pi = yi p. Amagat’s law: V = ΣVi with Vi = yi,V V at common T, p. These are mixture-specific relations. Boyle’s and Charles’s laws apply to a single species’ p–V–T variation; applying them to a mixture without specifying composition or partial properties is not meaningful as a standalone “mixture law.” Step-by-Step Solution: Identify mixture laws: Dalton and Amagat are designed for mixtures.Recognize Boyle/Charles are single-component relations.Therefore, the items “not applicable to mixture (as mixture laws)” are Boyle’s & Charles’s laws. Verification / Alternative check: Standard thermodynamics texts present Dalton and Amagat under “ideal-gas mixtures,” but treat Boyle/Charles as single-gas limiting behaviors. Why Other Options Are Wrong: Dalton and Amagat explicitly apply to mixtures; “none of these” is false; Graham’s law concerns rates of effusion, not a mixture property law. Common Pitfalls: Assuming “ideal gas” automatically makes the mixture obey every single-gas formula without composition awareness. Final Answer: Boyle's law & Charles's law (as pure-gas relations for the mixture)

Question 9

Property estimation tools: The Othmer chart is primarily used for estimating which thermal property from boiling/temperature data?

Accepted Answer

Introduction / Context: The Othmer chart is a classic graphical correlation used in chemical engineering to estimate latent heats of vaporization from readily available boiling or temperature data when detailed thermophysical tables are unavailable. Recognizing what a named chart estimates helps avoid misapplication in design problems. Given Data / Assumptions: Othmer charts relate temperature/boiling data to approximate latent heats. The options listed focus on surface/solution phenomena (mixing, wetting, adsorption), which are different categories of enthalpy change. Concept / Approach: Heats of mixing, wetting, and adsorption are interfacial or solution enthalpies requiring different measurements or correlations. The Othmer approach targets vaporization enthalpy (lambda), a bulk phase-change property. Since “latent heat of vaporization” is not among options a–c, the correct choice is “none of these.” Step-by-Step Solution: Identify the property tied to Othmer charts: latent heat of vaporization.Compare with listed options: none matches vaporization heat.Therefore, select “none of these.” Verification / Alternative check: Handbooks show Othmer-type correlations linking lambda with normal boiling point or reduced temperature, confirming the intended use. Why Other Options Are Wrong: Mixing, wetting, and adsorption heats arise from molecular interactions at interfaces or in solutions and are not what Othmer charts predict. Common Pitfalls: Confusing any “heat” correlation with solution phenomena; keep phase-change enthalpies distinct from interfacial enthalpies. Final Answer: none of these

Question 10

Latent heat–boiling point relation: Which named relation links the latent heat of vaporization to the normal boiling point for many non-associating liquids?

Accepted Answer

Introduction / Context: Trouton’s rule states that the entropy of vaporization at the normal boiling point of many non-associating liquids is approximately constant, about 85–88 J/mol·K. This empirical regularity provides a simple link between latent heat and boiling point via ΔHvap ≈ Tb * ΔSvap. Given Data / Assumptions: Liquid is non-associating (no strong hydrogen bonding). Property evaluated at the normal boiling point. Use as an estimate, not an exact law. Concept / Approach: Because ΔSvap at Tb is roughly constant for many liquids, ΔHvap scales approximately with Tb. Antoine equation correlates vapour pressure vs temperature, not directly ΔHvap; Kopp’s rule estimates heat capacity from atomic contributions; Kistyakowsky is not the relevant common name here. Thus, the correct named relation is Trouton’s rule. Step-by-Step Solution: Recall ΔSvap ≈ constant at Tb.Therefore, ΔHvap ≈ Tb * constant.Select Trouton’s rule. Verification / Alternative check: Comparing liquids like benzene, toluene, hexane shows ΔHvap/Tb values clustering around Trouton’s constant, except for hydrogen-bonded liquids like water. Why Other Options Are Wrong: Antoine is Pvap–T; Kopp is for heat capacity; “Clapeyron integral” is a general thermodynamic relation, not the empirical rule asked. Common Pitfalls: Applying Trouton’s rule to strongly associating liquids leads to significant errors. Final Answer: Trouton's rule

Question 11

Temperature dependence of heat capacity: As temperature rises (within ordinary ranges), how does the heat capacity of a typical substance tend to change?

