Question 1

Vapor terminology: A vapor whose partial pressure is less than its saturation (equilibrium) vapor pressure at the same temperature is called a __________ vapor.

Accepted Answer

Introduction / Context: Precise terminology in phase equilibria helps avoid confusion in design of condensers, evaporators, and dryers. The relation between a vapor’s actual partial pressure and the saturation vapor pressure at the same temperature classifies the state as saturated, superheated, or supersaturated. Given Data / Assumptions: Single-component vapor or a vapor in a gas mixture at temperature T. Saturation pressure p_sat(T) is the equilibrium vapor pressure at T. Partial pressure p_v of the vapor is known and p_v < p_sat(T). Concept / Approach: When p_v = p_sat(T), the vapor is saturated and in equilibrium with its liquid. If p_v < p_sat(T) at the same temperature, the vapor has “excess temperature” relative to its saturation state: it is superheated. Supersaturated refers to a metastable state where p_v > p_sat(T) (or, equivalently, where relative humidity > 100% in gas–vapor mixtures), which is unstable and tends to condense upon nucleation. Step-by-Step Solution: Compare actual partial pressure p_v to p_sat(T).Given p_v < p_sat(T).Therefore, the vapor is superheated.Select “superheated.” Verification / Alternative check: On T–s or P–h diagrams, states to the right of the saturated vapor line at the same pressure are superheated; equivalently, at constant T, a vapor with lower pressure than saturation is superheated. Why Other Options Are Wrong: Saturated: requires p_v = p_sat(T). Supersaturated: implies p_v > p_sat(T), the opposite case. Wet: indicates a mixture containing liquid droplets, not appropriate here. None of these: incorrect; the standard term is superheated. Common Pitfalls: Confusing “supersaturated” with “superheated”; supersaturation is an overpressure (or overconcentration) state prone to condensation, whereas superheating is an underpressure state relative to saturation. Final Answer: superheated

Question 2

Cox chart used in distillation design (especially for petroleum hydrocarbons): which variables are plotted?

Accepted Answer

Introduction / Context: Cox charts are convenient nomographs for estimating vapor pressures and boiling points, widely used in petroleum fractionation design and quick column sizing. They provide near-linear relationships over practical ranges for homologous series, simplifying interpolation without full Antoine parameters. Given Data / Assumptions: Hydrocarbon or similar organic compounds in the relevant temperature range. Chart is constructed so that data align approximately as straight lines. Used for boiling-point–vapor-pressure correlation and component comparison. Concept / Approach: On a Cox chart, the ordinate is typically log10 of vapor pressure, and the abscissa is temperature (often °C). For many petroleum hydrocarbons, plotting log(vapor pressure) against temperature yields roughly straight lines, enabling interpolation/extrapolation of boiling temperatures at specified pressures or vice versa. This is distinct from the Clausius-type plots of log P vs. 1/T, which are common but not the Cox construction referenced in refinery practice. Step-by-Step Solution: Identify axes: y-axis = log10(P); x-axis = temperature.Recognize purpose: obtain boiling point at a given P, or P at a given T, by straight-line interpolation.Select the option that states “temperature vs. log(vapor pressure).”Therefore, choose option (a). Verification / Alternative check: Comparing with standard refinery references shows Cox charts plotted as log P versus T, with families of curves for related compounds that are approximately linear across relevant ranges. Why Other Options Are Wrong: Vapor pressure vs. temperature without the log scale yields curved plots, less linear. log(temperature) variants are not standard Cox constructions. Common Pitfalls: Confusing Cox charts with Clausius–Clapeyron (log P vs. 1/T) or Antoine-based linearizations; the Cox chart’s linearity advantage for hydrocarbons lies in log P versus T. Final Answer: temperature vs. log(vapor pressure)

Question 3

Stoichiometry of alumina formation: If 1.5 moles of O2 react with aluminium to form Al2O3, what mass of aluminium (atomic weight = 27) is consumed?

