Question 1

General (ideal) gas equation in mass form: Select the correct form of the ideal gas equation for a fixed mass m at absolute temperature T.

Accepted Answer

Introduction / Context: The ideal gas equation is the cornerstone relation connecting pressure, volume, temperature, and amount of gas. Using the mass-based form is common in mechanical engineering calculations for compressors, nozzles, and air-standard cycles. Given Data / Assumptions: Ideal gas behavior. Specific gas constant R is on a per-mass basis. p is absolute pressure, v is total volume occupied by mass m. Concept / Approach: The molar form is pV = nR_uT, with n moles and universal constant R_u. Converting to mass m with specific gas constant R = R_u/M gives the mass form: pV = mRT. Rearranging for specific volume v_hat = V/m yields pv_hat = RT, often used in property tables and cycle work. Step-by-Step Solution: Start from pV = nR_uT.Use n = m/M and R = R_u/M ⇒ pV = mRT.Therefore the correct compact form with total volume v is pv = mRT. Verification / Alternative check: Dimensional check: [p][V] = energy; [m][R]*[T] has the same dimensions, confirming consistency. Why Other Options Are Wrong: pv = RTm and pv = (RT)m: Same as option A in value but unconventional ordering; however, the standard, unambiguous expression is pv = mRT. pvm = C: Suggests a constant product unrelated to temperature; incorrect. Common Pitfalls: Mixing molar and mass bases; forgetting that R is gas-specific whereas R_u is universal. Final Answer: pv = mRT

Question 2

Absolute zero pressure versus temperature: At what temperature can absolute zero pressure (perfect vacuum) be attained?

Accepted Answer

Introduction / Context: Absolute pressure is referenced to a perfect vacuum, while temperature scales like Celsius and Kelvin describe thermal energy states. It is tempting to connect “zero pressure” with a particular temperature, but they are different physical concepts and should not be conflated. Given Data / Assumptions: Absolute zero pressure means no gas molecules present. Absolute zero temperature is 0 K; Celsius values require conversion. Real systems cannot reach perfect vacuum or absolute zero temperature in practice. Concept / Approach: Pressure is due to molecular impacts; temperature measures average kinetic energy of molecules. While lowering temperature typically reduces pressure at fixed volume and mass, zero pressure requires removing all molecules (vacuum), not merely cooling to some particular temperature listed among the options. Therefore, none of the given temperatures guarantees zero absolute pressure. Step-by-Step Solution: Recognize that 0°C and 273 K are ordinary finite temperatures.At any finite temperature with gas present, pressure is greater than zero.Zero pressure needs absence of gas (vacuum), not a specific T from the list. Verification / Alternative check: From pV = mRT, with m > 0 and V finite, p is proportional to T; only if m = 0 (no molecules) can p be zero at any temperature, confirming the independence of a specific temperature value. Why Other Options Are Wrong: 0°C, 273°C, 273 K: None corresponds to perfect vacuum; these are arbitrary temperatures. Common Pitfalls: Confusing absolute zero temperature (0 K) with zero absolute pressure; they are distinct limits and neither is practically attainable in the strict sense. Final Answer: none of these

Question 3

Heat capacity ratio question: The ratio γ = cp/cv (specific heats at constant pressure and constant volume) is always __________ one for a gas.

Accepted Answer

Introduction / Context: The heat capacity ratio γ = cp/cv is a key parameter in isentropic flow, speed of sound, and adiabatic relations. Understanding its magnitude relative to unity is essential for predicting expansion/compression behavior and wave propagation in gases. Given Data / Assumptions: Ideal or calorically perfect gas for conceptual reasoning. Definitions: cp is heat per unit mass per degree at constant pressure; cv at constant volume. For a gas, cp = cv + R where R is the specific gas constant. Concept / Approach: Because cp = cv + R and R is positive for any gas, cp must exceed cv, hence γ = cp/cv > 1. Typical values are around 1.4 for diatomic gases (air at moderate temperatures) and around 1.67 for monatomic gases. Values decrease slightly with increasing temperature as vibrational modes become active in real gases, but remain greater than one. Step-by-Step Solution: Start with cp − cv = R > 0.Then cp > cv ⇒ cp/cv > 1.Therefore γ is greater than one for gases. Verification / Alternative check: For air at room temperature: cp ≈ 1.004 kJ/kg·K, cv ≈ 0.718 kJ/kg·K ⇒ γ ≈ 1.4 > 1, matching common textbook values. Why Other Options Are Wrong: Equal to one: Would imply R = 0 which is impossible for a gas. Less than one: Contradicts cp − cv = R > 0. Common Pitfalls: Confusing γ with polytropic index n; γ is a property ratio, whereas n depends on process path. Final Answer: greater than

Question 4

Comparing Otto and Diesel cycles at the same compression ratio: Which statement is correct about their thermal efficiencies for a fixed compression ratio?

