Question 1

Carnot Efficiency – Effect of Highest Temperature on Performance Statement check: 'By decreasing the highest temperature in the Carnot cycle, its efficiency is increased.' Determine its correctness.

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Introduction / Context:The Carnot cycle sets the theoretical upper limit of heat-engine efficiency: eta_C = 1 − T_L/T_H, where T_H is the high-temperature reservoir and T_L is the low-temperature reservoir. Understanding parameter effects avoids design misconceptions.Given Data / Assumptions:Reversible cycle exchanging heat only at T_H and T_L.Absolute temperatures (Kelvin scale).No internal irreversibilities or pressure losses.Concept / Approach:From eta_C = 1 − T_L/T_H, efficiency increases when T_H increases or when T_L decreases (or both). Therefore, decreasing the highest temperature T_H while keeping T_L fixed decreases eta_C, not increases it. The given statement is false unless T_L decreases proportionally more, which is not stated.Step-by-Step Solution:Write eta_C = 1 − T_L/T_H.Differentiate conceptually: ∂eta_C/∂T_H = +T_L/T_H^2 > 0 for fixed T_L.Hence, lowering T_H lowers eta_C.Conclude the statement is false as written.Verification / Alternative check:Numerical example: T_L = 300 K. If T_H drops from 1200 K to 900 K, eta_C falls from 1 − 300/1200 = 0.75 to 1 − 300/900 = 0.667.Why Other Options Are Wrong:Adding regeneration does not change the Carnot limit; the law holds for any working substance, not only ideal gases.Common Pitfalls:Confusing effects of increasing temperature difference ΔT with separately changing T_H or T_L; using Celsius instead of Kelvin.Final Answer:False

Question 2

Free (Joule) Expansion of an Ideal Gas – Energy Transfers and State Change During free expansion of an ideal gas into a vacuum, there is no boundary work and no heat transfer; internal energy remains constant and temperature is unchanged.

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Introduction / Context:Free expansion (Joule expansion) is a classic thought/bench experiment where a gas expands into a vacuum. It is invaluable for clarifying the roles of work, heat, and internal energy in non-quasi-equilibrium processes.Given Data / Assumptions:Ideal gas expands into an evacuated volume (no external resistance).Adiabatic, rigid container boundaries aside from the partition removal.No shaft work or other interactions.Concept / Approach:With no external pressure opposing expansion, boundary work is zero: W = ∫ p_ext dV = 0 (p_ext = 0). If the vessel is insulated, Q = 0. For an ideal gas, internal energy depends only on temperature; since ΔU = Q − W = 0, we have ΔU = 0 and therefore ΔT = 0. Pressure and specific volume change, but temperature does not for an ideal gas.Step-by-Step Solution:Remove partition between gas and vacuum.No resisting pressure → W1-2 = 0.Insulated system → Q1-2 = 0.First Law: ΔU = Q − W = 0 → dU = 0; for ideal gas, T2 = T1.Verification / Alternative check:Joule's vacuum bulb experiment confirms no temperature change for ideal gases. Real gases may cool or warm slightly (Joule–Thomson effects), but at low pressures the ideal result is approached.Why Other Options Are Wrong:Any claim of expansion work on surroundings contradicts p_ext = 0; asserting heat transfer violates the insulated assumption; stating ΔU ≠ 0 contradicts First Law for this setup.Common Pitfalls:Confusing free expansion with throttling (which imposes flow and maintains h constant, not u); assuming temperature must drop whenever gas expands (not true here for ideal gas).Final Answer:all of these

Question 3

Closed-Cycle Gas Turbine – Ideal Compressor Process In an ideal closed-cycle gas turbine (Brayton cycle), the compressor process is modeled as isentropic (adiabatic and reversible).