Accepted Answer

Introduction / Context: Heat capacity measures how much heat is needed to raise temperature by one degree. For most solids and liquids over moderate temperature ranges, Cp increases slightly with temperature as additional vibrational or configurational modes become accessible. Many gases also show increasing Cp with temperature due to activation of vibrational degrees of freedom. Given Data / Assumptions: Moderate pressures; no phase changes. Ordinary temperature spans used in engineering data. We are asking for the general trend, not exceptions. Concept / Approach: Empirical correlations (e.g., Shomate or polynomial Cp equations) typically show positive temperature coefficients for Cp in common ranges. While there are exceptions near transitions or decomposition, the general statement “Cp increases with T” is valid and widely used for preliminary calculations. Step-by-Step Solution: Consider common data tables for water, air, metals.Cp(T) generally exhibits a gentle upward slope with T.Therefore, select “increases.” Verification / Alternative check: Plotting Cp vs T for air (200–1000 K) or for water (liquid, 0–100 °C) confirms a mild increase. Why Other Options Are Wrong: “Decreases” and “unchanged” contradict typical data; “depends” is too vague for the standard generalization taught at this level. Common Pitfalls: Confusing specific cases (e.g., near critical points) with general behavior in standard ranges. Final Answer: increases

Question 12

Thermochemistry basics: The standard heat of reaction depends primarily on which state variable(s)?

Accepted Answer

Introduction / Context: The heat (enthalpy) of reaction reflects the enthalpy difference between products and reactants at a specified temperature and pressure. Kirchhoff’s law shows how the standard reaction enthalpy changes with temperature via the difference in heat capacities of products and reactants. Given Data / Assumptions: Ideal-gas or condensed phases away from strong compression effects. Reference to standard states (usually 1 bar) unless otherwise noted. Negligible pressure dependence of enthalpy for condensed phases and ideal gases. Concept / Approach: For ideal gases and most condensed phases, enthalpy is primarily a function of temperature. Pressure effects on enthalpy are small (zero for ideal gases). Therefore, the heat of reaction at standard pressure varies with T according to ΔH°(T2) = ΔH°(T1) + ∫(ΔCp) dT. Pressure may affect equilibrium and volume work but not the enthalpy change itself in the ideal approximation. Step-by-Step Solution: State ΔH depends on temperature through heat capacities.For ideal gases, ∂H/∂p = 0, so pressure has no direct effect.Therefore select “temperature.” Verification / Alternative check: Thermodynamic identities confirm enthalpy’s T dependence; Joule–Thomson effects are separate phenomena and small under standard conditions. Why Other Options Are Wrong: Pressure alone does not change ΔH° materially for ideal gases/condensed phases; “both” overstates; “neither” understates; “composition only” ignores the clear T dependence. Common Pitfalls: Confusing equilibrium position (pressure dependent) with reaction enthalpy (temperature dependent). Final Answer: temperature

Question 13

Ideal-gas identity: For an ideal gas, the difference between heat capacities at constant pressure and constant volume (Cp − Cv) equals what?

Accepted Answer

Introduction / Context: The Mayer relation links the constant-pressure and constant-volume heat capacities of an ideal gas: Cp − Cv = R. This identity follows from the ideal-gas equation of state combined with definitions of enthalpy and internal energy. Given Data / Assumptions: Ideal-gas behavior. Standard thermodynamic definitions: H = U + pV; for ideal gas, U = U(T) and H = H(T). Concept / Approach: Starting with dH = Cp dT and dU = Cv dT, and using H = U + pV and pV = RT per mole, differentiation yields Cp − Cv = R. This holds independent of the number of degrees of freedom for an ideal gas (which instead affects numerical values of Cp and Cv individually and thus gamma = Cp/Cv). Step-by-Step Solution: Per mole, pV = RT (ideal gas).H = U + pV → dH = dU + d(pV) = Cv dT + R dT.But dH = Cp dT, so Cp dT = Cv dT + R dT.Therefore, Cp − Cv = R. Verification / Alternative check: For a monatomic ideal gas: Cv = 3R/2, Cp = 5R/2; difference equals R, confirming the relation. Why Other Options Are Wrong: Fractions like R/2, multiples like 2R or 3R, and gamma*R are not general identities. Common Pitfalls: Mixing up Cp/Cv (gamma) with Cp − Cv. Gamma varies with molecular structure; Cp − Cv stays equal to R for any ideal gas. Final Answer: R

Question 14

Unit conversion check: One bar (1 bar) is approximately equal to how many atmospheres?

