Register to reply

Hi allWhy does a molten metal solidify with decrease in

by pukb
Tags: allwhy, decrease, metal, molten, solidify
Share this thread:
pukb
#1
Dec31-11, 08:35 AM
P: 92
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?
Phys.Org News Partner Engineering news on Phys.org
DIY glove-based tutor indicates muscle-memory potential
Tricorder XPRIZE: 10 teams advance in global competition to develop consumer-focused diagnostic device
Study shows local seismic isolation and damping methods provide optimal protection for essential computing equipment
Astronuc
#2
Jan1-12, 05:50 PM
Admin
Astronuc's Avatar
P: 21,880
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.
pukb
#3
Jan2-12, 02:38 AM
P: 92
But the question is exactly that. Why does it solidify at certain temperatures?
Cant it remain in liquid phase below melting point?

bm0p700f
#4
Jan2-12, 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.
bm0p700f
#5
Jan2-12, 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.
bm0p700f
#6
Jan2-12, 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.
uby
#7
Jan3-12, 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.
pukb
#8
Jan3-12, 11:00 PM
P: 92
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.
uby
#9
Jan4-12, 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 short-range order (on the molecular length scale) but lack long-range order (in the order of molecules) unlike solids. This lack of long-range 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 corner-shared 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.
Siress
#10
Jan11-12, 07:58 PM
P: 7
Quote Quote by Astronuc View Post
METALS tend to form along crystalline planes.

To make amorphous materials, particularly in metals, usually requires rapid solidification.
Fixed that for you. Be aware, however, that this is misleading. Crystallinity is merely a subset of a few types of solids; or more rigorously defined as a description of a microstructure with long-range order (theoretically infinite LRO). For instance, there exist crystalline metals, polymers, ceramics, et al.

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.

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?
Heat of fusion, or more generally just latent heat, is (assuming you know the definition already here) actually "hidden." It's not used in the chemical bonding, but is rather a form of potential energy associated with separating atoms.


Register to reply

Related Discussions
Chitosan Solidification Chemistry 0
Energy resultant of Solidification Point Of Ethanol Introductory Physics Homework 1
Solidification of H2O due to pressure Atomic, Solid State, Comp. Physics 0
Phase field simulation of alloy solidification ? Materials & Chemical Engineering 0