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Introduction / Context:Cycle modeling requires selecting ideal processes for major components. The Brayton compressor is ideally isentropic, providing a benchmark for defining isentropic efficiency and comparing to real polytropic compression behavior.Given Data / Assumptions:Closed Brayton loop with compressor, heater, turbine, and cooler.Idealizations: adiabatic, reversible compression and expansion.Working fluid approximated as an ideal gas with nearly constant properties.Concept / Approach:An isentropic compressor minimizes required work for a given pressure ratio in the ideal limit. Real compressors are polytropic with efficiencies less than 100%, but the isentropic model is the reference used in performance calculations and definitions (e.g., eta_c = isentropic work / actual work).Step-by-Step Solution:Specify desired pressure ratio r_p across the compressor.Model the process as adiabatic and reversible → isentropic.Relate temperature rise using T2/T1 = r_p^{(gamma−1)/gamma} for ideal gas.Use this to compute compressor work and overall cycle efficiency.Verification / Alternative check:T-s diagrams show vertical (constant entropy) lines for ideal compression/expansion. Real data deviate rightward due to irreversibilities.Why Other Options Are Wrong:Isothermal compression is not Brayton ideal; 'polytropic' describes real behavior but not the ideal assumption; 'none' and 'isobaric' are incorrect for a compressor.Common Pitfalls:Using isentropic relations outside their validity (large property variation); confusing stage polytropic efficiency with overall isentropic efficiency.Final Answer:isentropically

Question 4

Thermal Efficiency of an Ideal Simple Brayton (Gas Turbine) Cycle For an ideal gas turbine working on the simple Brayton cycle with perfect intercooling/recuperation absent, the thermal efficiency in terms of pressure ratio r_p and specific heat rat…

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Introduction / Context:The Brayton (gas-turbine) cycle is characterized by two isentropic and two constant-pressure processes. A widely used closed-form expression relates its ideal thermal efficiency to the compressor pressure ratio and the specific heat ratio gamma, enabling quick preliminary design and performance estimates.Given Data / Assumptions:Simple, ideal Brayton cycle (no regeneration, intercooling, or reheat).Isentropic compressor and turbine; isobaric heat addition and rejection.Ideal-gas behavior with constant gamma.Concept / Approach:Using the isentropic temperature–pressure relation T2/T1 = r_p^{(gamma−1)/gamma} and T3/T4 = r_p^{(gamma−1)/gamma}, one finds that the thermal efficiency depends only on r_p and gamma: eta_th = 1 − 1 / r_p^{(gamma−1)/gamma}. This shows efficiency increases with pressure ratio for the ideal case (up to limits when real effects are included).Step-by-Step Solution:Apply isentropic relations across compressor and turbine to relate temperatures to r_p.Write Q_in = m * Cp * (T3 − T2) and Q_out = m * Cp * (T4 − T1).Form eta_th = 1 − Q_out/Q_in and substitute the isentropic relations.Reduce to eta_th = 1 − 1 / r_p^{(gamma−1)/gamma}.Verification / Alternative check:An equivalent temperature-based form is eta_th = 1 − (T4 − T1)/(T3 − T2); substituting the isentropic equalities collapses it to the pressure-ratio expression above. Both are correct under ideal assumptions.Why Other Options Are Wrong:Option (b) omits the isentropic links; (c) is not a definition of thermal efficiency; (d) is correct in temperature form but only after invoking isentropic equalities—since the question asks for the standard r_p–gamma form, (a) is the most explicit; (e) is a mechanical efficiency-like ratio, not thermal efficiency.Common Pitfalls:Applying the ideal formula at high turbine inlet temperatures without accounting for variable properties and component efficiencies; confusing cycle thermal efficiency with component isentropic efficiencies.Final Answer:eta_th = 1 − 1 / r_p^((gamma − 1)/gamma)

Question 5

Gas Laws and Internal Energy — Interpreting Boyle’s Law Evaluate the statement: “Boyle’s law states that the change of internal energy of a perfect (ideal) gas is directly proportional to the change of temperature.” State whether this statement is c…

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Introduction / Context:This question checks conceptual clarity between two different facts in thermodynamics: (1) Boyle’s law and (2) the temperature dependence of internal energy for an ideal gas. Mixing them up is a common error in early coursework.Given Data / Assumptions:Ideal (perfect) gas behavior is assumed.Standard definitions: p = pressure, V = volume, T = absolute temperature, U = internal energy.No phase change or chemical reactions are involved.Concept / Approach:Boyle’s law states that for a fixed mass of an ideal gas at constant temperature (isothermal condition), p * V = constant. The statement in the prompt instead refers to internal energy U being a function only of temperature for an ideal gas, which is another well-known result often attributed to Joule’s law/ideal-gas property: U = U(T). That property is true, but it is not Boyle’s law. Therefore, attributing the internal-energy statement to Boyle’s law is incorrect.Step-by-Step Solution:Recall Boyle’s law: at constant T, p * V = constant (isothermal process).Recall ideal-gas internal energy: dU depends only on dT (Joule’s result for ideal gases).Compare: the prompt equates internal-energy dependence with Boyle’s law, which is a mismatch.Conclusion: The statement is incorrect; it confuses two separate results.Verification / Alternative check:On a p–V diagram, a Boyle’s-law path is an isotherm (rectangular hyperbola). Internal energy in an ideal gas remains constant along any isotherm because T is constant; however, that is a consequence of U(T), not the definition of Boyle’s law itself.Why Other Options Are Wrong:Choosing “Agree” would wrongly attribute the U–T dependence to Boyle’s law. Boyle’s law makes no direct statement about U.Common Pitfalls:Conflating names: Boyle’s (p–V at constant T), Charles’ (V–T at constant p), Gay–Lussac’s (p–T at constant V), and Joule’s statement on internal energy for an ideal gas.Final Answer:Disagree

