Question 1

A dry gas of volume V (litres) at temperature T has its pressure doubled from 1 kgf/cm^2 to 2 kgf/cm^2 at constant temperature. What will be the new gas volume?

Accepted Answer

Introduction / Context: Isothermal compression or expansion of gases is governed by Boyle’s law for ideal behavior. This relationship is used routinely in sizing compressors, storage vessels, and performing quick back-of-the-envelope checks. Given Data / Assumptions: Initial pressure P1 = 1 kgf/cm^2; final pressure P2 = 2 kgf/cm^2. Temperature constant (isothermal process). Gas assumed ideal and dry. Concept / Approach: Boyle’s law states P * V = constant for a given mass of ideal gas at constant temperature. Therefore, P1 V1 = P2 V2, so V2 = (P1/P2) * V1. Doubling the pressure halves the volume. Step-by-Step Solution: Write law: P1 V1 = P2 V2.Insert values: (1) * V = (2) * V2.Solve: V2 = V / 2.Conclude the new volume equals one-half of the original. Verification / Alternative check: A pressure–volume hyperbola confirms inverse proportionality: doubling P halves V. Why Other Options Are Wrong: 2V or V^2: contradict inverse relationship. V/4: would correspond to quadrupling the pressure, not doubling. Unchanged: only true if pressure remained constant. Common Pitfalls: Mixing absolute and gauge units; ensure consistent pressure basis when applying gas laws. Final Answer: V/2

Question 2

During a first-order phase transition such as melting, vaporisation, or sublimation at equilibrium, which specific thermodynamic property remains unchanged for the substance across the coexisting phases?

Accepted Answer

Introduction / Context: Phase changes involve latent heat and discontinuities in certain properties. Understanding which property is continuous across coexisting phases is fundamental for phase-equilibrium calculations, Clapeyron relations, and chemical potential concepts. Given Data / Assumptions: Consider a pure substance at equilibrium between two phases (e.g., liquid–vapour) at given T and P. Phase change occurs reversibly at these conditions. Concept / Approach: At phase equilibrium, the chemical potential of the substance is the same in each phase. For a pure substance, the chemical potential equals the molar Gibbs free energy, g. Thus, g_liquid = g_vapour at the coexistence T and P. In contrast, molar enthalpy (h), internal energy (u), entropy (s), and specific volume (v) undergo finite jumps corresponding to latent heat and volume change. Step-by-Step Solution: State equilibrium condition: chemical potentials equal in all coexisting phases.Use identity: chemical potential μ = molar Gibbs free energy g for a pure substance.Conclude: g is equal for both phases during the phase change at equilibrium.Recognize other properties (h, u, s, v) change by finite amounts (latent heat, volume jump). Verification / Alternative check: The Clapeyron equation relates discontinuities (Δh, Δv) while enforcing equality of μ or g across phases, reinforcing that g is the continuous potential. Why Other Options Are Wrong: Enthalpy/internal energy/entropy: change by latent heat terms. Specific volume: typically changes, often sharply at vaporisation. Common Pitfalls: Confusing continuity of Gibbs free energy (equal values across phases) with its derivatives; remember that first derivatives such as entropy and volume are discontinuous in first-order transitions. Final Answer: Gibbs free energy (molar)

Question 3

At N.T.P., 11.2 litres of ozone (O3) are contained in a vessel. How many atoms of oxygen are present? (Use Avogadro’s number 6.02 × 10^23 and N.T.P. molar volume 22.4 L/mol.)

Accepted Answer

Introduction / Context: Counting particles from gas volumes at standard conditions is a staple of stoichiometry. Here we translate a gas volume of ozone into a count of oxygen atoms, reinforcing the link between molar volume, moles of molecules, and constituent atoms. Given Data / Assumptions: N.T.P. molar volume = 22.4 L per mole of ideal gas. Avogadro’s number NA = 6.02 × 10^23 mol^−1. Species: ozone, O3 (three oxygen atoms per molecule). Concept / Approach: First convert litres of gas at N.T.P. to moles of O3 using the molar volume. Then multiply by 3 to get moles of oxygen atoms, and finally multiply by Avogadro’s number to obtain the number of atoms. Step-by-Step Solution: Compute moles of O3: n_O3 = 11.2 L / 22.4 L·mol^−1 = 0.5 mol.Convert to moles of O atoms: n_O(atoms) = 3 * n_O3 = 3 * 0.5 = 1.5 mol.Compute number of atoms: N = 1.5 * 6.02 × 10^23 = 9.03 × 10^23 atoms.Match to the closest option: 9.03 × 10^23. Verification / Alternative check: Dimensional consistency: litres cancel, leaving moles and then atoms. The factor of 3 accounts for triatomic ozone. Why Other Options Are Wrong: 3.01 × 10^22: too small by a factor of 30. 6.02 × 10^23: equals one mole of atoms; our sample has 1.5 moles of atoms. 1.20 × 10^24: corresponds to 2 moles of atoms, not 1.5. 9.03 × 10^24: an order of magnitude too large. Common Pitfalls: Accidentally treating ozone as diatomic oxygen or using the wrong molar volume (be consistent about N.T.P. vs S.T.P. definitions). Final Answer: 9.03 × 10^23

