Question 1

Path-independence of state properties: A system goes from state 1 to state 3 via path 1–2–3 with Q = +100 kJ and W = +60 kJ. Along an alternative path 1–4–3, the work is W = +20 kJ. What heat Q is absorbed along 1–4–3 (kJ)?

Accepted Answer

Introduction / Context: This problem tests recognition that internal energy is a state property and therefore path independent. While heat and work depend on the path, their difference (for a closed system neglecting KE/PE changes) equals the change in internal energy between the same end states. Given Data / Assumptions: Closed system; sign convention: heat to system positive, work by system positive. Path 1–2–3: Q_123 = +100 kJ, W_123 = +60 kJ. Path 1–4–3: W_143 = +20 kJ; Q_143 = ? Concept / Approach: First law for a closed system: ΔU = Q − W. Since states 1 and 3 are the same regardless of path, ΔU_123 = ΔU_143. Compute ΔU from the known path, then solve for the unknown heat on the alternative path using the same ΔU value. Step-by-Step Solution: Compute ΔU from 1–2–3: ΔU = Q − W = 100 − 60 = +40 kJ.Apply first law to 1–4–3: ΔU = Q − W.Substitute ΔU = +40 kJ and W = +20 kJ: 40 = Q − 20.Solve for Q: Q = 40 + 20 = +60 kJ. Verification / Alternative check: Check consistency: Larger work on the first path (60 kJ) required larger heat input (100 kJ) to reach the same ΔU; with smaller work (20 kJ) on the second path, proportionally smaller heat (60 kJ) is needed to maintain ΔU = +40 kJ. Why Other Options Are Wrong: −140, −80, −40 kJ: Incorrect sign/magnitude; they would imply a different ΔU, violating state-function path independence. Common Pitfalls: Using Q + W instead of Q − W; forgetting the sign convention (work by system positive). Final Answer: +60 kJ

Question 2

Nature of a throttling (Joule–Thomson) process: Classify the throttling process in terms of thermodynamic reversibility.

Accepted Answer

Introduction / Context: Throttling (Joule–Thomson) is a flow through a restriction such as a valve or porous plug with a significant pressure drop and negligible shaft work and heat transfer. It is ubiquitous in refrigeration, natural gas pressure letdown, and control valves. Given Data / Assumptions: Steady flow through a throttling device with large pressure drop. No shaft work, negligible kinetic and potential energy changes (often), and negligible heat transfer to surroundings. Real fluid with frictional effects in the restriction. Concept / Approach: Throttling is inherently dissipative because of viscous/frictional effects and non-equilibrium within the restriction. Entropy increases across the valve, and the process cannot be reversed without additional work input and reorganization. Enthalpy remains approximately constant (h1 ≈ h2) for ideal throttling, but irreversibility is indicated by Δs > 0 for most fluids. Step-by-Step Solution: Model device as adiabatic with no shaft work.Energy equation reduces to h1 ≈ h2 (neglecting KE/PE changes).Recognize strong internal friction and mixing → entropy generation.Conclude process is irreversible. Verification / Alternative check: On an h–p diagram, throttling follows an isenthalpic line with a rise in entropy for typical fluids at moderate pressures, confirming irreversibility. Why Other Options Are Wrong: Reversible/quasi-static: Contradicted by entropy generation and internal dissipation. “Reversible or irreversible”: Ambiguous; throttling in practice is not reversible. Common Pitfalls: Confusing “adiabatic and no shaft work” with “reversible”; adiabatic does not imply reversible. Final Answer: irreversible

Question 3

Identifying the correct law: For a perfect (ideal) gas, the change in internal energy is directly proportional to the change in temperature. Which named law states this relationship?