Accepted Answer

Introduction / Context: Engineering pressures may be quoted in bar or atmospheres. A quick conversion is essential for sanity checks and calculations. One atmosphere is 1.01325 bar, so 1 bar is about 0.987 atm—“approximately 1” for most practical problems. Given Data / Assumptions: 1 atm = 101.325 kPa; 1 bar = 100 kPa. Approximation acceptable for the question’s intent. Concept / Approach: Compute the ratio atm/bar or bar/atm: 1 bar / 1 atm = 100 / 101.325 ≈ 0.987, which is essentially 1 to two significant figures. Therefore, selecting “1” is the intended approximate equality. Step-by-Step Solution: Use kPa definitions.Take ratio: 100 / 101.325 ≈ 0.987.Therefore, 1 bar ≈ 1 atm (approx). Verification / Alternative check: Converting common pressures (e.g., 5 bar ≈ 4.93 atm) confirms the near-equality. Why Other Options Are Wrong: 10, 100, 1000 are orders of magnitude off; 0.1 would correspond to 0.1 atm, not 1 bar. Common Pitfalls: Confusing bar with mbar (millibar) or atm with “at” (technical atmosphere) in older literature. Final Answer: 1

Question 15

Water vapour pressure at the boiling point: At 100 °C, the equilibrium vapour pressure of pure water is closest to which of the following values?

Accepted Answer

Introduction / Context: By definition, at its normal boiling point (100 °C), water’s vapour pressure equals 1 atmosphere. Converting 1 atm into different pressure units is a frequent exam task and a useful field skill. Given Data / Assumptions: 1 atm ≈ 760 mm Hg ≈ 76 cm Hg ≈ 1.01325 bar. Mercury density ≈ 13.6 g/cm^3 (giving the 13.6 factor relating water vs mercury columns). Concept / Approach: Match 1 atm to the listed forms. 76 cm of mercury is equivalent to 760 mm of mercury, which equals approximately 1 atm at 0 °C calibration conditions. Therefore, “76 cm of Hg” is the correct choice among the options provided. Step-by-Step Solution: Recognize boiling point definition: Pvap = 1 atm.Convert 1 atm to a mercury column: 760 mm Hg = 76 cm Hg.Select “76 cm of Hg.” Verification / Alternative check: 1.013 bar also equals 1 atm; however it is not one of the offered correct forms here (unless explicitly listed). “76 cm Hg” matches exactly. Why Other Options Are Wrong: 100 N/m^2 is only 1 Pa, far too small; 13.6 cm Hg is 0.18 atm; 760 mm water column is much less than 1 atm due to water’s lower density; 1.013 bar would be correct numerically but is not the option chosen by the exam’s framing. Common Pitfalls: Mixing up water vs mercury manometer columns; remember the density factor 13.6. Final Answer: 76 cm of Hg

Question 16

Clausius–Clapeyron applicability: For deriving/using the Clausius–Clapeyron equation in its common integrated form for liquid–vapour systems, which assumptions are made?

Accepted Answer

Introduction / Context: The Clausius–Clapeyron equation relates vapour pressure to temperature via the latent heat of phase change. Its commonly used integrated form assumes certain simplifications that make the mathematics tractable and the result accurate over moderate temperature ranges. Given Data / Assumptions: Liquid–vapour equilibrium for a pure substance. Moderate pressures, away from the critical point. Latent heat treated as weakly temperature dependent over the interval of interest. Concept / Approach: Starting from the Clapeyron equation, dP/dT = ΔHvap / (T ΔV), the integrated form commonly used is ln P = −ΔHvap/R * (1/T) + constant. To reach this, two main assumptions are made: (1) the vapour behaves ideally so that Vvap ≈ RT/P, and (2) the liquid molar volume is negligible compared to the vapour volume so that ΔV ≈ Vvap. These lead directly to the integrated logarithmic relation. Step-by-Step Solution: Begin with dP/dT = ΔHvap /(T (Vvap − Vliq)).Assume Vliq ≪ Vvap ⇒ ΔV ≈ Vvap.Assume ideal vapour ⇒ Vvap = RT/P.Substitute and integrate to obtain ln P vs 1/T. Verification / Alternative check: Comparing with Antoine correlations shows good agreement over limited temperature ranges, validating the assumptions for many substances away from extremes. Why Other Options Are Wrong: Choosing only one assumption yields a less accurate or non-integrable simple form; “neither” is incorrect; the alternative in option e does not produce the standard integrated expression. Common Pitfalls: Applying the integrated form near the critical region or at high pressures where the assumptions break down. Final Answer: both (a) & (b)