Question 4

Unit conversion check: 1 Pascal (Pa), the SI unit of pressure, equals how many newtons per square metre (N/m^2)?

Accepted Answer

Introduction / Context: Consistent SI unit handling prevents serious design and safety errors. Pressure appears in fluid mechanics, thermodynamics, and transport operations; the Pascal is the SI coherent unit. Given Data / Assumptions: Definition of Pascal: 1 Pa = 1 N acting uniformly over 1 m^2. Derived from base SI units: kg, m, s. Concept / Approach: By definition, pressure = force / area. The SI unit of force is the newton (N), and area is in square metres (m^2). Hence, 1 Pa = 1 N / m^2 exactly. Larger units include kPa (10^3 Pa) and MPa (10^6 Pa). Step-by-Step Solution: Write formula: P = F / A.Insert units: Pa = N / m^2.Conclude 1 Pa equals 1 N per m^2 by definition.Recognize scaling: 1 kPa = 1000 N/m^2. Verification / Alternative check: Dimensional analysis from base units shows Pa = kg·m^−1·s^−2, matching N/m^2. Why Other Options Are Wrong: 10, 0.1, 1000, 0.01 N/m^2: incorrect scaling; these correspond to deca-, deci-, kilo-, centi- multiples not aligned with the base definition. Common Pitfalls: Confusing Pa with bar: 1 bar = 10^5 Pa = 100 kPa. Final Answer: 1

Question 5

Redox fundamentals: during an oxidation process, which of the following decreases for the species being oxidised?

Accepted Answer

Introduction / Context: Redox (reduction–oxidation) reactions are ubiquitous in corrosion, electrochemistry, metabolism, and industrial processes. A clear operational definition avoids confusion when balancing reactions or interpreting half-reactions. Given Data / Assumptions: Oxidation defined as loss of electrons (OIL RIG: Oxidation Is Loss, Reduction Is Gain). Oxidation number is a bookkeeping device that typically increases upon oxidation. Concept / Approach: When a species is oxidised, it donates electrons; therefore, its electron count decreases. Its oxidation number increases accordingly. The number of ions in solution is not a defining criterion for oxidation; ions may increase, decrease, or remain unchanged depending on the reaction context. Step-by-Step Solution: Apply definition: oxidation → loss of electrons.Infer consequence: electron count associated with that species decreases.Check oxidation number: it rises (not decreases) during oxidation.Recognize ion count is not a general indicator for oxidation state changes. Verification / Alternative check: Half-reaction examples: Fe^2+ → Fe^3+ + e^− shows one electron lost; oxidation number of Fe increases from +2 to +3 while electrons decrease by 1 for the species. Why Other Options Are Wrong: Oxidation number decreases: opposite of what occurs in oxidation. Number of ions: not a universal consequence of oxidation. All of these: cannot be true since (b) is false. Common Pitfalls: Equating oxidation with oxygen addition only; the electron-transfer definition is more general and applies broadly. Final Answer: number of electrons (electron count associated with the species)

Question 6

Colligative properties of dilute solutions depend entirely on what primary factor?