Accepted Answer

Introduction / Context: Simple stoichiometric calculations underpin material balances and reagent planning. Aluminium reacts with oxygen to form aluminium oxide (alumina), and the balanced equation provides the mole ratios needed to convert oxygen consumption into aluminium usage. Given Data / Assumptions: Balanced reaction: 4 Al + 3 O2 → 2 Al2O3 (equivalently, 2 Al + 1.5 O2 → Al2O3). Moles of O2 consumed: 1.5 mol. Atomic weight of Al: 27 g/mol. Concept / Approach: Use the stoichiometric coefficients from the balanced equation to relate moles of oxygen to moles of aluminium, then multiply by the molar mass to find the mass of aluminium consumed. Step-by-Step Solution: From 2 Al + 1.5 O2 → Al2O3, 1.5 mol O2 corresponds to 2 mol Al.Moles of Al required = 2 mol.Mass of Al = moles × molar mass = 2 × 27 g/mol = 54 g.Therefore, aluminium consumed = 54 g. Verification / Alternative check: Using the 4:3:2 stoichiometry also yields Al:O2 = 4:3 in moles; 1.5 mol O2 × (4/3) = 2 mol Al, as above. Why Other Options Are Wrong: 27 g would correspond to 1 mol Al, not enough for 1.5 mol O2. 5.4 g and 2.7 g are off by factors of 10. 81 g exceeds the stoichiometric requirement. Common Pitfalls: Failing to balance the reaction correctly or using atomic instead of molecular stoichiometry for oxygen (O2 vs. O). Final Answer: 54 g

Question 4

Degrees Brix (°Bx) for sugar solutions: which of the following formulas defines the Brix scale in terms of specific gravity G?

Accepted Answer

Introduction / Context: Degrees Brix (°Bx) is a widely used measure in food, beverage, and fermentation industries. It is defined as the mass percentage of sucrose in a sucrose–water solution. While hydrometers or refractometers are often used to estimate °Bx from density or refractive index, the formal definition is not a specific gravity formula but rather a weight percent specification tied to sucrose content. Given Data / Assumptions: G denotes specific gravity at a defined reference temperature. Hydrometer scales (e.g., °Baumé, °API) have simple algebraic relations to G. Brix is defined as % (w/w) sucrose; empirical correlations to G exist but are not the definition. Concept / Approach: By definition, 1 °Bx = 1 g sucrose per 100 g solution. Any relationship converting specific gravity to °Bx is an empirical correlation valid over a range and must be referenced to temperature and solution composition (sucrose-only). The choices given match other hydrometer scales: °Baumé for heavy liquids uses 145 − 145/G; °API uses 141.5/G − 131.5; and other linear forms are not the Brix definition. Therefore, none of the listed formulas is the definition of °Bx. Step-by-Step Solution: Recall the definition: °Bx = mass % sucrose.Compare with options: each gives a specific algebraic function of G.Identify known scales: option (c) is °Baumé (heavy liquids); option (e) is °API (petroleum).Conclude that none of the given formulas defines the Brix scale. Verification / Alternative check: Industry references define °Bx via sucrose mass fraction; tables and polynomial fits convert G to °Bx, but no single simple linear formula universally defines it across concentrations and temperatures. Why Other Options Are Wrong: (400/G) − 400 and 200(G − 1): not standard definitions for °Bx. 145 − (145/G): corresponds to °Baumé for heavy liquids, not °Brix. (141.5/G) − 131.5: °API gravity (petroleum), not °Brix. Common Pitfalls: Assuming °Brix is a hydrometer scale like °Baumé or °API. Remember: °Brix is a composition definition; instruments infer it via correlations from density or refractive index. Final Answer: none of these

Question 5

Combustion thermodynamics: The maximum adiabatic flame temperature attainable when a fuel burns in air is __________ the maximum adiabatic flame temperature when the same fuel burns in pure oxygen at the same initial conditions.

Accepted Answer

Introduction / Context: Adiabatic flame temperature is the theoretical maximum temperature a reacting mixture can reach when combustion proceeds adiabatically (no heat loss to surroundings) and at specified initial conditions. It is a key design parameter for furnaces, burners, gas turbines, and safety analysis in chemical engineering. Given Data / Assumptions: Same fuel, same initial temperature and pressure in both cases. Complete combustion assumed. Adiabatic conditions (no heat loss), negligible kinetic and potential energy changes. Comparison between two oxidizers: air (contains ~21% O2 and ~79% N2 by volume) versus pure oxygen (100% O2). Concept / Approach: The adiabatic flame temperature is found from an energy balance: chemical heat release (lower or higher heating value depending on water phase) is converted into sensible heating of the combustion products. When air is used, a large quantity of inert nitrogen is present. This nitrogen is heated along with the products, absorbing a significant portion of the released energy and thereby lowering the peak temperature. With pure oxygen, there is no nitrogen ballast; fewer total moles are heated, resulting in a higher adiabatic flame temperature. Step-by-Step Solution: Write energy balance: heat of reaction = sum of sensible heats of products.Case 1 (air): products include CO2, H2O, and substantial N2 that must be heated.Case 2 (O2): products are mainly CO2 and H2O; no nitrogen ballast to heat.For the same heat release, adding inert N2 increases heat capacity of the mixture, therefore the final adiabatic temperature is lower with air. Verification / Alternative check: Look up tabulated adiabatic flame temperatures: for common fuels (methane, hydrogen, CO), values in pure oxygen exceed values in air by several hundred kelvin, confirming the reasoning. Why Other Options Are Wrong: Higher than: contradicts the inert-dilution effect of nitrogen in air.Same as: would imply no effect of nitrogen ballast, which is false.Not related to: adiabatic flame temperature is directly related to oxidizer composition. Common Pitfalls: Confusing real flames (with heat losses, dissociation) with ideal adiabatic values; forgetting dissociation at very high temperatures can also lower the temperature but does not reverse the comparison trend. Final Answer: lower than