Accepted Answer

Introduction / Context: For ideal gases, internal energy is a function of temperature only. This foundational statement links microscopic molecular energy to macroscopic thermal behavior and is attributed to Joule's experiments and conclusions, commonly summarized as Joule's law for ideal gases. Given Data / Assumptions: Working substance is a perfect (ideal) gas. Internal energy is denoted by u; temperature by T. Specific heats are finite and positive. Concept / Approach: The law states: for an ideal gas, u = u(T) only. Hence any change in internal energy depends solely on temperature change, not on pressure or volume separately. Mathematically, du = cv * dT for an ideal gas. If temperature does not change, internal energy does not change, regardless of volume or pressure changes alone. Step-by-Step Solution: Recognize the ideal-gas property: u = u(T) ⇒ du = cv * dT.If T increases by ΔT, internal energy change Δu = cv * ΔT (direct proportionality).Therefore the correct named law is Joule's law. Verification / Alternative check: Free expansion of an ideal gas shows no temperature change and thus no change in internal energy despite a large volume change, confirming u depends on T only. Why Other Options Are Wrong: Boyle's law: P * V = constant at constant T.Charles' law: V ∝ T at constant P.Gay-Lussac law: P ∝ T at constant V.Avogadro's law: Equal volumes at same T and P contain equal numbers of molecules. Common Pitfalls: Assuming internal energy depends on pressure or volume for an ideal gas; that is true for real gases but not for ideal gases. Final Answer: Joule's law

Question 4

Thermodynamic pathway terminology: When a system moves from one equilibrium state to another, the sequence of intermediate states followed is called a ________.

Accepted Answer

Introduction / Context: Thermodynamics distinguishes between states (points) and the path taken between states. Correctly naming the path concept is essential for applying path-dependent quantities like heat and work versus state functions like internal energy and entropy. Given Data / Assumptions: Two distinct equilibrium states are defined for a closed or open system. Quasi-equilibrium description allows a continuous path of states. Heat and work depend on the path, not only on end states. Concept / Approach: The term for the path of successive equilibrium states is a thermodynamic process. A thermodynamic cycle is a special process that returns to the initial state. Laws of thermodynamics govern relations between properties and transfers but are not the path itself. Step-by-Step Solution: Define initial and final equilibrium states (state 1 and state 2).List the continuous sequence of intermediate states.Identify this sequence as the thermodynamic process connecting state 1 to state 2. Verification / Alternative check: On P–v or T–s diagrams, the curve between two points is the process path; enclosing a loop forms a cycle. Why Other Options Are Wrong: Thermodynamic law: a principle (e.g., First/Second law), not a path.Thermodynamic cycle: requires returning to the starting state.State function: a property defined by state, not a path. Common Pitfalls: Confusing a process with a law or with a specific device model; a process is generic and device-independent. Final Answer: thermodynamic process

Question 5

Sign of entropy change: For a gas (system) that absorbs heat, what is the sign of the system's entropy change?

Accepted Answer

Introduction / Context: Entropy indicates the degree of energy dispersion and is central to the Second law. Determining the sign of entropy change for heating or cooling helps assess feasibility and direction of processes in both closed and open systems. Given Data / Assumptions: System is a gas receiving heat. Reference is the system's entropy change, not the surroundings. Temperature is positive on an absolute scale. Concept / Approach: For a reversible path, ds = δQ_rev / T. If the system absorbs heat, δQ_rev > 0, and with T > 0, ds > 0. For irreversible heating, entropy change is evaluated via any reversible path between the same states; it remains positive because entropy is a state function determined by end states only. Step-by-Step Solution: Choose an equivalent reversible path to connect the same end states.Apply ds = δQ_rev / T with δQ_rev > 0.Conclude Δs_system > 0 (positive) for heat absorbed. Verification / Alternative check: For ideal gases at constant pressure: Δs = cp * ln(T2/T1) − R * ln(P2/P1). With heating at modest pressure change, ln(T2/T1) > 0 dominates, giving Δs > 0. Why Other Options Are Wrong: Negative: corresponds to net heat rejection by the system.Positive or negative: sign is not ambiguous when net heat is absorbed.Zero for any adiabatic: only reversible adiabatic has Δs = 0; general adiabatic may increase entropy.Undefined: entropy change is well-defined for state changes. Common Pitfalls: Mixing up system and surroundings; surroundings often lose heat and may have a negative local Δs while total entropy change remains nonnegative. Final Answer: positive

Question 6

Classifying properties: Which of the following is an extensive property of a thermodynamic system?