Accepted Answer

Introduction / Context: Colligative properties—such as boiling-point elevation, freezing-point depression, vapour-pressure lowering, and osmotic pressure—are invaluable for determining molar masses and understanding solution behavior in food, pharma, and chemical processes. Given Data / Assumptions: Dilute, ideally behaving solutions. Non-electrolytes compared with electrolytes (which dissociate into more particles). Solvent amount held constant when comparing effects. Concept / Approach: Colligative effects depend solely on the number of solute particles relative to the amount of solvent, not on the chemical identity of those particles. Hence, two different non-electrolytes at the same molality cause the same freezing-point depression. Electrolytes produce larger effects when they dissociate into multiple ions, effectively increasing the particle count (van ’t Hoff factor i). Step-by-Step Solution: Identify the property class: colligative → particle-number dependent.Relate to concentration: use molality or mole fraction as the controlling variable.Account for dissociation/association via van ’t Hoff factor i.Conclude dependence on number of particles, not chemical nature. Verification / Alternative check: Experimental plots of ΔT_f versus molality for many solutes collapse when corrected by i, confirming particle-number control. Why Other Options Are Wrong: Chemical composition/constitution/bond strength: influence non-colligative properties but not the magnitude of ideal colligative effects at equal particle counts. None of these: incorrect because particle number is the defining factor. Common Pitfalls: Using molarity instead of molality when temperature changes; remember that molality is temperature-independent and preferred for colligative calculations. Final Answer: number of solute molecules (or particles) present per amount of solvent

Question 7

In basic acid–base terminology, what pH range correctly characterizes an alkaline (basic) aqueous solution at 25°C?

Accepted Answer

Introduction / Context: Understanding how pH reflects the acidity or basicity of water based solutions is fundamental in chemistry, environmental engineering, and bioprocess control. pH is a logarithmic measure of hydrogen ion activity. At 25°C, a neutral solution has pH 7, acidic solutions are below 7, and alkaline (basic) solutions are above 7. This question checks whether you can correctly map qualitative terms like alkaline to their quantitative pH interval. Given Data / Assumptions: Temperature reference is 25°C, where water autoionization constant Kw is 1.0 * 10^-14. Neutral water has pH = 7.00 at 25°C. Alkalinity corresponds to excess hydroxide relative to hydrogen ions. Concept / Approach: By definition, pH = -log10[H+]. When [H+] equals 1.0 * 10^-7 mol L^-1, pH is 7 (neutral). If [H+] decreases below 10^-7, the pH increases above 7. Such solutions contain a higher proportion of hydroxide ions than hydrogen ions and are called alkaline or basic. Conversely, if [H+] increases above 10^-7, pH falls below 7 and the solution is acidic. These thresholds are standard at 25°C and are widely used in analytical and industrial practice. Step-by-Step Solution: Identify the neutral reference: pH = 7 at 25°C.Define acidic vs basic: acidic if pH < 7, basic if pH > 7.Match term alkaline with pH > 7.Select the choice that states pH > 7. Verification / Alternative check: Because pOH = 14 - pH at 25°C, an alkaline solution has pOH < 7, confirming that [OH-] > [H+], which is equivalent to pH > 7. Why Other Options Are Wrong: 7: Identifies neutrality, not alkalinity.< 7: This is the acidic region.Constant over a wide range: pH varies with composition; no wide constant region defines alkalinity. Common Pitfalls: Confusing neutrality with alkalinity; ignoring temperature dependence of Kw outside 25°C; assuming logarithmic scale changes are linear in concentration space. Final Answer: > 7

Question 8

At N.T.P. (0°C, 1 atm), which statement about ideal-gas molar volumes and air density is incorrect?

Accepted Answer

Introduction / Context: Gas property facts at reference conditions are frequent quick checks in chemical engineering calculations. N.T.P. is commonly taken as 0°C and 1 atm. Ideal gases at N.T.P. occupy a well known molar volume, and conversions between gram moles, kilogram moles, and pound moles are standard. The density of real air at N.T.P., however, is not exactly 1 g per litre. This item tests attention to canonical values and unit consistency. Given Data / Assumptions: N.T.P. reference: 0°C (273.15 K) and 1 atm. Molar volume of an ideal gas at N.T.P.: about 22.414 L per mol. Dry air density at 0°C, 1 atm is about 1.293 kg m^-3 (i.e., 1.293 g L^-1). Concept / Approach: Use ideal gas molar volume to validate statements about gram moles, kilogram moles, and pound moles. Then compare the quoted dry air density to the accepted value. The statements giving 22.4 litres per gram mole and 22.4 m^3 per kilogram mole are consistent. For 1 lb-mol, the corresponding volume at 0°C is approximately 359 ft^3 (as the 379.5 ft^3 value corresponds to 60°F, not 32°F). The given air density of 1 g L^-1 is too low compared with the standard 1.293 g L^-1; therefore it is the incorrect statement. Step-by-Step Solution: Check molar volume: 22.414 L mol^-1 at N.T.P.Convert to kg-mol: 22.414 m^3 per kmol ≈ 22.4 m^3 per kg-mol.Convert to lb-mol at 0°C: ≈ 359 ft^3 per lb-mol (consistent).Compare dry air density: expected ≈ 1.293 g L^-1, not 1 g L^-1. Verification / Alternative check: Using PV = nRT at 273.15 K and 1 atm reproduces 22.414 L mol^-1. Standard air composition yields a molar mass ~28.97 g mol^-1, giving density 28.97 g per 22.414 L ≈ 1.293 g L^-1. Why Other Options Are Wrong: They are correct statements at N.T.P. and match accepted reference values for ideal gases. Common Pitfalls: Mixing N.T.P. with S.T.P. or 60°F base conditions; rounding aggressively; using 1 g L^-1 as a rough mnemonic instead of the accurate 1.293 g L^-1. Final Answer: Density of dry air at N.T.P. is 1 g per litre.