Accepted Answer

Introduction / Context: The ideal Otto and Diesel cycles model spark-ignition and compression-ignition engines, respectively. For a fair comparison at the same compression ratio, the difference in the mode of heat addition leads to different efficiencies. Recognizing which cycle is more efficient under this condition is a staple concept in thermodynamics. Given Data / Assumptions: Air-standard analysis with constant specific heats. Equal compression ratio r for both cycles. Otto: heat addition at constant volume; Diesel: heat addition at constant pressure with cut-off ratio ρ > 1. Concept / Approach: Otto cycle efficiency depends only on compression ratio: η_Otto = 1 − 1/r^(γ−1). Diesel efficiency also depends on cut-off ratio ρ and is always lower than Otto's at the same r because part of Diesel's heat addition occurs at larger specific volumes (lower average temperatures of heat addition). Therefore, for an equal r, η_Otto > η_Diesel. Step-by-Step Solution: Fix r for both cycles.Otto: heat at constant volume → higher peak and average temperature of heat addition.Diesel: constant-pressure addition over finite cut-off ρ → lower average temperature of heat addition.Hence η_Otto > η_Diesel for the same r. Verification / Alternative check: Plugging typical numbers (γ ≈ 1.4, r = 8–10) into standard formulas shows Otto’s η several points higher than Diesel’s unless ρ approaches 1 (which would trend toward Otto behavior). Why Other Options Are Wrong: Same efficiency: False because Diesel's dependence on ρ reduces η. Diesel more efficient at same r: Opposite to theory. Efficiency independent of compression ratio: Both cycles increase η with r. Common Pitfalls: Confusing real-engine trends (Diesel engines often have higher r) with ideal-cycle comparisons at the same r. Final Answer: For a given compression ratio, Otto cycle is more efficient than Diesel cycle.

Question 5

Heating a gas at constant pressure (isobaric heating) In classical thermodynamics for a gas in a piston–cylinder or similar device, when the gas is heated while the pressure is maintained constant, what happens to its state variables?

Accepted Answer

Introduction / Context: An isobaric (constant-pressure) process is common in piston–cylinder devices with a sliding weight or regulated outlet. Understanding how temperature and volume respond to heat addition at constant pressure is foundational for analyzing compressors, heaters, and gas turbine components. Given Data / Assumptions: Gas contained in a device that can expand to keep pressure constant. Heat is added to the gas. Ideal-gas behavior used for intuition; real gases follow the same qualitative trend for ordinary conditions because cp > 0. Concept / Approach: At constant pressure, the First law on a per–unit mass basis gives: δq = dh for ideal gases, where h = h(T). Thus heat addition increases enthalpy and therefore temperature. From the ideal-gas relation p * v = R * T at fixed p, any increase in T requires a proportional increase in specific volume v. Hence both T and v rise when heat is supplied isobarically. Step-by-Step Solution: Hold pressure constant: p = const.Add heat: δq > 0 ⇒ for ideal gas, dh = δq ⇒ T increases because h = h(T).Use p * v = R * T with p constant ⇒ v ∝ T ⇒ volume increases. Verification / Alternative check: On a T–v diagram at fixed pressure, isobars slope upward; heating moves the state to higher T and higher v. In practical devices, the piston moves outward while the gas warms, confirming both increases. Why Other Options Are Wrong: Only temperature or only volume increases: Inconsistent with p * v = R * T at constant p.No change or temperature decrease: Violates positive cp and heat addition effects for gases. Common Pitfalls: Confusing isobaric with isochoric processes. At constant volume, temperature rises but volume does not; at constant pressure, both temperature and volume rise together. Final Answer: both temperature and volume will increase

Question 6

Standard atmosphere equivalences at mean sea level Select the correct consolidated statement for the standard atmospheric pressure at sea level (International Standard Atmosphere).