Question 6

Mixed ideal gases and basic solution concepts: Identify the single incorrect statement among the following (assume ideal-gas behavior and standard molar volume 22.4 L/mol at N.T.P., so that 1 kmol occupies 22.4 Nm^3).

Accepted Answer

Introduction / Context: This problem checks fluency with multiple foundational ideas: normality and dilution in solutions, Dalton’s law for gas mixtures, Avogadro’s law for gas volumes at N.T.P., and mole-to-volume conversions. Mastering these enables quick screening of statements for consistency. Given Data / Assumptions: N.T.P. molar volume: 22.4 L/mol; therefore 1 kmol occupies 22.4 Nm^3. Ideal-gas behavior and Dalton’s law for partial pressures. Equal weights of O2 and CH4 are mixed with no reaction. Ozone O3 contains three oxygen atoms per molecule. Concept / Approach: For gases, partial pressure fraction equals mole fraction (at fixed T,V). For solutions, normality scales inversely with dilution factor. At N.T.P., volume is proportional to moles; converting “atoms” in O3 to “molecules” divides the atom count by 3 before using the molar volume. Step-by-Step Solution: (a) Dilution: dividing solute per liter by 10 reduces normality to N/10 → correct.(b) Equal weights of O2 and CH4: molar masses are 32 (O2) and 16 (CH4). If masses are equal, n_O2 : n_CH4 = (m/32):(m/16) = 1:2. Hence y_O2 = 1/3, not 1/2; oxygen contributes one-third of total pressure → statement is incorrect.(c) 9.034 × 10^23 oxygen atoms in O3 correspond to 9.034e23/3 = 3.011e23 molecules ≈ 0.5 mol; volume = 0.5 × 22400 c.c = 11200 c.c → correct.(d) 1 kmol occupies 22.4 Nm^3 at N.T.P. → correct by definition. Verification / Alternative check: Compute mole fractions explicitly for part (b) to reaffirm y_O2 = 0.333… and y_CH4 = 0.666…, so the oxygen partial pressure is one-third of the total. Why Other Options Are Wrong: They are not wrong; each is consistent with basic definitions and calculations. Common Pitfalls: Equating equal masses with equal moles; misreading N.T.P. molar volume as 22400 Nm^3 instead of 22.4 Nm^3; forgetting to divide by 3 when converting oxygen atoms to O3 molecules. Final Answer: When equal weights of O2 and CH4 are mixed at room temperature, oxygen exerts one-half of the total pressure.

Question 7

Raoult’s law prerequisites: Raoult’s law states that the equilibrium vapor pressure exerted by a component in a solution is proportional to its mole fraction in the liquid phase. This generalization assumes which of the following about the solution …

Accepted Answer

Introduction / Context: Raoult’s law underpins many introductory vapor–liquid equilibrium (VLE) treatments. It works best for ideal or nearly ideal solutions—typically those with similar molecular sizes and intermolecular forces—so that deviations from ideality are small. Given Data / Assumptions: Binary or multicomponent liquid solutions in equilibrium with their vapor. No chemical reaction or association between unlike molecules. Non-polar or weakly polar species of comparable size. Concept / Approach: Raoult’s law states p_i = x_i * p_i^sat. This relation follows from the ideal-solution model, which assumes the partial molar properties mirror those of the pure liquids at the same temperature. Ideal-solution behavior is expected when molecular interactions between like and unlike species are similar, and when molecules have comparable sizes and shapes. Strong polarity, hydrogen bonding, or specific association leads to nonideal behavior and the need for activity coefficients. Step-by-Step Solution: Identify conditions for ideality: similar size, similar interaction strength, and absence of complexation or association.Match with options: (a), (b), and (c) each reflect these conditions.Therefore, all three are required for the best applicability of Raoult’s law. Verification / Alternative check: Mixing chemically similar hydrocarbons (e.g., hexane–heptane) typically obeys Raoult’s law closely. In contrast, polar–nonpolar mixtures (e.g., ethanol–hexane) show large deviations and require activity-coefficient models. Why Other Options Are Wrong: Selecting only one or two assumptions ignores the others that also materially affect ideality. Common Pitfalls: Assuming Raoult’s law applies to strongly associating systems; forgetting that activity coefficients (γ_i) correct for nonideality: p_i = x_i * γ_i * p_i^sat. Final Answer: All of (a), (b), and (c).

