Metallurgical Thermodynamics And Kinetics — Part 2 MCQs with Explanations
Practice high-yield Metallurgical Engineering MCQs on Metallurgical Thermodynamics And Kinetics — Part 2 with worked academic solutions for PAEC, NESCOM, HIT, FPSC.
Questions
40
Difficulty Level
Exam-Level
Target Exams
PAEC, NESCOM, HIT...
Estimated Time
48 mins
Exam-Level PAEC NESCOM HIT FPSC
Verified for Commission Syllabus • Oct 14, 2026
40
Verified Practice Questions
Interactive test mode with detailed academic worked solutions
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🎉 Set complete! All 40 questions answered.
Question 1 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
The weight % of nitrogen (N 2 ) in liquid iron can be expressed as:
Correct Answer: Option A — $$K{\left( {P{N_2}} \right)^{\frac{1}{2}}}$$
Detailed Academic Explanation:
$$K{\left( {P{N_2}} \right)^{\frac{1}{2}}}$$ is the correct answer to “The weight % of nitrogen (N 2 ) in liquid iron can be expressed as:”. This question is standard in competitive examination syllabi for Metallurgical Thermodynamics And Kinetics.
💡Key Conceptual Anchor:Metallurgical Thermodynamics And Kinetics
Question 2 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
The ratio of fugacity of the substance to the fugacity in its standard state
Correct Answer: Option D — Both A and C
Detailed Academic Explanation:
Both A and C is the correct answer to “The ratio of fugacity of the substance to the fugacity in its standard state”. This question is standard in competitive examination syllabi for Metallurgical Thermodynamics And Kinetics.
💡Key Conceptual Anchor:Metallurgical Thermodynamics And Kinetics
Question 3 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
Melting point of a metal is 1356 K. When the liquid metal is undercooled to 1256 K, the free energy change for solidification, ΔG L-S = - 1000 J mol -1 . On the other hand, if the liquid metal is undercooled to 1200 K, the free energy change (in J mol -1 ) for solidification is
Correct Answer: Option C — -1550
Detailed Academic Explanation:
-1550 is the correct answer to “Melting point of a metal is 1356 K. When the liquid metal is undercooled to 1256 K, the free energy change for solidification, ΔG L-S = - 1000 J mol -1 . On the other hand, if the liquid metal is undercooled to 1200 K, the free energy change (in J mol -1 ) for solidification is”. This question is standard in competitive examination syllabi for Metallurgical Thermodynamics And Kinetics.
💡Key Conceptual Anchor:Metallurgical Thermodynamics And Kinetics
Question 4 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
For a spontaneous natural process at constant temperature and pressure, the free energy of the system always
Correct Answer: Option A — increases
Detailed Academic Explanation:
increases is the correct answer to “For a spontaneous natural process at constant temperature and pressure, the free energy of the system always”. This question is standard in competitive examination syllabi for Metallurgical Thermodynamics And Kinetics.
💡Key Conceptual Anchor:Metallurgical Thermodynamics And Kinetics
Question 5 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
The rate of reaction for first order reaction
Correct Answer: Option B — Rate = K[A]
Detailed Academic Explanation:
Rate = K[A] is the correct answer to “The rate of reaction for first order reaction”. This question is standard in competitive examination syllabi for Metallurgical Thermodynamics And Kinetics.
💡Key Conceptual Anchor:Metallurgical Thermodynamics And Kinetics
Question 6 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
Decomposition of calcium carbonate occurs as: CaCO 3 = CaO + CO 2 : ΔG 0 T = 17710 - 158T J. mol -1 What will be the decomposition temperature (in K) of CaCO 3 , if the activity of CaO is 0.5 and the pressure is 3 atmospheres?
Correct Answer: Option A — 142.7
Detailed Academic Explanation:
142.7 is the correct answer to “Decomposition of calcium carbonate occurs as: CaCO 3 = CaO + CO 2 : ΔG 0 T = 17710 - 158T J. mol -1 What will be the decomposition temperature (in K) of CaCO 3 , if the activity of CaO is 0.5 and the pressure is 3 atmospheres”. This question is standard in competitive examination syllabi for Metallurgical Thermodynamics And Kinetics.