Question 6

Process Identification — If p * V Remains Constant, Is the Process Isothermal? A thermodynamic process for a fixed mass of gas is carried out such that the product of pressure and volume (p * V) remains constant throughout. Is it correct to call thi…

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Introduction / Context:Correctly naming a process from observed relationships between properties is a fundamental skill. For an ideal gas, the relationship p * V = constant is the hallmark of an isothermal process for a fixed mass, directly from the ideal-gas equation p * V = m * R * T.Given Data / Assumptions:Fixed mass of ideal gas; gas constant R is constant.Product p * V is maintained constant during the process.Quasi-static or otherwise well-defined thermodynamic states.Concept / Approach:For an ideal gas: p * V = m * R * T. If p * V remains constant for a given m and R, then T must remain constant. A constant temperature process is, by definition, an isothermal process. Heating or expansion (mentioned colloquially) can occur while T stays constant if heat transfer balances boundary work.Step-by-Step Solution:Write ideal-gas relation: p * V = m * R * T.Given p * V = constant with fixed m, R ⇒ T = constant.A process at constant T is called isothermal ⇒ statement is correct.Verification / Alternative check:On a p–V diagram, an isotherm is a rectangular hyperbola (p = C/V). Any path that follows p * V = constant traces an isothermal curve for an ideal gas.Why Other Options Are Wrong:“No” would imply that constant p * V does not imply constant T, which contradicts the ideal-gas relation.Common Pitfalls:Assuming that “heating” always raises temperature; in isothermal expansion, added heat equals work done, keeping T unchanged.Final Answer:Yes

Question 7

Gas Properties — Name of the Temperature Where the Extrapolated Gas Volume Becomes Zero On the absolute temperature scale, the temperature at which the volume of an ideal gas extrapolates to zero is called:

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Introduction / Context:Understanding absolute temperature and its physical meaning is central to thermodynamics and kinetic theory. Extrapolating gas law behavior to zero volume identifies a key reference point on the Kelvin scale.Given Data / Assumptions:Ideal-gas behavior at low densities.Linear V–T relationship at constant pressure: V ∝ T.Use of the absolute (Kelvin) temperature scale.Concept / Approach:Charles’ law indicates V1/T1 = V2/T2 at constant pressure for an ideal gas. Extrapolating this straight line to V = 0 yields T = 0 K, called the absolute zero temperature. At absolute zero, ideal-gas pressure and volume vanish in the limiting sense, consistent with minimal thermal motion.Step-by-Step Solution:Start from V ∝ T (constant pressure, fixed mass).Extrapolate to V = 0 ⇒ T = 0 K.Therefore the correct term is “absolute zero temperature.”Verification / Alternative check:Absolute zero is the zero point of the Kelvin scale; thermodynamic temperature is defined so that Carnot efficiency depends only on temperature ratio T_cold/T_hot.Why Other Options Are Wrong:“Absolute temperature” is the general term for temperature measured from 0 K; “absolute scale of temperature” is the name of the scale, not the specific point.Common Pitfalls:Confusing the scale (Kelvin) with the specific zero point; applying real-gas deviations at high pressures where linearity breaks down.Final Answer:absolute zero temperature

Question 8

Series Carnot Engines — Finding the Intermediate Temperature Two reversible (Carnot) heat engines are arranged in series. Engine 1 operates between 1600 K (hot reservoir) and T2 K; Engine 2 operates between T2 K and 400 K. If the two engines have eq…

