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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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Question 1 of 40
★ PAEC ★ NESCOM

The weight % of nitrogen (N 2 ) in liquid iron can be expressed as:

Question 2 of 40
★ PAEC ★ NESCOM

The ratio of fugacity of the substance to the fugacity in its standard state

Question 3 of 40
★ 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

Question 4 of 40
★ PAEC ★ NESCOM

For a spontaneous natural process at constant temperature and pressure, the free energy of the system always

Question 5 of 40
★ PAEC ★ NESCOM

The rate of reaction for first order reaction

Question 6 of 40
★ 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?

Question 7 of 40
★ 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

Question 8 of 40
★ PAEC ★ NESCOM

G = U + PV - TS. Then which one of the following is correct?

Question 9 of 40
★ 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)

Question 10 of 40
★ PAEC ★ NESCOM

When there is existence of uniform temperature throughout the system is called

Question 11 of 40
★ PAEC ★ NESCOM

The Daltons law of partial pressure can be written as

Question 12 of 40
★ PAEC ★ NESCOM

When two phases α and β in an alloy are in thermodynamic equilibrium, then

Question 13 of 40
★ 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

Question 14 of 40
★ 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.

Question 15 of 40
★ PAEC ★ NESCOM

For an ideal gas, Cp - Cv is

Question 16 of 40
★ 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

Question 17 of 40
★ PAEC ★ NESCOM

Uphill diffusion means diffusion from

Question 18 of 40
★ PAEC ★ NESCOM

Select the correct plot of Gibbs free energy (G) Vs. Temperature (T) for a single component system from the following:

Question 19 of 40
★ PAEC ★ NESCOM

The part of the universe excluding the system is known as

Question 20 of 40
★ 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

Question 21 of 40
★ PAEC ★ NESCOM

The formulae for change in entropy

Question 22 of 40
★ PAEC ★ NESCOM

Critical value of the Gibbs energy of nucleation at equilibrium temperature is

Question 23 of 40
★ PAEC ★ NESCOM

The constant term "2" in mathematical expression for phase rule is stands for

Question 24 of 40
★ 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

Question 25 of 40
★ 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

Question 26 of 40
★ PAEC ★ NESCOM

The chemical potential of a component 1 in a solution is given by, μi =

Question 27 of 40
★ PAEC ★ NESCOM

Isothermal compressibility of a material is given by

Question 28 of 40
★ PAEC ★ NESCOM

The process in which all small steps are in equilibrium, so that entire process is in equilibrium is

Question 29 of 40
★ 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.

Question 30 of 40
★ PAEC ★ NESCOM

What will be the degree of freedom for single phase system of binary system?

Question 31 of 40
★ PAEC ★ NESCOM

Which one of the following is not an intensive property?

Question 32 of 40
★ PAEC ★ NESCOM

Activation energy of a chemical reaction, homogeneous or heterogeneous, is graphically estimated from a plot between

Question 33 of 40
★ 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.

Question 34 of 40
★ PAEC ★ NESCOM

Which of the following partial derivative is equal to $${\left( {\frac{{\partial S}}{{\partial P}}} \right)_T}?$$

Question 35 of 40
★ PAEC ★ NESCOM

Heat capacity at constant pressure is defined by

Question 36 of 40
★ PAEC ★ NESCOM

In thermodynamics, the law of conservation of energy is expressed in the form of

Question 37 of 40
★ PAEC ★ NESCOM

Which of the following phase will have lowest entropy

Question 38 of 40
★ PAEC ★ NESCOM

The first law of thermodynamics can be stated as:

Question 39 of 40
★ 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

Question 40 of 40
★ PAEC ★ NESCOM

The sequence of processes which return back to its initial point is called

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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.

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