Accepted Answer

Introduction / Context: Engineering calculations often require converting standard atmospheric pressure between multiple unit systems. Being fluent with these equivalences avoids mistakes in sizing pumps, compressors, and instrumentation calibrations. Given Data / Assumptions: Standard atmosphere (mean sea level): approximately 101325 Pa. Mercury column at 0°C has density supporting 760 mm at 1 atm. Bar is a convenient non-SI metric unit used in practice. Concept / Approach: Key exact or conventional values used in engineering: 1 atm ≈ 101325 N/m^2 (Pa) ≈ 1.013 × 10^5 N/m^2. This equals approximately 1.013 bar (since 1 bar = 10^5 Pa) and corresponds to a mercury column of 760 mm under standard conditions. Therefore, the grouped option stating “all of these” is correct. Note that 1.013 × 10^6 N/m^2 would be 10.13 bar, which is not standard atmospheric pressure. Step-by-Step Solution: Recognize 1 atm ≈ 101325 Pa = 1.013 × 10^5 N/m^2.Convert to bar: 101325 Pa ÷ 10^5 Pa/bar ≈ 1.013 bar.Recall barometric equivalence: 760 mm Hg at 0°C ≈ 1 atm. Verification / Alternative check: Instrument calibration tables and ISO/ISA standards list these same equivalences, validating the consolidated choice. Why Other Options Are Wrong: 1.013 × 10^6 N/m^2: Ten times larger than 1 atm. Common Pitfalls: Mixing bar (10^5 Pa) with Pascal and kiloPascal; remember 1 bar = 100 kPa (exact by definition). Final Answer: all of these

Question 7

Fuel chemistry – principal elemental constituents For common fossil fuels used in engines and furnaces, which elements are the principal constituents determining heating value and combustion products?

Accepted Answer

Introduction / Context: Combustion analysis and emission predictions begin with understanding elemental composition. Most conventional liquid and gaseous fuels are hydrocarbons, so identifying the dominant elements is essential for stoichiometry and calorific value calculations. Given Data / Assumptions: Fuels include coal-derived gases, petroleum liquids, and natural gas. Trace elements (sulphur, nitrogen, oxygen) may be present but in smaller amounts. Heating value correlates strongly with carbon–hydrogen content. Concept / Approach: Hydrocarbons consist of hydrogen and carbon. On combustion, they yield CO2 and H2O as primary products, with heating value driven by the formation of strong bonds (CO2, H2O). Minor sulphur leads to SO2/SO3; fuel-bound oxygen slightly reduces net heating value but does not dominate composition for most fossil fuels. Step-by-Step Solution: Recognize that gasoline, diesel, LPG, and natural gas are hydrocarbon mixtures.Infer that C and H are principal elements.Select option stating carbon and hydrogen. Verification / Alternative check: Fuel standards (ASTM/EN) report ultimate analyses with carbon and hydrogen as the largest mass fractions for petroleum fuels and natural gas, confirming the choice. Why Other Options Are Wrong: Pairs including oxygen, sulphur, nitrogen, or inert gases do not represent the dominant elemental makeup in typical fossil fuels. Common Pitfalls: Overemphasizing trace sulphur or oxygenated additives; these affect emissions or octane/cetane but are not the principal constituents. Final Answer: carbon and hydrogen

Question 8

Simple Brayton (gas turbine) cycle – determinants of work ratio In a simple, open Brayton cycle without regeneration or intercooling, the work ratio (net work / turbine work) primarily depends on which of the following parameters?

Accepted Answer

Introduction / Context: The work ratio measures how much of the turbine's output is available as net work after driving the compressor. It is a key performance indicator for gas turbines and is sensitive to temperature limits and pressure ratio choices. Given Data / Assumptions: Simple Brayton cycle: compressor → combustor → turbine. Fixed component efficiencies for qualitative discussion. No regeneration, intercooling, or reheat. Concept / Approach: Turbine work and compressor work scale with absolute temperature levels. Increasing turbine inlet temperature (maximum cycle temperature) raises turbine work more than compressor work, improving work ratio. Lowering minimum cycle temperature (compressor inlet) increases density and reduces compressor work per unit mass flow, improving the ratio. Pressure ratio alters both compressor and turbine specific works; there exists an optimum pressure ratio for peak specific work and efficiency at given Tmax and Tmin. Therefore, all listed parameters influence the work ratio significantly. Step-by-Step Solution: Raise Tmax: increases turbine specific work → higher work ratio.Lower Tmin: reduces compressor specific work → higher work ratio.Vary pressure ratio: changes both stage enthalpy rises/drops, affecting numerator and denominator. Verification / Alternative check: T–s diagrams show larger turbine area minus compressor area as Tmax increases or Tmin decreases; parametric studies confirm sensitivity to pressure ratio as well. Why Other Options Are Wrong: Single-parameter claims ignore the coupled dependence of Brayton-cycle work terms on both temperature limits and pressure ratio. Common Pitfalls: Focusing only on pressure ratio while neglecting inlet cooling or turbine inlet temperature constraints imposed by materials. Final Answer: all of these

Question 9

Otto cycle efficiency – dependence on specific heats Does the air-standard efficiency of the Otto (constant-volume heat addition) cycle depend on the specific heats of the working substance?

Accepted Answer

Introduction / Context: Air-standard cycle analysis expresses thermal efficiency in terms of compression ratio and the ratio of specific heats (gamma). Understanding this dependence informs material choices and mixture effects in spark-ignition engine modeling. Given Data / Assumptions: Air-standard assumptions, constant properties for simplicity. Compression ratio r fixed by engine geometry. Gamma = cp / cv. Concept / Approach: The ideal Otto-cycle efficiency is η = 1 − 1 / r^(γ−1). Thus efficiency depends on r and γ. Because γ = cp / cv derives from specific heats, any change in cp or cv (and thus γ) alters η. Real gases have temperature-dependent heat capacities, making γ vary with temperature; the dependence persists even in more refined models. Step-by-Step Solution: Recall η_Otto = 1 − 1 / r^(γ−1).Note γ depends on specific heats of the working substance.Therefore η is a function of both r and γ → depends on specific heats. Verification / Alternative check: Comparing dry air (γ ≈ 1.4) to products-rich mixtures (γ lower) at the same r shows reduced ideal efficiency when γ decreases. Why Other Options Are Wrong: “No” or temperature-only claims contradict the standard efficiency formula; compression ratio and γ govern the ideal result. Common Pitfalls: Assuming efficiency depends only on r; γ changes with composition and temperature, subtly shifting η even at fixed r. Final Answer: Yes

Question 10

Isochoric heating (constant volume) – energy and work effects When a gas is heated at constant volume, which statement best describes where the supplied heat goes and what happens to the system properties?

Accepted Answer

Introduction / Context: Isochoric (constant-volume) processes occur in rigid vessels and also model heat rejection/addition legs in ideal cycles. Properly accounting for energy and work clarifies why cp and cv differ for gases. Given Data / Assumptions: Closed system gas in a rigid container (dv = 0). Heat is supplied to the gas. Ideal-gas relation used for intuition; qualitative result holds broadly. Concept / Approach: The First law for a closed system is ΔU = Q − W. With constant volume, boundary work W_b = ∫ p dv = 0. Therefore, any heat added (Q > 0) increases internal energy: ΔU = Q. For an ideal gas, u = u(T), so increasing internal energy directly increases temperature and pressure (since p = R * T / v and v is fixed). No external boundary work is done because volume does not change. Step-by-Step Solution: Set dv = 0 ⇒ W_b = ∫ p dv = 0.Apply First law: ΔU = Q − W_b = Q.For ideal gas, u = u(T) ⇒ ΔU > 0 ⇒ temperature rises; pressure rises because p ∝ T at fixed v. Verification / Alternative check: Calorimetry in rigid bombs demonstrates temperature rise with heat input and no boundary work; measured cv relates directly to the temperature change at constant volume. Why Other Options Are Wrong: External work during expansion does not occur at constant volume.Statements implying internal energy reduction contradict the First law for Q > 0 with W_b = 0. Common Pitfalls: Confusing pressure rise with work: pressure can rise without boundary work if volume is fixed. Final Answer: increases the internal energy of the gas and increases the temperature of the gas

Question 11

Gas constant R – relation to specific heats For a calorically perfect ideal gas, the gas constant R is equal to which of the following combinations of the mass-specific heat capacities?

