
#1
Dec3111, 08:35 AM

P: 90

Hi all
Why does a molten metal solidify with decrease in temperature. or in other words why is a closed packed structure more stable than an amorphous liquid melt. I have read that the free energy below melting point is lesser for solid state than that of for liquid state at the same temperature.But the question that is troubling me is why is free energy in solid state lesser? 



#2
Jan112, 05:50 PM

Admin
P: 21,634

Liquid to solid. Most elements do this, but at different temperatures.
Solids tend to form along crystalline planes. To make amorphous materials, particularly in metals, usually requires rapid solidification. Metal oxides would more readily form amorphous compounds, particularly where different cation species are involved. 



#3
Jan212, 02:38 AM

P: 90

But the question is exactly that. Why does it solidify at certain temperatures?
Cant it remain in liquid phase below melting point? 



#4
Jan212, 11:59 AM

P: 128

Hi allWhy does a molten metal solidify with decrease in
Supercooling liquids is possible. Water can be cooled to about 253K without it freezing but it has to be clean water free of impurities. A liquid freezes becuase it nucleates at a point. Nucleation occurs due to impurities, surface texture of container e.t.c and it becomes energetically favourable to change state.
A good treatment of this is in "Phase Transformations of Metals and Alloys" by Porter and Eastling. 



#5
Jan212, 12:14 PM

P: 128

Page 10 of this book desribe this.
"In dealing with pahse transformations we are often concerned with the difference in free energy between two phases at temperatures away from the equilibrium temperature. For example, if a liquid is undercooled by a ΔT below T[itex]_{m}[/itex] beofre it solidifies, solidification will be accompanied by a decrease in free energy [itex]\Delta[/itex]G. This free energy decrease provised the driving force for solidification" At temp T [itex]\Delta[/itex]G = [itex]\Delta[/itex]H T[itex]\Delta[/itex]S [itex]\Delta[/itex]H = H[itex]^{L}[/itex]  H[itex]^{S}[/itex] [itex]\Delta[/itex]S = S[itex]^{L}[/itex]  S[itex]^{S}[/itex] At eqilibrium melting temperature T[itex]_{m}[/itex] the free energies of the liquid and solid are the same. So [itex]\Delta[/itex]G = 0 [itex]\Delta[/itex]S = L/T[itex]_{m}[/itex] If the liquids is supercooled by even a small ammount then it become energically favourable for the liquid to become a solid and realese latent heat. 



#6
Jan212, 12:37 PM

P: 128

Page 187 goes into more detail.
When a particle of solid nucleates it's free energy will depend on its size [itex]\Delta[/itex]G = [itex]\frac{4}{3}[/itex][itex]\pi[/itex]r[itex]^{3}[/itex][itex]\Delta[/itex]G[itex]_{v}[/itex] + 4[itex]\pi[/itex]r[itex]^{2}[/itex][itex]\gamma[/itex][itex]_{sl}[/itex] Subscript r and v represent radius and volume terms and [itex]\gamma[/itex] is surface energy of the interface. There is a critical radius r[itex]^{*}[/itex]. If r < r[itex]^{*}[/itex] then the system can lower it free energy dissolution of the solid. If r > r[itex]^{*}[/itex] then the system can lower its free energy by growing the solid. [itex]\Delta[/itex]G[itex]_{v}[/itex] = L[itex]_{v}[/itex]ΔT/T[itex]_{m}[/itex] [itex]\Delta[/itex]T is the undercooling temperature. r[itex]^{*}[/itex] = (2[itex]\gamma^{3}_{SL}[/itex]T[itex]^{2}_{m}/L_{v})(\frac{1}{ΔT^{2}}[/itex]) So if you could measure this critical radius you could determine the melting point for the undercooling required. 



#7
Jan312, 10:58 AM

P: 176

If you agree that the formation of a chemical bond tends to lower free energy (leaving aside the details of enthalpy and entropy), then a state which maximizes the bond density will tend to be the most stable. Solids are more close packed than liquids (i.e., they have a higher bond density), thus they tend to represent a lower energy state.
Solids melt when the thermal energy of its constituent atoms/molecules overcome the bond energy. (This is why, near the melting point of most materials, the product of the coefficient of thermal expansion and absolute temperature gives a strain on the order of ten percent  much more and the solid is no longer stable  the bond length becomes too great!) Once you understand this, a more interesting question is why does the solid assume the crystal symmetry that it does? (i.e., why aren't all solids close packed) But I suspect this is a little beyond what you're looking for. 



#8
Jan312, 11:00 PM

P: 90

I agree that free energy decreases when a chemical bond is formed. Is it due to release of heat of fusion and decrease in entropy? In what form was the excess energy (heat of fusion) stored in liquid state?
The question which uby has mentioned was troubling me from some time. I have read that all metals are not FCC/HCP because it depends on electronic structure of atoms. But I am not fully satisfied. 



#9
Jan412, 01:02 PM

P: 176

pukb: Entire theories of bonding have been created whose primary purpose are to explain the observed structures and properties of materials. Solid state physics, physical and molecular chemistry, etc. Unfortunately, there isn't a very concise explanation and you will need to endure a lot of reading to fully grasp these theories. I myself am not fully versed in anything but the most introductory aspects of most of these theories.
A couple quick points: generally, the entropic contribution to the total free energy of condensed phases is small compared to the enthalpic contribution  this is not to be confused with the contribution to the free energy of solution/mixing/etc., where there can often be zero enthalpic contribution. Most liquids exhibit shortrange order (on the molecular length scale) but lack longrange order (in the order of molecules) unlike solids. This lack of longrange order is manifested in a lower bond density. The greater the bond density, the lower the overall free energy. For example, [SiO4] tetrahedra in a silicate liquid are mostly cornershared with each other (allowing for many degrees of freedom and an increased free volume) while in the solid they can exhibit much more rigid bonds along faces and/or edges depending on the crystal structure. 



#10
Jan1112, 07:58 PM

P: 7

As far as solidification is concern, it's an energetically driven phenomenon that, in the most refined way known, is detailed above by bm0p700f from the exact textbook that I, too, favor. 


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