Question 1

Entropy as a state function — choosing appropriate independent variables: For a simple compressible substance, entropy can be expressed as a function of which independent variables?

Accepted Answer

Introduction / Context:Entropy s is a property (state function). For simple compressible substances (including ideal gases), the thermodynamic state is fully specified by two independent intensive variables. Choosing the proper pair is foundational for property calculations and tables.Given Data / Assumptions:Single-phase, simple compressible substance.Equilibrium conditions (uniform properties).Fixed composition.Concept / Approach:Because s is a state property, it can be expressed as s = s(p, T) or s = s(T, v) (and other valid pairs like s(p, v) with constitutive closure). “Heat” and “work” are path-dependent interactions, not properties; they cannot serve as state coordinates. The choice of independent variables affects convenience: gases often use (T, p), while tables for water may use (T, v) or (T, p).Step-by-Step Solution:Recognize that two independent variables determine the state for a simple compressible substance.Identify valid pairs: (p, T) and (T, v) are both sufficient to fix state and hence s.Eliminate invalid pair: (Q, W) are not properties and depend on the process path.Therefore, the most complete correct option is “either (p, T) or (T, v).”Verification / Alternative check:Property relations (e.g., Maxwell relations) and equations of state allow expressing s in different coordinate pairs; steam tables frequently tabulate s against T and p.Why Other Options Are Wrong:Options A and B are each partially correct but not uniquely most complete; option C is incorrect because Q and W are not state variables.Common Pitfalls:Assuming a unique pair is required; in reality, any independent pair plus an equation of state suffices for a simple compressible substance.Final Answer:either (p, T) or (T, v) for a simple compressible substance

Question 2

Avogadro’s law — statement check: “Equal volumes of all gases, at the same temperature and pressure, contain equal numbers of molecules.” Is this statement (Avogadro’s law) correct?

Accepted Answer

Introduction / Context:Avogadro’s law underpins the concept of molar volume of gases and the equivalence between volume ratios and mole ratios under identical temperature and pressure. It is a cornerstone for stoichiometry and gas-mixture calculations.Given Data / Assumptions:Ideal-gas behavior at the same temperature and pressure.Equal volumes are being compared.Fixed overall conditions across samples.Concept / Approach:If T and p are equal, the ideal-gas law pV = nRT shows V ∝ n. Therefore, equal volumes correspond to equal moles n, which implies equal numbers of molecules (since molecules = n * N_A, with N_A Avogadro’s number). This is the content of Avogadro’s law.Step-by-Step Solution:Write ideal-gas relation: pV = nRT with T and p equal for both samples.Rearrange: V = (nR/ p) * T ⇒ V ∝ n.Hence equal V implies equal n and therefore equal number of molecules.Verification / Alternative check:Molar volume at standard temperature and pressure (STP) is approximately 22.414 L per mole (idealized), illustrating the law’s practical use.Why Other Options Are Wrong:Marking “False” would contradict the ideal-gas model and widespread experimental validation under dilute conditions.Common Pitfalls:Extending the law to high-pressure or highly non-ideal regimes without corrections; forgetting it applies to gases, not liquids/solids.Final Answer:True

Question 3

Molecular mass — correct modern definition: “The molecular mass of a substance is the number of times a molecule is heavier than the hydrogen atom.” Evaluate this statement for correctness under the modern standard.

Accepted Answer

Introduction / Context:Precise definitions matter in chemistry and thermodynamics. The historical hydrogen-based scale has been superseded; today, molecular (or relative molecular) mass is referenced to the carbon-12 isotope standard. This distinction affects how we define atomic and molecular masses in SI.Given Data / Assumptions:Modern IUPAC definition using the carbon-12 scale.Relative molecular mass is dimensionless; molar mass uses g/mol or kg/mol.Hydrogen exists as isotopes (protium, deuterium, tritium), complicating a hydrogen-based reference.Concept / Approach:The modern standard defines the atomic mass unit (u) so that the mass of a carbon-12 atom is exactly 12 u. Relative molecular mass M_r of a molecule is the ratio of its mass to 1/12 of the mass of a carbon-12 atom, not to the mass of a hydrogen atom. Therefore, the statement comparing to hydrogen is outdated and incorrect today.Step-by-Step Solution:Identify the claim: “heavier than the hydrogen atom.”Recall standard: 1 u corresponds to 1/12 of carbon-12 mass; molecular masses are sums of atomic relative masses on this scale.Conclude that the hydrogen comparison does not define molecular mass under modern conventions.Verification / Alternative check:Periodic tables list relative atomic masses on the carbon-12 scale; calculated molecular masses (e.g., H2O ≈ 18.015) are consistent with this standard, not with hydrogen-based scaling.Why Other Options Are Wrong:Marking “True” would accept an obsolete definition and conflict with SI/IUPAC standards.Common Pitfalls:Confusing molar mass (kg/mol) with relative molecular mass (dimensionless); citing historical hydrogen scale without noting its replacement.Final Answer:False