Question 9

Raoult's law: under which condition does the proportionality between a component's vapor pressure and its liquid-phase mole fraction fail to apply?

Accepted Answer

Introduction / Context: Raoult's law describes ideal solutions where each component's partial vapor pressure equals its mole fraction times the pure-component vapor pressure. Ideal behavior requires similar intermolecular interactions between like and unlike species. Many real mixtures deviate from this ideality, especially when specific interactions or reactions occur upon mixing. Knowing when Raoult's law breaks down is essential for accurate phase-equilibrium calculations and for selecting activity-coefficient models. Given Data / Assumptions: Temperature and pressure are moderate so vapor behaves nearly ideally. We are evaluating liquid-phase nonideality conditions. Key assumptions for Raoult's law include comparable molecular size, nonpolarity, and no preferential surface enrichment. Concept / Approach: When unlike molecules interact much more strongly or weakly than like molecules, or when chemical association or reaction occurs, the solution is nonideal. Chemical association between unlike molecules (e.g., hydrogen bonding, complex formation, acid–base reactions) alters escaping tendencies, so vapor pressures are not proportional to simple mole fractions. In such cases, activity coefficients depart from unity, and Raoult's law must be corrected using models like Margules, Wilson, NRTL, or UNIQUAC. Step-by-Step Solution: Recall Raoult's law requirement: similar interactions and no reaction on mixing.Identify condition that violates this: chemical combination or association between unlike molecules.Conclude Raoult's law does not apply under that condition.Select the option describing molecular association/chemical combination. Verification / Alternative check: Classical examples include acetone–chloroform (specific H-bonding) and acid–base pairs, all showing strong deviations from Raoult's law and requiring activity coefficients substantially different from 1. Why Other Options Are Wrong: No surface concentration, similar size, and comparable attractions are exactly the ideal-solution assumptions under which Raoult's law is valid. Common Pitfalls: Assuming any nonideality invalidates Raoult's law entirely; mild nonideality can be handled by activity coefficients, but strong association requires more advanced models. Final Answer: There is chemical combination or molecular association between unlike molecules on mixing.

Question 10

Closed, rigid vessel heating: Three moles of N2 at 30°C are heated at constant volume to 250°C. With Cv = 20.8 J mol^-1 °C^-1, what total heat is required (neglect vessel heat capacity)?

Accepted Answer

Introduction / Context: This problem is a straightforward application of constant-volume heating of an ideal gas. In a rigid vessel, the volume does not change, so boundary work is zero. The heat added equals the change in internal energy, which is n * Cv * ΔT for ideal gases. Mastering such calculations is essential for quick energy balances in reactors and closed vessel heating steps. Given Data / Assumptions: Gas: nitrogen, treated as ideal with temperature-independent heat capacities over the range. n = 3 mol, T1 = 30°C, T2 = 250°C. Cv = 20.8 J mol^-1 °C^-1, Cp = 29.1 J mol^-1 °C^-1 (Cp is not used here). Rigid vessel implies constant volume, so W = 0. Concept / Approach: For an ideal gas at constant volume, ΔU = n * Cv * ΔT and Q = ΔU because there is no boundary work. Temperature change is computed on the Celsius scale because only differences matter (ΔT is the same in K and °C). Substitute the given values to find Q. Step-by-Step Solution: Compute temperature rise: ΔT = 250 - 30 = 220 °C.Write the relation at constant volume: Q = n * Cv * ΔT.Insert numbers: Q = 3 * 20.8 * 220 J.Evaluate product: 3 * 20.8 = 62.4; 62.4 * 220 = 13728 J.Select the option that matches 13728 J. Verification / Alternative check: As a check, if Cp had been used accidentally, the heat would be higher (3 * 29.1 * 220 ≈ 19194 J), which does not match the constant-volume constraint and confirms that Cv is the correct property here. Why Other Options Are Wrong: 19206 J: close to using Cp instead of Cv.4576 J and 12712 J: result from arithmetic or unit mistakes, or using wrong ΔT. Common Pitfalls: Using Cp instead of Cv; converting temperatures to Kelvin for initial and final values but forgetting that ΔT is unchanged; including boundary work in a rigid vessel where it is zero. Final Answer: 13728 J