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Introduction / Context:Carnot engines provide an ideal benchmark. When two reversible engines are placed in series with equal performance metrics (equal efficiency or identical heat input/output ratios), we can relate their temperature ratios to find the unknown intermediate temperature.Given Data / Assumptions:Engine 1: T_hot1 = 1600 K, T_cold1 = T2.Engine 2: T_hot2 = T2, T_cold2 = 400 K.Both engines are reversible (Carnot) and have equal efficiencies.Concept / Approach:For a Carnot engine, efficiency eta = 1 - T_cold/T_hot. Equal efficiencies imply 1 - T2/1600 = 1 - 400/T2, which directly yields T2.Step-by-Step Solution:Set eta1 = eta2 ⇒ 1 - T2/1600 = 1 - 400/T2.Rearrange ⇒ T2/1600 = 400/T2.Multiply both sides by 1600*T2 ⇒ T2^2 = 1600 * 400 = 640000.Take positive root: T2 = sqrt(640000) = 800 K.Verification / Alternative check:Check efficiencies: eta1 = 1 - 800/1600 = 0.5; eta2 = 1 - 400/800 = 0.5. Both equal 50%, confirming the result.Why Other Options Are Wrong:1000 K, 1200 K, or 1400 K would lead to unequal efficiencies (not both 50%).Common Pitfalls:Equating heat transfers instead of efficiencies without a consistent basis; forgetting to use absolute temperatures.Final Answer:800 K

Question 9

Gas Turbine Metrics — Definition of Work Ratio In a simple gas turbine (Brayton) plant, the work ratio is defined as which of the following expressions, where W_t is turbine work and W_c is compressor work?

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Introduction / Context:Besides thermal efficiency, gas-turbine designers track the work ratio to see how much of the turbine’s output is actually available as net work after driving the compressor. A higher work ratio indicates a larger fraction of the turbine work remains for useful output.Given Data / Assumptions:Ideal Brayton components for definition; real losses are not needed to define the metric.Single-shaft or coupled turbine-compressor arrangement.W_t and W_c are per unit mass of working fluid.Concept / Approach:Net work W_net = W_t - W_c. Work ratio wr is defined as W_net divided by turbine work output: wr = (W_t - W_c) / W_t. This reveals how much of the turbine’s gross power escapes the internal demand of the compressor.Step-by-Step Solution:Define W_net = W_t - W_c.Define wr = W_net / W_t.Substitute W_net ⇒ wr = (W_t - W_c) / W_t.Verification / Alternative check:In practice, simple gas turbines often have wr in the 0.3–0.5 range at design point; improvements like intercooling/recuperation alter W_c and W_t, affecting wr accordingly.Why Other Options Are Wrong:(W_t - W_c)/W_c, W_t/W_c, and W_c/W_t describe other ratios (net-to-compressor, turbine-to-compressor, compressor-to-turbine) and are not the standard definition of work ratio.Common Pitfalls:Confusing work ratio with back work ratio (defined as W_c/W_t), which quantifies the compressor’s fraction of turbine work.Final Answer:(W_t - W_c) / W_t

Question 10

Fuel Technology — Is Coke Oven Gas Produced by Burning Coke? Judge the statement: “Coke oven gas is obtained by burning coke in an oven.” Indicate whether this is correct or incorrect and recall the actual source of coke oven gas.

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Introduction / Context:Coke oven gas is an important by-product fuel in iron and steel plants. Knowing how it is produced helps distinguish it from producer gas or blast furnace gas and prevents misconceptions about its composition and calorific value.Given Data / Assumptions:Industrial coke production from coking (destructive distillation) of coal in absence of air.By-products: coke, tar, ammonia liquor, benzol, and coke oven gas.Controlled heating without combustion within sealed ovens.Concept / Approach:In a coke oven, coal is heated in the absence of air to drive off volatile matter. The evolved gases and vapors, after cooling and cleaning, constitute coke oven gas (rich in H2, CH4, CO, and other light hydrocarbons). It is not generated by burning coke; combustion would consume fuel rather than produce this gas stream.Step-by-Step Solution:Identify the industrial operation: destructive distillation (carbonization) of coal.Recognize the gaseous by-product: coke oven gas.Conclude the statement is incorrect because the gas is not created by burning coke.Verification / Alternative check:Plant flow diagrams show raw gas leaving ovens to by-product recovery units; combustion occurs in separate heating flues, not within the charge space.Why Other Options Are Wrong:“Agree” confuses production (by coking coal) with utilization (burning the gas or coke elsewhere).Common Pitfalls:Mixing up different industrial gases: producer gas (air blown over hot coke), water gas (steam over hot coke), and coke oven gas (from coal carbonization).Final Answer:Disagree