Accepted Answer

Introduction / Context: Thermodynamic properties are either intensive (independent of system size) or extensive (scale with the amount of substance). Correct classification is essential for setting up balances and for forming specific (intensive) properties by normalization. Given Data / Assumptions: System can be subdivided into identical parts. Properties considered: Pressure, Volume, Temperature, Density, Specific volume. Concept / Approach: Extensive properties are additive. If we combine two identical subsystems, total value doubles. Volume adds directly (V_total = V1 + V2), making it extensive. By contrast, pressure, temperature, density, and specific volume are intensive; they do not simply add when systems are combined at equilibrium. Step-by-Step Solution: Test additivity: combine two equal systems.Volume doubles → extensive.Pressure/temperature remain unchanged at equilibrium → intensive.Density and specific volume are ratios of extensive properties → intensive. Verification / Alternative check: Dimensional reasoning: volume scales with length^3 of the system size, confirming extensivity; intensive properties are size-independent. Why Other Options Are Wrong: Pressure and temperature are equilibrating variables, not additive.Density and specific volume are derived intensives (mass-normalized or volume-normalized quantities). Common Pitfalls: Confusing total energy (extensive) with specific energy (intensive); always check if the property adds when systems are merged. Final Answer: Volume

Question 7

What does the Second law define most fundamentally among thermodynamic quantities?

Accepted Answer

Introduction / Context: The First law introduces energy conservation and defines internal energy differences. The Second law introduces directionality and the concept of entropy, enabling statements about feasibility and efficiency beyond mere energy accounting. Given Data / Assumptions: Closed or open systems exchanging heat and work. Thermal reservoirs at absolute temperatures. Reversible and irreversible processes considered. Concept / Approach: The Second law defines entropy (S) through inequalities and equalities relating heat transfer and absolute temperature. For reversible processes, dS = δQ_rev / T. For any process, ΔS_total ≥ 0. This definition allows quantifying irreversibility and setting the maximum efficiency of heat engines and refrigerators. Step-by-Step Solution: State the reversible relation: dS = δQ_rev / T.Extend to general processes: ΔS_system + ΔS_surroundings ≥ 0.Use S to derive limits, e.g., η_Carnot = 1 − Tc/Th. Verification / Alternative check: Entropy balance predictions match real observations (e.g., mixing increases entropy, heat flows from hot to cold spontaneously). Why Other Options Are Wrong: Heat and work are modes of energy transfer defined operationally, not by the Second law.Internal energy is defined via the First law and state equations.Temperature scale can be related to the Second law, but the property most directly defined is entropy. Common Pitfalls: Equating the Second law only with heat-engine limits; its deeper role is defining entropy and irreversibility. Final Answer: entropy

Question 8

Identify the property: Which thermodynamic property of a working substance changes in proportion to reversible heat transfer and is used to measure irreversibility?

Accepted Answer

Introduction / Context: Among thermodynamic properties, one uniquely links heat transfer to temperature in a way that quantifies directional constraints and irreversibility. That property is entropy, central to the Second law and energy quality analysis. Given Data / Assumptions: Reversible heat transfer δQ_rev at absolute temperature T. Property S with differential dS = δQ_rev / T for reversible processes. General inequality ΔS_total ≥ 0 for all real processes. Concept / Approach: Entropy increases when reversible heat is added (dS = δQ_rev / T). In any real (irreversible) process, entropy production is positive. Enthalpy and internal energy are energy measures; they do not directly quantify irreversibility. External energy is not a standard thermodynamic property. Step-by-Step Solution: Use the defining relation: dS = δQ_rev / T.Identify that S tracks reversible heat interaction and sets limits for engines and refrigerators.Select entropy as the correct property. Verification / Alternative check: Isentropic (ΔS = 0) idealizations represent reversible adiabatic steps, confirming entropy's role as the irreversibility yardstick. Why Other Options Are Wrong: Enthalpy/internal energy measure energy content but do not by themselves indicate reversibility.External energy and specific heat are either non-standard or material parameters, not the irreversibility measure. Common Pitfalls: Confusing zero heat transfer (adiabatic) with zero entropy change; only reversible adiabatic is isentropic. Final Answer: entropy

Question 9

Combustion stoichiometry check: One kilogram of sulphur (S) reacts with one kilogram of oxygen (O2) to yield two kilograms of sulphur dioxide (SO2). Is this statement correct?