Question 11

If a solution has pH = 3 at 25°C, what is its hydroxide ion concentration [OH-] in mol L^-1?

Accepted Answer

Introduction / Context: Relating pH and pOH is a staple skill in aqueous chemistry. At 25°C, water self ionization is quantified by Kw = [H+][OH-] = 1.0 * 10^-14. Given pH, you can determine hydrogen ion concentration and then hydroxide ion concentration using Kw. This type of conversion is common in environmental monitoring, bioprocess pH control, and analytical calculations. Given Data / Assumptions: pH = 3.00 at 25°C. Kw at 25°C equals 1.0 * 10^-14. Activities are approximated by concentrations for dilute solutions. Concept / Approach: Compute [H+] from pH, then use Kw to compute [OH-]. Specifically, [H+] = 10^-pH = 10^-3 mol L^-1. Then [OH-] = Kw / [H+] = 10^-14 / 10^-3 = 10^-11 mol L^-1. Alternatively, use pH + pOH = 14, so pOH = 11 and [OH-] = 10^-11 mol L^-1. Both approaches are equivalent at 25°C. Step-by-Step Solution: Calculate hydrogen ion concentration: [H+] = 10^-3 mol L^-1.Apply the water ion product: Kw = [H+][OH-] = 10^-14.Solve for hydroxide: [OH-] = 10^-14 / 10^-3 = 10^-11 mol L^-1.Match to the correct option: 10^-11. Verification / Alternative check: Using pOH: pOH = 14 - 3 = 11, hence [OH-] = 10^-11 mol L^-1, confirming the result. Why Other Options Are Wrong: 10^-10: corresponds to pOH = 10 and pH = 4, not pH 3.10^-3: this is [H+], not [OH-], for pH 3.10^-4: would imply pOH = 4 and pH = 10, a basic solution. Common Pitfalls: Forgetting that pH + pOH = 14 applies at 25°C; mixing up [H+] and [OH-]; ignoring that the scale is logarithmic, so a change of 1 pH unit is a tenfold change in [H+]. Final Answer: 10^-11

Question 12

Unit conversions for mass: which of the following stated conversion formulas is incorrect?

Accepted Answer

Introduction / Context: Accurate engineering calculations rely on correct unit conversions. Tonne, metric ton, U.S. short ton, and long (imperial) ton are frequently confused. This question asks you to identify the incorrect statement among several common mass conversions encountered in process design and supply chain calculations. Given Data / Assumptions: 1 tonne (metric ton) = 1000 kg exactly. 1 kilogram = 2.20462 pounds (lbs). 1 U.S. short ton = 2000 lbs ≈ 907.1847 kg ≈ 0.907 tonne. 1 long ton (ton) = 2240 lbs ≈ 1016.0469 kg ≈ 1.016 tonnes. Concept / Approach: Convert the tonne to pounds using 1 kg = 2.20462 lbs. Then validate the other conversions. The first statement incorrectly lists 22.046 lbs for a tonne, which is short by a factor of 100. Correct is approximately 2204.6 lbs. The other statements align with standard references within rounding tolerance. Step-by-Step Solution: Compute pounds per tonne: 1000 kg * 2.20462 lbs kg^-1 ≈ 2204.62 lbs.Compare with the given value 22.046 lbs → incorrect by two orders of magnitude.Validate short ton: 2000 lbs ≈ 907 kg → 0.907 tonne and ≈ 0.893 of a long ton (2000/2240).Validate long ton: 2240 lbs ≈ 1016 kg ≈ 1.016 tonnes and 2240/2000 = 1.12 U.S. tons. Verification / Alternative check: Cross check with any reputable unit-conversion table or dimensional analysis tool to confirm the numeric factors. Why Other Options Are Wrong: They are actually correct within normal engineering rounding, so they are not the incorrect formulas. Common Pitfalls: Dropping zeros in conversions; mixing mass (tonne) and force (ton-force); confusing long and short ton definitions. Final Answer: 1 tonne = 1000 kg = 22.046 lbs

Question 13

Hess's law application: The total enthalpy change (net heat) for a chemical process is independent of which factor?