Question 11

Stoichiometry (Gaseous Volumes) — Methane Combustion Products Consider complete combustion of methane (CH4) with oxygen. At the same temperature and pressure (ideal-gas behavior), is the statement correct that 1 m^3 of methane requires 2 m^3 of oxyg…

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Introduction / Context:Volume (or mole) ratios for reacting gases follow directly from balanced chemical equations when temperature and pressure are the same before and after reaction and all species are considered in the gaseous state. This is widely used in burner design and flue-gas calculations.Given Data / Assumptions:Balanced reaction for complete combustion of methane.Ideal-gas behavior; equal T and p conditions.Water is considered as vapor (steam) in the products.Concept / Approach:The balanced equation is: CH4 + 2 O2 → CO2 + 2 H2O (vapor). Because equal gas volumes represent equal moles at the same T and p, the coefficients provide the volume ratios directly: 1 : 2 → 1 : 2.Step-by-Step Solution:Write stoichiometry: CH4 + 2 O2 → CO2 + 2 H2O.Match volumes to coefficients: 1 m^3 CH4 requires 2 m^3 O2.Products: 1 m^3 CO2 and 2 m^3 H2O vapor (assuming steam, not liquid water).Thus the statement is correct under the stated conditions.Verification / Alternative check:On a molar (or volumetric) basis, air requirement can also be inferred: 1 m^3 CH4 needs 2 m^3 O2, which corresponds to 2 / 0.21 ≈ 9.52 m^3 of air theoretically, confirming consistency.Why Other Options Are Wrong:“Incorrect” would only be justified if water were condensed to liquid (not a gas) or if T, p were not matched; the prompt explicitly allows water or steam, with the gaseous case implied.Common Pitfalls:Forgetting that the simple volume–coefficient mapping holds only for gases at the same T and p; ignoring the nitrogen in air when calculating flue-gas volumes.Final Answer:Correct

Question 12

State Variables of a Perfect (Ideal) Gas — Which Variables Govern Its Physical Properties? Identify the set of variables that control the thermodynamic state and physical properties of a perfect (ideal) gas in equilibrium.

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Introduction / Context:State description for a simple compressible substance (such as an ideal gas) requires identifying independent variables that uniquely fix the state. For ideal gases, the classic triad is pressure, volume, and temperature.Given Data / Assumptions:Simple, single-phase ideal gas.Equilibrium (no significant gradients).Fixed composition (molar mass constant).Concept / Approach:The ideal-gas equation p * V = m * R * T relates pressure p, volume V, and temperature T for a fixed mass m and gas constant R. Any two independent variables determine the third. Together, they govern density, specific volume, internal energy (through T), and other derived properties.Step-by-Step Solution:Recognize the equation of state: p * V = m * R * T.For fixed m and R, two independent variables among p, V, T determine the state.Hence, the full set p, V, and T control the gas’s physical state and properties.Verification / Alternative check:Experimental P–V–T surfaces for gases confirm that specifying two variables (e.g., p and T) fixes specific volume v via v = R * T / p.Why Other Options Are Wrong:Each of A, B, or C alone is incomplete; all three together constitute the governing variables.Common Pitfalls:Using Celsius rather than Kelvin for ideal-gas calculations; forgetting composition must be fixed for R to be constant.Final Answer:all of these

Question 13

Gauge vs Absolute Pressure — Converting a Vessel Reading A pressure gauge on a vessel reads 25 bar (gauge). If atmospheric pressure is 1.03 bar, determine the absolute pressure inside the vessel. Take g = 9.81 m/s^2 (not directly needed).