Accepted Answer

Introduction / Context: The link between the gas constant R and the specific heats cp and cv is central to ideal-gas thermodynamics. Remembering the correct algebraic relation prevents frequent mistakes in property calculations and cycle analysis. Given Data / Assumptions: Ideal gas with constant specific heats (calorically perfect). Mass-specific heats cp and cv. Definition of gamma: γ = cp / cv. Concept / Approach: For ideal gases, enthalpy h = cp * T and internal energy u = cv * T. The enthalpy–internal energy relation is h − u = R * T, which rearranges to (cp − cv) * T = R * T. Cancel T to obtain the fundamental identity R = cp − cv. This holds on both a per–unit-mass and per–mole basis (with appropriate R values). Step-by-Step Solution: Start from h = u + p * v and p * v = R * T (ideal gas).Differentiate with respect to T: dh/dT − du/dT = R.Recognize dh/dT = cp and du/dT = cv ⇒ R = cp − cv. Verification / Alternative check: Measured values for air at room temperature: cp ≈ 1.005 kJ/kg·K, cv ≈ 0.718 kJ/kg·K, difference ≈ 0.287 kJ/kg·K, matching R for air. Why Other Options Are Wrong: Sum/product/ratio/square do not satisfy the identity derived from h − u and the ideal-gas equation of state. Common Pitfalls: Confusing γ = cp / cv with R = cp − cv; both appear often but represent distinct relations. Final Answer: difference of two specific heats

Question 12

Comparing manufactured gases – highest calorific value Among the following manufactured/industrial fuel gases, which typically has the highest calorific value per unit volume at standard conditions?

Accepted Answer

Introduction / Context: Manufactured gases differ widely in composition (H2, CO, CH4, N2, CO2) and therefore in calorific value (CV). Knowing their relative CV is important for burner design, engine suitability, and pipeline distribution planning. Given Data / Assumptions: Typical compositions without enrichment. Standard temperature and pressure for volumetric CV comparison. Historical industry definitions of coal gas, producer gas, Mond gas, and blast furnace gas. Concept / Approach: Coal gas (town gas) produced from coal carbonization contains a significant fraction of hydrogen, methane, and light hydrocarbons, leading to a relatively high CV. Producer gas and Mond gas are nitrogen-diluted (air used in gasification), with lower CV. Blast furnace gas contains substantial CO2 and N2 with modest CO, yielding the lowest CV among those listed. Hence, coal gas ranks highest on a volumetric basis among the options provided. Step-by-Step Solution: Identify gas origins: carbonization (coal gas) vs air–steam gasification (producer/Mond) vs metallurgical off-gas (BFG).Link higher fractions of CH4/H2/CO to higher CV; inert dilution lowers CV.Select coal gas as having the highest CV among the listed options. Verification / Alternative check: Representative CV (approximate, MJ/Nm^3): coal gas ~ 18–20; producer gas ~ 4–6; Mond gas ~ 5–7; blast furnace gas ~ 3–4. Rankings confirm the choice. Why Other Options Are Wrong: Producer/Mond/BFG have substantial nitrogen or carbon dioxide dilution, lowering CV.Coke oven gas can be high CV but is not among the original set in the source list; the comparative context here points to coal gas as the typical highest CV versus producer/Mond/BFG. Common Pitfalls: Confusing coal gas with natural gas; natural gas (mostly methane) has even higher CV, but it is not a manufactured gas in this context. Final Answer: Coal gas

Question 13

Stoichiometric combustion of ethylene (C2H4) – products per kilogram Consider the complete (stoichiometric) combustion of 1 kg of ethylene, C2H4. It produces 22/7 kg of carbon dioxide. What is the corresponding mass of water (liquid or steam) formed?

