- #1

- 325

- 3

## Main Question or Discussion Point

Say we had some solid and it's being heated up towards it's melting point. It reaches that temperature, and then an infinitesimal increase in temperature T will cause the crystal to break apart. It loses its ordered structure and so the entropy S has increased.

If I'm not mistaken, any spontaneous process occurs if it will allow a system to lower its free energy G, ie. reduce the useful energy it has available to do work, where it is defined as

G = U + pV - TS = H - TS, where enthalpy H = U + pV.

similarly the change in free energy is [tex] \Delta G = \Delta U + p\Delta V - T\Delta S [/tex].

So, when the entropy change [tex] \Delta S [/tex] is positive, the [tex] -T\Delta S [/tex] term is negative. That means if [tex]|-T\Delta S| > |\Delta H| [/tex] the change in free energy [tex] \Delta G [/tex] will be negative and melting will indeed happen, right?

My first issue is with this enthalpy term, H. I'm stuggling to picture how it will change. First we have the internal energy U, which is the Coulomb interaction between neighbouring molecules, right? I'm imagining this as an attractive potential, so when the solid melts it increases the separation of its molecules, which must mean an increase in potential energy. Thus U increases, which is a positive contribution to H.

Then we have the pV term. This is the mechanical work a system does on its environment, right? So by melting (if we aren't dealing with water), the volume of the stuff that was initially in the solid state (ie. our "system"?) has increased, which is a further positive increase to [tex] \Delta H [/tex]. It has expanded out, exerting a force on the fluid initially surrounding it (its "environment"?) So, if I'm correct about everything so far, melting involves an increase in enthalpy, right? But it only proceeds if the free energy loss of the [tex] -T\Delta S [/tex] term outweighs it?

This is the opposite of what I keep reading. For example, on wikipedia,

So what's going on here, where am I getting mixed up?

I'm also confused over what happens when going the other way, cooling a liquid down to its melting point, which is defined as the point where the free energies of liquid & solid are the same. So, if we started forming a solid from a liquid, we have decreased the entropy of the molecules involved as they have taken on a more ordered arrangement. This negative increase in entropy would give a positive [tex] \Delta G[/tex], so why should this process ever occur?

If I'm not mistaken, any spontaneous process occurs if it will allow a system to lower its free energy G, ie. reduce the useful energy it has available to do work, where it is defined as

G = U + pV - TS = H - TS, where enthalpy H = U + pV.

similarly the change in free energy is [tex] \Delta G = \Delta U + p\Delta V - T\Delta S [/tex].

So, when the entropy change [tex] \Delta S [/tex] is positive, the [tex] -T\Delta S [/tex] term is negative. That means if [tex]|-T\Delta S| > |\Delta H| [/tex] the change in free energy [tex] \Delta G [/tex] will be negative and melting will indeed happen, right?

My first issue is with this enthalpy term, H. I'm stuggling to picture how it will change. First we have the internal energy U, which is the Coulomb interaction between neighbouring molecules, right? I'm imagining this as an attractive potential, so when the solid melts it increases the separation of its molecules, which must mean an increase in potential energy. Thus U increases, which is a positive contribution to H.

Then we have the pV term. This is the mechanical work a system does on its environment, right? So by melting (if we aren't dealing with water), the volume of the stuff that was initially in the solid state (ie. our "system"?) has increased, which is a further positive increase to [tex] \Delta H [/tex]. It has expanded out, exerting a force on the fluid initially surrounding it (its "environment"?) So, if I'm correct about everything so far, melting involves an increase in enthalpy, right? But it only proceeds if the free energy loss of the [tex] -T\Delta S [/tex] term outweighs it?

This is the opposite of what I keep reading. For example, on wikipedia,

*"...melting occurs because the entropy, S, gain in your system by spatial randomization of the molecules has overcome the enthalpy, H,***loss**due to breaking the crystal packing forces."So what's going on here, where am I getting mixed up?

I'm also confused over what happens when going the other way, cooling a liquid down to its melting point, which is defined as the point where the free energies of liquid & solid are the same. So, if we started forming a solid from a liquid, we have decreased the entropy of the molecules involved as they have taken on a more ordered arrangement. This negative increase in entropy would give a positive [tex] \Delta G[/tex], so why should this process ever occur?

Last edited: