Question 1

Ideal Gas Turbine (Brayton) Plant – Assumed Nature of Compression and Expansion Under the ideal-cycle model, the compressor and turbine processes in a gas-turbine plant are treated as isentropic (adiabatic and reversible). Select the correct assumpt…

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Introduction / Context:To analyze gas-turbine (Brayton) performance, we adopt an ideal model with simplified, reversible processes. These assumptions enable closed-form relations between temperatures and pressure ratio and a compact expression for thermal efficiency. This question focuses on the assumed character of compression and expansion in that ideal model.Given Data / Assumptions:Simple Brayton cycle without regeneration, intercooling, or reheat.Negligible pressure losses in heat exchangers and passages.Calorically perfect gas with nearly constant gamma.Concept / Approach:In the ideal Brayton cycle, the compressor and turbine are modeled as adiabatic and reversible, i.e., isentropic. This yields T2/T1 = r_p^((gamma−1)/gamma) for the compressor and T3/T4 = r_p^((gamma−1)/gamma) for the turbine (with r_p the pressure ratio). These relations underpin the well-known efficiency formula eta = 1 − 1/r_p^((gamma−1)/gamma).Step-by-Step Solution:Specify component models: compressor and turbine both adiabatic and reversible.Invoke isentropic temperature-pressure relations for ideal gases.Relate temperatures to r_p and gamma for performance predictions.Conclude the correct assumption is “isentropic”.Verification / Alternative check:Real machines are polytropic with efficiencies below 100%; the ideal isentropic benchmark is used to define and compute those efficiencies and to bound performance.Why Other Options Are Wrong:Isothermal compression/expansion is not representative for Brayton hardware; “polytropic” describes real behavior, not the ideal assumption; “isobaric” processes apply to the heater/cooler (heat addition/rejection), not the turbomachinery.Common Pitfalls:Applying isentropic relations at very high temperatures without accounting for variable properties; misusing absolute vs. gauge pressures when computing r_p.Final Answer:isentropic

Question 2

Thermochemical Conversion of Wood – Product of Low-Temperature, Limited-Air Heating When wood is heated with a limited supply of air to a temperature not less than about 280°C (pyrolysis/low-temperature carbonisation), the principal solid fuel produ…

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Introduction / Context:Thermochemical conversion of biomass includes pyrolysis, gasification, and combustion. Under oxygen-limited heating of wood, volatiles are driven off and a carbon-rich solid residue remains. Identifying the name and conditions of this product is basic knowledge for fuel technology and renewable energy systems.Given Data / Assumptions:Wood heated with limited air (pyrolytic conditions).Temperature not less than about 280°C, rising through 300–500°C for typical charcoal making.Batch or continuous retort/klins used; volatiles may be recovered as by-products.Concept / Approach:In low-oxygen heating (destructive distillation), cellulose, hemicellulose, and lignin thermally decompose. The process releases tars, methanol, acetic acid, and light gases, leaving a porous carbonaceous solid. The conventional name for the solid residue from wood is wood charcoal. It has high fixed carbon, low volatile content (relative to the original wood), and good adsorptive properties. It differs from coke, which is produced by high-temperature carbonisation of coal, not wood.Step-by-Step Solution:Specify conditions: limited air and T ≥ 280°C initiate wood pyrolysis.Recognize products: gases + condensable tars + carbon-rich solid.Name the solid product from wood: wood charcoal.Differentiate from coal-derived coke and manufactured briquettes.Verification / Alternative check:Industrial and traditional charcoal production commonly operates in the 300–500°C range, yielding biochar/charcoal; analytical proximate/ultimate analyses confirm high fixed carbon and low ash for good hardwood charcoals.Why Other Options Are Wrong:Coke derives from coal at ~900–1100°C; bituminous coal and peat are geological materials, not products of wood pyrolysis; briquetted coal is a manufactured form factor, not a pyrolysis product of wood.Common Pitfalls:Using “coke” generically for any carbonized solid; overlooking the lower temperature regime and biomass origin that define charcoal.Final Answer:wood charcoal

Question 3

Entropy change when heat is removed from a gas Consider a closed system of an ideal (perfect) gas. If heat is removed from the gas (i.e., the system rejects heat), is the change in the system's entropy necessarily positive?