💡Key Conceptual Anchor:Metallurgical Thermodynamics And Kinetics
Question 7 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
Roasting of a metallic sulfide MS in pure oxygen is carried out at a particular temperature where the probable stable solid phases are M, MS, MO and MSO 4 . In case the system attains equilibrium and thus satisfies phase rule, it is possible to obtain a combination of the following solid phases in the roasted product. P. M, MS and MO Q. M, MS, MO and MSO 4 R. MSO 4 , MS and MO S. MS and MO
Correct Answer: Option B — P or, Q or, R
Detailed Academic Explanation:
P or, Q or, R is the correct answer to “Roasting of a metallic sulfide MS in pure oxygen is carried out at a particular temperature where the probable stable solid phases are M, MS, MO and MSO 4 . In case the system attains equilibrium and thus satisfies phase rule, it is possible to obtain a combination of the following solid phases in the roasted product. P. M, MS and MO Q. M, MS, MO and MSO 4 R. MSO 4 , MS and MO S. MS and MO”. This question is standard in competitive examination syllabi for Metallurgical Thermodynamics And Kinetics.
💡Key Conceptual Anchor:Metallurgical Thermodynamics And Kinetics
Question 8 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
G = U + PV - TS. Then which one of the following is correct?
$${\left( {\frac{{\partial V}}{{\partial T}}} \right)_p} = - {\left( {\frac{{\partial S}}{{\partial P}}} \right)_T}$$ is the correct answer to “G = U + PV - TS. Then which one of the following is correct”. This question is standard in competitive examination syllabi for Metallurgical Thermodynamics And Kinetics.
💡Key Conceptual Anchor:Metallurgical Thermodynamics And Kinetics
Question 9 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
In case of homogenous nucleation, the critical edge length for a cubed shaped nucleus is (Y : Energy per unit area of interface between the product and the parent phase; Δg : Gibbs free energy change per unit volume)
Correct Answer: Option A — $$ - \frac{{4Y}}{{\Delta g}}$$
Detailed Academic Explanation:
$$ - \frac{{4Y}}{{\Delta g}}$$ is the correct answer to “In case of homogenous nucleation, the critical edge length for a cubed shaped nucleus is (Y : Energy per unit area of interface between the product and the parent phase; Δg : Gibbs free energy change per unit volume)”. This question is standard in competitive examination syllabi for Metallurgical Thermodynamics And Kinetics.
💡Key Conceptual Anchor:Metallurgical Thermodynamics And Kinetics
Question 10 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
When there is existence of uniform temperature throughout the system is called
Correct Answer: Option A — Thermal equilibria
Detailed Academic Explanation:
Thermal equilibria is the correct answer to “When there is existence of uniform temperature throughout the system is called”. This question is standard in competitive examination syllabi for Metallurgical Thermodynamics And Kinetics.
💡Key Conceptual Anchor:Metallurgical Thermodynamics And Kinetics
Question 11 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
The Daltons law of partial pressure can be written as
Correct Answer: Option A — $${X_i} = \frac{{{P_i}}}{{{P_T}}}$$
Detailed Academic Explanation:
$${X_i} = \frac{{{P_i}}}{{{P_T}}}$$ is the correct answer to “The Daltons law of partial pressure can be written as”. This question is standard in competitive examination syllabi for Metallurgical Thermodynamics And Kinetics.
💡Key Conceptual Anchor:Metallurgical Thermodynamics And Kinetics
Question 12 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
When two phases α and β in an alloy are in thermodynamic equilibrium, then
Correct Answer: Option D — $$G_i^\alpha = G_i^\beta $$
Detailed Academic Explanation:
$$G_i^\alpha = G_i^\beta $$ is the correct answer to “When two phases α and β in an alloy are in thermodynamic equilibrium, then”. This question is standard in competitive examination syllabi for Metallurgical Thermodynamics And Kinetics.