Accepted Answer

Introduction / Context: Sulphur oxidation is a standard stoichiometric reaction in combustion and gas treatment. Verifying mass ratios builds comfort with balancing reactions and predicting flue-gas composition for desulphurization or sulfuric acid production trains. Given Data / Assumptions: Reaction considered: S + O2 → SO2 (complete oxidation to dioxide). Atomic masses: S = 32, O = 16. Ideal, complete conversion without side reactions. Concept / Approach: Balance by moles: 1 mol S (32 kg/kmol) reacts with 1 mol O2 (32 kg/kmol) to form 1 mol SO2 (64 kg/kmol). Thus, per kilogram basis, 1 kg S needs 1 kg O2 and produces 2 kg SO2. While further oxidation to SO3 can occur in catalytic processes, the primary combustion product without catalyst is SO2. Step-by-Step Solution: Write reaction: S + O2 → SO2.Compute masses: 32 + 32 → 64 (kg per kmol basis).Scale per kg: 1 kg S + 1 kg O2 → 2 kg SO2. Verification / Alternative check: Mass conservation and standard oxidation supports the same ratio; flue-gas analyzers report SO2 as the primary sulfur oxide absent catalytic conversion. Why Other Options Are Wrong: Dependence on excess air or pressure does not change the stoichiometric mass ratio for the ideal reaction.SO3 formation requires specific catalytic conditions; it is not the default product in simple combustion. Common Pitfalls: Confusing oxygen atom mass with O2 molecular mass; always use O2 = 32 for combustion stoichiometry. Final Answer: True

Question 10

Coal testing: Ultimate analysis of coal determines the percentage of which constituents?

Accepted Answer

Introduction / Context: Coal characterization uses proximate and ultimate analyses. Ultimate analysis provides elemental composition essential for combustion calculations, emission estimates, and boiler design (air requirement, flue-gas composition, SOx/NOx potential). Given Data / Assumptions: Standard ultimate analysis reports mass percentages of C, H, N, S, O, and ash. Moisture may be reported separately or included depending on reporting basis. Laboratory procedures follow standardized methods (e.g., ASTM). Concept / Approach: Ultimate analysis breaks coal into elements: carbon, hydrogen, nitrogen, sulphur, oxygen (by difference), and ash. Therefore, options naming carbon; hydrogen and nitrogen; and sulphur and ash are each part of ultimate analysis. The combined option “all of these” is correct. Step-by-Step Solution: Identify ultimate-analysis elements: C, H, N, S, O, ash.Match options: each listed group belongs to the ultimate analysis.Select the inclusive answer. Verification / Alternative check: Engineering manuals and standards confirm the same constituent list for ultimate analysis used in boiler and environmental calculations. Why Other Options Are Wrong: Any single subset alone is incomplete; moisture alone is proximate analysis content, not the full ultimate analysis. Common Pitfalls: Confusing proximate (moisture, volatile matter, fixed carbon, ash) with ultimate (elemental) analysis. Final Answer: all of these

Question 11

Carnot cycle efficiency relationship For a heat engine operating on the Carnot cycle between a highest temperature T1 and a lowest temperature T2 (absolute units), which formula gives the thermal efficiency?

Accepted Answer

Introduction / Context: The Carnot cycle sets the theoretical upper limit of efficiency for any heat engine operating between two thermal reservoirs. Its efficiency depends only on the reservoir temperatures when expressed in absolute units (Kelvin). Given Data / Assumptions: Reversible cycle between hot reservoir at T1 and cold reservoir at T2. No internal irreversibilities; quasi-static processes. Temperatures are absolute (Kelvin) and T1 > T2 > 0. Concept / Approach: From entropy balance for a reversible cycle, Q1/T1 = Q2/T2. Efficiency η = 1 − (Q2/Q1) = 1 − (T2/T1). This simple ratio highlights why reducing T2 or increasing T1 raises the maximum possible efficiency, albeit material limits constrain T1 in practice. Step-by-Step Solution: Write reversible entropy relation: Q1/T1 = Q2/T2.Express efficiency: η = 1 − Q2/Q1.Combine to obtain η = 1 − T2/T1. Verification / Alternative check: Plugging T2 → 0 gives η → 1, and T2 → T1 gives η → 0, matching physical expectations and limiting cases. Why Other Options Are Wrong: Forms with 1 − T1/T2 or (T2 − T1)/T2 invert temperatures incorrectly.η = T2/T1 is not an efficiency expression.(T1 − T2)/T2 overestimates efficiency and can exceed 1. Common Pitfalls: Using Celsius instead of Kelvin; only absolute temperatures are valid in the formula. Final Answer: η = 1 − T2/T1

Question 12

Ironmaking practice: Which fuel is predominantly used in the blast furnace to produce pig iron from iron ore?

Accepted Answer

Introduction / Context: The blast furnace reduces iron oxides to iron using a carbonaceous fuel that also supplies both heat and reducing gases. Industrial practice relies on a specific metallurgical fuel for permeability, strength, and reactivity under furnace conditions. Given Data / Assumptions: Operation at high temperatures with strong gas flow and burden load. Need for stable furnace permeability and minimal fines generation. Fuel must provide both carbon (as CO generator) and structural support. Concept / Approach: Hard coke (metallurgical coke) is the standard blast furnace fuel. Its high carbon content, mechanical strength, low volatile matter, and porosity allow it to support the burden while reacting with oxygen and CO2 to form CO, the principal reducing gas. Alternatives such as soft coke or raw coals lack the necessary strength and would degrade, impairing permeability; limited pulverised coal injection is used as a supplemental reductant/energy source but does not replace coke's structural role. Step-by-Step Solution: Identify operational needs: heat, reduction, and bed support.Match properties: hard coke meets strength and reactivity requirements.Select hard coke as the principal fuel used. Verification / Alternative check: Modern furnaces employ pulverised coal injection rates but still maintain a substantial coke rate to preserve furnace permeability and hearth conditions. Why Other Options Are Wrong: Soft coke and raw bituminous coal disintegrate, causing channeling and pressure drop issues.Pulverised coal is an auxiliary injectant, not the sole bed-supporting fuel.Charcoal can be used in small/experimental furnaces but is not predominant industrially. Common Pitfalls: Assuming coal injection replaces coke entirely; structural and permeability requirements prevent full substitution in conventional blast furnaces. Final Answer: hard coke

Question 13

Gas laws and isobaric heating When a gas is heated while the pressure is held constant (isobaric process), the governing ideal-gas relationship is known as:

Accepted Answer

Introduction / Context: In introductory thermodynamics and physical chemistry, several “gas laws” describe how an ideal gas behaves when one state variable is held constant. This question targets the scenario of heating at constant pressure (an isobaric process), asking which named relationship correctly governs the volume–temperature change. Given Data / Assumptions: Ideal-gas behavior is assumed so that p * V = n * R * T applies. Process condition: pressure p is held constant (isobaric heating). Amount of gas n is fixed (closed mass). Concept / Approach: For a fixed amount of an ideal gas at constant pressure, the volume is directly proportional to absolute temperature: V ∝ T. This proportionality is historically called Charles' law. By contrast, Boyle's law refers to isothermal changes (p * V = constant), Gay-Lussac's law often denotes the p–T relation at constant volume, and Avogadro's law relates volume to moles at fixed p and T. Step-by-Step Solution: Start with p * V = n * R * T.At constant p and fixed n, rearrange to V = (n * R / p) * T.Hence V / T = constant, which is exactly Charles' law.Therefore, the governing law for isobaric heating is Charles' law. Verification / Alternative check: Graphing V versus T (in kelvin) at constant pressure gives a straight line through the origin. Laboratory piston–cylinder demonstrations show that as the gas is heated at fixed external pressure, the piston rises proportionally with T, consistent with Charles' law. Why Other Options Are Wrong: Boyle's law: p * V = constant holds when T is constant, not when p is constant. Gay-Lussac law: commonly states p ∝ T at constant V, which is not the present constraint. Avogadro's law: V ∝ n at fixed p and T; unrelated to heating at constant p with fixed n. Dalton's law: about partial pressures in mixtures, not single-gas isobaric heating. Common Pitfalls: Using Celsius instead of kelvin breaks the direct proportionality V ∝ T. Always work with absolute temperature for gas laws. Final Answer: Charles' law

Question 14

Power units and conversions One kilowatt (kW) equals which of the following in base S.I. units?

Accepted Answer

Introduction / Context: Power quantifies the rate at which work is done or energy is transferred. The S.I. unit of power is the watt (W), defined as one joule per second. Kilowatt is a common multiple used in electrical and mechanical engineering. This question checks your ability to convert between kW and the fundamental S.I. units newton, metre, and second. Given Data / Assumptions: 1 W = 1 J/s. 1 J = 1 N·m. 1 kW = 1000 W. Concept / Approach: Combine the definitions systematically. Since 1 W = 1 J/s and 1 J = 1 N·m, it follows that 1 W = 1 N·m/s. Therefore, multiplying by 1000 gives 1 kW = 1000 N·m/s. This representation is universally valid for mechanical and electrical power, independent of the conversion path. Step-by-Step Solution: Start: 1 W = 1 J/s.Use: 1 J = 1 N·m → 1 W = 1 N·m/s.Scale: 1 kW = 1000 W → 1000 N·m/s.Thus, select 1000 N·m/s as the correct equivalence. Verification / Alternative check: Electrical: P = V * I; 1 kW could be 1000 W at 250 V and 4 A, yet in base units the rate of energy transfer is still 1000 J every second, i.e., 1000 N·m/s. Why Other Options Are Wrong: 1 N·m/s is 1 W, not 1 kW. 100 N·m is work/energy, not power (missing per-second). 1 × 10^6 N·m/s corresponds to 1 MW, not 1 kW. 1000 J is energy, not a rate. Common Pitfalls: Confusing energy (joule) with power (joule per second). Always check dimensions include time in the denominator for power. Final Answer: 1000 N·m/s

Question 15

Closed-cycle gas turbines and the Brayton–Joule model State whether the following is correct: “A closed-cycle gas turbine operates on the Joule (Brayton) cycle under air-standard assumptions.”

Accepted Answer

Introduction / Context: Gas turbines—open or closed—are commonly analyzed with the Brayton (also called Joule) cycle: isentropic compression, constant-pressure heat addition, isentropic expansion, and constant-pressure heat rejection. The question asks if this framework applies to closed-cycle configurations. Given Data / Assumptions: Air-standard or ideal-gas modeling is used for cycle description. Closed cycle recirculates a working fluid through external heat exchangers. Compression and expansion are treated as isentropic in the ideal model. Concept / Approach: Both open- and closed-cycle gas turbines follow the same Brayton–Joule sequence in idealized analysis. The difference lies in how heat is supplied and rejected: open cycles burn fuel directly, while closed cycles use external heaters/coolers and may include recuperation. Despite these hardware differences, the thermodynamic cycle remains Brayton. Step-by-Step Solution: Identify the four ideal processes: isentropic compression, constant-pressure heating, isentropic expansion, constant-pressure cooling.Recognize that closing the loop with heat exchangers does not change the process sequence.Therefore, the statement is correct: a closed-cycle gas turbine operates on the Joule (Brayton) cycle. Verification / Alternative check: Textbook T–s and p–v diagrams for closed Brayton cycles mirror those of open cycles; only the placement of combustor vs. external heater differs. Why Other Options Are Wrong: “Only helium” or “only without a recuperator” adds unnecessary conditions; the cycle identification is independent of working fluid or recuperation. Rankine is the vapor power cycle (phase change), not applicable to gas turbines. Common Pitfalls: Equating “closed cycle” with a different thermodynamic cycle; closure refers to fluid recirculation, not to a change in the ideal process sequence. Final Answer: True

Question 16

Perpetual motion machines – identifying the violated law Evaluate the statement: “A machine that violates the first law of thermodynamics is called a perpetual motion machine of the second kind.”

Accepted Answer

Introduction / Context: The terminology of perpetual motion machines (PMMs) distinguishes between violations of the first and second laws of thermodynamics. This question checks whether you can correctly associate the violated principle with the PMM “kind.” Given Data / Assumptions: First law: conservation of energy. Second law: directionality/entropy, impossibility of perfect conversion or spontaneous cold-to-hot transfer. PMM terminology is standard in thermodynamics texts. Concept / Approach: A perpetual motion machine of the first kind (PMM1) violates the first law by creating energy from nothing or delivering net work without energy input. A perpetual motion machine of the second kind (PMM2) respects energy conservation but violates the second law, for example by converting all heat from a single reservoir to work or transferring heat from cold to hot with no other effect. Step-by-Step Solution: Identify the claimed violation: first law (energy conservation).Recall definitions: PMM1 ↔ first-law violation; PMM2 ↔ second-law violation.Therefore, labeling a first-law violator as a PMM2 is incorrect. Verification / Alternative check: Classic examples of PMM2 include Kelvin–Planck and Clausius violations (100% heat-to-work engines; spontaneous cold-to-hot transfer). None of these create energy; they contravene entropy-based limitations. Why Other Options Are Wrong: “Correct” and conditional variants misclassify the violation. System type (open/closed) and scale (microscopic) do not change the law being violated. Common Pitfalls: Confusing “perpetual motion” generically with the specific PMM1 vs. PMM2 classification. Always tie PMM1 to energy creation and PMM2 to second-law violations. Final Answer: Incorrect

Question 17

Heat quantity definitions – raising the temperature of water The amount of heat required to raise the temperature of 1 kg of water by 1 K (approximately 1 °C) is called:

Accepted Answer

Introduction / Context: Historically, thermal energy was often measured in calories. Although the S.I. unit of energy is the joule, legacy thermal units still appear in engineering practice and older data sheets. This question asks you to identify the traditional name for the amount of heat that raises 1 kg of water by approximately 1 K. Given Data / Assumptions: Water mass considered: 1 kg. Temperature rise: 1 K (or 1 °C over the relevant range). Standard pressure conditions assumed. Concept / Approach: By definition, 1 kilocalorie (kcal) is the amount of heat required to raise the temperature of 1 kilogram of water by 1 °C. In S.I. units, 1 kcal ≈ 4.1868 kJ. While modern standards use joules exclusively, the historical definition clearly labels this quantity as one kilocalorie. Step-by-Step Solution: Recall caloric definitions: 1 cal raises 1 g of water by 1 °C; 1 kcal raises 1 kg by 1 °C.Match the problem statement: 1 kg of water, 1 K rise → fits 1 kilocalorie.Therefore choose “kilocalorie.” Verification / Alternative check: Convert to S.I.: 1 kcal ≈ 4186.8 J. Using c_p,water ≈ 4.186 kJ/(kg·K), the heat for 1 kg and 1 K is Q = m * c_p * ΔT ≈ 1 kg * 4.186 kJ/(kg·K) * 1 K = 4.186 kJ, consistent with 1 kcal. Why Other Options Are Wrong: Specific heat capacity of water is a property (≈ 4.186 kJ/(kg·K)), not a unit name for a fixed quantity of heat. Joule is the S.I. unit; the numerical amount would be ~4186.8 J, but the named historical quantity is “kilocalorie.” Btu and watt-hour are different energy units (1 Btu heats 1 lb of water by 1 °F; 1 Wh = 3600 J). Common Pitfalls: Mixing up the unit name with the material property. The question asks for the name of the amount of heat, not the value of c_p. Final Answer: kilocalorie

Question 18

Universal (molar) gas constant – relation to specific gas constant The universal gas constant R_u (per mole) is related to a gas’s specific gas constant R (per unit mass) by which product?

Accepted Answer

Introduction / Context: Two common forms of the gas constant appear in thermodynamics: R_u, the universal (molar) gas constant with units J/(mol·K), and R, the specific gas constant with units J/(kg·K). Converting between them requires the molar (molecular) mass of the gas. Given Data / Assumptions: R_u ≈ 8.314 kJ/(kmol·K) or 8.314 J/(mol·K). R depends on the particular gas: R = R_u / M. M is the molar (molecular) mass in kg/kmol or kg/mol consistent with units. Concept / Approach: The relationship is R_u = M * R, where M is the molar mass of the gas and R is its specific gas constant on a per-mass basis. This follows directly from writing the ideal-gas equation either as p * V = n * R_u * T (molar form) or p * v = R * T (specific form) and substituting n = m/M. Step-by-Step Solution: Start with p * V = n * R_u * T and n = m / M.Divide by mass m to get p * v = (R_u / M) * T.Identify R = R_u / M → rearrange to R_u = M * R.Therefore, the correct product is molecular mass times specific gas constant. Verification / Alternative check: For air: M ≈ 28.97 kg/kmol, R ≈ 0.287 kJ/(kg·K). Product M * R ≈ 28.97 * 0.287 ≈ 8.314 kJ/(kmol·K) = R_u, confirming the relation. Why Other Options Are Wrong: Atomic mass pertains to single atoms; many gases are diatomic or polyatomic; the correct parameter is molar mass of the gas species. Products with c_p or c_v do not produce R_u; instead, for ideal gases c_p − c_v = R (not R_u). “Molar mass divided by c_p” is dimensionally inconsistent. Common Pitfalls: Mixing per-mole and per-mass quantities. Always ensure unit consistency when converting between R and R_u. Final Answer: molecular mass * specific gas constant (R)