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Introduction / Context:Entropy quantifies the energy dispersal per unit temperature and sets limits on heat-to-work conversion. This question examines the sign of entropy change when a gas rejects heat. Understanding the distinction between entropy transfer with heat and entropy generation due to irreversibility is crucial for thermodynamics and refrigeration problems.Given Data / Assumptions:Closed system containing a fixed mass of ideal gas.Heat is removed from the system (Q Temperature is measured on the Kelvin scale.Process may be reversible or irreversible unless otherwise specified.Concept / Approach:The entropy change of a system over any process can be written as ΔS_system = ∫(δQ_rev/T) + S_gen, where S_gen ≥ 0 accounts for irreversibilities. For a reversible path at temperature T, rejecting heat (δQ_rev Step-by-Step Solution:State the relation: ΔS_system = ∫(δQ_rev/T) + S_gen.For heat removal at representative temperature T: ∫(δQ_rev/T) is negative.For a reversible cooling step: S_gen = 0, hence ΔS_system For an irreversible cooling step: ΔS_system = negative quantity + S_gen; sign depends on magnitude of S_gen.Therefore, the blanket statement “entropy change is positive when heat is removed” is incorrect.Verification / Alternative check:Cooling water or gas in a reversible heat exchanger reduces the fluid’s entropy, while the surroundings gain entropy by |Q|/T_surroundings. The total entropy of the universe increases or stays the same, but the system’s entropy typically decreases upon heat removal.Why Other Options Are Wrong:Option A claims always positive—contradicted by reversible cooling. Option C claims always zero—false except for special contrived paths. Option D ignores the path dependence of heat and the role of T. Option E is partially insightful but still does not make the original unconditional statement true; the correct evaluation of the stem is “No”.Common Pitfalls:Confusing system entropy change with total entropy generation; assuming “heat out” must increase entropy due to the second law (it does not for the system; it does for the surroundings or the total).Final Answer:No, entropy change is generally negative when heat is removed at a given temperature

Question 4

Fuel classification used in power generation In fuel technology, the term “steam coal” (used for raising steam in boilers) generally refers to which type of coal?

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Introduction / Context:Coal terminology can be confusing because it mixes rank (lignite, sub-bituminous, bituminous, anthracite), coking properties (coking vs. non-coking), and form (lump vs. pulverised). “Steam coal” is a commercial term commonly used in power generation markets to denote fuels intended for steam-raising in boilers.Given Data / Assumptions:Steam coal is sold primarily for boiler firing to generate steam and electricity.Emphasis is on calorific value, ash, volatile matter, and non-coking behavior.Coking properties are not required for power boilers.Concept / Approach:Steam coal is typically non-coking bituminous coal selected for combustion characteristics rather than metallurgical coke production. Pulverisation is a preparation method applied to many ranks and is not a distinct coal type. Brown coal (lignite) is lower rank with higher moisture and often termed “lignite,” not “steam coal” in strict trade usage, although it can be used in power stations.Step-by-Step Solution:Differentiate “rank” (bituminous, lignite, anthracite) from “coking property.”Steam coal is purchased for boilers; coking behavior is unnecessary and often undesirable.Hence, steam coal aligns with non-coking bituminous coal specifications.Verification / Alternative check:Power-plant fuel contracts and standards (for example, specifying volatile matter, ash, sulfur, and calorific value) identify non-coking bituminous grades as typical “steam coal.”Why Other Options Are Wrong:Pulverised coal (A) is a physical form, not a coal rank/type. Brown coal (B) is lignite and usually named as such. Coking bituminous (C) is reserved for metallurgical coking. Anthracite (E) is a distinct high-rank fuel and not commonly called steam coal in trade.Common Pitfalls:Equating “pulverised coal” with a coal type; conflating coking and non-coking uses.Final Answer:Non-coking bituminous coal used for thermal power (steam-raising)

Question 5

Energy conservation principle — identify the correct law “Energy can neither be created nor destroyed, but it can be transformed from one form to another.” This fundamental statement is known as:

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Introduction / Context:Every thermodynamic analysis begins with energy accounting. The principle that total energy is conserved while changing forms is the bedrock of engineering calculations in power plants, refrigeration, engines, and process equipment.Given Data / Assumptions:Macroscopic (classical) thermodynamics framework.Energy transfer modes: heat and work.Systems may be closed or open; formulation adapts accordingly.Concept / Approach:The First Law of Thermodynamics expresses energy conservation. In a closed system, ΔE_system = Q_in − W_out. In a steady-flow open system (control volume), the rate form includes enthalpy flow and kinetic/potential energy changes but still conserves total energy. The law allows interconversion among forms (thermal, mechanical, electrical, chemical) without loss of total amount.Step-by-Step Solution:Identify the quoted statement: conservation with transformation allowed.Map to thermodynamic law: the First Law.Eliminate other options: Zeroth law (thermal equilibrium), Second law (direction/entropy), kinetic theory (microscopic model).Verification / Alternative check:Practical examples include electric heaters (work → heat), heat engines (heat → work), and generators (mechanical → electrical). All obey energy balance relationships.Why Other Options Are Wrong:Zeroth law defines temperature via equilibrium; second law sets limits on conversion efficiency; kinetic theory describes molecular behavior, not energy conservation law.Common Pitfalls:Assuming the First Law guarantees 100% heat-to-work conversion—this is restricted by the Second Law.Final Answer:First law of thermodynamics

Question 6

Atomic mass — evaluate the definition “The atomic mass of an element is the number of times the atom of that element is heavier than the hydrogen atom.” Is this definition correct under the modern standard?

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Introduction / Context:Atomic and molecular masses are foundational for stoichiometry, gas laws, and thermodynamic property evaluation. Historically, hydrogen was used as a reference, but modern standards use the carbon-12 scale to avoid ambiguity from hydrogen isotopes and measurement issues.Given Data / Assumptions:IUPAC standard defines the atomic mass unit (u) via carbon-12.Relative atomic mass (dimensionless) and molar mass (kg/mol) are related but distinct concepts.Isotopic composition influences standard atomic weights.Concept / Approach:The modern definition sets the mass of a carbon-12 atom to exactly 12 u. Relative atomic mass of an element is the weighted average of isotopic masses relative to 1/12 of carbon-12, not to hydrogen. Therefore, the statement tying atomic mass directly to a hydrogen atom is outdated and incorrect today.Step-by-Step Solution:Recognize the claim: comparison to hydrogen atom as reference.Recall the accepted standard: carbon-12 provides the reference for the atomic mass unit.Conclude the statement is false under the modern convention.Verification / Alternative check:Periodic tables list relative atomic masses on the carbon-12 scale (e.g., O ≈ 15.999, Fe ≈ 55.845) derived from isotopic distributions, confirming the standard.Why Other Options Are Wrong:Options C and D mischaracterize references; option E mixes a true factor (isotopes) with the wrong reference definition.Common Pitfalls:Confusing “atomic mass unit reference” with “arbitrary hydrogen reference”; mixing molar mass and relative atomic mass.Final Answer:False

Question 7

Kinetic theory of gases — mean translational kinetic energy per kilomole For an ideal gas at absolute temperature T, what is the mean translational kinetic energy per kilogram-mole (kmol) of gas? (Ru denotes the universal gas constant.)

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Introduction / Context:The kinetic theory links temperature to molecular motion. For monatomic ideal gases (and for the translational component of any ideal gas), the average translational kinetic energy depends linearly on absolute temperature. Converting per-molecule results to per-kilomole uses the universal gas constant Ru.Given Data / Assumptions:Ideal gas behavior.Translational degrees of freedom are three in three-dimensional space.Ru is the universal gas constant per kilomole.Concept / Approach:Per molecule, mean translational kinetic energy is (3/2) * k_B * T. Multiplying by Avogadro’s number N_A converts to per mole: (3/2) * N_A * k_B * T = (3/2) * R, where R is the gas constant per mole. Per kilomole, the corresponding constant is Ru, hence energy becomes (3/2) * Ru * T.Step-by-Step Solution:Start with E_trans,per molecule = (3/2) * k_B * T.Multiply by Avogadro’s number: E_trans,per mole = (3/2) * R * T.Per kmol, replace R with Ru: E_trans,per kmol = (3/2) * Ru * T.Therefore, the correct choice is 1.5 * Ru * T.Verification / Alternative check:Dimensional check: Ru has units kJ/(kmol·K) in engineering units, so energy per kmol is kJ/kmol as expected for a temperature-proportional energy.Why Other Options Are Wrong:RuT underestimates by a factor of 1.5; 2RuT and 3RuT overestimate; 0.5Ru*T is not supported by kinetic theory.Common Pitfalls:Mixing molar (R) with universal (Ru) constants; confusing total kinetic energy with internal energy for polyatomic gases—here we refer specifically to translational component that scales as (3/2) * Ru * T per kmol.Final Answer:1.5 * Ru * T

Question 8

Diesel (air-standard) cycle — effect of cut-off ratio on efficiency Consider the ideal Diesel cycle at a fixed compression ratio and fixed specific heat ratio. How does the thermal efficiency vary as the cut-off ratio increases?