💡Key Conceptual Anchor:Metallurgical Thermodynamics And Kinetics
Question 13 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
Consider an ideal solution of component A and B. The entropy of mixing per mole of an alloy containing 50 wt% B, is
Correct Answer: Option A — R In 2
Detailed Academic Explanation:
R In 2 is the correct answer to “Consider an ideal solution of component A and B. The entropy of mixing per mole of an alloy containing 50 wt% B, is”. This question is standard in competitive examination syllabi for Metallurgical Thermodynamics And Kinetics.
💡Key Conceptual Anchor:Metallurgical Thermodynamics And Kinetics
Question 14 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
The hydrogen content of steel in equilibrium with hydrogen gas at 1 bar pressure is 28 ppm at some temperature. Hydrogen content in the metal at the same temperature gets reduced at 1 ppm, when the equilibrium PH 2 changes to . . . . . . . . bar.
Correct Answer: Option B — $$\frac{1}{{28}}$$
Detailed Academic Explanation:
$$\frac{1}{{28}}$$ is the correct answer to “The hydrogen content of steel in equilibrium with hydrogen gas at 1 bar pressure is 28 ppm at some temperature. Hydrogen content in the metal at the same temperature gets reduced at 1 ppm, when the equilibrium PH 2 changes to . . . . . . . . bar.”. This question is standard in competitive examination syllabi for Metallurgical Thermodynamics And Kinetics.
💡Key Conceptual Anchor:Metallurgical Thermodynamics And Kinetics
Question 15 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
For an ideal gas, Cp - Cv is
Correct Answer: Option A — R
Detailed Academic Explanation:
R is the correct answer to “For an ideal gas, Cp - Cv is”. This question is standard in competitive examination syllabi for Metallurgical Thermodynamics And Kinetics.
💡Key Conceptual Anchor:Metallurgical Thermodynamics And Kinetics
Question 16 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
At 1200°C, the standard Gibbs energy of thermal decomposition of one mole of wustite into Fe and O 2 is 168 J. The corresponding dissociation pressure (in atm.) is
Correct Answer: Option B — 1.22 × 10-12
Detailed Academic Explanation:
1.22 × 10-12 is the correct answer to “At 1200°C, the standard Gibbs energy of thermal decomposition of one mole of wustite into Fe and O 2 is 168 J. The corresponding dissociation pressure (in atm.) is”. This question is standard in competitive examination syllabi for Metallurgical Thermodynamics And Kinetics.
💡Key Conceptual Anchor:Metallurgical Thermodynamics And Kinetics
Question 17 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
Uphill diffusion means diffusion from
Correct Answer: Option A — a lower concentration and lower chemical potential to higher concentration and higher chemical potential
Detailed Academic Explanation:
a lower concentration and lower chemical potential to higher concentration and higher chemical potential is the correct answer to “Uphill diffusion means diffusion from”. This question is standard in competitive examination syllabi for Metallurgical Thermodynamics And Kinetics.
💡Key Conceptual Anchor:Metallurgical Thermodynamics And Kinetics
Question 18 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
Select the correct plot of Gibbs free energy (G) Vs. Temperature (T) for a single component system from the following:
Correct Answer: Option B — Q
Detailed Academic Explanation:
Q is the correct answer to “Select the correct plot of Gibbs free energy (G) Vs. Temperature (T) for a single component system from the following:”. This question is standard in competitive examination syllabi for Metallurgical Thermodynamics And Kinetics.
💡Key Conceptual Anchor:Metallurgical Thermodynamics And Kinetics
Question 19 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
The part of the universe excluding the system is known as
Correct Answer: Option A — the surrounding
Detailed Academic Explanation:
the surrounding is the correct answer to “The part of the universe excluding the system is known as”. This question is standard in competitive examination syllabi for Metallurgical Thermodynamics And Kinetics.
