Q: Purpose of a bypass stream in process flowsheets: Why is a bypass line often included in chemical process operations?

Introduction / Context: Bypass lines are common in industrial flowsheets. They allow a fraction of a stream to circumvent a unit operation and later be mixed back, enabling operators to fine-tune properties like temperature, concentration, and flow without disturbing upstream or downstream steady states. Given Data / Assumptions: The bypass line splits a feed, sending part around a unit (e.g., heater, cooler, reactor) and later recombining. Steady-state process; standard control objectives (temperature or composition). Concept / Approach: A bypass provides a simple, fast-acting manipulation variable. For example, mixing a cooled stream with a warm bypass stream sets the final temperature precisely. Similarly, a bypass around a separator can trim concentration. It is primarily a control and flexibility tool; it does not inherently increase conversion or yield. Step-by-Step Reasoning: Introduce split: F = F_bypass + F_through_unit.Manipulate split ratio to reach target outlet T or composition x.Controller adjusts bypass valve to correct deviations quickly. Verification / Alternative check: Heat balance on a mixer with a bypassed heater shows the outlet temperature depends on the bypass fraction, demonstrating controllability without changing heater duty or upstream flows. Why Other Options Are Wrong: Improves conversion / increases yield: Not inherently; reaction performance is dictated by kinetics, residence time, and conditions inside the reactor, not by external bypass piping. Reduces pressure drop by default: Pressure behavior depends on hydraulics; a bypass can even add valves and fittings. None of these: Incorrect because process control benefits are real and well-established. Common Pitfalls: Assuming a bypass boosts reaction conversion; ignoring that bypass manipulation can affect residence time distribution if used improperly. Final Answer: Facilitates better control of the process (e.g., temperature or composition trimming)

Q: Steady-state mass balance: For a system operating at steady state, when does the simple relation “inputs = outputs” hold true?

Introduction / Context: Mass balances distinguish between total mass and component mass. At steady state, accumulation is zero. Even with reactions, total mass is conserved, though individual species balances include generation or consumption terms. Given Data / Assumptions: System at steady state (d/dt of holdup = 0). No leaks or unaccounted losses; total mass is conserved. Chemical reactions may occur internally. Concept / Approach: Total mass balance at steady state is: total mass in = total mass out. Component balances are: in − out + generation − consumption = 0. Reactions affect components, not the conservation of total mass (assuming no nuclear reactions). Step-by-Step Solution: Total: Σṁ_in = Σṁ_out (steady state, closed to leaks).Component i: ṁ_in,i − ṁ_out,i + r_i = 0, where r_i accounts for reaction stoichiometry.Thus “inputs = outputs” holds for the total mass even with reaction. Verification / Alternative check: Consider A → B in a CSTR at steady state. Although A decreases and B increases, the sum of mass flow rates remains constant if density changes are accounted for in volumetric terms. Why Other Options Are Wrong: Without reaction only / only without reaction and without losses: Too restrictive; the total mass relation is broader. None of these / 100% conversion requirement: Not relevant to total mass conservation. Common Pitfalls: Confusing total mass balance with species balances; overlooking leaks or evaporative losses which would violate the simple equality. Final Answer: With chemical reaction (total mass basis, no accumulation)

Q: Physical adsorption versus chemisorption: Which statement correctly contrasts physisorption with chemisorption?

Introduction / Context: Adsorption processes are central to separations and catalysis. Two limiting mechanisms are physical adsorption (physisorption) and chemical adsorption (chemisorption), which differ in bonding, energetics, and capacity. Given Data / Assumptions: Physisorption arises from van der Waals forces; multilayer adsorption is possible (BET behavior). Chemisorption involves specific chemical bonding; typically monolayer-limited (Langmuir behavior). Typical heats: physisorption ~ 20–40 kJ/mol; chemisorption often much larger. Concept / Approach: Because physisorption is weaker and non-specific, it can build multiple layers, often leading to a higher total uptake per mass at suitable temperatures. Chemisorption is stronger, often activated, and limited to surface sites, reducing capacity to about one monolayer. Step-by-Step Reasoning: Physisorption: weak forces, low activation barrier, possible multilayer → higher overall capacity.Chemisorption: strong bonds, activated process, site-specific → lower capacity (monolayer).Therefore, the statement about higher quantity adsorbed per unit mass fits physisorption. Verification / Alternative check: BET isotherms show increasing uptake with relative pressure due to multilayer formation, confirming higher capacities for physisorption at lower temperatures. Why Other Options Are Wrong: Surface-reaction control, high activation energy, very large heat of adsorption, strong bonds: all characterize chemisorption, not physisorption. Common Pitfalls: Assuming higher heat means better separation; confusing monolayer (Langmuir) with multilayer (BET) models. Final Answer: Quantity adsorbed per unit adsorbent mass is generally higher (multilayer possible)

Q: Kinetic theory of gases core idea: According to the kinetic theory, which statement is correct?