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Introduction / Context:The Diesel cycle differs from the Otto cycle by adding heat at (approximately) constant pressure over a finite cut-off. The cut-off ratio (rc) is the ratio of cylinder volumes at the end and start of heat addition. Understanding how rc affects efficiency is essential in engine thermodynamics.Given Data / Assumptions:Air-standard Diesel cycle with ideal gas and constant specific heats.Compression ratio r is fixed.No pumping, friction, or heat losses in the ideal model.Concept / Approach:Diesel efficiency can be written in terms of r, k, and rc. For fixed r and k, increasing rc prolongs heat addition at higher specific volumes, lowering the mean effective temperature of heat addition relative to the isentropic temperature rise. This reduces thermal efficiency compared with smaller rc, approaching Otto behavior only as rc → 1.Step-by-Step Solution:Recognize: as rc increases (longer constant-pressure heat addition), expansion begins later at higher volume.The average temperature at which heat is supplied decreases relative to the temperature rise achievable with earlier expansion.Consequently, the cycle’s efficiency decreases with increasing rc for constant r and k.Verification / Alternative check:Plot η_Diesel vs rc for a fixed r (e.g., r = 16, k = 1.4). Curves show monotonic decrease in η as rc increases above 1.Why Other Options Are Wrong:Option A contradicts standard Diesel analysis; option C ignores rc’s role; option D has no basis in the ideal model; option E is true only in the limiting case rc → 1.Common Pitfalls:Confusing the practical increase of real Diesel efficiency due to higher r (compared with SI engines) with the theoretical effect of rc when r is held fixed.Final Answer:Efficiency decreases with increasing cut-off ratio

Question 9

Flow processes — gas expansion through a minute aperture When an ideal (perfect) gas is forced through an aperture of minute dimensions or a porous plug, the process is best described as which of the following?

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Introduction / Context:Gas expansions can occur in different ways: through turbines (doing work), through nozzles (kinetic energy rise), or through restrictions such as valves and porous plugs (throttling). Correctly naming the process is vital for enthalpy and temperature change predictions in refrigeration and gas processing.Given Data / Assumptions:Aperture or porous plug of minute dimensions causing significant pressure drop.Steady-flow, adiabatic device; no shaft work.Negligible changes in kinetic and potential energies.Concept / Approach:For throttling (Joule–Thomson) processes, the steady-flow energy equation reduces to h1 ≈ h2 (enthalpy remains approximately constant). Temperature may change depending on the gas and inlet conditions (via Joule–Thomson coefficient). This differs from free expansion, which is a non-flow, sudden expansion into vacuum with no defined inlet/outlet states for steady flow.Step-by-Step Solution:Identify device: minute aperture/porous plug with pressure drop ⇒ throttling.Apply steady-flow energy balance with Q ≈ 0, W ≈ 0, ΔKE ≈ 0, ΔPE ≈ 0 ⇒ h2 ≈ h1.Conclude process name: Joule–Thomson (throttling) expansion.Verification / Alternative check:In refrigeration cycles using gas (e.g., Bell–Coleman alternatives for liquids use expansion valves), throttling is standard for liquids; for gases, temperature change depends on inversion temperature and initial state.Why Other Options Are Wrong:Isothermal and adiabatic reversible (isentropic) imply different energy changes; free expansion is unrestrained into vacuum, not through a tiny aperture; nozzle expansion produces large velocity rises, not enthalpy-constant throttling.Common Pitfalls:Confusing free expansion with throttling; assuming temperature must always drop—ideal gas shows no temperature change in throttling because h depends only on T.Final Answer:Throttling process (Joule–Thomson expansion)

Question 10

General gas process relation — identify the most general law For quasi-static expansion or compression of a gas between two equilibrium states, which relation represents the general polytropic form that encompasses special cases like isothermal and …