💡Key Conceptual Anchor:Metallurgical Thermodynamics And Kinetics
Question 20 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
Reaction between A and B results in an intermediate complex AB* which leads to the final product AB as, A + B ⟶ AB* ⟶ AB. Collision rate theory views the rate as dependent on
Correct Answer: Option A — frequency of breakdown of intermediate product
Detailed Academic Explanation:
frequency of breakdown of intermediate product is the correct answer to “Reaction between A and B results in an intermediate complex AB* which leads to the final product AB as, A + B ⟶ AB* ⟶ AB. Collision rate theory views the rate as dependent on”. This question is standard in competitive examination syllabi for Metallurgical Thermodynamics And Kinetics.
💡Key Conceptual Anchor:Metallurgical Thermodynamics And Kinetics
Question 21 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
The formulae for change in entropy
Correct Answer: Option D — All are correct
Detailed Academic Explanation:
All are correct is the correct answer to “The formulae for change in entropy”. This question is standard in competitive examination syllabi for Metallurgical Thermodynamics And Kinetics.
💡Key Conceptual Anchor:Metallurgical Thermodynamics And Kinetics
Question 22 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
Critical value of the Gibbs energy of nucleation at equilibrium temperature is
Correct Answer: Option B — Infinite
Detailed Academic Explanation:
Infinite is the correct answer to “Critical value of the Gibbs energy of nucleation at equilibrium temperature is”. This question is standard in competitive examination syllabi for Metallurgical Thermodynamics And Kinetics.
💡Key Conceptual Anchor:Metallurgical Thermodynamics And Kinetics
Question 23 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
The constant term "2" in mathematical expression for phase rule is stands for
Correct Answer: Option D — both A and B
Detailed Academic Explanation:
both A and B is the correct answer to “The constant term "2" in mathematical expression for phase rule is stands for”. This question is standard in competitive examination syllabi for Metallurgical Thermodynamics And Kinetics.
💡Key Conceptual Anchor:Metallurgical Thermodynamics And Kinetics
Question 24 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
Consider a reaction with activation energy of 8.314 kJ/mol that takes place at 300 K. If the reaction rate is to be tripled, the temperature of the reaction should be
Correct Answer: Option B — 447.5 K
Detailed Academic Explanation:
447.5 K is the correct answer to “Consider a reaction with activation energy of 8.314 kJ/mol that takes place at 300 K. If the reaction rate is to be tripled, the temperature of the reaction should be”. This question is standard in competitive examination syllabi for Metallurgical Thermodynamics And Kinetics.
💡Key Conceptual Anchor:Metallurgical Thermodynamics And Kinetics
Question 25 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
During the reduction of an oxide of hydrogen gas, it is observed that the rate of reaction increases by almost three fold due to a slight increases in temperature. The most likely rate controlling step is
Correct Answer: Option D — combined mass transfer of H2 and H2O
Detailed Academic Explanation:
combined mass transfer of H2 and H2O is the correct answer to “During the reduction of an oxide of hydrogen gas, it is observed that the rate of reaction increases by almost three fold due to a slight increases in temperature. The most likely rate controlling step is”. This question is standard in competitive examination syllabi for Metallurgical Thermodynamics And Kinetics.
💡Key Conceptual Anchor:Metallurgical Thermodynamics And Kinetics
Question 26 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
The chemical potential of a component 1 in a solution is given by, μi =
Correct Answer: Option D — $$\left( {\frac{{\delta G}}{{\delta {n_1}}}} \right)P,V,{n_2},{n_3}...$$
Detailed Academic Explanation:
$$\left( {\frac{{\delta G}}{{\delta {n_1}}}} \right)P,V,{n_2},{n_3}...$$ is the correct answer to “The chemical potential of a component 1 in a solution is given by, μi =”. This question is standard in competitive examination syllabi for Metallurgical Thermodynamics And Kinetics.
💡Key Conceptual Anchor:Metallurgical Thermodynamics And Kinetics
Question 27 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
Isothermal compressibility of a material is given by
$$ - \frac{1}{V}{\left( {\frac{{\partial V}}{{\partial P}}} \right)_T}$$ is the correct answer to “Isothermal compressibility of a material is given by”. This question is standard in competitive examination syllabi for Metallurgical Thermodynamics And Kinetics.