Question 8

Boiling and molecular energies: A pure liquid is in equilibrium with its vapor exactly at its boiling point. On average, molecules in the liquid phase and in the vapor phase have equal __________.

Accepted Answer

Introduction / Context: Thermal equilibrium between phases at the boiling point is a cornerstone of phase-change thermodynamics. Understanding which molecular-average properties are equal clarifies what is controlled by temperature and what is controlled by phase structure. Given Data / Assumptions: Pure substance at its boiling point. Liquid and vapor phases coexisting at equilibrium (no net evaporation or condensation). Uniform temperature across both phases. Concept / Approach: Average translational kinetic energy of molecules in any idealized phase depends solely on absolute temperature. Since both phases are at the same temperature at equilibrium, molecules in liquid and vapor have the same average kinetic energy. However, potential energy and total energy differ due to distinct intermolecular configurations and binding energies in the phases. Intermolecular forces are stronger in liquids; in vapors, molecules are much farther apart on average. Step-by-Step Solution: Equal temperature at equilibrium implies equal average molecular kinetic energy.Potential energy differs because of different average separations and bonding environments.Total energy (kinetic + potential) therefore differs between phases. Verification / Alternative check: From kinetic theory: average translational kinetic energy per molecule is (3/2) k_B T (idealized); same T in both phases → same kinetic energy. Why Other Options Are Wrong: Intermolecular forces of attraction: stronger in liquids; not equal.Potential energy: lower (more negative) in liquids due to stronger interactions; not equal.Total energy: sum differs because potential energy differs. Common Pitfalls: Equating equal temperature with equal total energy; forgetting that phase changes involve latent heat which reflects potential-energy changes at constant temperature. Final Answer: kinetic energy

Question 9

Thermal unit conversion check: Select the correct conversion between British Thermal Units and calories (small-c).

Accepted Answer

Introduction / Context: Accurate energy-unit conversions are essential in energy balances, heat-exchanger design, and utility billing. Two common non-SI units are the British Thermal Unit (BTU) and the calorie (small-c, defined relative to heating water). Given Data / Assumptions: Use the common engineering approximation: 1 BTU ≈ 252 cal (small calories). Be aware that 1 kcal (Cal, food calorie) = 1000 cal. Concept / Approach: Standard reference conversions are: 1 BTU = 1055 J (approx), and 1 cal = 4.186 J. Combining these gives 1 BTU / 1 cal ≈ 1055 / 4.186 ≈ 252. Hence the correct direct conversion is 1 BTU ≈ 252 cal. Step-by-Step Solution: Start from SI: 1 cal = 4.186 J.Also: 1 BTU ≈ 1055 J.Compute ratio: 1055 / 4.186 ≈ 252 → 1 BTU ≈ 252 cal. Verification / Alternative check: Equivalently, 1 kcal ≈ 3.968 BTU. Multiplying 252 cal by 3.968×10^-3 BTU/cal returns approximately 1 BTU, confirming consistency. Why Other Options Are Wrong: 453.6 calories: likely confusion with mass conversions (1 lb = 453.6 g); not an energy relation.1 calorie = 252 BTU and 1 calorie = 453.6 BTU invert the ratio and are off by orders of magnitude. Common Pitfalls: Mixing calories (cal) with kilocalories (kcal); confusing pound-mass conversion with energy conversion; rounding inconsistently. Final Answer: 1 BTU = 252 calories

Question 10

Colligative property relationship: For dilute solutions, osmotic pressure is directly proportional to which of the following quantities?