Introduction / Context: The kinetic theory of gases provides a microscopic explanation of macroscopic gas properties, relating temperature and pressure to molecular motion and collisions. Given Data / Assumptions: Ideal gas model: point particles, elastic collisions, negligible intermolecular forces except during collisions. Thermal energy is entirely kinetic in this model. Concept / Approach: In kinetic theory, absolute temperature T is proportional to the average translational kinetic energy of gas molecules. Collisions are perfectly elastic (no net energy loss), and the volume of molecules is negligible relative to the container volume. Step-by-Step Reasoning: For an ideal monatomic gas: average kinetic energy per molecule ∝ T.Collisions conserve kinetic energy and momentum (elastic).Molecular volume is neglected in the ideal model; real-gas deviations are handled by equations like van der Waals. Verification / Alternative check: Gas temperature rises when average molecular speed increases; pressure at constant volume rises accordingly, consistent with kinetic interpretations. Why Other Options Are Wrong: Energy loss in collisions or inelastic ideal-gas collisions contradicts the model. Appreciable molecular volume violates the point-particle assumption of ideal gases. “None of these” is invalid because option (c) is correct. Common Pitfalls: Extending ideal behavior to high pressures/low temperatures where real-gas effects and finite molecular size matter. Final Answer: Absolute temperature measures the average kinetic energy of molecules

Q: Ideal solutions and Raoult’s law: For an ideal binary solution, how does the total vapor pressure vary with composition (mole fraction)?

Introduction / Context: Understanding vapor–liquid equilibrium (VLE) begins with ideal solutions, where Raoult’s law applies. This provides a linear dependence of partial pressures, and hence total pressure, on composition. Given Data / Assumptions: Ideal solution behavior: each component obeys Raoult’s law. Binary mixture at fixed temperature. Concept / Approach: Raoult’s law states p_i = x_i * p_i^sat(T). The total pressure is P_total = Σ p_i = x_1 p_1^sat + x_2 p_2^sat. Because x_2 = 1 − x_1, P_total is a linear function of x_1 (and of x_2). Step-by-Step Solution: Write P_total(x_1) = x_1 p_1^sat + (1 − x_1) p_2^sat.Rearrange: P_total = p_2^sat + x_1 (p_1^sat − p_2^sat).This is a straight line in x_1 with slope (p_1^sat − p_2^sat). Verification / Alternative check: Graphical VLE diagrams for ideal pairs (e.g., benzene–toluene) show linear total pressure lines at constant temperature. Why Other Options Are Wrong: Inverse, exponential, parabolic, or negligible change: Do not match Raoult’s law behavior for ideal solutions. Common Pitfalls: Mixing up ideal solution (Raoult) with ideal gas behavior; forgetting temperature must be held constant for the linear relation. Final Answer: Linearly

Q: Trend of latent heat near the critical point: How does the heat of vaporisation change with pressure, and what happens at the critical point?

Introduction / Context: Phase-change energetics are vital for boilers, condensers, and power cycles. The latent heat of vaporisation reflects the energy required to separate liquid and vapor phases; its behavior with pressure reveals what happens as a fluid approaches the critical state. Given Data / Assumptions: Pure fluid undergoing equilibrium vaporisation along the saturation curve. Critical point: liquid and vapor phases become indistinguishable. Concept / Approach: The Clausius–Clapeyron relation shows latent heat λ ∝ T * Δv * (dP_sat/dT). As pressure increases toward the critical pressure, the specific volume difference Δv = v_g − v_l shrinks. Consequently, λ decreases and must reach zero at the critical point where Δv → 0. Step-by-Step Reasoning: Increase pressure → move up saturation curve → liquid and vapor densities converge.As Δv decreases, λ correspondingly decreases.At the critical point, Δv = 0 → λ = 0. Verification / Alternative check: Steam tables show the latent heat of water declining with pressure and vanishing at the critical point (~22.06 MPa), validating the trend. Why Other Options Are Wrong: Increase or remain constant: Contradicted by data and theory. Only “becomes zero” without noting the continuous decrease omits the trend. Common Pitfalls: Confusing latent heat with sensible heat; assuming a constant latent heat across pressures. Final Answer: Both decreases with pressure and becomes zero at the critical point

Question 1

Equilibrium in multicomponent, multiphase systems: Which approach cannot be used to calculate equilibrium relations?

Accepted Answer

Introduction / Context: Phase and reaction equilibria require relations (e.g., K-values, activity coefficients) to specify how components distribute among phases. Several approaches exist: measurement, empirical correlations, and theoretical models. The phase rule, however, serves a different purpose. Given Data / Assumptions: Gibbs phase rule: F = C − P + 2 (for nonreactive systems). Goal: Obtain actual equilibrium relationships (e.g., y vs. x, distribution ratios), not just degrees of freedom. Concept / Approach: The phase rule determines the number of independent intensive variables that must be fixed to define the state. It does not give numerical equilibrium relations. Those must come from experiments, empirical correlations (e.g., Antoine, Wilson, NRTL), or theoretical equations (e.g., EOS with mixing rules). Step-by-Step Solution: Identify need: equilibrium relations (K_i, γ_i, φ_i).Match methods: experimental, empirical, theoretical models all provide such relations.Exclude phase rule: provides F (degrees of freedom), not the relations themselves. Verification / Alternative check: In VLE design, phase rule tells you how many variables to set (e.g., T and P), but K-values still require data or models. Why Other Options Are Wrong: Experimental data, empirical equations, theoretical equations all directly yield or fit equilibrium relations. Common Pitfalls: Assuming the phase rule can replace property models; it cannot provide numerical partitioning. Final Answer: phase rule

Question 2

Latent heat of vaporisation: Identify the correct overall trend with temperature and pressure, and its value at the critical point.