💡Key Conceptual Anchor:Metallurgical Thermodynamics And Kinetics
Question 28 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
The process in which all small steps are in equilibrium, so that entire process is in equilibrium is
Correct Answer: Option C — quasi-static process
Detailed Academic Explanation:
quasi-static process is the correct answer to “The process in which all small steps are in equilibrium, so that entire process is in equilibrium is”. This question is standard in competitive examination syllabi for Metallurgical Thermodynamics And Kinetics.
💡Key Conceptual Anchor:Metallurgical Thermodynamics And Kinetics
Question 29 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
pure nickel melts at 1726K at 10 5 Pa (1 atm.) pressure, ΔV (solid-liquid) = 0.26 × 10 -6 m 3 . mol -1 and the heat of fusion, ΔHf = 18000 J.mol -1 . The melting point of pure nickel, when acted upon by a pressure of 10 8 Pa is . . . . . . . . K.
Correct Answer: Option B — 1723.5
Detailed Academic Explanation:
1723.5 is the correct answer to “pure nickel melts at 1726K at 10 5 Pa (1 atm.) pressure, ΔV (solid-liquid) = 0.26 × 10 -6 m 3 . mol -1 and the heat of fusion, ΔHf = 18000 J.mol -1 . The melting point of pure nickel, when acted upon by a pressure of 10 8 Pa is . . . . . . . . K.”. This question is standard in competitive examination syllabi for Metallurgical Thermodynamics And Kinetics.
💡Key Conceptual Anchor:Metallurgical Thermodynamics And Kinetics
Question 30 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
What will be the degree of freedom for single phase system of binary system?
Correct Answer: Option B — 2
Detailed Academic Explanation:
2 is the correct answer to “What will be the degree of freedom for single phase system of binary system”. This question is standard in competitive examination syllabi for Metallurgical Thermodynamics And Kinetics.
💡Key Conceptual Anchor:Metallurgical Thermodynamics And Kinetics
Question 31 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
Which one of the following is not an intensive property?
Correct Answer: Option C — volume
Detailed Academic Explanation:
volume is the correct answer to “Which one of the following is not an intensive property”. This question is standard in competitive examination syllabi for Metallurgical Thermodynamics And Kinetics.
💡Key Conceptual Anchor:Metallurgical Thermodynamics And Kinetics
Question 32 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
Activation energy of a chemical reaction, homogeneous or heterogeneous, is graphically estimated from a plot between
Correct Answer: Option D — In k versus $$\frac{1}{T}$$
Detailed Academic Explanation:
In k versus $$\frac{1}{T}$$ is the correct answer to “Activation energy of a chemical reaction, homogeneous or heterogeneous, is graphically estimated from a plot between”. This question is standard in competitive examination syllabi for Metallurgical Thermodynamics And Kinetics.
💡Key Conceptual Anchor:Metallurgical Thermodynamics And Kinetics
Question 33 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
2cu(s) + 0.5 O 2 (g) = Cu 2 O(s); ΔG° = - 162200 + 69.24 T, J 2Cu(l) + 0.5 O 2 (g) = CU 2 O(s); ΔG° = - 1188300 + 88.48 T, J The molar free energy/change at 1300 K for the transformation of solid Cu to liquid Cu will be . . . . . . . . J.
Correct Answer: Option C — 544
Detailed Academic Explanation:
544 is the correct answer to “2cu(s) + 0.5 O 2 (g) = Cu 2 O(s); ΔG° = - 162200 + 69.24 T, J 2Cu(l) + 0.5 O 2 (g) = CU 2 O(s); ΔG° = - 1188300 + 88.48 T, J The molar free energy/change at 1300 K for the transformation of solid Cu to liquid Cu will be . . . . . . . . J.”. This question is standard in competitive examination syllabi for Metallurgical Thermodynamics And Kinetics.