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Introduction / Context:Pressure instruments often indicate gauge pressure, which excludes atmospheric pressure. Many thermodynamic relations, however, require absolute pressure. Correctly converting between the two is a basic, critical skill.Given Data / Assumptions:Gauge pressure p_g = 25 bar.Atmospheric pressure p_atm = 1.03 bar.Absolute pressure p_abs = p_g + p_atm for a standard gauge referenced to atmosphere.Concept / Approach:By definition, p_abs = p_g + p_atm. Negative gauge readings would imply vacuum; here the gauge is positive, so we add p_atm to obtain the absolute pressure.Step-by-Step Solution:Write p_abs = p_g + p_atm.Substitute values: p_abs = 25 + 1.03.Compute: p_abs = 26.03 bar.Verification / Alternative check:As a quick check, absolute pressure must exceed gauge pressure by about 1 bar near sea level; 26.03 bar is reasonable.Why Other Options Are Wrong:23.97 bar would be gauge minus atmosphere (vacuum case); 25 bar is gauge, not absolute; 34.81 bar has no basis in the given data.Common Pitfalls:Confusing absolute and gauge references; mixing different units (Pa, kPa, bar) without consistent conversion.Final Answer:26.03 bar

Question 14

Cycle Identification — Select the Incorrect Statement Among the following cycle descriptions, choose the single incorrect statement concerning the idealized processes in standard power cycles.

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Introduction / Context:Recognizing the signature process sets of common thermodynamic cycles is essential for analyzing engines and gas turbines. This question asks you to identify the incorrect description among several standard cycles.Given Data / Assumptions:Ideal, reversible cycles are considered for identification.No real-world losses or component inefficiencies are implied.Canonical textbook definitions are used.Concept / Approach:The Otto cycle (spark-ignition) has two isentropic (compression/expansion) and two constant-volume (heat addition/rejection) processes. The Ericsson cycle has two isothermal and two constant-pressure processes with regeneration. The Joule (Brayton) cycle, however, comprises two isentropic and two constant-pressure processes, not constant volume. Therefore, the statement assigning constant-volume processes to the Joule cycle is incorrect.Step-by-Step Solution:Check Otto: CV + CV + two isentropes ⇒ correct.Check Ericsson: isothermals + CP with regeneration ⇒ correct.Check Joule/Brayton: should be CP + CP + two isentropes; the given statement says CV, which is wrong.Therefore, option A is the incorrect statement.Verification / Alternative check:Component mapping: compressor/turbine are isentropic; combustor/exhaust are nearly constant-pressure → matches Brayton, confirming option A is wrong.Why Other Options Are Wrong:Option D (“all of the above”) is not correct because B and C are true; only A is incorrect.Common Pitfalls:Confusing Otto (CV heat addition) with Brayton (CP heat addition); assuming “all of the above” when only one statement is actually wrong.Final Answer:A Joule (Brayton) cycle consists of two constant volume and two isentropic processes.

Question 15

Combustion stoichiometry for water formation: If 1 kg of hydrogen burns completely with 8 kg of oxygen, what total mass of liquid water is produced?

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Introduction / Context:This problem checks your understanding of basic combustion stoichiometry and conservation of mass. When hydrogen reacts with oxygen to form water, the chemical equation and molar masses determine the exact product mass, which is crucial in boiler calculations, fuel–air ratios, and emissions estimates.Given Data / Assumptions:Chemical reaction: 2H2 + O2 → 2H2O (complete combustion, no side products).Masses given: 1 kg of hydrogen reacts with 8 kg of oxygen.Assume ideal, complete conversion to liquid water with no losses.Concept / Approach:The law of conservation of mass states that total mass of reactants equals total mass of products. The balanced equation shows a 1:1 mole ratio between H2 and H2O, but masses differ because molecular weights differ. Here, however, the problem already supplies the exact reacting masses, allowing a simple mass balance without converting to moles (though both routes agree).Step-by-Step Solution:Write total reactant mass: m_reactants = m_H2 + m_O2 = 1 kg + 8 kg = 9 kg.Conservation of mass: m_products = m_reactants for a closed, complete process.Hence, total water produced: m_H2O = 9 kg.(Molar check) Molecular masses: H2 = 2, O2 = 32, H2O = 18. Per stoichiometry, 4 kg H2 + 32 kg O2 → 36 kg H2O; scaling by 1/4 gives 1 kg H2 + 8 kg O2 → 9 kg H2O.Verification / Alternative check:The molar method confirms the mass-balance method. Both demonstrate that all hydrogen and oxygen supplied in the given ratio end up in the water product, with no unreacted species for the ideal case.Why Other Options Are Wrong:1 kg of water: Ignores oxygen's mass contribution.7 kg or 8 kg of water: Do not match total reactant mass; violate mass conservation.Common Pitfalls:Confusing mole and mass ratios, or forgetting to include oxygen's mass in the final product. Also, do not subtract; in combustion of elements, product mass equals the sum of reactant masses.Final Answer:9 kg of water

Question 16

Absolute (vacuum) pressure concept: Absolute zero pressure corresponds to which physical condition or location?

