This set of Phase Transformation Multiple Choice Questions & Answers (MCQs) focuses on “Solidification – Nucleation in Pure Metal”.

1. Consider a given volume of liquid at a temperature T below Tm with a free energy G’. If some of the atoms of the liquid cluster together to form a small sphere of solid, the free energy of the system will change to G, given by________ (Assume V1 volume of the solid sphere, V2 volume of liquid, A is the solid/liquid interfacial area, γ solid/liquid interface free energy, G1 and G2 are the free energies per unit volume of solid and liquid respectively).

a) V1G1 + V2G2 + Aγ

b) V1G2 + V2G1 + Aγ

c) V1G1 + V2G2 – Aγ

d) V1G2 + V2G1 – Aγ

View Answer

Explanation: If some of the atoms of the liquid cluster together to form a small sphere of solid, the free energy of the system will change to G2 is given by the formula V1G1 + V2G2 + Aγ and this is a part of homogeneous nucleation and the formation of the solid results in the change in the free energy which is given by ΔG = G1-G2.

2. If the volume of the solid sphere and the volume of the liquid is given as V1 and V2 respectively, G1 and G2 are the free energies per unit volume of solid and liquid respectively. The free energy of the system without any solid present is given by_______

a) G = (V1+V2)G2

b) G = V1G1 + V2G2

c) G = V1G2 + V2G1

d) G = (V1+V2)G1

View Answer

Explanation: The free energy of the system without any solid present is given by the formula G = (V1+V2)*G2 and hence the formation of solid will directly lead to a change in free energy, G*-G where the value of G* is given as V1G1 + V2G2 + Aγ.

3. If the latent heat of fusion is given as L and the melting temperature of that substance is given as x. For an undercooling ΔT, the change in free energy is given as _________

a) ΔG = (L*ΔT)/x

b) ΔG = L*ΔT * x

c) ΔG = L*x/ΔT

d) ΔG = L/ (ΔT*x)

View Answer

Explanation: The change in free energy is given as ΔG = (L*ΔT)/x. Below Tm, ΔGv is positive so that the free energy change associated with the formation of a small volume of solid has a negative contribution due to the lower free energy of a bulk solid, but there is also a positive contribution due to the creation of a solid/liquid interface.

4. Let’s take the equilibrium melting temperature of a fluid ‘A’ as Tm and which is cooled below its equilibrium melting temperature (Tm) then there is always possibility of a driving force F for solidification and it might be expected that the liquid phase of ‘A’ would spontaneously solidify.

a) True

b) False

View Answer

Explanation: The above mentioned statement is not always true there are some exceptions. Let’s take the case of nickel, under suitable conditions nickel can be undercooled (or super cooled) to 250 K below Tm (1453 degree Celsius) and held there indefinitely without any transformation occurring.

5. The free energy change associated with homogeneous nucleation of a sphere (assume the radius as 5m) has an interfacial energy term and a volume free energy. Calculate the value of the interfacial term if the interfacial energy is given as (1/п) kJ/m^{2}Kmol?

a) 1000 kJ/Kmol

b) 10 kJ/Kmol

c) 1 kJ/Kmol

d) 100 kJ/Kmol

View Answer

Explanation: The interfacial energy term is directly proportional to the square of the radius and the related equation is the product of the surface area and the interfacial energy and is given as 4пr

^{2}γ.

6. Calculate the volume free energy associated with homogenous nucleation of sphere if the radius of the sphere is given as 5m and the ΔGv is given as (3/п) kJ/m^{3}Kmol?

a) 500 kJ/Kmol

b) 50 kJ/Kmol

c) 5 kJ/Kmol

d) 1000 kJ/Kmol

View Answer

Explanation: Here when it comes to the interface term is directly proportional to r

^{2}and the volume free energy is directly proportional to r

^{3}here in order to calculate the volume free energy we require the value of r and ΔGv and the related equation is given as 4пr

^{3}ΔGv.

7. For a given undercooling there is a certain radius r*, which is associated with a maximum excess free energy. If r< r* the system can lower its free energy by_____

a) Creating a void

b) Maintaining its volume

c) Transferring liquid

d) Dissolution of solid

View Answer

Explanation: Actually, the interface term is directly proportional to r

^{2}and the volume free energy is directly proportional to r

^{3}and hence the creation of small particles of solid always leads to a free energy increase. It is this increase that is able to maintain the liquid phase in a metastable state almost indefinitely at temperatures below Tm. If r< r* the system can lower its free energy by dissolution of solid.