Accepted Answer

Introduction / Context: Latent heat of vaporisation is the energy required to convert saturated liquid to saturated vapour at a given pressure/temperature. Its variation with T and P is foundational in boiler design, refrigeration, and power cycles. Given Data / Assumptions: Considering a pure component undergoing liquid–vapour phase change. Equilibrium transitions along the saturation curve up to the critical point. Concept / Approach: As temperature increases toward the critical temperature, the enthalpy difference between saturated liquid and vapour shrinks because liquid and vapour properties converge. Similarly, at higher pressures along the saturation curve, the two phases become more alike. At the critical point, the phases are indistinguishable, so latent heat is zero. Step-by-Step Solution: Increase T (along saturation): Δh_vap decreases.Increase P (along saturation): Δh_vap also generally decreases.At critical point: Δh_vap = 0 due to identical phases. Verification / Alternative check: Steam tables show h_g − h_f decreasing as T approaches the critical temperature (for water, near 374 °C), confirming the trend. Why Other Options Are Wrong: Each individual statement (a), (b), (c) is correct; therefore the combined choice “all (a), (b) & (c)” is appropriate. Common Pitfalls: Assuming latent heat is constant or increases with temperature; in reality it declines to zero at the critical point. Final Answer: all (a), (b) & (c)

Question 3

Pressure units: 1 torr is equal to how many millimetres of mercury (mm Hg)?

Accepted Answer

Introduction / Context: Pressure units arise across vacuum systems, distillation columns, and laboratory measurements. Torr and mm Hg are closely related historical units, and recognizing their equivalence avoids confusion during unit conversions. Given Data / Assumptions: Definition: 1 torr was historically defined as exactly 1 mm Hg. Modern standards set 1 torr = 133.322 Pa and 1 mm Hg (at 0 °C) ≈ 133.322 Pa, preserving practical equivalence. Concept / Approach: Because both units trace to the same physical mercury column height at standard gravity and temperature, they are effectively identical for routine engineering use, though Pascal is preferred in SI contexts. Step-by-Step Solution: Use definition: 1 torr = 1 mm Hg.Optional SI mapping: 1 torr ≈ 133.322 Pa; 760 torr ≈ 1 atm. Verification / Alternative check: Check with standard conversions: 760 mm Hg = 760 torr ≈ 101325 Pa (1 atm), confirming equivalence. Why Other Options Are Wrong: 0.1 or 10 or 1000 mm Hg would imply inconsistent scaling with atmospheric pressure and are not supported by definitions. Common Pitfalls: Confusing “torr” with “mTorr” (millitorr); remember 1 mTorr = 0.001 torr = 0.001 mm Hg. Final Answer: 1

Question 4

Steam distillation concept check: Assume benzene and water are mutually insoluble. Their normal boiling points are 80.1°C (benzene) and 100°C (water). At a total pressure of 1 atm, what is the boiling point of a liquid–liquid mixture of benzene and …

Accepted Answer

Introduction / Context: Immiscible (mutually insoluble) liquid mixtures such as benzene and water exhibit a special boiling behavior exploited in steam distillation. Instead of following Raoult’s law for ideal, miscible solutions, the total vapor pressure is the sum of the pure-component vapor pressures, allowing the mixture to boil below the normal boiling point of either component. Given Data / Assumptions: Benzene and water are assumed insoluble in each other (two liquid phases). Normal boiling points: benzene = 80.1°C; water = 100°C at 1 atm. Total system pressure = 1 atm; boiling occurs when total vapor pressure equals 1 atm. Concept / Approach: For immiscible liquids, the total vapor pressure P_total at a given temperature equals P_benzene^sat(T) + P_water^sat(T). Boiling begins when P_total reaches the external pressure. Because both components contribute to P_total, this condition is met at a temperature lower than the normal boiling point of either pure liquid. Step-by-Step Solution: Identify condition: boil when P_benzene^sat(T_boiling) + P_water^sat(T_boiling) = 1 atm.Since each vapor pressure rises with temperature, their sum reaches 1 atm at a temperature below 80.1°C, because neither component alone must achieve 1 atm.Therefore, T_boiling(mixture) < min(80.1°C, 100°C) = 80.1°C. Verification / Alternative check: Typical steam-distillation temperatures for benzene–water at 1 atm are around the high 60s to low 70s °C, clearly below 80.1°C, confirming the principle. Why Other Options Are Wrong: 80.1°C or 100°C: Applies to pure-component boiling, not an immiscible pair. Greater than 80.1°C but less than 100°C, or greater than 100°C: Contrary to the additive vapor pressure rule for immiscible mixtures. Common Pitfalls: Confusing immiscible behavior with ideal-solution Raoult’s law; forgetting that both components contribute independently to total vapor pressure. Final Answer: Less than 80.1°C

Question 5

Purpose of a bypass stream in process flowsheets: Why is a bypass line often included in chemical process operations?