Accepted Answer

Introduction / Context:Absolute pressure is measured relative to a perfect vacuum, not to atmospheric pressure. Understanding what “zero absolute pressure” truly means prevents mistakes when converting between absolute and gauge pressures and when applying gas laws and vacuum technology concepts.Given Data / Assumptions:Absolute zero pressure means no molecules (or no molecular motion impacting the boundary) in the space considered.Sea-level atmospheric pressure is a positive absolute pressure.Temperatures must be on an absolute scale (Kelvin) to reason correctly.Concept / Approach:Pressure at the microscopic level arises from molecular impacts on surfaces, which are related to molecular momentum transfer. A perfect vacuum has no molecules and hence no momentum exchange, giving zero absolute pressure. Real locations like sea level or the earth's center do not correspond to zero absolute pressure. Temperature statements must also be physically meaningful; there is no such thing as −273 K (Kelvin is non-negative).Step-by-Step Solution:Define absolute zero pressure: p_abs = 0 when there are no molecular impacts.Rule out sea level: p_abs ≈ 101.325 kPa at sea level, not zero.Rule out −273 K: Kelvin scale cannot be negative; absolute zero is at 0 K, not −273 K.Rule out center of earth: Enormous pressures exist due to overburden, not zero.Verification / Alternative check:From kinetic theory, pressure is proportional to number density * mean squared molecular speed * molecular mass. Setting number density to zero yields zero pressure, matching the vacuum concept.Why Other Options Are Wrong:At sea level: Clearly nonzero absolute pressure.At −273 K: Non-physical temperature on the Kelvin scale.At earth's center: Extremely high pressure due to gravity and overlying mass.Common Pitfalls:Confusing gauge pressure zero (which occurs roughly at atmospheric pressure) with absolute zero pressure; mixing Celsius/Kelvin offsets when discussing absolute zero temperature versus vacuum pressure.Final Answer:when molecular momentum of the system becomes zero

Question 17

Extensive property definition (thermodynamics): In the context of properties of a system, what is true about an extensive property?

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Introduction / Context:Thermodynamic properties are categorized as intensive or extensive. This classification is fundamental for formulating balances, scaling systems, and interpreting property tables. The wording in some options can be misleading, so it is important to rely on the precise definition.Given Data / Assumptions:Extensive properties depend on the size or mass of the system.Examples: mass, volume, total energy, total entropy, total internal energy.Intensive properties are independent of system size (e.g., temperature, pressure, density).Concept / Approach:An extensive property doubles when the system size is doubled at the same state. None of the listed options A–C are proper definitions; they misuse process words (heat transferred, work done, energy utilized) instead of property words. Heat and work are path functions, not properties, and “utilised” is not a property concept. Therefore, the only correct choice among those provided is “none of these.”Step-by-Step Solution:Identify correct definition: property that scales with system size → extensive.Evaluate options: A–C talk about transfers/uses during a process, not state properties.Conclude “none of these”.Verification / Alternative check:Doubling a system: two identical tanks each with mass m and volume V together have mass 2m and volume 2V, confirming mass and volume are extensive.Why Other Options Are Wrong:Heat/work “transferred/done”: Not properties; they are path-dependent interactions, not state descriptors.“Energy utilised”: Vague process description, not a property.Common Pitfalls:Confusing properties (state functions) with interactions (path functions); assuming “energy” alone must be intensive—note that total energy is extensive while specific energy (per mass) is intensive.Final Answer:none of these

Question 18

Open system (control volume) identification: Which statement correctly defines an open system in thermodynamics?