Accepted Answer

Introduction / Context: Stoichiometric combustion problems convert a fuel's chemical formula into precise product and reactant masses. Ethylene (C2H4) is a common unsaturated hydrocarbon; balancing its combustion gives fixed product ratios independent of air excess. Given Data / Assumptions: Chemical formula: C2H4. Complete combustion to CO2 and H2O. Basis: 1 kg of fuel; standard atomic masses C = 12, H = 1, O = 16. Concept / Approach: Balance the reaction on a molar basis, then convert to mass per kilogram of fuel. Combustion equation: C2H4 + 3 O2 → 2 CO2 + 2 H2O. From this, two moles of CO2 and two moles of H2O are formed per mole of C2H4 burned. Convert moles to masses and scale by the molar mass of ethylene. Step-by-Step Solution: Molar masses: M_C2H4 = 212 + 41 = 28 kg/kmol; M_CO2 = 44 kg/kmol; M_H2O = 18 kg/kmol.Per 1 kmol fuel: products are 2 kmol CO2 and 2 kmol H2O.Masses per 1 kmol fuel: CO2 = 244 = 88 kg; H2O = 218 = 36 kg.Per kilogram fuel, divide by 28: CO2 = 88/28 = 22/7 kg; H2O = 36/28 = 9/7 kg. Verification / Alternative check: The given CO2 yield 22/7 kg per kg fuel matches the calculation, confirming consistent stoichiometry. Therefore, the water formed must be 9/7 kg per kg fuel. Why Other Options Are Wrong: 11/7, 7/4, 11/4, 5/7 do not satisfy the mass ratios implied by the balanced equation; only 9/7 aligns with 2 moles of water per mole of ethylene. Common Pitfalls: Using oxygen requirement data to infer products directly; products are fixed by fuel composition under complete combustion, independent of excess air. Final Answer: 9/7

Question 14

Charles’ law – volume–temperature relation at constant pressure According to Charles’ law, the volume of a given mass of a perfect gas varies how with its absolute temperature when the absolute pressure remains constant?

Accepted Answer

Introduction / Context: Charles’ law is one of the classical gas laws, explaining how volume depends on temperature when pressure is held constant. It is foundational for understanding thermal expansion of gases and calibrating constant-pressure thermometers. Given Data / Assumptions: Fixed mass of an ideal (perfect) gas. Constant absolute pressure. Temperature measured on an absolute scale (K). Concept / Approach: The ideal-gas equation p * V = n * R * T reduces at constant p and n to V ∝ T. Thus if absolute temperature doubles, volume doubles; if temperature decreases, volume decreases in the same proportion. This direct proportionality is Charles’ law. Step-by-Step Solution: Start from p * V = n * R * T.Hold p and n constant ⇒ V = (n * R / p) * T.Conclude V ∝ T on an absolute (Kelvin) scale. Verification / Alternative check: Experimental plots of V versus T at constant p produce straight lines extrapolating toward V → 0 near T → 0 K, consistent with the law. Why Other Options Are Wrong: Indirect, exponential, logarithmic, or independent relations do not follow from the ideal-gas law under the stated constraint. Common Pitfalls: Using Celsius instead of Kelvin; direct proportionality is valid only with absolute temperature. Near condensation or very high pressures, real gases deviate, but the ideal relation remains a good model over modest ranges. Final Answer: directly

Question 15

Industrial fuels – common choice for cement kilns and metallurgical heating In industrial practice, which fuel is mostly used as the primary heat source in the cement industry (rotary kilns) and widely in high-temperature metallurgical processes (e.…

Accepted Answer

Introduction / Context: Cement rotary kilns and many metallurgical furnaces demand intense, controllable heat release, clean flame characteristics, and good temperature uniformity. Selecting the right industrial fuel impacts flame temperature, kiln throughput, clinker quality, and emissions. Given Data / Assumptions: Cement plants commonly fire rotary kilns requiring stable, high-temperature flames. Metallurgical operations (e.g., reheating and some reduction furnaces) require efficient, controllable combustion. Fuel should be suitable for burner injection and air–fuel mixing. Concept / Approach: Pulverised coal (finely ground) is pneumatically conveyed and burned through specially designed kiln burners. Fine particle size ensures rapid ignition and complete combustion, enabling high flame temperatures and good heat transfer. Its logistics, cost, and burner compatibility make it a mainstream choice in cement and many metallurgical applications. Step-by-Step Solution: Identify operational needs: high, controllable heat release in a continuous process.Match fuel handling: fine coal can be metered, blown, and mixed with air efficiently.Consider alternatives: coke and wood charcoal are solid lumps, less suitable for burner atomisation; fuel oil/gas are alternatives but not “mostly used” in many regions historically.Conclude: pulverised coal aligns best with typical kiln and furnace firing systems. Verification / Alternative check: Standard kiln burner designs, coal mills, and inline calciner systems are engineered around pulverised coal firing; retrofits may also accept petcoke or alternative fuels, but pulverised coal remains the benchmark. Why Other Options Are Wrong: Coke/bituminous coke: primarily a reductant in blast furnaces; not commonly pulverised for kiln main firing. Wood charcoal: inconsistent supply and reactivity; not mainstream for cement kilns. Fuel oil only: used in some plants, but not the “mostly used” traditional choice. Common Pitfalls: Confusing metallurgical reductants (lump coke) with burner-grade, pneumatically conveyed fuels. Burner systems need fine, reactive fuel for flame shaping. Final Answer: pulverised coal