Question 19

Simple Brayton cycle efficiency – dependence on pressure ratio Does the ideal (air-standard) efficiency of a simple gas-turbine cycle depend on the compressor pressure ratio?

Accepted Answer

Introduction / Context: The simple Brayton (Joule) cycle models gas turbines with two isentropic processes and two constant-pressure processes. A hallmark result is that the air-standard thermal efficiency depends on the pressure ratio across the compressor–turbine pair. Given Data / Assumptions: Air-standard model with constant specific heats. Isentropic compression and expansion. No intercooling, reheating, or regeneration in the “simple” cycle. Concept / Approach: The ideal Brayton efficiency may be written as η = 1 − (1 / r_p)^((γ − 1)/γ), where r_p is the pressure ratio and γ = c_p / c_v. This expression shows a direct dependence on r_p: increasing pressure ratio (within limits) increases η, up to an optimum when real effects are considered. Step-by-Step Solution: Relate temperature ratios to pressure ratio for isentropic stages: T2/T1 = r_p^((γ − 1)/γ).Express heat addition and rejection at constant pressure using temperature differences.Form η = 1 − Q_out/Q_in and simplify to obtain dependence on r_p.Conclude: ideal efficiency increases with r_p per the stated formula. Verification / Alternative check: Plotting η versus r_p with γ ≈ 1.4 shows monotonic increase for the ideal case, matching standard textbook curves for Brayton cycles without component losses. Why Other Options Are Wrong: “No” contradicts the closed-form efficiency expression. Intercooling or regeneration alter efficiency but are not prerequisites for dependence on r_p. Non-idealities change the numeric optimum but not the fundamental dependence. Common Pitfalls: Confusing ideal-cycle results with real-engine optima, which also depend on turbine inlet temperature limits and component efficiencies. Final Answer: Yes

Question 20

No-heat-transfer processes A process in which the working substance neither receives nor rejects heat to the surroundings during compression or expansion is called:

Accepted Answer

Introduction / Context: Thermodynamic processes are often idealized by constraints on heat transfer or temperature. This question asks you to name the process that proceeds with zero heat exchange between the system and its surroundings. Given Data / Assumptions: No heat crosses the boundary: Q = 0 during the process. Compression or expansion of a gas is implied. Work interactions may still occur. Concept / Approach: When Q = 0 for the entire process, it is termed adiabatic. If, in addition, the process is reversible, it is isentropic; however, the term “adiabatic” alone specifies only the heat-transfer condition. An isothermal process, by contrast, maintains constant temperature and generally requires heat transfer to offset compression/expansion work. Step-by-Step Solution: Apply the first law in differential form: δQ = dU + δW.Set δQ = 0 for an adiabatic process → dU = −δW (internal energy change balances boundary work).Therefore, the correct name for the process is “adiabatic process.” Verification / Alternative check: In rapid (well-insulated) compressions, temperature rises because work is converted to internal energy without heat rejection—classic adiabatic behavior observed in compression-ignition engines. Why Other Options Are Wrong: Isothermal requires heat exchange to maintain constant temperature during volume change. “Polytropic with n = 1” is isothermal for ideal gases, not adiabatic. “Hyperbolic process” is an older term for p * V = constant (isothermal). Isobaric maintains pressure, not heat flow. Common Pitfalls: Equating “adiabatic” with “isentropic” unconditionally; irreversibilities make many adiabatic processes non-isentropic. Final Answer: adiabatic process

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