Question 17

Psychrometry application (saturated air at 20 °C): A rigid vessel of volume 1000 m^3 contains air that is saturated with water vapour at a total pressure of 100 kPa and a temperature of 20 °C. Given that the saturation vapour (partial) pressure of w…

Accepted Answer

Introduction / Context: In psychrometry and gas–vapour calculations, the water vapour in a closed, saturated air–water system can be quantified using the ideal-gas relationship applied to the vapour phase. At a given temperature, the partial pressure of water vapour equals the saturation vapour pressure, independent of the total pressure. This question asks for the mass of water vapour in a large rigid vessel at 20 °C when the air is saturated. Given Data / Assumptions: Vessel volume V = 1000 m^3 (rigid). Total pressure P_total = 100 kPa at T = 20 °C. Saturation vapour pressure of water at 20 °C: p_sat = 2.34 kPa. Water vapour behaves ideally (reasonable at 20 °C and low partial pressure). Specific gas constant for water vapour: R_v ≈ 461.5 J/kg·K. Concept / Approach: For a saturated mixture, the partial pressure of water vapour equals the saturation pressure at the specified temperature. The vapour mass then follows from the ideal-gas equation in the form m = p_v * V / (R_v * T_abs). Only the vapour partial pressure is used, not the total pressure, because Dalton’s law applies and each component obeys the ideal-gas law with its own partial pressure. Step-by-Step Solution: Convert temperature to absolute: T_abs = 20 + 273.15 = 293.15 K.Use p_v = p_sat = 2.34 kPa = 2340 Pa.Apply m = p_v * V / (R_v * T_abs).Compute m ≈ (2340 Pa * 1000 m^3) / (461.5 J/kg·K * 293.15 K) ≈ 17.3 kg.Round to the nearest option: 17 kg. Verification / Alternative check: A quick order-of-magnitude check: at a few kPa and room temperature, ideal-gas density ρ ≈ p / (R_v * T) ≈ 2340 / (461.5 * 293) ≈ 0.0173 kg/m^3; multiplying by 1000 m^3 gives ≈ 17.3 kg, confirming the result. Why Other Options Are Wrong: 20, 25, 34 kg: these overestimate the vapour mass; they would imply higher p_v or lower T than given. 12 kg: underestimates the mass; inconsistent with the ideal-gas calculation at the stated saturation pressure. Common Pitfalls: Using total pressure instead of vapour partial pressure; forgetting to convert °C to K; mixing up the universal gas constant with the specific gas constant for water vapour; rounding too early. Final Answer: 17

Question 18

Expressing dilute sulfuric acid strength as normality: An aqueous solution contains 2.45% by weight H2SO4 and has specific gravity 1.011 (at the stated conditions). Calculate the acid strength expressed in normality (equivalents per litre).

Accepted Answer

Introduction / Context: Normality (N) is commonly used for acid–base titrations and process dosing when the stoichiometric equivalents matter. For sulfuric acid (H2SO4), which is diprotic, the normality is twice the molarity if both protons are titratable. Here we convert a weight percent solution with known density (specific gravity) into normality directly on a per-litre basis. Given Data / Assumptions: Mass fraction of H2SO4 = 2.45% w/w. Specific gravity (density) ≈ 1.011 g/mL ⇒ 1011 g per litre of solution. Equivalent weight of H2SO4 for acid–base reactions ≈ 98.08 / 2 ≈ 49 g/eq. Assume full 2-equivalent acidity of H2SO4 in this dilute range. Concept / Approach: Compute grams of H2SO4 present in one litre of solution from the given mass fraction and density, then divide by the equivalent weight to obtain equivalents per litre (normality). Alternatively, find molarity first and multiply by 2 for a diprotic acid; both routes should coincide for dilute solutions where full dissociation to 2 H+ is assumed in stoichiometric calculations. Step-by-Step Solution: Mass of 1 L solution = 1011 g.Mass of H2SO4 per L = 0.0245 * 1011 ≈ 24.77 g.Equivalents per L = (24.77 g) / (49 g/eq) ≈ 0.5055 eq/L.Therefore, Normality ≈ 0.5055 N.(Check via molarity: 24.77/98.08 ≈ 0.2525 mol/L; N = 2 * 0.2525 ≈ 0.505.) Verification / Alternative check: Independent routes (equivalent-weight method and molarity×2) agree within rounding, validating the computation and the assumption of full two-proton acidity for H2SO4 in such dilute solutions. Why Other Options Are Wrong: 0.2500, 0.2528: these represent molarity, not normality, effectively missing the factor of 2 for diprotic acid. 0.5000: rounded too aggressively; slightly low compared with the precise density-based calculation. 0.3000: not supported by either molarity or equivalent-weight calculations. Common Pitfalls: Confusing molarity with normality; ignoring the density when converting weight percent to per-litre basis; using molecular weight instead of equivalent weight for polyprotic acids. Final Answer: 0.5055

Question 19

Handy conversion: 1 cubic metre (1 m^3) is approximately equal to how many litres or cubic feet?

Accepted Answer

Introduction / Context: Unit conversions between SI and Imperial systems are routine in design, data sheets, and vendor specifications. Remembering a few anchor conversions improves accuracy and speed. One such anchor is the relation between cubic metres, litres, and cubic feet, used in tank sizing, ventilation, and gas metering. Given Data / Assumptions: 1 m = 100 cm; 1 L = 1 dm^3 = 0.001 m^3. 1 m^3 = 1000 L exactly (by definition). 1 ft = 0.3048 m exactly, hence 1 m^3 = (1 / 0.3048)^3 ft^3 ≈ 35.3147 ft^3. Concept / Approach: Use exact definitions for base units to compute derived volume conversions. When selecting from approximate multiple-choice options, choose the closest value to the exact conversion while rejecting clearly incorrect litre values that differ by orders of magnitude. Step-by-Step Solution: Compute litres: 1 m^3 = 1000 L (exact).Compute cubic feet: 1 m^3 ≈ 35.3147 ft^3.Among the options, only ≈ 35 ft^3 matches.Therefore, the best approximation is 35 ft^3. Verification / Alternative check: Reverse check: 35 ft^3 * (0.0283168 m^3/ft^3) ≈ 0.991 m^3, close to 1 m^3; the small difference is due to rounding of “35”. Why Other Options Are Wrong: 28 L and 4.5 L: off by a factor of ~36 and ~222 respectively; 1 m^3 is much larger (1000 L). 4.5 ft^3: too small; about one-eighth of a cubic metre is ~4.3 ft^3, not 1 m^3. 100 L: still an order of magnitude too small. Common Pitfalls: Confusing litres with cubic decimetres; mixing length and volume conversions; misremembering 1 ft = 0.3 m without squaring/cubing when moving to area/volume. Final Answer: 35 ft^3

Question 20

Specific heat capacity units (SI): Identify the correct SI unit for Cp, the specific heat capacity at constant pressure.

Accepted Answer

Introduction / Context: Specific heat capacity at constant pressure, Cp, quantifies the energy required to raise the temperature of a unit mass of a substance by one kelvin at constant pressure. Correct units matter in energy balances, heat-exchanger design, and transient heating calculations across process engineering. Given Data / Assumptions: Cp refers to a per-mass property (specific), not per-mole or per-volume. SI coherent units: energy in joules (J), temperature in kelvin (K), mass in kilograms (kg). Concept / Approach: By definition, Cp = (∂h/∂T)_p with h in J/kg (specific enthalpy) and T in K. Therefore, the unit becomes (J/kg)/K = J/kg·K. Distinguish this from thermal conductivity and heat flux units, which also include W and length dimensions but represent different physical properties. Step-by-Step Solution: Start with Cp definition: energy per mass per temperature rise at constant pressure.Energy unit: joule (J). Mass unit: kilogram (kg). Temperature unit: kelvin (K).Combine: Cp has unit J/(kg·K) = J/kg·K. Verification / Alternative check: In molar form, Cp,m would be J/(mol·K) or kJ/(kmol·K). The presence of “per mass” vs “per mole” distinguishes specific vs molar heat capacity. Why Other Options Are Wrong: W/m^2·K and W/m·K: these are heat transfer coefficient and thermal conductivity units, not heat capacity. J/m^3·K: a volumetric heat capacity, not specific (per mass) Cp. kJ/kmol·K: a molar heat capacity unit, not specific mass-based Cp. Common Pitfalls: Confusing Cp with k (thermal conductivity) or h (heat transfer coefficient); using molar units when data or balances are on a mass basis. Final Answer: J/kg·K

Stoichiometry Questions