💡Key Conceptual Anchor:Metallurgical Thermodynamics And Kinetics
Question 34 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
Which of the following partial derivative is equal to $${\left( {\frac{{\partial S}}{{\partial P}}} \right)_T}?$$
$$ - {\left( {\frac{{\partial V}}{{\partial T}}} \right)_P}$$ is the correct answer to “Which of the following partial derivative is equal to $${\left( {\frac{{\partial S}}{{\partial P}}} \right)_T}?$$”. This question is standard in competitive examination syllabi for Metallurgical Thermodynamics And Kinetics.
💡Key Conceptual Anchor:Metallurgical Thermodynamics And Kinetics
Question 35 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
Heat capacity at constant pressure is defined by
Correct Answer: Option C — $${\left( {\frac{{\partial H}}{{\partial T}}} \right)_p}$$
Detailed Academic Explanation:
$${\left( {\frac{{\partial H}}{{\partial T}}} \right)_p}$$ is the correct answer to “Heat capacity at constant pressure is defined by”. This question is standard in competitive examination syllabi for Metallurgical Thermodynamics And Kinetics.
💡Key Conceptual Anchor:Metallurgical Thermodynamics And Kinetics
Question 36 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
In thermodynamics, the law of conservation of energy is expressed in the form of
Correct Answer: Option B — first law of thermodynamics
Detailed Academic Explanation:
first law of thermodynamics is the correct answer to “In thermodynamics, the law of conservation of energy is expressed in the form of”. This question is standard in competitive examination syllabi for Metallurgical Thermodynamics And Kinetics.
💡Key Conceptual Anchor:Metallurgical Thermodynamics And Kinetics
Question 37 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
Which of the following phase will have lowest entropy
Correct Answer: Option A — solid
Detailed Academic Explanation:
solid is the correct answer to “Which of the following phase will have lowest entropy”. This question is standard in competitive examination syllabi for Metallurgical Thermodynamics And Kinetics.
💡Key Conceptual Anchor:Metallurgical Thermodynamics And Kinetics
Question 38 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
The first law of thermodynamics can be stated as:
Correct Answer: Option A — dE = δQ - δW
Detailed Academic Explanation:
dE = δQ - δW is the correct answer to “The first law of thermodynamics can be stated as:”. This question is standard in competitive examination syllabi for Metallurgical Thermodynamics And Kinetics.
💡Key Conceptual Anchor:Metallurgical Thermodynamics And Kinetics
Question 39 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
Integral enthalpy of mixing (in J/mol) of liquid (Cu, Zn) solution can be approximated by ΔH m mix = - 19250 X Cu X Zn The corresponding partial molar enthalpy of mixing (in J/mol) for Cu is
Correct Answer: Option D — - 19250 x2zn
Detailed Academic Explanation:
- 19250 x2zn is the correct answer to “Integral enthalpy of mixing (in J/mol) of liquid (Cu, Zn) solution can be approximated by ΔH m mix = - 19250 X Cu X Zn The corresponding partial molar enthalpy of mixing (in J/mol) for Cu is”. This question is standard in competitive examination syllabi for Metallurgical Thermodynamics And Kinetics.
💡Key Conceptual Anchor:Metallurgical Thermodynamics And Kinetics
Question 40 of 40
★ PAEC
★ NESCOM
★ PAEC
★ NESCOM
The sequence of processes which return back to its initial point is called
Correct Answer: Option C — cyclic process
Detailed Academic Explanation:
cyclic process is the correct answer to “The sequence of processes which return back to its initial point is called”. This question is standard in competitive examination syllabi for Metallurgical Thermodynamics And Kinetics.
💡Key Conceptual Anchor:Metallurgical Thermodynamics And Kinetics
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💡Examiner's Advisory for Metallurgical Engineering
In competitive commission exams (FPSC, PPSC, KPPSC, SPSC, CSS, PMS), questions are frequently phrased with distractor options designed to catch rote learners. Whenever you answer an MCQ, read the Detailed Academic Explanation and Conceptual Anchor to understand why the remaining options are incorrect.