Accepted Answer

Introduction / Context: Bypass lines are common in industrial flowsheets. They allow a fraction of a stream to circumvent a unit operation and later be mixed back, enabling operators to fine-tune properties like temperature, concentration, and flow without disturbing upstream or downstream steady states. Given Data / Assumptions: The bypass line splits a feed, sending part around a unit (e.g., heater, cooler, reactor) and later recombining. Steady-state process; standard control objectives (temperature or composition). Concept / Approach: A bypass provides a simple, fast-acting manipulation variable. For example, mixing a cooled stream with a warm bypass stream sets the final temperature precisely. Similarly, a bypass around a separator can trim concentration. It is primarily a control and flexibility tool; it does not inherently increase conversion or yield. Step-by-Step Reasoning: Introduce split: F = F_bypass + F_through_unit.Manipulate split ratio to reach target outlet T or composition x.Controller adjusts bypass valve to correct deviations quickly. Verification / Alternative check: Heat balance on a mixer with a bypassed heater shows the outlet temperature depends on the bypass fraction, demonstrating controllability without changing heater duty or upstream flows. Why Other Options Are Wrong: Improves conversion / increases yield: Not inherently; reaction performance is dictated by kinetics, residence time, and conditions inside the reactor, not by external bypass piping. Reduces pressure drop by default: Pressure behavior depends on hydraulics; a bypass can even add valves and fittings. None of these: Incorrect because process control benefits are real and well-established. Common Pitfalls: Assuming a bypass boosts reaction conversion; ignoring that bypass manipulation can affect residence time distribution if used improperly. Final Answer: Facilitates better control of the process (e.g., temperature or composition trimming)

Question 6

Steady-state mass balance: For a system operating at steady state, when does the simple relation “inputs = outputs” hold true?

Accepted Answer

Introduction / Context: Mass balances distinguish between total mass and component mass. At steady state, accumulation is zero. Even with reactions, total mass is conserved, though individual species balances include generation or consumption terms. Given Data / Assumptions: System at steady state (d/dt of holdup = 0). No leaks or unaccounted losses; total mass is conserved. Chemical reactions may occur internally. Concept / Approach: Total mass balance at steady state is: total mass in = total mass out. Component balances are: in − out + generation − consumption = 0. Reactions affect components, not the conservation of total mass (assuming no nuclear reactions). Step-by-Step Solution: Total: Σṁ_in = Σṁ_out (steady state, closed to leaks).Component i: ṁ_in,i − ṁ_out,i + r_i = 0, where r_i accounts for reaction stoichiometry.Thus “inputs = outputs” holds for the total mass even with reaction. Verification / Alternative check: Consider A → B in a CSTR at steady state. Although A decreases and B increases, the sum of mass flow rates remains constant if density changes are accounted for in volumetric terms. Why Other Options Are Wrong: Without reaction only / only without reaction and without losses: Too restrictive; the total mass relation is broader. None of these / 100% conversion requirement: Not relevant to total mass conservation. Common Pitfalls: Confusing total mass balance with species balances; overlooking leaks or evaporative losses which would violate the simple equality. Final Answer: With chemical reaction (total mass basis, no accumulation)

Question 7

Physical adsorption versus chemisorption: Which statement correctly contrasts physisorption with chemisorption?

Accepted Answer

Introduction / Context: Adsorption processes are central to separations and catalysis. Two limiting mechanisms are physical adsorption (physisorption) and chemical adsorption (chemisorption), which differ in bonding, energetics, and capacity. Given Data / Assumptions: Physisorption arises from van der Waals forces; multilayer adsorption is possible (BET behavior). Chemisorption involves specific chemical bonding; typically monolayer-limited (Langmuir behavior). Typical heats: physisorption ~ 20–40 kJ/mol; chemisorption often much larger. Concept / Approach: Because physisorption is weaker and non-specific, it can build multiple layers, often leading to a higher total uptake per mass at suitable temperatures. Chemisorption is stronger, often activated, and limited to surface sites, reducing capacity to about one monolayer. Step-by-Step Reasoning: Physisorption: weak forces, low activation barrier, possible multilayer → higher overall capacity.Chemisorption: strong bonds, activated process, site-specific → lower capacity (monolayer).Therefore, the statement about higher quantity adsorbed per unit mass fits physisorption. Verification / Alternative check: BET isotherms show increasing uptake with relative pressure due to multilayer formation, confirming higher capacities for physisorption at lower temperatures. Why Other Options Are Wrong: Surface-reaction control, high activation energy, very large heat of adsorption, strong bonds: all characterize chemisorption, not physisorption. Common Pitfalls: Assuming higher heat means better separation; confusing monolayer (Langmuir) with multilayer (BET) models. Final Answer: Quantity adsorbed per unit adsorbent mass is generally higher (multilayer possible)

Question 8

Kinetic theory of gases core idea: According to the kinetic theory, which statement is correct?