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Introduction / Context:Many real compression/expansion paths can be approximated by the polytropic relation p * v^n = C. This flexible model reduces to familiar special cases for particular n values and is widely used in compressor and expander performance analysis.Given Data / Assumptions:Simple compressible gas, quasi-static path between two states.Ideal-gas equation of state may be used for interpretation.Constant specific heats over the range (for adiabatic special case).Concept / Approach:The polytropic law p * v^n = C includes: n = 1 (isothermal, p * v = C), n = γ (reversible adiabatic, p * v^γ = C), and other n values modeling heat transfer during compression/expansion. It is more general than any one special case and fits measured data by choosing n appropriately.Step-by-Step Solution:Recognize that option B is the equation of state (not a process relation).Option A is the specific isothermal case (n = 1).Option D is the adiabatic special case and does not represent all real processes.Thus, the general process relation sought is p * v^n = C.Verification / Alternative check:Experimental compressor/expander data often fit a constant n over a pressure ratio range, validating the polytropic model for engineering prediction.Why Other Options Are Wrong:Equations for specific cases (A, D) lack generality; the ideal-gas equation (B) is a state equation; (E) is Charles’ law at constant pressure.Common Pitfalls:Using γ indiscriminately for non-adiabatic processes; confusing process path laws with equations of state.Final Answer:p * v^n = C

Question 11

Nature of heat and work in thermodynamics — select the correct statement Which of the following statements correctly characterizes heat and work in thermodynamics?

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Introduction / Context:Correctly distinguishing properties from interactions is essential. State properties (like pressure, temperature, internal energy) belong to the system at an instant, whereas heat and work are modes of energy transfer across the boundary during a process.Given Data / Assumptions:System boundaries are well defined.Processes are considered between equilibrium states.Sign conventions for heat/work are standard (engineering or physics, as long as consistent).Concept / Approach:Heat and work are not contained in a system; they occur only while a process is taking place and always involve the boundary. Consequently, they are path functions—dependent on the process path, not solely on initial and final states—unlike properties, which are point functions determined by state variables alone.Step-by-Step Solution:Evaluate A: Boundary phenomena — correct, since no “heat in a system” exists except as energy transfer across the boundary.Evaluate B: Energy in transit — correct; heat/work are the means by which energy crosses boundaries.Evaluate C: Path functions — correct; different paths between same states can yield different Q and W values.Therefore, all three statements are true ⇒ select “all of the above.”Verification / Alternative check:In cyclic processes returning to the same state, ΔE = 0 but ∮δQ and ∮δW need not be zero and often are nonzero, demonstrating their path dependence.Why Other Options Are Wrong:Option E contradicts the fundamental definition; heat and work are not properties.Common Pitfalls:Speaking of “heat content” of a system—more precise terms are internal energy, enthalpy, or specific heat capacity.Final Answer:all of the above

Question 12

Refrigeration cycles — reversed Joule (Brayton) identification The reversed form of the Joule (Brayton) cycle used for air refrigeration (gas-cycle refrigeration) is commonly known as:

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Introduction / Context:Refrigeration cycles can be realized by reversing power cycles. When the Joule (Brayton) gas turbine cycle is run in reverse, it forms a gas-cycle refrigeration system widely known in aviation and early refrigeration history as the Bell–Coleman cycle.Given Data / Assumptions:Working fluid is a gas (often air) treated approximately as ideal.Key components: compressor, heat exchanger(s), expander/turbine.Reversed operation relative to the power-producing Brayton cycle.Concept / Approach:In the reversed Brayton (Bell–Coleman) cycle, air is compressed, rejected heat to the surroundings at high pressure, expanded in a turbine to a lower temperature, and then absorbs heat from the refrigerated space before recompression. The performance metric is the coefficient of performance based on desired cooling effect over net work input.Step-by-Step Solution:Identify the forward power cycle: Joule/Brayton (compressor → heat addition at constant pressure → turbine → heat rejection at constant pressure).Reverse the sequence for refrigeration: compressor → reject heat → turbine expansion to produce cooling → absorb heat at low temperature.Recognize the historical name: Bell–Coleman cycle.Verification / Alternative check:Aircraft air-cycle cooling packs use the same principle: compressed bleed air is cooled, expanded through a turbine to produce low-temperature air for cabin cooling.Why Other Options Are Wrong:Carnot is the theoretical maximum efficiency cycle; Rankine is a vapor cycle; Stirling is an external-combustion isothermal/isochoric cycle; Gifford–McMahon is a cryogenic cooler using a displacer and valves, not the reversed Brayton arrangement.Common Pitfalls:Confusing reversed Brayton with reversed Rankine (vapor-compression) commonly used in household refrigeration; mixing gas-cycle and vapor-cycle terminology.Final Answer:Bell–Coleman cycle