Accepted Answer

Introduction / Context: Osmotic pressure is a classic colligative property, meaning it depends on the number of solute particles rather than their identity (for ideal dilute solutions). It connects directly to other colligative phenomena such as vapor-pressure lowering, boiling-point elevation, and freezing-point depression. Given Data / Assumptions: Dilute, ideal solutions (van’t Hoff law applies). Non-electrolyte solute unless otherwise specified; for electrolytes, use the van’t Hoff factor i. Constant solvent properties aside from colligative effects. Concept / Approach: The van’t Hoff equation for ideal dilute solutions is π = i C R T, where π is osmotic pressure, C is molar concentration of solute, R is the gas constant, and T is absolute temperature. Vapor-pressure lowering Δp_v / p_v^0 ≈ x_solute (mole fraction), and for dilute solutions, C is proportional to x_solute at fixed solvent amount. Hence osmotic pressure is proportional to solute concentration and to absolute temperature, and it correlates linearly with the relative lowering of vapor pressure. Step-by-Step Solution: Relate π to concentration and temperature: π ∝ C and π ∝ T.Relate π to vapor-pressure lowering: both π and Δp_v are proportional to the number of solute particles (colligative), yielding a direct proportionality under dilute conditions.Therefore, all three statements hold in the ideal, dilute limit. Verification / Alternative check: Derivations from chemical potential equality across a semipermeable membrane lead to πV = n_s R T, consistent with van’t Hoff’s law and the proportionalities stated. Why Other Options Are Wrong: Picking only one or two ignores the complete set of proportionalities valid for dilute, ideal solutions. Common Pitfalls: Applying these relations to concentrated or strongly nonideal solutions without activity corrections; neglecting ionization (van’t Hoff factor i) for electrolytes. Final Answer: All of (a), (b), and (c)

Question 11

pH comparison and acidity scale: Two aqueous solutions are labeled A1 (pH = 2) and A2 (pH = 6). Which statement best quantifies their relative acidity/alkalinity?

Accepted Answer

Introduction / Context: The pH scale is logarithmic: pH = −log10[H+]. Each one-unit change in pH corresponds to a tenfold change in hydrogen-ion concentration. Understanding this helps compare the strength of acidity or basicity between solutions quickly and precisely. Given Data / Assumptions: Solution A1: pH = 2 → [H+] = 10^−2 mol/L. Solution A2: pH = 6 → [H+] = 10^−6 mol/L. Both are aqueous solutions at comparable conditions; activity effects neglected for simplicity. Concept / Approach: Relative acidity is determined by the ratio of hydrogen-ion concentrations. A difference of 4 pH units means a factor of 10^4 between [H+] values. Lower pH corresponds to higher [H+], hence higher acidity. Step-by-Step Solution: Compute [H+] for A1: 10^−2.Compute [H+] for A2: 10^−6.Take the ratio: (10^−2) / (10^−6) = 10^(−2 + 6) = 10^4 = 10,000.Therefore, A1 is 10,000 times more acidic (greater [H+]) than A2. Verification / Alternative check: The pOH or hydroxide concentration would show the inverse trend: A2 is more basic than A1 by the same factor, consistent with pH + pOH ≈ 14 at 25°C for dilute aqueous solutions. Why Other Options Are Wrong: “A2 is more acidic”: contradicts the pH values (higher pH means less acidic).“A1 is very slightly acidic”: pH 2 corresponds to strong acidity, not slight.“Both (a) and (c)”: includes a false statement, so the combined option is incorrect. Common Pitfalls: Forgetting the logarithmic nature of pH; thinking a 1-unit change is small when it is a tenfold change in [H+]; mixing up acidity with alkalinity. Final Answer: A1 is 10,000 times more acidic than A2.

Question 12

Kinetic theory link: For molecular speeds in a gas, the characteristic speed (e.g., root-mean-square speed) varies with absolute temperature T as __________.

Accepted Answer

Introduction / Context: The kinetic theory of gases relates molecular speed distributions to temperature and molecular mass. The root-mean-square (rms) speed, most probable speed, and average speed all scale with the square root of temperature for a given gas. Given Data / Assumptions: Ideal-gas behavior; negligible intermolecular potential effects for the scaling argument. Fixed molar mass M for the gas under consideration. Concept / Approach: The rms speed is u_rms = √(3 R T / M) for a gas of molar mass M. Both the average speed and the most probable speed have similar √T dependence with different constants (u_avg = √(8 R T / (π M)), u_mp = √(2 R T / M)). Therefore, any representative molecular speed scales as the square root of absolute temperature, not linearly with T. Step-by-Step Solution: Recall u_rms = √(3 R T / M).Hold M and R constant for a given gas.Therefore, u_rms ∝ √T. Verification / Alternative check: Plotting measured speed versus temperature on log–log axes gives a slope of 0.5, confirming the square-root dependence. Why Other Options Are Wrong: Proportional to T or T^2: would vastly overpredict speed increases with temperature.Inversely proportional to T: contradicts the fundamental kinetic theory relation. Common Pitfalls: Confusing pressure dependence with temperature scaling; assuming linearity out of habit; ignoring molecular mass in the complete expression. Final Answer: directly proportional to √T

Question 13

Gas-phase nonideality in VLE modeling: In vapor–liquid equilibrium calculations, which of the following does not represent gas-phase deviation from ideal-gas behavior?