Accepted Answer

Introduction / Context: Hess's law states that the enthalpy change of a process depends only on the initial and final states of the system, not on the pathway taken. This powerful idea allows engineers and chemists to compute heats of reaction by combining tabulated steps, even if the overall reaction cannot be measured directly. Identifying what enthalpy change is independent of helps avoid conceptual errors in energy balances. Given Data / Assumptions: State function: enthalpy H depends on state variables (T, P, composition, phase). Pathway can be arbitrary sequences of reactions or steps. Initial and final states are clearly defined. Concept / Approach: Because enthalpy is a state function, ΔH depends only on initial and final states. Therefore, ΔH is independent of the number or nature of intermediate steps. However, ΔH does depend on conditions such as temperature, pressure, and phases of reactants and products, because those define the initial and final states. Any option suggesting independence from these state-defining parameters would be incorrect. Step-by-Step Solution: Identify the property: enthalpy is a state function.State what follows: ΔH is path independent.Find the option representing the path details: number of intermediate reactions.Select that option as the independent factor. Verification / Alternative check: Construct hypothetical cycles that begin and end at the same state; the sum of ΔH around a closed loop is zero, confirming path independence and the validity of Hess's law. Why Other Options Are Wrong: Temperature, pressure, and physical states define the state of the system, so ΔH changes when they change. Common Pitfalls: Confusing path independence of ΔH with path independence of heat q in general processes; only under constant pressure with negligible kinetic and potential energy changes does heat equal ΔH. Final Answer: Number of intermediate chemical reactions involved

Question 14

Osmosis and colligative properties: The osmotic pressure of a solution increases when which variable is decreased (other factors constant)?

Accepted Answer

Introduction / Context: Osmotic pressure is a colligative property proportional to the solute particle concentration in a solution. It matters in biological systems, membrane separations, and food processing. Understanding how manipulation of volume, concentration, and temperature affects osmotic pressure helps in designing desalination, dialysis, and concentration processes. Given Data / Assumptions: For dilute solutions, van 't Hoff relation applies: π = i * M * R * T. M is molarity = moles of solute per litre of solution. i and R are taken as constants for the same solute and system; T is constant unless otherwise noted. Concept / Approach: Holding solute moles and temperature constant, decreasing the solution volume increases molarity M. Because π is directly proportional to M, osmotic pressure increases as volume decreases. Decreasing solute concentration would lower M and therefore lower π. Decreasing temperature would also lower π. Only a decrease in volume, with moles fixed, raises π in this list of options. Step-by-Step Solution: Write π = i * M * R * T for dilute solutions.Fix i, R, T and number of solute moles.Note that M = n / V; decreasing V increases M.Conclude that π increases when V decreases. Verification / Alternative check: Membrane operations observe higher osmotic back-pressure as solutions concentrate during water removal, consistent with π rising as volume falls at constant solute moles. Why Other Options Are Wrong: Solute concentration: decreasing it reduces π, not increases it.Temperature: lowering T reduces π in the van 't Hoff equation.None of these: incorrect because volume decrease clearly increases π under constant moles. Common Pitfalls: Confusing mass fraction with molarity; ignoring that π depends on particle concentration, not just mass of solute; forgetting the temperature factor in π for nonisothermal systems. Final Answer: Volume

Question 15

Ideal-gas change: If a gas's absolute temperature is doubled while its pressure is simultaneously reduced to one-half, how does its volume change?

Accepted Answer

Introduction / Context: Combined gas law questions test the ability to relate pressure, volume, and temperature for an ideal gas. Engineers constantly use these relations for compressor sizing, storage calculations, and reactor volume changes. Recognizing proportionalities quickly can save time and avoid algebraic errors. Given Data / Assumptions: Ideal gas behavior applies. Initial state: P1, V1, T1. Final state: P2 = P1/2, T2 = 2*T1. Gas amount n is constant. Concept / Approach: The ideal gas law rearranged gives V ∝ T/P for fixed n. Apply the specified scalings to find the multiplicative effect on volume. The linearity of volume with respect to T and inverse proportionality with respect to P allows a quick ratio method without solving for absolute values. Step-by-Step Solution: Start with V ∝ T/P for fixed n.Compute factor: (T2/T1) / (P2/P1) = (2) / (1/2) = 4.Therefore, V2 = 4 * V1.Select the option indicating a fourfold increase. Verification / Alternative check: Plug into PV = nRT twice: P1V1/T1 = P2V2/T2 → V2 = V1 * (T2/T1) * (P1/P2) = V1 * 2 * 2 = 4 V1, confirming the ratio reasoning. Why Other Options Are Wrong: No change or doubling: underestimate the combined effect of increased T and decreased P.One-fourth of original: inverts the correct proportionality. Common Pitfalls: Mixing up absolute and relative temperature (always use absolute temperature); applying Celsius differences directly to proportional relations; forgetting that halving pressure increases volume for a fixed amount of gas. Final Answer: Fourfold increase