Question 4

Charles’ law — thermal expansion coefficient at constant pressure: According to Charles’ law, all perfect gases change in volume by what fraction of their original volume at 0 °C for each 1 °C rise in temperature at constant pressure?

Accepted Answer

Introduction / Context:Charles’ law defines the proportional expansion of an ideal gas with temperature at constant pressure. Expressing the change per degree near 0 °C leads to the classical fraction used in early engineering texts and helps build intuition for the absolute temperature scale.Given Data / Assumptions:Ideal-gas behavior.Reference volume is the volume at 0 °C (273 K).Small temperature increments around this point at constant pressure.Concept / Approach:Charles’ law states V ∝ T at constant p. Using absolute temperature, a 1 K rise from 273 K to 274 K increases volume by approximately 1/273 of the 0 °C volume. Hence, per 1 °C change near 0 °C, the fractional change is about 1/273 of the original volume at 0 °C.Step-by-Step Solution:Start with V1 / T1 = V2 / T2 at constant p.Let T1 = 273 K (0 °C), T2 = 274 K; then V2/V1 = 274/273.Hence ΔV / V1 = (274 − 273)/273 = 1/273 per degree Celsius at 0 °C.Verification / Alternative check:Over larger ranges, V scales linearly with T: at 100 °C (373 K), V ≈ (373/273) * V_0, consistent with the same proportionality.Why Other Options Are Wrong:Fractions 1/27, 1/93, and 1/173 are not consistent with the absolute zero offset and would imply incorrect zero-volume intercepts.Common Pitfalls:Using Celsius directly in proportional formulas; forgetting the need to convert to Kelvin to maintain linear proportionality.Final Answer:1/273 of the original volume

Question 5

Thermal efficiency comparison at identical peak conditions: For air-standard cycles operating between the same maximum pressure and the same maximum temperature, which cycle is more efficient?

Accepted Answer

Introduction / Context:This question probes a classic comparison among ideal air-standard cycles (Otto, Diesel, and Dual) when the upper limits of temperature and pressure are fixed. Understanding which cycle is more efficient under identical peak conditions clarifies how the mode of heat addition (constant volume vs constant pressure) affects thermal efficiency.Given Data / Assumptions:Air-standard assumptions with constant specific heats.Same maximum temperature T_max and same maximum pressure p_max for the cycles compared.Internally reversible compression and expansion (isentropic legs).Concept / Approach:For identical T_max and p_max, the position of state points forces different compression ratios and different amounts of heat addition at various specific volumes. The Diesel cycle, with constant-pressure heat addition, tends to distribute heat at relatively lower average specific volume than the Otto cycle would under the same peak limits, resulting in a higher average temperature of heat addition for Diesel in this constrained comparison. Consequently, for fixed p_max and T_max, η_Diesel > η_Otto, while the Dual cycle efficiency generally lies between them for most parameter choices.Step-by-Step Solution:Fix T_max and p_max for both cycles to ensure comparable limits.Note that, to meet these peaks, the Otto cycle must have a different compression ratio than the Diesel cycle.Evaluate effect of heat-addition path: constant volume (Otto) vs constant pressure (Diesel) on mean temperature of heat addition.Conclude that, under the stated constraints, the Diesel cycle achieves higher thermal efficiency than the Otto cycle.Verification / Alternative check:Graphical p–V or T–s comparisons with identical peaks show the Diesel rectangle/area achieving a larger net work for the same heat input. Textbook orderings commonly report Diesel > Dual > Otto for fixed maximum pressure and heat input; with both p_max and T_max constrained, Diesel remains more efficient than Otto.Why Other Options Are Wrong:Otto more efficient: True only when compression ratio is fixed (not the case here).Dual more efficient than both: Typically lies between Diesel and Otto for comparable constraints.Dual less efficient than both: Too strong; depends on parameters and is not generally valid.Common Pitfalls:Confusing comparisons at fixed compression ratio with those at fixed peak conditions; assuming conclusions for one constraint automatically carry over to the other.Final Answer:Diesel cycle is more efficient than Otto cycle

Question 6

Air-standard efficiency order at constant maximum pressure and the same heat input: Arrange the cycles by efficiency (highest to lowest).

Accepted Answer

Introduction / Context:Efficiency comparisons among Otto, Diesel, and Dual cycles depend on what constraints are held fixed. This item focuses on the case where the maximum pressure is the same and the total heat added is the same for all cycles—an important design-stage comparison for peak-limited machines such as engines and gas generators.Given Data / Assumptions:Air-standard model with constant specific heats.Identical maximum pressure p_max for all cycles considered.Same total heat input Q_in across cycles.Reversible compression/expansion (isentropic legs).Concept / Approach:Under fixed p_max and Q_in, the manner of heat addition changes the average temperature at which heat is supplied. Diesel adds heat at constant pressure over a range of volumes and tends to yield a higher mean effective temperature of heat addition than the Otto cycle in this constrained comparison. The Dual cycle splits heat addition between constant volume and constant pressure, and its efficiency typically falls between Diesel and Otto. Hence, the general order is Diesel > Dual > Otto.Step-by-Step Solution:Set p_max equal and Q_in equal for all cycles.Compare heat-addition modes: CV (Otto), CP (Diesel), mixed (Dual).Assess average temperature of heat addition and resulting net work for each cycle.Conclude the efficiency ranking: Diesel > Dual > Otto.Verification / Alternative check:Standard thermodynamics texts illustrate this ordering with p–V/T–s diagrams and parametric studies; as the cut-off ratio approaches 1, Dual trends toward Otto; as the CV fraction decreases, Dual trends toward Diesel, remaining intermediate.Why Other Options Are Wrong:Otto > Diesel > Dual or Dual-first orders: These contradict the typical ranking for equal p_max and Q_in.Common Pitfalls:Carrying over the “same compression ratio” result (Otto > Diesel) to the “same maximum pressure and heat input” case; the constraint change reverses the order.Final Answer:Diesel cycle > Dual cycle > Otto cycle

Question 7

Ideal Diesel cycle composition: How many and what types of thermodynamic processes constitute the air-standard Diesel cycle?

Accepted Answer

Introduction / Context:The Diesel cycle is the idealized model for compression-ignition engines. Knowing the nature of each process in the cycle is essential for writing energy equations, deriving efficiency formulas, and interpreting p–V and T–s diagrams.Given Data / Assumptions:Air-standard analysis with constant specific heats.Internally reversible compression and expansion.Heat addition at constant pressure; heat rejection at constant volume.Concept / Approach:The Diesel cycle consists of four processes: (1) isentropic compression, (2) constant-pressure heat addition, (3) isentropic expansion, and (4) constant-volume heat rejection. This combination distinguishes Diesel from Otto (constant-volume heat addition) and Dual (split CV then CP addition).Step-by-Step Solution:Identify compression: 1→2 isentropic.Heat addition: 2→3 at constant pressure.Expansion: 3→4 isentropic.Heat rejection: 4→1 at constant volume.Verification / Alternative check:Plotting on p–V shows a vertical up-right CP line for 2→3 and a horizontal CV line for 4→1; T–s exhibits an isentropic pair with s constant during 1→2 and 3→4.Why Other Options Are Wrong:Two CV + two isentropic: That is the Otto cycle, not Diesel.Two CP + two isentropic: Rejects constant-volume rejection.Two CV + two isothermal: Not an air-standard engine cycle.Common Pitfalls:Assuming both heat interactions are constant pressure; confusing Diesel with Dual, where heat addition starts at CV and then continues at CP.Final Answer:one constant pressure, one constant volume, and two isentropic

Question 8

Kelvin–Planck statement of the second law: Is it possible to construct a cyclic heat engine whose sole effect is to convert all absorbed heat into work?

Accepted Answer

Introduction / Context:This question targets the Kelvin–Planck statement of the second law of thermodynamics, which places a fundamental limit on the performance of heat engines operating in cycles.Given Data / Assumptions:A cyclic device exchanging heat and work with surroundings.Sole purpose: convert heat into work with no other effect.No allowance for additional reservoirs or changes elsewhere.Concept / Approach:The Kelvin–Planck statement asserts that it is impossible to devise a cyclic engine that converts all heat received from a single thermal reservoir entirely into work. Some heat must be rejected to a lower-temperature reservoir. Thus, 100% conversion in a cycle violates the second law.Step-by-Step Solution:Define a cyclic engine with heat input Q_in.If W_out = Q_in and Q_out = 0, efficiency would be 100%.Second law forbids this for any real or ideal cyclic device.Verification / Alternative check:Carnot's theorem: Even the ideal reversible engine (Carnot) has efficiency η = 1 − T_c/T_h 0 unless T_c = 0 K (unattainable).Why Other Options Are Wrong:Possible / possible with an ideal gas / possible at absolute zero: Each contradicts the second law or invokes unattainable conditions.Common Pitfalls:Confusing one-shot energy conversion (non-cyclic) with cyclic operation; misinterpreting the role of multiple reservoirs.Final Answer:impossible

Question 9

Entropy trend of the universe: Is the entropy of the universe continually increasing according to the second law?

Accepted Answer

Introduction / Context:The second law of thermodynamics, in its broad cosmological interpretation, states that the total entropy of the universe (system plus surroundings) does not decrease. This idea underlies irreversibility and the direction of natural processes.Given Data / Assumptions:Universe modeled as system + surroundings.Real processes involve irreversibilities (friction, mixing, heat transfer across finite ΔT, chemical reaction, etc.).Entropy is a state property; changes are path-independent.Concept / Approach:For any real process, ΔS_universe = ΔS_system + ΔS_surroundings ≥ 0. Equality holds only for the idealized limit of a reversible process. Thus, as processes unfold spontaneously, the universe’s entropy increases or, in the reversible limit, remains constant—never decreases.Step-by-Step Solution:Write the entropy balance for combined system and surroundings.Recognize sources of entropy generation (internal irreversibility).Conclude monotonic non-decreasing behavior of universe entropy.Verification / Alternative check:Clausius inequality for any cycle, ∮(δQ/T) ≤ 0, is equivalent to ΔS_universe ≥ 0 for real processes, confirming the directionality of time in thermodynamics.Why Other Options Are Wrong:Incorrect: Violates the second law.True only for closed systems / reversible cycles: The statement applies generally; reversible cycles give ΔS_universe = 0, not Common Pitfalls:Interpreting “always increasing” as strictly increasing in every infinitesimal step; the rigorous statement is “never decreases,” with equality for reversible processes.Final Answer:Correct

Question 10

Behavior of superheated vapor: A superheated vapor most closely behaves like which idealized substance?

Accepted Answer

Introduction / Context:When a vapor is heated significantly above its saturation temperature at a given pressure, it is called superheated. Engineers often approximate superheated vapors as ideal (perfect) gases to simplify property evaluations and process calculations.Given Data / Assumptions:Vapor well above saturation line at the working pressure.Moderate pressures where intermolecular forces are relatively weak.Use of standard correlations or tables when necessary.Concept / Approach:Far from the saturation dome, compressibility factors approach unity and the ideal-gas equation pV = mRT becomes an adequate approximation. Thus, a superheated vapor behaves similarly to a perfect gas, allowing the use of cp, cv, and R with minimal error for many engineering calculations (though property tables remain more accurate near saturation or at very high pressures).Step-by-Step Solution:Identify the state as superheated, not saturated or wet.Apply ideal-gas approximations when Z ≈ 1.Use perfect-gas relations for processes (e.g., ΔU = mcvΔT, ΔH = mcpΔT) as first estimates.Verification / Alternative check:Comparison of superheated-steam table values with ideal-gas estimates at moderate pressure shows small deviations, validating the approximation away from the dome.Why Other Options Are Wrong:Air/steam/ordinary gas: These labels are either specific substances or vague terms; the intended idealized behavior is the perfect gas model.Common Pitfalls:Using ideal-gas relations too close to saturation or at very high pressures, where real-gas effects become significant.Final Answer:perfect gas

Question 11

Fuel preparation terminology: Coal that is first dried and then ground to a fine powder in a pulverizing mill is called what?