Question 16

Stoichiometry of carbon monoxide combustion by mass One kilogram of carbon monoxide (CO) requires 4/7 kg of oxygen (O2) for complete oxidation. The complete reaction produces which mass of carbon dioxide (CO2)?

Accepted Answer

Introduction / Context: Combustion stoichiometry connects mass of reactants to mass of products using balanced chemical equations and molar masses. This problem tests your ability to translate a known oxygen requirement into the corresponding product mass. Given Data / Assumptions: Reaction: 2 CO + O2 → 2 CO2 (complete oxidation). Given mass ratio: 1 kg CO needs 4/7 kg O2. Molar masses: CO = 28 kg/kmol, O2 = 32 kg/kmol, CO2 = 44 kg/kmol. Concept / Approach: Use the balanced equation to derive proportional mass relationships. For 2 mol CO (56 kg) reacting with 1 mol O2 (32 kg), products are 2 mol CO2 (88 kg). Scale linearly to 1 kg CO to find the produced CO2 mass. Step-by-Step Solution: From 56 kg CO → 88 kg CO2.Therefore, per 1 kg CO: CO2 produced = 88/56 kg = 11/7 kg.Given O2 requirement per 1 kg CO is 32/56 = 4/7 kg, which is consistent with the statement.Hence the correct product mass is 11/7 kg CO2. Verification / Alternative check: Mass conservation: Inputs 1 kg CO + 4/7 kg O2 = (7/7 + 4/7) kg = 11/7 kg total; output is 11/7 kg CO2, confirming balance. Why Other Options Are Wrong: 11/3 kg: too large; violates mass conservation for the given inputs. 7/3 kg or 8/3 kg of CO: lists CO as a product; complete oxidation produces CO2, not CO. 1 kg CO2: ignores added oxygen mass. Common Pitfalls: Forgetting to use molar mass ratios or neglecting oxygen contribution to product mass. Final Answer: 11/7 kg of carbon dioxide gas

Question 17

Calorific values – relationship between higher and lower heating values Is the following statement correct? “The lower calorific value (LCV) of a fuel may be obtained by adding the latent heat of the steam formed during combustion to the higher calo…

Accepted Answer

Introduction / Context: Higher and lower calorific values differ in whether the water formed during combustion is condensed and its latent heat recovered. Many exam questions probe this sign convention. Given Data / Assumptions: HCV assumes products are cooled to condense water and recover latent heat. LCV assumes water remains as vapour and latent heat is not recovered. Fuels containing hydrogen produce water upon combustion. Concept / Approach: The relationship is: HCV = LCV + latent heat of water formed (per unit fuel). Equivalently, LCV = HCV − latent heat associated with product water. Therefore, adding latent heat to HCV to “obtain LCV” is incorrect; you subtract that heat to go from HCV to LCV. Step-by-Step Solution: Define LH_vap = latent heat of vaporisation of water in products.State identity: LCV = HCV − LH_vap (per unit fuel considering water formed).Interpret the prompt: it says “adding” LH_vap to HCV → would give a value larger than HCV, contradicting LCV < HCV.Thus the statement is false. Verification / Alternative check: Numerical examples: Natural gas HCV ≈ LCV + 10%–11% (due to water condensation). LCV is lower because latent heat is not recovered in typical exhaust conditions. Why Other Options Are Wrong: Any “Yes” option reverses the sign convention. “LCV is always greater than HCV” is the opposite of reality; HCV ≥ LCV. Common Pitfalls: Confusing which value includes condensation. Remember: HCV counts the latent heat; LCV does not. Final Answer: No

Question 18

Atomic mass comparison – how many times heavier is an oxygen atom than a hydrogen atom? Based on relative atomic masses (H ≈ 1, O ≈ 16), the oxygen atom is how many times heavier than the hydrogen atom?