Question 16

Real gases approximate ideal behavior best under which conditions of pressure and temperature?

Accepted Answer

Introduction / Context: Ideal-gas behavior is a useful approximation when intermolecular forces and molecular sizes are negligible compared with the average spacing of molecules. Recognizing when this holds is crucial for deciding whether PV = nRT is acceptable or whether real-gas equations of state (e.g., van der Waals, Peng–Robinson) are required for accuracy. Given Data / Assumptions: Real gases deviate due to finite molecular volume and attractive/repulsive forces. Deviation decreases when the molecules are far apart and kinetic energy dominates. Critical region and saturation conditions promote nonideality. Concept / Approach: At low pressure, the average distance between molecules is large, so excluded volume and pairwise attractions contribute less to the equation of state. At high temperature, molecules have higher kinetic energy, which reduces the relative importance of attractive interactions. The combination of low P and high T therefore minimizes nonideality and makes the ideal gas law most reliable. Step-by-Step Solution: Relate nonideality to intermolecular forces and finite size.Identify conditions that minimize these effects: low P (reduces density) and high T (increases kinetic energy).Conclude that low pressure and high temperature approach ideal behavior.Select the corresponding option. Verification / Alternative check: Compressibility factor Z approaches 1 for many gases as P → 0 and T » Tc; generalized charts show Z ≈ 1 under these conditions. Why Other Options Are Wrong: High P and/or low T push gases toward condensation and large deviations from ideality. Common Pitfalls: Using the ideal gas law near saturation lines or the critical point; ignoring strong polarity or association that causes deviations even at moderate conditions. Final Answer: Low pressure and high temperature

Question 17

Distillation practice: In a fractionating column, refluxing (returning part of the condensed distillate to the top of the column) is a recycling operation primarily intended to achieve what outcome?

Accepted Answer

Introduction / Context: Reflux is central to binary and multicomponent distillation. By returning a portion of condensed overhead liquid to the top tray or packing, we create internal liquid–vapour contact that sharpens the separation. This question tests whether you understand that the primary objective of reflux is to enrich the more volatile component in the overhead product, not merely to save heat or increase total yield. Given Data / Assumptions: A conventional fractionating column with trays or packing. Distillate is partially returned as reflux; the remaining is withdrawn as product. Steady operation near design reflux ratio. Concept / Approach: Reflux increases the liquid flow down the column, improving mass transfer on each stage. The counter-current contact between rising vapour (rich in light key) and descending liquid (leaner in light key) increases the number of effective equilibrium stages, thereby increasing distillate purity for a given column height. Energy effects are secondary; heat is usually supplied by the reboiler and removed by the condenser. Step-by-Step Solution: Identify the function of reflux: enhance rectification by internal mass transfer.Link to composition: overhead liquid returns to wash heavier components downward.Outcome: higher mole fraction of light key in vapour/distillate (product enrichment).Hence, select “product enrichment.” Verification / Alternative check: McCabe–Thiele analysis shows that increasing reflux ratio moves the operating lines closer to the diagonal, increasing theoretical stages or allowing higher purity at fixed stages—both signify product enrichment. Why Other Options Are Wrong: Heat conservation is not the primary aim; yield enhancement depends on material balance and cut points, not reflux alone; “none” is incorrect; pressure stabilization is handled by venting/controls, not reflux per se. Common Pitfalls: Confusing overall recovery (yield) with composition (purity). Reflux chiefly improves purity; recovery is set by cut location and flow splits. Final Answer: product enrichment

Question 18

Vapour pressure trend: For liquids of similar chemical nature at a fixed temperature, how does vapour pressure vary with increasing molecular weight?