Accepted Answer

Introduction / Context:Large steam generators and some industrial furnaces use coal in a finely divided form to increase surface area, promote rapid combustion, and improve flame stability. Correct terminology helps avoid confusion with other prepared fuels such as briquettes or coke.Given Data / Assumptions:Coal is first dried to reduce moisture.Then crushed and ground to a fine powder in a pulverizer.Combustion occurs in a burner designed for dust-firing.Concept / Approach:The product is called pulverized coal (or pulverized fuel). The small particle size enables near-complete burnout within milliseconds in suspension firing, leading to high combustion efficiency and easier control. This is distinct from briquetted coal, which is made by compacting fine coal with a binder into larger lumps, and from wood charcoal, which is a carbon-rich solid derived from wood, not coal.Step-by-Step Solution:Identify the processing steps: drying → grinding to fine powder.Recognize the intended firing method: suspension firing in boilers.Name the product: pulverized coal (pulverized fuel).Verification / Alternative check:Boiler manuals and combustion texts describe PF (pulverized-fuel) firing systems with mills, classifiers, burners, and cyclones or ESPs—terminology matches the process described.Why Other Options Are Wrong:Briquetted coal: Agglomerated, not powdered; used for domestic or small industrial firing.Wood charcoal: Derived from wood via pyrolysis; not coal.Bituminous coal (lumped): Raw size-graded coal, not pulverized.Common Pitfalls:Assuming any prepared coal is a briquette; ignoring moisture control prior to pulverizing which is vital for mill performance and flame stability.Final Answer:pulverized coal (pulverized fuel)

Question 12

Process identification by p*v behavior: A gas is heated or expanded such that the product p * v remains constant throughout the process. What is this process called?

Accepted Answer

Introduction / Context:Many thermodynamic processes are classified by simple functional relations among p, v, and T. Recognizing the label associated with p * v = constant helps with quick diagramming and integration of work terms.Given Data / Assumptions:Throughout the process, p * v = constant for the working gas.Ideal-gas assumption may or may not be invoked; the definition of the process is algebraic.No additional constraints (like heat transfer conditions) are specified.Concept / Approach:The general polytropic relation is p * v^n = constant. When n = 1, the p–v curve is a rectangular hyperbola, and the process is called a hyperbolic process. For an ideal gas, an isothermal process also satisfies p * v = m * R * T = constant if temperature is constant; thus, for ideal gases, a hyperbolic process with n = 1 coincides with isothermal. However, the broader, name-by-equation classification is “hyperbolic.”Step-by-Step Solution:Write the observed law: p * v = C.Compare with polytropic form p * v^n = C → n = 1.Therefore, classify as a hyperbolic (n = 1) process.Verification / Alternative check:On a p–v plot, p = C/v gives a rectangular hyperbola; the area under the curve ∫p dv = C * ln(v2/v1) is the characteristic work expression for n = 1.Why Other Options Are Wrong:Isothermal: True for ideal gases with constant T, but the strict name tied to p * v = constant is “hyperbolic.”Adiabatic: Follows p * v^γ = constant for an ideal gas; n = γ ≠ 1.Polytropic: Too general; n can be any real number, the specific case n = 1 is hyperbolic.Common Pitfalls:Assuming p * v = constant automatically implies no heat transfer; that is only guaranteed for ideal-gas isothermal processes, not for real gases or non-ideal conditions.Final Answer:hyperbolic process

Question 13

Fuel property definition: The amount of heat released by the complete combustion of 1 kg of a fuel is called what?

Accepted Answer

Introduction / Context:Calorific value (heating value) is a fundamental property used to size burners, boilers, and engines, and to estimate emissions per unit of energy delivered. It is defined per unit mass for solid and liquid fuels and per unit volume or mole for gaseous fuels.Given Data / Assumptions:Combustion is complete, with products cooled to a reference state.Basis: 1 kg of fuel for solids/liquids.Two conventions exist: higher and lower heating values.Concept / Approach:The general term “calorific value” refers to the heat released by complete combustion per kilogram of fuel. Subtypes include Higher Calorific Value (HHV), which assumes water in products is condensed (latent heat recovered), and Lower Calorific Value (LCV/LHV), which assumes water remains vapor (latent heat not recovered). Both are specific cases of the broader “calorific value.”Step-by-Step Solution:Define calorific value generically: CV = heat released / mass of fuel.Recognize HHV and LHV as reporting conventions.Select “calorific value” as the correct umbrella term.Verification / Alternative check:Laboratory bomb calorimeter tests measure HHV directly; LHV is computed by subtracting the latent heat of vaporization corresponding to the formed water.Why Other Options Are Wrong:Heat energy: Vague; not a property name.Lower calorific value / higher calorific value: These are specific variants, not the general definition asked.Common Pitfalls:Comparing fuels using mixed bases (wet vs dry, HHV vs LHV) without conversion; misinterpreting volumetric values of gases as directly comparable to mass-based values of solids.Final Answer:calorific value

Question 14

Charcoal production temperature check: Is wood charcoal produced by heating wood at only 100°C?

Accepted Answer

Introduction / Context:Charcoal is a carbon-rich solid produced by the pyrolysis (destructive distillation) of wood or biomass. The temperature regime is crucial to drive off volatiles and produce a stable, high-carbon product suitable for metallurgical, cooking, or filtration uses.Given Data / Assumptions:Heating of wood in limited or no oxygen.Target product: wood charcoal, not merely dried wood.Industrial practice rather than laboratory micro-samples.Concept / Approach:At about 100°C, wood only dries (free moisture removal). True pyrolysis starts substantially higher. Charcoal production typically occurs around 450–600°C (and can be higher depending on desired properties), at which hemicellulose, cellulose, and lignin decompose, releasing volatiles and leaving a porous carbon matrix. Therefore, the statement that charcoal is produced at 100°C is false.Step-by-Step Solution:Recognize drying range: roughly 25–150°C removes moisture.Onset of pyrolysis: hundreds of °C; sustained carbonization at ≈ 450–600°C.Conclude that 100°C is insufficient for charcoal production.Verification / Alternative check:Industrial retorts and traditional kilns operate well above 400°C; product analysis (fixed carbon content, volatile matter) confirms the need for higher temperatures to achieve charcoal quality.Why Other Options Are Wrong:Yes/Only in vacuum/Only with catalysts: None overcome the fundamental requirement that significant pyrolysis temperatures are necessary; vacuum or catalysts do not enable full carbonization at 100°C.Common Pitfalls:Equating “drying” with “carbonizing”; assuming that removing water alone creates charcoal.Final Answer:No

Question 15

Reversible vs. irreversible heat engines In classical thermodynamics, what is the correct statement about the efficiencies of reversible engines compared with irreversible engines operating between the same two heat reservoirs?

Accepted Answer

Introduction / Context:Heat-engine efficiency is a cornerstone concept in thermodynamics. The Carnot framework compares any engine against an ideal reversible benchmark operating between two fixed temperature reservoirs. This question checks whether you recognize the unique property of reversible engines and how their efficiency compares with that of irreversible (real) engines.Given Data / Assumptions:Two thermal reservoirs at fixed temperatures Th (hot) and Tc (cold).Engines may be reversible (idealized, no dissipation) or irreversible (real, with friction, gradients, losses).All engines considered operate in cycles between the same Th and Tc.Concept / Approach:The Carnot theorem states two key results: (1) No engine operating between given reservoirs can be more efficient than a reversible engine operating between the same reservoirs; and (2) All reversible engines operating between the same two reservoirs have the same efficiency, which depends only on Th and Tc. Therefore, reversibility defines an upper bound that real engines can only approach but never exceed.Step-by-Step Solution:Recall Carnot efficiency: eta_Carnot = 1 − Tc/Th (temperatures on an absolute scale).Property: Any reversible engine between Th and Tc must match eta_Carnot.Irreversible engines have internal entropy production ⇒ lower efficiency: eta_irrev Thus, all reversible engines share the same efficiency; irreversible engines cannot equal or exceed it.Verification / Alternative check:Consider connecting a reversible engine to drive a reversible refrigerator between the same reservoirs. If two reversible engines had different efficiencies, one could form a perpetual motion of the second kind by cycling heat and work contradictorily—impossible by the Second law—thus they must have identical efficiencies.Why Other Options Are Wrong:Same efficiency for reversible and irreversible: violates Carnot’s inequality.Irreversible engines have maximum efficiency: false; irreversibility reduces efficiency.All engines are designed reversible: real designs can only approximate reversibility.Reversible less efficient than irreversible: contradicts the Second law.Common Pitfalls:Confusing “highly efficient” with “reversible.” Reversible is an idealization with zero entropy production; practical engines always incur losses, so their efficiencies depend on design, but all lie below the reversible limit.Final Answer:All the reversible engines have the same efficiency.

Question 16

Gas-law nomenclature at constant pressure The statement “All perfect gases change in volume by 1/273 of their original volume at 0°C for every 1°C change in temperature at constant pressure” refers to which named law?

Accepted Answer

Introduction / Context:Classical gas laws describe limiting behaviors of ideal gases when one state variable is held fixed. Recognizing the correct eponym ensures you apply the right proportionality for experiments and calculations involving constant-pressure heating or cooling.Given Data / Assumptions:Perfect (ideal) gas behavior.Pressure held constant.Reference condition at 0°C (273.15 K).Concept / Approach:Charles’ law states that, at constant pressure, the volume of a given mass of gas is directly proportional to its absolute temperature. Expressed differentially near 0°C, the fractional change in volume per 1°C is approximately 1/273 of the volume at 0°C, reflecting linear expansion on the Kelvin scale. Boyle’s law pertains to P–V inverse proportionality at constant temperature; Gay-Lussac’s commonly names P–T proportionality at constant volume (terminology conventions vary regionally), while Joule’s law relates internal energy of an ideal gas to temperature only.Step-by-Step Solution:At constant pressure, V ∝ T (absolute).Near 0°C, ΔV/V0 ≈ ΔT/273 → 1°C change gives ≈ 1/273 of V0.Thus the stated rule is Charles’ law.Verification / Alternative check:From the ideal-gas equation P * V = n * R * T, with P and n fixed, V/T is constant. This yields the linear relation underlying the 1/273 figure at 0°C.Why Other Options Are Wrong:Boyle: constant T, not constant P.Gay-Lussac: commonly P–T at constant V, not V–T at constant P.Joule: internal energy independence from volume for ideal gases.Avogadro: equal volumes at the same T and P contain equal numbers of molecules.Common Pitfalls:Memorizing names without the associated constraints (which variable is held constant). Always tie the eponym to the controlled variable.Final Answer:Charles' law

Question 17

Town (coal) gas composition – main constituents Coal gas (also called town gas) produced by coal carbonisation primarily contains which major components?

Accepted Answer

Introduction / Context:Manufactured “town gas” historically fueled lighting and heating before widespread natural gas. Knowing its composition is key for understanding calorific value, combustion air requirements, and safety considerations (toxicity and explosion limits).Given Data / Assumptions:Coal gas is obtained via destructive distillation (carbonisation) of coal.Typical downstream cleanup removes tars and ammonia but leaves a fuel-rich mixture.Qualitative composition comparison only is required.Concept / Approach:Coal gas commonly contains substantial hydrogen (often the largest fraction), carbon monoxide, and light hydrocarbons such as methane and small olefins. Carbon dioxide and nitrogen may be present in smaller amounts, depending on process and cleanup. This mixture gives coal gas a higher calorific value than air-diluted producer or Mond gases, but the presence of CO makes it toxic.Step-by-Step Solution:Identify origin: carbonisation → release of H2, CO, hydrocarbons.Recognize minor constituents: CO2, N2, H2S, water vapor traces.Conclude the statement describing H2, CO, and hydrocarbons as main components is correct.Verification / Alternative check:Representative analyses show H2 and CH4 together with CO forming the bulk of combustible species, consistent with common engineering texts on manufactured gases.Why Other Options Are Wrong:“Disagree” options conflict with well-documented historical compositions.Attributing dominance to nitrogen/carbon dioxide confuses coal gas with blast furnace or flue gases.Common Pitfalls:Confusing coal gas with producer gas (air–steam gasification) which is far more nitrogen-diluted and has lower calorific value.Final Answer:Agree

Question 18

Polytropic index for adiabatic (isentropic) behavior For a perfect gas undergoing a reversible adiabatic process, the polytropic relation p * v^n = C holds with n equal to which parameter?

Accepted Answer

Introduction / Context:Polytropic processes unify many paths with a single exponent n in p * v^n = constant. Recognizing special n values helps you quickly select formulas for work, heat, and temperature changes, especially for adiabatic compression or expansion of gases.Given Data / Assumptions:Perfect (ideal) gas.Reversible adiabatic (isentropic) process: no heat transfer and no internal dissipation.Constant specific heats for simplicity so that γ = cp/cv is constant.Concept / Approach:For a reversible adiabatic process of an ideal gas, p * v^γ = constant, T * v^(γ−1) = constant, and p^(1−γ) * T^γ = constant all hold. These are derived from the First law with Q = 0 and the ideal-gas equation, leading to the identification that the polytropic exponent n coincides with γ for isentropic behavior.Step-by-Step Solution:Start: Q = 0 for adiabatic; reversible implies ds = 0 for ideal gas.Combine du = cv dT with p dv + v dp = R dT.Eliminate T to obtain p * v^γ = constant ⇒ n = γ.Verification / Alternative check:On a log–log P–v plot, the slope of an isentropic line is −γ. Empirical compressor maps approximate this behavior at high efficiencies.Why Other Options Are Wrong:n = 0: isobaric; n = 1: isothermal for ideal gases; n → ∞: isochoric; n = −1: not the isentropic case.Common Pitfalls:Confusing any adiabatic with isentropic; irreversibilities change the effective exponent to a value different from γ in polytropic modeling.Final Answer:γ