8. Calculate the critical radius if the ΔGv and γ values are given as 5 kJ/m^{3}Kmol and 2.5 kJ/m^{2}Kmol respectively? (Homogeneous nucleation)

a) 4m

b) 2m

c) 1m

d) 0.5m

View Answer

Explanation: Critical radius can be calculate if we know the values of interfacial energy and ΔGv and the related equation is as 2γ/ΔGv and when we substitute the respective values we get r* as 2*2.5/5=1m.

9. Calculate the ΔG’ (critical) if the ΔGv and γ values are given as √Π kJ/m^{3}Kmol and 3kJ/m^{2}Kmol respectively? (Homogenous nucleation)

a) 144 kJ/Kmol

b) 81 kJ/Kmol

c) 125 kJ/Kmol

d) 16 kJ/Kmol

View Answer

Explanation: Here in this case the ΔG’ (critical) can be calculated using the formula ΔG’ (critical) = 16Πγ

^{3}/3(ΔGv)

^{2}. Here if we substitute the respective value we get the required solution and its vlue is given as 144 kJ/Kmol.

10. Calculate the critical radius, if the interface energy is given as 3kJ/m^{2}Kmol and the undercooling is given as 300K (ΔT)? (Assume the latent heat of fusion per unit volume as L and the melting point temperature (Tm) as 150K).

a) 3/L

b) 6/L

c) 6L

d) 3L

View Answer

Explanation: Here in this case the critical radius using the formula (2γ/L)*(Tm/ΔT), where γ denotes the interface so if the substitute the respective values in the formula we get, (2*3/L)*(0.5) =3/L.

11. If the liquid contains x atoms per unit volume, calculate the number of clusters that have reached the critical size (Y), if the temperature is taken at T (K) and the critical value of change in free energy at homogenous nucleation is given as ΔG’?

a) Y = x*exp(-ΔG’/RT)

b) Y = x*exp(-ΔG’/kT)

c) Y = x*exp(ΔG’/RT)

d) Y = x*exp(ΔG’/kT)

View Answer

Explanation: Here the number of clusters the reach the critical size is dependent on the temperature and the critical value of free energy and the related formula is given as Y =x*exp (-ΔG’/kT) where k represents the Boltzmann constant.

12. Calculate the value of ΔGv if the interfacial energy is given as 3kJ/mm^{2}Kmol and the critical radius is given as 1.5mm?

a) 2 kJ/mm^{3}Kmol

b) 4 kJ/mm^{3}Kmol

c) 5 kJ/mm^{3}Kmol

d) 3 kJ/mm^{3}Kmol

View Answer

Explanation: The value of ΔGv can be calculated using the critical radius and the interfacial energy and the related formula is given by twice the interfacial energy divided by the critical radius gives the value of ΔGv and here when we substitute the required value, we get b as the answer.

13. Value of shape factor of heterogeneous nucleation if Θ is given as 60°?

a) 5/32

b) 5/16

c) 5/4

d) 5/8

View Answer

Explanation: The value of shape factor is given by s(Θ) = (2+cosΘ)*(1-cosΘ)

^{2}/4 and hence when we substitute the value of Θ as 60, we get the value of cos(Θ) as 0.5 and further when we substitute the value in the above mentioned equation we get the required solution.

14. In some metallic systems the liquid can be rapidly cooled to temperatures below the so-called glass transition temperature without the formation of crystalline solid.

a) False

b) True

View Answer

Explanation: Since atomic jumps from the liquid on to the cluster are thermally activated, fₒ will in fact diminish with decreasing temperature. In some metallic systems the liquid can be rapidly cooled to temperatures below the so-called glass transition temperature without the formation of crystalline solid. fₒ is very small at these temperatures and the supercooled liquid is a relatively stable metallic glass or amorphous metal.

15. Which among the following are true about the activation barrier of homogeneous and heterogeneous nucleation?

a) ΔG(hom) > ΔG(het)

b) ΔG(hom) << ΔG(het)

c) ΔG(hom) is always a constant but ΔG(het) varies with temperature

d) Nothing can be predicted

View Answer

Explanation: This happens because the activation barrier of heterogeneous nucleation is given as ΔG(het) = ΔG(hom)*s(Θ) and here the extra term is known as the shape factor and is dependent on the wetting angle hence this statement is proved.

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