Accepted Answer

Introduction / Context: Accurate VLE modeling requires accounting for nonideal behavior in both phases. Gas-phase deviations are often treated via equations of state or fugacity coefficients, while liquid-phase deviations are captured by activity coefficients. Knowing which tool belongs to which phase avoids modeling errors. Given Data / Assumptions: Nonreacting vapor–liquid equilibrium. Potential nonideality in either phase. Standard thermodynamic framework using fugacity to express chemical potential. Concept / Approach: The vapor-phase fugacity of component i is f_i^v = φ_i y_i P, where φ_i is the fugacity coefficient computed from an equation of state or virial method. The liquid-phase fugacity is f_i^l = γ_i x_i f_i^0 (or γ_i x_i p_i^sat for simple models), where γ_i is the activity coefficient from a liquid nonideality model (e.g., Wilson, NRTL, UNIQUAC). Thus, equations of state and φ are vapor-phase nonideality tools; γ belongs to the liquid phase. Step-by-Step Solution: Identify phase: (a) and (b) are vapor-phase corrections.Recognize that γ belongs to liquid-phase modeling.Therefore, the item that does not address gas-phase nonideality is the activity coefficient. Verification / Alternative check: Common practice: φ–φ methods (EoS for both phases) or γ–φ methods (activity model for liquid, EoS for vapor) illustrate the separation of roles. Why Other Options Are Wrong: Equation of state and fugacity coefficient specifically target vapor-phase nonideality; they are not incorrect. Common Pitfalls: Using γ for the gas phase; forgetting that γ approaches 1 for ideal liquids, while φ approaches 1 for ideal gases under low P and high T. Final Answer: Activity coefficient (γ) for the liquid phase

Question 14

In solution thermodynamics, which relation is obeyed by an ideal (Raoultian) liquid solution over the entire composition range?

Accepted Answer

Introduction / Context: An ideal liquid solution is a cornerstone model in chemical engineering thermodynamics. It assumes that intermolecular interactions between unlike molecules are the same as those between like molecules, so escaping tendencies scale directly with composition. Knowing which law characterizes such solutions helps in vapor–liquid equilibrium (VLE) calculations, distillation design, and solvent selection. Given Data / Assumptions: Binary or multicomponent liquid mixture with chemically similar components. Temperature and pressure are such that components behave ideally in the liquid phase. Vapor phase is near ideal so the basic form of the law applies. Concept / Approach: For an ideal liquid solution, the partial pressure of each component i is p_i = x_i * P_i^sat at the system temperature, where x_i is the liquid mole fraction and P_i^sat is the pure-component saturation pressure. This is Raoult's law. Other listed relations have different scopes: Boyle's law is for gases (P ∝ 1/V at constant T), Amagat's law concerns additive volumes of gas mixtures, and Trouton's rule gives approximate latent heat–boiling point trends for many liquids. Henry's law applies for dilute solutes in non-ideal (or solvent-dominated) systems, not for ideal solutions over the full composition range. Step-by-Step Solution: Identify the definition of an ideal liquid solution: activity coefficients ≈ 1 for all components.Match to the constitutive relation: p_i = x_i * P_i^sat → Raoult's law.Conclude that Raoult's law is the appropriate answer. Verification / Alternative check: Experimental VLE data of near-ideal pairs (e.g., benzene–toluene) closely follow Raoult's predictions, validating the model under similarity in molecular interactions. Why Other Options Are Wrong: Boyle's law and Amagat's law apply to gases, not liquids; Trouton's rule is an empirical boiling/latent heat correlation; Henry's law handles dilute, not full-range ideality. Common Pitfalls: Confusing Raoult's and Henry's domains; assuming ideality for highly dissimilar molecules. Final Answer: Raoult's law

Question 15

Raoult’s law design question: A non-volatile solute lowers a solvent’s vapor pressure by 10 mm Hg when its mole fraction is 0.2. To achieve a 20 mm Hg lowering under the same conditions, what should be the solvent mole fraction?