Question 19

Nature of throttling (Joule–Thomson) process Classify a throttling expansion (flow through a partially closed valve or porous plug) in terms of reversibility and flow type.

Accepted Answer

Introduction / Context:Throttling is ubiquitous in refrigeration, gas distribution, and pressure reduction stations. Understanding its thermodynamic character clarifies why enthalpy is (approximately) conserved while entropy increases, and why it is modeled as an irreversible steady-flow process.Given Data / Assumptions:Steady flow through a valve/orifice/porous plug.Negligible heat transfer and shaft work; kinetic/potential energy changes typically small.Large pressure drop across a localized restriction.Concept / Approach:In throttling, the control volume analysis (steady-flow energy equation) with Q ≈ 0, W_s ≈ 0, and negligible KE/PE changes yields h1 ≈ h2. Large pressure gradients, viscous dissipation, and mixing cause internal entropy generation, so the process is inherently irreversible. Because inlet and outlet conditions are steady, throttling is a steady-flow process even though the fluid experiences a non-equilibrium drop through the restriction.Step-by-Step Solution:Apply steady-flow energy equation: h1 + KE1 + PE1 + q = h2 + KE2 + PE2 + w_s.With q ≈ 0, w_s ≈ 0, KE/PE small ⇒ h1 ≈ h2 (enthalpy nearly conserved).Irreversibility: friction and turbulence cause entropy increase, so s2 > s1.Therefore: irreversible steady-flow expansion.Verification / Alternative check:Joule–Thomson coefficient experiments (μ_JT) measure temperature changes at constant enthalpy, confirming h ≈ const while entropy rises in real throttling valves.Why Other Options Are Wrong:Assuming ideal-gas exception: even ideal gases experience irreversible throttling (though ΔT ≈ 0 for an ideal gas because h = h(T)).Entropy decreases claim contradicts the Second law for adiabatic throttling of real fluids.Common Pitfalls:Confusing throttling with isentropic expansion in turbines; turbines produce work with near-reversible expansion, whereas throttling destroys availability without producing work.Final Answer:Correct

Question 20

Units for calorific value of gaseous fuels When reporting the calorific value (heating value) of a gaseous fuel at standard conditions, which unit expression is appropriate?

Accepted Answer

Introduction / Context:Heating value quantifies the chemical energy released per quantity of fuel. For gases, volume-based billing and metering at standard temperature and pressure (STP or specified base conditions) make volumetric units the norm in industry and codes.Given Data / Assumptions:Fuel in gaseous state measured by volume under defined base conditions.Higher or lower heating value may be specified; the unit form remains volumetric for gases.Standard conditions are assumed for comparability.Concept / Approach:For liquids and solids, calorific value is expressed per unit mass (kJ/kg). For gases, because custody transfer and metering are volumetric, values are given per unit volume, commonly kJ/m^3 (or MJ/Nm^3). Area-based units have no meaning for energy content, and bare kJ lacks a reference basis. Therefore, kJ/m^3 is the correct dimensional form among the options provided.Step-by-Step Solution:Identify fuel phase: gaseous → use volume basis.Select energy per volume units: kJ/m^3 (or MJ/Nm^3 in practice).Reject mass-based or area-based alternatives for gases.Verification / Alternative check:Utility specifications list natural gas HHV as, e.g., 38–42 MJ/Nm^3 or similar, confirming volumetric units are standard.Why Other Options Are Wrong:kJ/kg is appropriate for solids/liquids, not for gases in general trade.kJ and kJ/m^2 lack a physically meaningful basis here.kJ/mol·K is a heat capacity unit, not a calorific value.Common Pitfalls:Failing to state reference conditions (e.g., 0°C, 1 atm) when reporting volumetric heating values; these must be specified for accurate comparison.Final Answer:kJ/m^3

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