Accepted Answer

Introduction / Context:Thermodynamic analyses distinguish between closed systems (mass fixed) and open systems (control volumes) where mass can flow. Power plants, compressors, turbines, nozzles, pumps, and heat exchangers are typically modeled as open systems during steady operation.Given Data / Assumptions:Open systems allow mass transfer across the control surface.Energy can cross as heat and/or work.Steady or unsteady conditions may apply, but the definition does not depend on steadiness.Concept / Approach:An open system (control volume) exchanges both energy and mass with its surroundings. A closed system allows energy transfer as heat or work but not mass transfer. An isolated system allows neither energy nor mass transfer. Hence the only option that matches the definition of an open system is the one stating that heat, work, and mass may all cross the boundary.Step-by-Step Solution:Check mass transfer: required for an open system.Check energy transfer: heat/work exchanges commonly occur as well.Select the option mentioning both energy and mass crossing the boundary.Verification / Alternative check:Examples: air compressor (mass in and out, shaft work in, heat to surroundings), gas turbine (mass in/out, heat from combustion, shaft work out) all fit the open-system definition.Why Other Options Are Wrong:A: Describes a closed system.B: Mass crosses but claims no heat/work—too restrictive and not the defining criterion.D: Describes an isolated system.Common Pitfalls:Equating open system with steady flow only; they are related but not synonymous.Final Answer:both the heat and work as well as mass of the working substance crosses the boundary of the system

Question 19

Adiabatic expansion (ideal gas): Which expression correctly gives the work done during a reversible adiabatic expansion from state (p1, v1, T1) to (p2, v2, T2)? (R = gas constant, γ = cp/cv)

Accepted Answer

Introduction / Context:Adiabatic processes are central to compressors, turbines, and nozzles. For an ideal gas undergoing a reversible adiabatic (isentropic) expansion, several equivalent formulas exist for work. Selecting the correct sign and arrangement is essential for solving cycle problems accurately.Given Data / Assumptions:Ideal gas with constant specific heats (γ = cp/cv).Reversible adiabatic (no heat transfer, Q = 0; isentropic for ideal gas).Initial state (p1, v1, T1), final state (p2, v2, T2).Concept / Approach:For a reversible adiabatic process: pv^γ = constant and Q = 0. The first law for a closed system gives W = ∫p dv. Using the adiabatic relation, the integral reduces to an expression in end states: W = (p1v1 − p2v2)/(γ − 1). Equivalent temperature-based forms are W = mcv*(T1 − T2) and W = mR(T1 − T2)/(γ − 1). Correct signs reflect that T2 Step-by-Step Solution:Use pv^γ = C and W = ∫ p dv from v1 to v2.Integrate to obtain W = (p1v1 − p2v2)/(γ − 1).Check sign with T-drop: T1 > T2 so W > 0 for expansion.Verification / Alternative check:From the first law with Q = 0: ΔU = −W and ΔU = mcv*(T2 − T1) ⇒ W = mcv(T1 − T2), which is consistent with the pressure–volume form above.Why Other Options Are Wrong:(p2v2 − p1v1)/(γ − 1): Sign error.mR(T2 − T1)/(γ − 1): Sign error; should be T1 − T2.mcv(T2 − T1): Again sign error for expansion.Common Pitfalls:Forgetting sign conventions; mixing up γ − 1 with 1 − γ; assuming T2 > T1 in an expansion (true only with heat input, not adiabatic expansion).Final Answer:W = (p1v1 - p2v2) / (γ - 1)

Question 20

Gas cooled at constant pressure (qualitative behavior): When a gas is cooled at constant pressure, what happens to its temperature and volume?

Accepted Answer

Introduction / Context:Many real processes approximate constant-pressure behavior, such as heating or cooling in ducts and atmospheric processes. Understanding the qualitative change in volume with temperature at constant pressure is a direct application of the ideal gas law and is vital for interpreting P–V–T diagrams and process narratives.Given Data / Assumptions:Ideal gas behavior for qualitative trends.Pressure p held constant externally.No phase change within the temperature range considered.Concept / Approach:The ideal gas equation pV = mRT shows V ∝ T at fixed p and mass. Therefore, reducing T (cooling) directly reduces V at constant p. Temperature obviously decreases by definition of cooling. Thus, both the temperature and the volume decrease during constant-pressure cooling.Step-by-Step Solution:Hold p constant: pV = mRT ⇒ V = mR*T/p.Decrease T: T ↓ implies V ↓ proportionally.Hence, both T and V fall when cooling at constant p.Verification / Alternative check:On a T–V diagram at constant p, the process traces a straight line through the origin (for ideal gas), confirming direct proportionality between V and T.Why Other Options Are Wrong:Temperature increases while cooling: Contradiction in terms.Volume increases under cooling at constant p: Opposite of ideal gas relation.Both increase: Again contradicts cooling definition and gas law.Common Pitfalls:Confusing constant-pressure with constant-volume or constant-temperature processes; forgetting to hold mass constant when applying pV = mRT.Final Answer:both temperature and volume decreases

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