Accepted Answer

Introduction / Context: Vapor-pressure lowering by a non-volatile solute is a classic colligative property described by Raoult’s law for ideal or near-ideal solutions. Engineers use this proportionality for estimating boiling point elevation, designing evaporators, and checking composition from pressure data. Given Data / Assumptions: Initial vapor-pressure lowering ΔP_1 = 10 mm Hg at x_solute,1 = 0.2. Target vapor-pressure lowering ΔP_2 = 20 mm Hg. Same temperature and same solvent; non-volatile solute; ideal-solution assumption. Concept / Approach: For an ideal solution with a non-volatile solute, vapor-pressure lowering ΔP = x_solute * P^0, where P^0 is the pure-solvent saturation pressure at the temperature. Because P^0 is constant at fixed T, ΔP is directly proportional to the solute mole fraction. Thus, doubling ΔP requires doubling x_solute, and the solvent mole fraction is x_solvent = 1 − x_solute. Step-by-Step Solution: From ΔP ∝ x_solute, compute x_solute,2 = (ΔP_2 / ΔP_1) * x_solute,1 = (20/10)*0.2 = 0.4.Then x_solvent,2 = 1 − 0.4 = 0.6.Therefore, the solvent mole fraction should be 0.6. Verification / Alternative check: Using the more explicit Raoult relation: P_solvent = x_solvent * P^0 → ΔP = P^0 − x_solvent P^0 = (1 − x_solvent)P^0 = x_solute P^0. Doubling ΔP doubles x_solute, so the result is consistent. Why Other Options Are Wrong: 0.2 and 0.1 correspond to smaller solvent fractions than required; 0.4 would not double ΔP; 0.8 would halve ΔP relative to the initial case. Common Pitfalls: Confusing solute and solvent fractions; forgetting that P^0 is constant at constant temperature. Final Answer: 0.6

Question 16

Combined gas law reasoning: If the pressure of a gas is reduced to one-half and its absolute temperature is doubled, how does its volume change (amount of gas fixed)?

Accepted Answer

Introduction / Context: Many rapid process estimates in gas handling rely on the combined gas law. Understanding proportional relationships between pressure, temperature, and volume helps in sizing vessels, relief systems, and compressors when moles are constant. Given Data / Assumptions: Ideal-gas approximation holds. Amount of gas (n) is constant. P2 = 0.5 P1 and T2 = 2 T1; absolute temperatures used. Concept / Approach: The combined gas law states: P V / T = constant for a fixed amount of ideal gas. Therefore, V ∝ T/P at constant n. Compute the ratio V2/V1 by applying the given changes in T and P directly to this proportionality. Step-by-Step Solution: Write V2/V1 = (T2/P2) / (T1/P1).Substitute T2 = 2T1 and P2 = 0.5P1 → V2/V1 = (2T1/0.5P1) / (T1/P1) = (2/0.5) = 4.Hence, the volume increases by a factor of four. Verification / Alternative check: Use PV = nRT: If P halves and T doubles, the right-hand side doubles while the left-hand side has half the pressure; volume must quadruple to balance the equation. Why Other Options Are Wrong: One-quarter is the inverse of the correct factor; two times ignores the pressure change magnitude; no change or half-volume contradicts the combined proportionality. Common Pitfalls: Forgetting to use absolute temperature; mixing up direct and inverse proportionalities of T and P with V. Final Answer: increases four times

Question 17

Gas-mixture volumes: “The total volume of a gas mixture equals the sum of the pure-component volumes at the mixture pressure and temperature.” This statement corresponds to which law?

Accepted Answer

Introduction / Context: Two additive laws describe ideal gas mixtures: Dalton’s law of additive partial pressures and Amagat’s law of additive partial volumes. Correctly distinguishing them is crucial for multicomponent gas calculations and mixture property estimates. Given Data / Assumptions: Ideal or near-ideal gas behavior. Same temperature and pressure for components and mixture. Statement explicitly refers to additivity of volumes. Concept / Approach: Amagat’s law states that the total volume of a gas mixture equals the sum of the partial volumes that each component would occupy alone at the mixture temperature and pressure. Dalton’s law, in contrast, states that total pressure equals the sum of partial pressures at the mixture temperature and volume. Gay-Lussac’s and Avogadro’s laws concern proportionalities for pure gases, not mixture additivity rules. Step-by-Step Solution: Identify key phrase: “sum of pure component volumes.”Match with Amagat’s law of additive volumes.Select Amagat’s law as the correct option. Verification / Alternative check: For ideal gases, Amagat’s and Dalton’s laws can be shown to be consistent and lead to the same mixture relationships because PV = nRT applies component-wise. Why Other Options Are Wrong: Dalton’s law refers to pressures, not volumes; Gay-Lussac’s and Avogadro’s are single-gas proportionalities; Boyle’s is P–V inverse relationship for a pure gas at constant temperature. Common Pitfalls: Swapping Dalton’s and Amagat’s because both are “additive” statements; always note whether the additive variable is pressure or volume. Final Answer: Amagat's law

Question 18

Units check in mechanics and thermodynamics: Which of the following is not a unit of pressure?