Question 13

Free expansion of a gas (against vacuum): In classical thermodynamics, consider a free expansion where a gas initially confined by a partition expands into a vacuum after the partition is removed. For this process, the boundary (p*dV) work done by t…

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Introduction / Context:A free expansion is a classic thought experiment in thermodynamics. A gas suddenly expands into a vacuum with no opposing external pressure. Understanding why the boundary work term is zero clarifies differences between work, heat, and changes in internal energy, and prevents common misconceptions about “expansion implies work”.Given Data / Assumptions:Initial gas in one chamber; the other chamber is vacuum.Partition removed; expansion is rapid and unresisted.No external pressure opposing the expansion (p_ext = 0).Closed system; no shaft work, no moving boundary against resistance.Concept / Approach:Boundary work is defined as W_b = ∫ p_ext * dV. In free expansion, the external pressure p_ext equals zero throughout the expansion, so the integral gives zero independent of the change in volume or temperature. Internal energy may still change if heat exchange occurs with the surroundings, but for an ideal gas in an insulated container the internal energy (and temperature) remain the same.Step-by-Step Solution:Identify work term: W_b = ∫ p_ext dV.For free expansion, p_ext = 0 at all instants.Therefore W_b = ∫ 0 * dV = 0.Result holds regardless of ideal/real gas, as it depends on external pressure.Verification / Alternative check:A paddle wheel doing work or a resisting piston would be required for nonzero boundary work. Neither exists here; hence zero work is consistent with the definition and experiments.Why Other Options Are Wrong:False: Contradicts W_b definition.Only for ideal gases: Incorrect; the integral uses external pressure, not gas model.Zero only if temperature stays constant: Temperature behavior is irrelevant to W_b in free expansion.Common Pitfalls:Confusing “gas expands” with “system does work”. Work requires an opposing force; vacuum provides none.Final Answer:True

Question 14

Petroleum refining basics: In fractional distillation of crude oil, the gasoline (petrol) fraction typically distills over which approximate temperature range (°C)?

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Introduction / Context:Crude oil is separated into useful products by fractional distillation. Each fraction is characterized by a boiling range corresponding to hydrocarbon molecular weights. Recognizing the approximate range for petrol (gasoline) is essential for refinery overview questions and fuel property discussions.Given Data / Assumptions:Conventional atmospheric distillation column conditions.Typical refinery cut points; actual ranges vary with crude and operation.Gasoline (petrol) is a light fraction used in spark-ignition engines.Concept / Approach:In a standard atmospheric distillation, lighter fractions boil off first. Gasoline typically spans the low-to-moderate temperature region. Kerosene, gas oils, and residues boil at progressively higher temperatures. Therefore, among the ranges offered, the lightest interval (65° to 220°C) aligns with gasoline, while higher ranges correspond to kerosene (roughly 150–275°C), diesel/light gas oil (around 200–370°C), and heavier gas oils/residues beyond that.Step-by-Step Solution:Identify the lightest temperature window among the options.Associate lower boiling range with lighter hydrocarbons (C5–C10).Select 65° to 220°C as the gasoline cut.Verification / Alternative check:Petroleum textbooks and refinery data sheets commonly list gasoline cuts roughly 30–200°C or 40–205°C; 65–220°C remains a reasonable approximate range for generic MCQs.Why Other Options Are Wrong:220–345°C and 345–470°C: More typical of kerosene to gas oil cuts.470–550°C: Characteristic of vacuum gas oils/residual fractions, not petrol.Common Pitfalls:Assuming precise fixed temperatures; in practice, ranges overlap and are adjusted by product specifications.Final Answer:65° to 220°C

Question 15

Thermodynamic property concept: The heat-related energy stored within a gas that manifests as a rise in its temperature is known as what property?