Accepted Answer

Introduction / Context: The kinetic theory of gases provides a microscopic explanation of macroscopic gas properties, relating temperature and pressure to molecular motion and collisions. Given Data / Assumptions: Ideal gas model: point particles, elastic collisions, negligible intermolecular forces except during collisions. Thermal energy is entirely kinetic in this model. Concept / Approach: In kinetic theory, absolute temperature T is proportional to the average translational kinetic energy of gas molecules. Collisions are perfectly elastic (no net energy loss), and the volume of molecules is negligible relative to the container volume. Step-by-Step Reasoning: For an ideal monatomic gas: average kinetic energy per molecule ∝ T.Collisions conserve kinetic energy and momentum (elastic).Molecular volume is neglected in the ideal model; real-gas deviations are handled by equations like van der Waals. Verification / Alternative check: Gas temperature rises when average molecular speed increases; pressure at constant volume rises accordingly, consistent with kinetic interpretations. Why Other Options Are Wrong: Energy loss in collisions or inelastic ideal-gas collisions contradicts the model. Appreciable molecular volume violates the point-particle assumption of ideal gases. “None of these” is invalid because option (c) is correct. Common Pitfalls: Extending ideal behavior to high pressures/low temperatures where real-gas effects and finite molecular size matter. Final Answer: Absolute temperature measures the average kinetic energy of molecules

Question 9

Ideal solutions and Raoult’s law: For an ideal binary solution, how does the total vapor pressure vary with composition (mole fraction)?

Accepted Answer

Introduction / Context: Understanding vapor–liquid equilibrium (VLE) begins with ideal solutions, where Raoult’s law applies. This provides a linear dependence of partial pressures, and hence total pressure, on composition. Given Data / Assumptions: Ideal solution behavior: each component obeys Raoult’s law. Binary mixture at fixed temperature. Concept / Approach: Raoult’s law states p_i = x_i * p_i^sat(T). The total pressure is P_total = Σ p_i = x_1 p_1^sat + x_2 p_2^sat. Because x_2 = 1 − x_1, P_total is a linear function of x_1 (and of x_2). Step-by-Step Solution: Write P_total(x_1) = x_1 p_1^sat + (1 − x_1) p_2^sat.Rearrange: P_total = p_2^sat + x_1 (p_1^sat − p_2^sat).This is a straight line in x_1 with slope (p_1^sat − p_2^sat). Verification / Alternative check: Graphical VLE diagrams for ideal pairs (e.g., benzene–toluene) show linear total pressure lines at constant temperature. Why Other Options Are Wrong: Inverse, exponential, parabolic, or negligible change: Do not match Raoult’s law behavior for ideal solutions. Common Pitfalls: Mixing up ideal solution (Raoult) with ideal gas behavior; forgetting temperature must be held constant for the linear relation. Final Answer: Linearly

Question 10

Trend of latent heat near the critical point: How does the heat of vaporisation change with pressure, and what happens at the critical point?

Accepted Answer

Introduction / Context: Phase-change energetics are vital for boilers, condensers, and power cycles. The latent heat of vaporisation reflects the energy required to separate liquid and vapor phases; its behavior with pressure reveals what happens as a fluid approaches the critical state. Given Data / Assumptions: Pure fluid undergoing equilibrium vaporisation along the saturation curve. Critical point: liquid and vapor phases become indistinguishable. Concept / Approach: The Clausius–Clapeyron relation shows latent heat λ ∝ T * Δv * (dP_sat/dT). As pressure increases toward the critical pressure, the specific volume difference Δv = v_g − v_l shrinks. Consequently, λ decreases and must reach zero at the critical point where Δv → 0. Step-by-Step Reasoning: Increase pressure → move up saturation curve → liquid and vapor densities converge.As Δv decreases, λ correspondingly decreases.At the critical point, Δv = 0 → λ = 0. Verification / Alternative check: Steam tables show the latent heat of water declining with pressure and vanishing at the critical point (~22.06 MPa), validating the trend. Why Other Options Are Wrong: Increase or remain constant: Contradicted by data and theory. Only “becomes zero” without noting the continuous decrease omits the trend. Common Pitfalls: Confusing latent heat with sensible heat; assuming a constant latent heat across pressures. Final Answer: Both decreases with pressure and becomes zero at the critical point

Question 11

Solutions with the same osmotic pressure: What is the standard term for solutions that exhibit equal osmotic pressure at the same temperature?