Accepted Answer

Introduction / Context: Unit consistency prevents costly design mistakes. Pressure units commonly encountered include Pa (N/m^2), kPa, bar, atm, and torr. Recognizing non-pressure units avoids dimensional errors in calculations and software inputs. Given Data / Assumptions: We seek the item that is not dimensionally pressure. Standard SI and engineering units considered. Concept / Approach: Pressure has dimensions of force per area. SI base unit is pascal (Pa) = N/m^2. 1 kPa = 10^3 Pa; 1 bar = 10^5 Pa; 1 torr ≈ 133.322 Pa. A parsec is an astronomical unit of length (approximately 3.086 × 10^16 m), unrelated to pressure. Step-by-Step Solution: Check each option's dimensionality.kPa, bar, and N/m^2 are pressure; torr is also a pressure unit.Parsec is length; therefore, not a pressure unit. Verification / Alternative check: Convert 1 bar to Pa (10^5 Pa) and compare with kPa and N/m^2 to confirm they are consistent pressure units. Why Other Options Are Wrong: They are all legitimate pressure units used across disciplines (SI, metric, and vacuum technology). Common Pitfalls: Confusing torr (pressure) with toroidal geometry; assuming bar is non-SI and therefore not a pressure unit—it is non-SI but widely accepted for pressure. Final Answer: Parsec

Question 19

Heat capacity of solids vs. temperature: How does the atomic (molar) heat capacity of solid elements vary as temperature decreases?

Accepted Answer

Introduction / Context: Understanding how heat capacity varies with temperature is central to low-temperature operations, cryogenics, and solid-state process design. Classical Dulong–Petit behavior (approximately constant molar heat capacity near room temperature) breaks down at low temperatures, where quantum effects govern. Given Data / Assumptions: Pure solid elements; temperature decreasing from ambient toward cryogenic conditions. No phase transitions in the considered range. Concept / Approach: As temperature decreases, vibrational modes in a crystal are progressively “frozen out.” Debye theory predicts C_v ∝ T^3 at sufficiently low temperatures. Consequently, the atomic (molar) heat capacity decreases as temperature falls and tends toward zero in the limit of 0 K (absolute zero). It does not approach zero at 0°C; that option confuses Celsius with Kelvin. Step-by-Step Solution: Recall low-temperature limit: C_v ≈ a * T^3 for many solids → decreasing function of T.At intermediate/room temperatures, C_v approaches classical values but still generally drops as T is lowered.Hence, the correct qualitative trend is “decreases.” Verification / Alternative check: Handbook curves for copper, aluminum, and silicon all show C_p and C_v diminishing with T, approaching zero as T → 0 K, validating the trend. Why Other Options Are Wrong: “Increases” contradicts empirical and theoretical behavior; “unchanged” ignores temperature dependence; “approaches zero at 0°C” is incorrect (should be at 0 K); the “first increases then decreases” pattern is not the general law for elemental solids. Common Pitfalls: Confusing Celsius and Kelvin scales; assuming Dulong–Petit holds at all temperatures. Final Answer: decreases

Question 20

Counting molecules in a micro-drop: A drop of water weighs 0.018 g. It contains 6.023 × 10^? molecules. What is the missing exponent?

Accepted Answer

Introduction / Context: Back-of-the-envelope mole–molecule conversions are invaluable in reaction engineering and analytical chemistry. Here, we estimate the number of molecules in a tiny water drop using molar mass and Avogadro’s number. Given Data / Assumptions: Mass of water drop = 0.018 g. Molar mass of water = 18 g/mol. Avogadro’s number ≈ 6.023 × 10^23 molecules/mol. Concept / Approach: Number of moles n = mass / molar mass. Number of molecules N = n × N_A. Because 0.018 g is a thousandth of 18 g, the number of moles is 0.001 mol, i.e., 10^-3 mol. Multiply by 6.023 × 10^23 to obtain 6.023 × 10^20 molecules. Step-by-Step Solution: n = 0.018 g / 18 g·mol^-1 = 0.001 mol.N = n × N_A = 0.001 × 6.023 × 10^23 = 6.023 × 10^20 molecules.Therefore, the missing exponent is 20. Verification / Alternative check: Scale reasoning: 1 mol has 6.023 × 10^23 molecules; 10^-3 mol has 10^-3 of that count → exponent 23 − 3 = 20, consistent with the calculation. Why Other Options Are Wrong: 10^23 corresponds to 1 mol, not 0.001 mol; 10^26 and 10^19 or lower deviate by orders of magnitude; 10^17 is far too small. Common Pitfalls: Using 18 kg/kmol incorrectly; forgetting to convert mass to moles; misplacing powers of ten. Final Answer: 10^20

Stoichiometry Questions