Accepted Answer

Introduction / Context: Relative atomic masses allow quick comparisons of atomic weights and are foundational in stoichiometry and chemical balancing. Given Data / Assumptions: Hydrogen standard atomic mass ≈ 1. Oxygen standard atomic mass ≈ 16. Use approximate integer values suitable for basic calculations. Concept / Approach: The question asks for a ratio: mass(O) / mass(H). Using standard relative masses, this ratio is 16/1 = 16. Step-by-Step Solution: Assign relative masses: H = 1, O = 16.Compute ratio = 16 / 1.Result = 16 times. Verification / Alternative check: In water H2O, molecular mass = 2*1 + 16 = 18, further reinforcing oxygen’s much larger atomic mass relative to hydrogen. Why Other Options Are Wrong: 2, 4, 8, 12 do not match standard atomic mass ratios. Common Pitfalls: Confusing atomic number with atomic mass; atomic number counts protons, not mass units. Final Answer: 16

Question 19

Process classification – are isothermal and adiabatic processes necessarily reversible? Is it correct to regard isothermal processes and adiabatic processes as always reversible, or can they be reversible or irreversible depending on how they are co…

Accepted Answer

Introduction / Context: Thermodynamic processes are classified both by path constraints (isothermal, adiabatic) and by reversibility. These classifications are independent; a process may satisfy a constraint yet still be irreversible. Given Data / Assumptions: Isothermal: temperature held constant. Adiabatic: no heat transfer. Reversibility depends on absence of friction, infinitesimal gradients, and quasi-equilibrium. Concept / Approach: Reversibility is a quality of the path, not merely the constraint. An isothermal compression with finite temperature difference heat transfer or friction is irreversible. An adiabatic expansion across a throttle is irreversible; a reversible adiabatic process (isentropic) requires no dissipative effects. Step-by-Step Solution: Identify constraint (isothermal or adiabatic).Evaluate path characteristics: presence of friction, turbulence, finite ΔT heat transfer, unrestrained expansion → irreversibility.If all dissipations are eliminated and process is quasi-static, the same constraint can be satisfied reversibly.Therefore, such processes can be reversible or irreversible. Verification / Alternative check: Textbook examples: reversible adiabatic (isentropic) compression vs. throttling (adiabatic but irreversible); reversible isothermal expansion of an ideal gas vs. isothermal mixing/compression with friction (irreversible). Why Other Options Are Wrong: “Reversible process” or “irreversible process” alone overgeneralise. “None of these” ignores established thermodynamic distinctions. Common Pitfalls: Equating “adiabatic” with “isentropic” without the reversibility requirement; isentropic = adiabatic + reversible. Final Answer: reversible or irreversible process

Question 20

Solid fuel preparation – definition of briquetted coal Confirm the statement: “When finely ground coal is moulded under pressure with or without a binding material, the resulting fuel is called briquetted coal.”

Accepted Answer

Introduction / Context: Briquetting converts fine coal or coal dust into larger agglomerates for easier handling, transportation, and combustion. The definition focuses on forming under pressure, often with binders. Given Data / Assumptions: Feed: finely ground coal or fines. Process: moulding or pressing under pressure. Binders: optional (e.g., molasses, pitch, bentonite) depending on coal rank and process. Concept / Approach: Briquetted coal refers to densified, shaped products (briquettes) produced from fines. The essence is mechanical compaction (and sometimes thermal curing) with or without a binder to achieve strength and durability. Step-by-Step Solution: Identify raw material: coal fines unsuitable for direct use.Apply compaction/moulding under pressure (with/without binder).Product obtained: briquetted coal — easier to store, transport, and burn.Hence, the statement matches the standard definition. Verification / Alternative check: Industry practices (roller presses, piston presses) produce pillow or oval briquettes consistent with this definition. Why Other Options Are Wrong: Restricting to specific coal ranks or additives is unnecessary; multiple feedstocks and binders can be used. Common Pitfalls: Confusing “briquettes” with “coke” (destructive distillation) or “pellets” from ore fines; processes and products are different. Final Answer: Yes

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