Accepted Answer

Introduction / Context: Osmosis drives solvent across semipermeable membranes from low to high solute chemical potential. In bioprocessing and medicine, matching osmotic pressures prevents cell lysis or plasmolysis; such solutions are called isotonic. Given Data / Assumptions: Comparison at the same temperature through a semipermeable membrane. Solute does not cross the membrane; solvent can. Concept / Approach: Osmotic pressure π for dilute solutions is given by π = i * C * R * T (van’t Hoff relation), where i is the dissociation factor. Two solutions with equal π at the same T are isotonic. Step-by-Step Reasoning: Identify equal π → no net solvent flow across the membrane.Hence, tonicity is matched → solutions are isotonic. Verification / Alternative check: In intravenous therapy, 0.9% saline is roughly isotonic with blood plasma, demonstrating the concept practically. Why Other Options Are Wrong: Dilute/ideal/saturated/supersaturated describe concentration, nonideality, or phase equilibrium, not equality of osmotic pressure. Common Pitfalls: Confusing isotonic with isosmotic (equal osmolarity but different membrane permeabilities can change effective tonicity). Final Answer: Isotonic

Question 12

Vapour pressure correlation name check: The empirical equation log10 P = A − B/(t + C) for pure components is known as the ________ equation.

Accepted Answer

Introduction / Context: Correlations for pure-component vapor pressures are used in VLE design and simulation. A widely used empirical fit is the Antoine equation, convenient over moderate temperature ranges. Given Data / Assumptions: Equation form: log10 P = A − B/(t + C). P is often in mmHg or bar depending on parameterization; t is temperature (commonly °C). Concept / Approach: Recognize the canonical form and its name. The Antoine equation is a three-parameter modification of the Clausius–Clapeyron relation that improves fitting accuracy across limited ranges. Step-by-Step Reasoning: Match form: log10 P vs 1/(t + C) with constants A, B, C → Antoine.Other listed items are unrelated: Kopp’s and Trouton’s are rules of thumb; Kistyakowsky is a different correlation family; Redlich–Kwong is an equation of state. Verification / Alternative check: Process simulators store Antoine constants for thousands of compounds, confirming the naming convention. Why Other Options Are Wrong: They do not correspond to the given functional form for vapor pressure. Common Pitfalls: Using Antoine constants outside their valid temperature range; mixing units for P without matching the fitted constants. Final Answer: Antoine equation

Question 13

Definition check: Heat of transition in phase/allotrope changes: Heat of transition is the heat evolved or absorbed when a substance is converted from which state to which?

Accepted Answer

Introduction / Context: Thermochemical quantities distinguish different types of transformations: fusion (melting), vaporisation, sublimation, and transition. Heat of transition specifically involves rearrangements between allotropic forms—distinct structural forms of the same element (e.g., graphite ⇄ diamond, rhombic ⇄ monoclinic sulfur). Given Data / Assumptions: Allotropy refers to structural variations at the same aggregate state (usually solid). Process occurs at constant pressure with measurable heat effect. Concept / Approach: Identify the transformation type: phase changes between solid, liquid, and vapor have their own heats (fusion, vaporisation, sublimation). A transition between two allotropic forms is neither simple melting nor boiling; the associated enthalpy change is the heat of transition. Step-by-Step Reasoning: Allotrope A (solid) ⇄ Allotrope B (solid) → heat absorbed or released = heat of transition.Examples: rhombic sulfur → monoclinic sulfur; white tin → gray tin. Verification / Alternative check: Thermochemistry tables list separate ΔH values for transition distinct from fusion or vaporisation, confirming the definition. Why Other Options Are Wrong: Vapor ↔ liquid/solid and solid → liquid are standard phase-change heats (vaporisation, condensation, sublimation, fusion) not “transition.” Liquid to supercritical fluid involves approaching the critical point, not an allotropic change. Common Pitfalls: Equating allotropic transitions with polymorphic phase changes between different aggregate states; mixing up heat-of-transition with latent heat of fusion. Final Answer: One allotropic form to another allotropic form

Question 14

At normal temperature and pressure (NTP), what is the mass (in grams) of 56 litres of ammonia gas (NH3)? Show understanding of molar volume and molar mass before selecting the value.

Accepted Answer

Introduction / Context: Gas stoichiometry questions at normal temperature and pressure (NTP) test your ability to combine the molar volume concept with molar mass. At NTP (or STP in many texts), one mole of an ideal gas occupies a standard molar volume, which lets us convert from a measured volume to moles and then to mass for any gas, including ammonia (NH3). Given Data / Assumptions: NTP molar volume = 22.4 L per mol (ideal-gas assumption). Given gas volume of NH3 = 56 L. Molar mass of NH3 = 14 (N) + 3*1 (H) = 17 g/mol. Gas behaves ideally for this calculation, which is the standard exam convention. Concept / Approach: Use two conversions in sequence: volume → moles using the molar volume; then moles → mass using the molar mass. Keep units explicit to avoid errors and round only at the end. Step-by-Step Solution: n = volume / molar volume = 56 L / 22.4 L·mol^-1 = 2.5 mol.Mass = n * molar mass = 2.5 mol * 17 g·mol^-1 = 42.5 g.Therefore, 56 L of NH3 at NTP has a mass of 42.5 g. Verification / Alternative check: You can scale from the known: 22.4 L weighs 17 g. 44.8 L would weigh 34 g; add another 11.2 L (half of 22.4 L) which weighs 8.5 g. Total 34 + 8.5 = 42.5 g, confirming the result without calculators. Why Other Options Are Wrong: 2.5 g: Confuses moles (2.5 mol) with grams. 56 g: Treats density as 1 g/L, which is not valid for NH3. 2800 g: An off-by-two-orders-of-magnitude error. 17 g: Mass of 22.4 L, not 56 L. Common Pitfalls: Using the wrong molar volume, ignoring units, or mixing up moles and grams. Always convert volumes to moles before applying molar mass. Final Answer: 42.5 g

Question 15

In analytical chemistry, a solution that resists significant pH change upon addition of small amounts of acid or base is called a what?