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Introduction / Context:When a gas is heated, its microscopic motions (translation, rotation, vibration) intensify. The energy associated with these microscopic degrees of freedom is tracked via a state property that dictates temperature changes and appears in the first law. Identifying this property correctly is fundamental for energy balance problems.Given Data / Assumptions:Closed system perspective for a simple compressible gas.No phase change; heating changes temperature and possibly volume/pressure.Microscopic interpretation via kinetic theory and molecular structure.Concept / Approach:Internal energy, U, is the energy contained within the system due to molecular motions and interactions. For ideal gases, changes in internal energy depend only on temperature and are quantified by ΔU = m * cv * ΔT. External energy is not a standard thermodynamic property; kinetic energy typically refers to macroscopic bulk motion; “molecular energy” is an informal phrase encompassing internal energy components but is not the standard property name used in balances.Step-by-Step Solution:Relate heating to temperature rise.Recognize ΔU as the property capturing microscopic energy changes.Conclude the correct term is internal energy.Verification / Alternative check:Applying the first law for a closed system: ΔU = Q − W (neglecting KE and PE changes). A positive heat input at constant volume leads directly to ΔU = m * cv * ΔT, confirming the role of internal energy.Why Other Options Are Wrong:External energy: Not a defined thermodynamic property.Kinetic energy: Refers to bulk motion of the system as a whole.Molecular energy: Vague wording; the formal property is internal energy.Common Pitfalls:Mixing up path functions (heat, work) with properties (U, H, T, p).Final Answer:internal energy

Question 16

Reversible adiabatic (isentropic) ideal-gas relation: For a reversible adiabatic process between states 1 and 2 of an ideal gas, which expression correctly gives T1/T2?

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Introduction / Context:Isentropic (reversible adiabatic) relations connect temperature, pressure, and volume changes of ideal gases and appear in nozzle, compressor, and turbine analyses. Choosing the correct exponent and arrangement is critical for accurate calculations.Given Data / Assumptions:Ideal gas with constant specific heats; γ = cp/cv.Process is adiabatic and reversible (isentropic).States 1 and 2 with (p1, v1, T1) and (p2, v2, T2).Concept / Approach:For an ideal gas under isentropic change, p * v^γ = constant and T * v^(γ − 1) = constant. From the latter, T1 * v1^(γ − 1) = T2 * v2^(γ − 1). Rearranging gives T1/T2 = (v2/v1)^(γ − 1). Alternative correct forms include T1/T2 = (p1/p2)^((γ − 1)/γ) and T2/T1 = (v1/v2)^(γ − 1). Only the volume-based expression in the offered list matches the correct relation exactly as written.Step-by-Step Solution:Start with T * v^(γ − 1) = constant.Form ratio: T1/T2 = (v2/v1)^(γ − 1).Confirm consistency with pressure form if needed.Verification / Alternative check:Combine pv^γ = const with ideal-gas law to obtain Tp^((1 − γ)/γ) = const, which is equivalent and yields the same temperature ratio when expressed in pressure terms.Why Other Options Are Wrong:(v1/v2)^(γ − 1): Reciprocal of the correct ratio.(p1/p2)^(γ − 1): Wrong exponent; should be ((γ − 1)/γ).(p2/p1)^((γ − 1)/γ): Reciprocal in pressure form.Common Pitfalls:Confusing exponents γ − 1 and (γ − 1)/γ; flipping state indices inadvertently.Final Answer:T1/T2 = (v2/v1)^(γ - 1)

Question 17

Atomic mass comparison: Relative to oxygen, the atomic mass of nitrogen is ________.

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Introduction / Context:Atomic masses are fundamental constants used in chemical stoichiometry, gas property evaluation, and conversion between molar and mass-based quantities. Comparing nitrogen and oxygen is common because air is mainly composed of N2 and O2.Given Data / Assumptions:Standard atomic masses: N ≈ 14, O ≈ 16 (on the 12C = 12 scale).Comparison is per atom, not per molecule.No isotope effects considered.Concept / Approach:Atomic mass of nitrogen (≈14) is smaller than that of oxygen (≈16). Therefore, on a molar basis, nitrogen-containing gases tend to have lower molar masses than comparable oxygen-containing gases, affecting densities and specific gas constants R = R_u/M.Step-by-Step Solution:Recall N: ~14 u; O: ~16 u.Compare directly: 14 Conclude: nitrogen has less atomic mass than oxygen.Verification / Alternative check:Air’s mean molar mass (~28.97 kg/kmol) reflects dominance of N2 (28) with O2 (32), consistent with nitrogen’s lower atomic mass.Why Other Options Are Wrong:Equal to/more than: Contradict tabulated atomic masses.Varies with temperature: Atomic mass is a constant; temperature affects state, not mass.Common Pitfalls:Confusing atomic mass with molecular mass (e.g., N2 = 28 vs O2 = 32).Final Answer:less than

Question 18

Universal gas constant (molar basis): Select the correct SI value and unit for the universal gas constant R_u.