Accepted Answer

Introduction / Context: Trends in vapour pressure are essential for predicting volatility, boiling points, and separation difficulty. For homologous or chemically similar liquids, intermolecular forces scale with size and structure. This question probes your understanding of the inverse relation between molecular weight and vapour pressure within a related chemical family at the same temperature. Given Data / Assumptions: Liquids are chemically similar (e.g., n-alkanes). Temperature is fixed. No association/strong hydrogen bonding anomalies are introduced. Concept / Approach: Heavier molecules typically have stronger dispersion forces and larger surface areas, making their molecules harder to escape into the vapour phase. Thus, at a given temperature, heavier members exhibit lower equilibrium vapour pressures. This explains why boiling points increase with molecular weight in a series like C5–C10 alkanes. Step-by-Step Solution: Consider a homologous series (e.g., n-alkanes).As molecular weight increases, cohesive forces increase.Higher cohesion lowers vapour pressure at the same temperature.Therefore, select “decreases.” Verification / Alternative check: Antoine constants and vapour–pressure tables confirm declining vapour pressure with increasing carbon number at fixed temperature. Why Other Options Are Wrong: “Increases” contradicts data; “remains unchanged” ignores molecular interactions; “increases linearly” is not supported; “no definite trend” is false for similar chemical families. Common Pitfalls: Assuming a universal rule across dissimilar chemistries; the trend holds best for closely related liquids. Final Answer: decreases

Question 19

Gas mixture partial pressures (Dalton’s law): An empty rigid vessel at 40°C is charged with 40 g methane (CH4) and 40 g oxygen (O2). What fraction of the total pressure is exerted by oxygen?

Accepted Answer

Introduction / Context: Dalton’s law states that each gas in a mixture exerts a partial pressure proportional to its mole fraction at a fixed temperature and volume. Converting masses to moles is crucial because pressure shares depend on moles, not mass. Given Data / Assumptions: T = 40°C, rigid vessel (volume constant). m(CH4) = 40 g; m(O2) = 40 g. Ideal-gas behaviour and homogeneous mixture. Concept / Approach: Partial pressure fraction of a component equals its mole fraction: yi = ni / ntotal. Compute moles for both species, then take the ratio for oxygen. Step-by-Step Solution: Molar masses: M(CH4) = 16 g/mol; M(O2) = 32 g/mol.n(CH4) = 40 / 16 = 2.5 mol.n(O2) = 40 / 32 = 1.25 mol.n(total) = 2.5 + 1.25 = 3.75 mol.y(O2) = n(O2) / n(total) = 1.25 / 3.75 = 1/3. Verification / Alternative check: Because equal masses of CH4 and O2 were used, the lighter gas contributes more moles. Therefore, oxygen’s fraction must be less than methane’s, consistent with 1/3 vs 2/3. Why Other Options Are Wrong: 1/2 assumes equal moles, not masses; 1/4 and 3/4 are inconsistent with the computed mole counts; 2/3 would be methane’s fraction, not oxygen’s. Common Pitfalls: Using mass fractions instead of mole fractions; forgetting to divide by the correct molar masses. Final Answer: 1/3

Question 20

Real-gas behaviour: The temperature at which the second virial coefficient (B) of a real gas becomes zero is known as the ________.

Accepted Answer

Introduction / Context: The virial equation of state expands real-gas behaviour in powers of 1/V or p, with the second virial coefficient B(T) capturing pairwise interactions. The temperature where B(T) = 0 is important because the gas shows nearly ideal compressibility over a moderate pressure range. Given Data / Assumptions: Definition of B(T) as a function of temperature alone for a given gas. Classical virial EoS context. Concept / Approach: At the Boyle temperature, Z = pV/RT approaches 1 with minimal slope at low pressures, since the first correction term (involving B) vanishes. This does not mean perfect ideality at all pressures; higher virial terms still matter at higher p. Step-by-Step Solution: Recall virial form: Z = 1 + B(T)p/RT + C(T)p^2/…Set B(T) = 0 → lowest-order deviation disappears.By definition, this T is the Boyle temperature. Verification / Alternative check: Compressibility-factor charts show that near the Boyle temperature, the Z–p curve is flat at low p. Why Other Options Are Wrong: Eutectic relates to solid–solid–liquid equilibria; boiling and critical temperatures are phase-change properties; consolute point applies to liquid–liquid miscibility. Common Pitfalls: Confusing Boyle temperature with Boyle’s law (isothermal ideality) at any temperature—real gases approximate it only near this special temperature at low p. Final Answer: Boyle temperature

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