Accepted Answer

Introduction / Context: Being able to name and recognize a solution that maintains an approximately constant pH is fundamental in biochemistry, pharmaceutics, and analytical chemistry. Such solutions are essential for enzyme assays, titrations, and many industrial processes where pH stability is critical. Given Data / Assumptions: The solution exhibits a reasonably constant pH when small amounts of acid or base are added. We are not discussing colloidal stability, ideality, or phase equilibria; the focus is pH control. Concept / Approach: A buffer solution is typically prepared from a weak acid and its conjugate base (or a weak base and its conjugate acid). The Henderson–Hasselbalch relationship describes how the ratio of conjugate base to acid controls pH. The capacity to neutralize small additions of H+ or OH- is the buffering action. Step-by-Step Solution: Identify the behavior: resistance to pH change under small acid/base addition.Match to definition: this is the defining property of a buffer solution.Therefore, the correct term is “Buffer solution”. Verification / Alternative check: Common examples include acetate buffers (acetic acid/acetate), phosphate buffers (various H2PO4-/HPO4^2- pairs), and Tris buffers in biochemistry. Each resists pH change within a characteristic pH range near the pKa of the weak acid/base pair. Why Other Options Are Wrong: Ideal/non-ideal solutions concern activity behavior, not pH stabilization. Colloidal solution relates to particle dispersion, not acid–base buffering. Azeotropic solution refers to constant-boiling mixtures, unrelated to pH control. Common Pitfalls: Assuming strong acid plus strong base mixture forms a buffer—true buffers require a weak component and its conjugate pair in appropriate ratios. Final Answer: Buffer solution

Question 16

Gas over water correction: The method used to calculate the amount of dry gas from a measured volume of moist gas (collected over water) is based on which gas law?

Accepted Answer

Introduction / Context: When a gas sample is collected over water, the measured pressure includes contributions from both the dry gas and water vapor. Accurately determining the dry gas volume requires correcting for the water vapor’s partial pressure. This is a staple calculation in physical chemistry and gas analysis. Given Data / Assumptions: Total pressure P_total = P_dry gas + P_water vapor. Temperature is known so that the saturation vapor pressure of water can be obtained. Ideal-gas behavior is assumed for the correction. Concept / Approach: Dalton’s law states that the total pressure of a gas mixture equals the sum of the partial pressures of its components. Thus, P_dry gas = P_total − P_water(vapor). Once P_dry gas is known, use the ideal gas relation to compute moles or corrected volume of the dry gas at the same conditions. Step-by-Step Solution: Measure P_total and gas volume at given T.Look up P_sat,water at that T.Compute P_dry = P_total − P_sat,water.Use P_dry * V / (R * T) to obtain moles of dry gas (or correct V accordingly). Verification / Alternative check: If the collection temperature is lowered, P_sat,water decreases, making the water-vapor correction smaller—consistent with Dalton’s framework and observed behavior. Why Other Options Are Wrong: Charles’s and Boyle’s laws relate pairs of state variables but not directly to mixture partial pressures. Avogadro’s hypothesis concerns equal volumes at equal T and P containing equal numbers of molecules. Graham’s law is about diffusion/effusion rates, not pressure correction. Common Pitfalls: Forgetting to subtract P_sat,water or using the wrong temperature for P_sat lookup will skew the dry-gas calculation. Final Answer: Dalton’s law of partial pressures

Question 17

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

Accepted Answer

Introduction / Context: Colligative properties include boiling point elevation, freezing point depression, relative lowering of vapor pressure, and osmotic pressure. They are central to solution thermodynamics and analytical determinations like molar mass measurements of solutes. Given Data / Assumptions: Dilute, ideal (or nearly ideal) solutions. Solute is non-volatile for vapor pressure considerations. Electrolytes may dissociate, thus changing the effective number of particles. Concept / Approach: Colligative effects are proportional to the number of solute particles in a given amount of solvent, not their identity. This is why electrolytes, which dissociate into multiple ions, show larger effects than nonelectrolytes at the same molality. The van ’t Hoff factor accounts for dissociation/association when necessary. Step-by-Step Solution: Identify property class: colligative → depends on particle count.Therefore, the determining factor is the number of solute molecules (or total particles) per amount of solvent.Hence option (c) correctly captures the dependence. Verification / Alternative check: Compare a 0.1 m glucose solution with a 0.1 m NaCl solution (near complete dissociation, i ≈ 2). The NaCl solution exhibits roughly twice the colligative effect, illustrating the role of particle number, not chemical nature. Why Other Options Are Wrong: (a), (b), and (e) relate to identity and interactions; in the ideal dilute limit, colligative properties ignore identity and depend only on count. (d) is invalid because a correct general statement exists in (c). Common Pitfalls: Ignoring dissociation/association (van ’t Hoff factor i) or using molarity instead of molality when temperature changes are involved in freezing/boiling problems. Final Answer: Number of solute molecules present per given amount of solvent

Question 18

Specific gravity scales: Which scale is used exclusively for liquids heavier than water?