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Introduction / Context:The universal gas constant R_u links macroscopic thermodynamic variables on a molar basis via pV = n R_u T. Correct numerical value and units are essential for accurate conversions between molar and mass-based formulations.Given Data / Assumptions:Molar basis: per kmol (engineering) or per mol (scientific).Common engineering unit: J/(kmol·K).R_u ≈ 8.314 J/(mol·K) = 8314 J/(kmol·K).Concept / Approach:Because 1 kmol = 1000 mol, multiply the molar value by 1000 to convert: 8.314 J/(mol·K) × 1000 = 8314 J/(kmol·K). Many textbooks denote kmol as “kgmole” in older notation, hence 8314 J/kgmole·K. The specific gas constant for a particular gas is R = R_u / M, where M is molar mass (kg/kmol).Step-by-Step Solution:Start with R_u = 8.314 J/(mol·K).Convert to per kmol: multiply by 1000 → 8314 J/(kmol·K).Match with provided unit convention: J/kgmole-K.Verification / Alternative check:Using R = R_u/M for air with M ≈ 28.97 kg/kmol gives R ≈ 287 J/(kg·K), a widely known value, confirming consistency.Why Other Options Are Wrong:8.314 J/kgmole-K, 83.14, 831.4: Off by factors of 10^3, 10^2, and 10, respectively.Common Pitfalls:Mixing molar and mass-specific constants; missing the 1000 factor when switching mol→kmol.Final Answer:8314 J/kgmole-K

Question 19

Cycle identification: Which ideal cycle consists of two constant-pressure (isobaric) processes and two isentropic processes?

Accepted Answer

Introduction / Context:Identifying cycles by their process sequence is crucial for analyzing gas turbines, internal combustion engines, and refrigeration systems. The air-standard Brayton (Joule) cycle models the simple gas turbine engine used in power generation and aviation.Given Data / Assumptions:Air-standard analysis, constant specific heats.Two isentropic legs (compressor and turbine).Two constant-pressure legs (combustor heat addition and cooler/stack heat rejection).Concept / Approach:The Brayton cycle sequence is: 1→2 isentropic compression, 2→3 constant-pressure heat addition, 3→4 isentropic expansion, 4→1 constant-pressure heat rejection. This distinguishes it from the Otto cycle (constant-volume heat addition), Diesel cycle (constant-pressure heat addition and constant-volume heat rejection), Carnot (isothermal and isentropic), and Stirling (isothermal with regeneration).Step-by-Step Solution:Map processes to components: compressor → combustor → turbine → exhaust/cooler.Check constraints: CP for heating/rejection, isentropic for compression/expansion.Conclude it is the Brayton (Joule) cycle.Verification / Alternative check:p–V or T–s diagrams show two vertical isentropes and two near-horizontal CP lines for Brayton, consistent with gas turbine operation.Why Other Options Are Wrong:Carnot: Isothermal + isentropic, not constant-pressure.Stirling: Isothermal with regeneration, not isentropic legs.Otto: Constant-volume heat addition and rejection at constant volume.Common Pitfalls:Confusing Brayton with Diesel; Diesel's heat rejection is at constant volume, not constant pressure.Final Answer:Brayton (Joule) cycle

Question 20

Pressure unit conversion: One bar corresponds to what value in SI base units (N/m^2)?

Accepted Answer

Introduction / Context:Bar is a convenient non-SI pressure unit frequently used in engineering practice, weather reports, and manufacturer datasheets. Converting between bar and pascals (N/m^2) is necessary for consistency in calculations and standards compliance.Given Data / Assumptions:1 Pa = 1 N/m^2 by definition.Exact definition: 1 bar = 100,000 Pa.Atmospheric pressure at sea level ≈ 1.01325 bar (approximate reference).Concept / Approach:By definition, 1 bar equals 10^5 pascals. This makes bar convenient for approximating atmospheric pressure while maintaining simple powers of ten in engineering problems.Step-by-Step Solution:Write conversion: 1 bar = 100,000 Pa.Replace Pa with N/m^2: 1 bar = 1 × 10^5 N/m^2.Select the matching option.Verification / Alternative check:Check sanity: 1 bar ≈ 0.987 atm; 1 atm = 101,325 Pa, close to 10^5 Pa, confirming the order of magnitude.Why Other Options Are Wrong:10^2, 10^3, 10^4 N/m^2: Off by several orders of magnitude.Common Pitfalls:Confusing bar with atm or using kPa incorrectly; remember 1 bar = 100 kPa exactly.Final Answer:1 × 10^5 N/m^2

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