Accepted Answer

Introduction / Context: Process engineers encounter several specific gravity scales in industry. Knowing which scale applies to which type of liquid simplifies interpretation of measurements in chemical plants and laboratories. Given Data / Assumptions: Comparing well-known empirical scales: Baumé, Twaddell, API, and Brix. Water reference temperature is standard for the scales’ definitions. Concept / Approach: The Twaddell scale is defined for liquids heavier than water only. Baumé has two related scales (one for lighter-than-water, one for heavier-than-water liquids). API gravity is widely used for petroleum fractions, generally lighter than water. Brix measures sugar content, not specific gravity directly for generic liquids. Step-by-Step Solution: Identify the scale restricted to heavy liquids: Twaddell.Confirm alternatives: Baumé is dual; API is for hydrocarbons (lighter); Brix is sugar concentration.Therefore, the correct choice is Twaddell scale. Verification / Alternative check: Lookup relations: Twaddell degrees are proportional to (SG − 1) for SG > 1, aligning with the “heavier than water” restriction. Why Other Options Are Wrong: Baumé: has heavy and light scales, not exclusive. API: oriented to petroleum liquids, which mostly have SG < 1. Brix: expresses percent sucrose equivalent; not a general SG scale. None of these: invalid given the correct Twaddell option. Common Pitfalls: Confusing Baumé’s dual definition with exclusivity, or mistaking API for a universal SG scale. Final Answer: Twaddell scale

Question 19

When a solute dissolves in a solvent, which of the following properties of the original solvent remains unchanged (if any)? Choose the best statement.

Accepted Answer

Introduction / Context: Adding a solute to a solvent changes the solution’s properties relative to the pure solvent. Engineers must anticipate these changes to design separations, heat transfer operations, and reaction systems accurately. Given Data / Assumptions: Dilute to moderate concentrations where colligative effects and transport-property changes are observable. Non-reactive solute unless stated otherwise. Concept / Approach: Key properties of the solvent are generally altered upon dissolution: vapor pressure decreases (Raoult’s law for nonvolatile solutes), boiling point elevates, freezing point depresses, viscosity and surface tension often change, and even specific heat of the resulting solution differs from that of the pure solvent. Therefore, the safest general statement is that none of these listed properties remain unchanged. Step-by-Step Solution: Check each property qualitatively after dissolution.Vapor pressure: decreases for nonvolatile solute → changed.Viscosity: typically increases (electrolytes, polymers) → changed.Specific heat: solution Cp differs from pure solvent → changed.Thus, “None of these” is correct. Verification / Alternative check: Practical examples: adding salt to water lowers vapor pressure (hence boiling point elevation) and increases viscosity; adding sugar changes Cp and surface tension. Data tables confirm these trends. Why Other Options Are Wrong: (a), (b), (c), (e): Each property is known to change upon dissolution in most real systems. Common Pitfalls: Assuming very dilute solutions behave identically to pure solvent; while changes may be small, they are generally nonzero and measurable. Final Answer: None of these

Question 20

Methods of crystallisation from solution: Which set of strategies can induce crystallisation of a solute from its solution?

Accepted Answer

Introduction / Context: Crystallisation is a cornerstone separation and purification technique in chemical and pharmaceutical processing. Engineers often control supersaturation by manipulating solvent removal, temperature, or solution composition to nucleate and grow crystals selectively. Given Data / Assumptions: Solute has finite solubility in the solvent. No chemical reaction; only physical separation by phase change. Concept / Approach: Crystallisation requires supersaturation. Three common routes are: (1) evaporating solvent to increase solute concentration; (2) cooling (or heating, for retrograde cases) to reduce solubility; and (3) adding a second solute/antisolvent to reduce solubility of the target (salting-out or antisolvent crystallisation). Step-by-Step Solution: Route (a): Evaporation increases concentration at fixed T → supersaturation.Route (b): Cooling most solutions lowers solubility → supersaturation forms.Route (c): Adding a soluble salt or antisolvent changes activity coefficients/solubility → supersaturation.Hence all listed methods are valid approaches. Verification / Alternative check: Industrial crystallisers (evaporative and cooling types) and antisolvent crystallisation in pharma validate each route. Process selection depends on thermodynamics, kinetics, and impurity profile. Why Other Options Are Wrong: (a), (b), (c) alone are incomplete—each is valid, but the best answer is that all apply. (e) is false because multiple standard methods exist. Common Pitfalls: Overcooling causing excessive nucleation and fines; evaporating too fast causing occluded impurities; using an inappropriate antisolvent leading to oiling out instead of crystallisation. Final Answer: All of the above (a), (b) and (c)

Stoichiometry Questions