Helmholtz Free Energy

  Рет қаралды 26,316

Physical Chemistry

Physical Chemistry

Күн бұрын

Пікірлер: 12
@kina4288
@kina4288 9 ай бұрын
You are always so clear with the presumptions (T and V in this case) which, when not explained prior to arriving to the solution, makes thermodynamic so confusing to student. 🎉
@h7opolo
@h7opolo Жыл бұрын
you're a good teacher. i like your stoic style. subbed.
@PhysicalChemistry
@PhysicalChemistry Жыл бұрын
A stoic thanks to you
@tajrinNoor
@tajrinNoor 7 ай бұрын
Great work👏👏👏👏
@madhavdua1246
@madhavdua1246 2 жыл бұрын
Thanks Sir 😇🙏
@PhysicalChemistry
@PhysicalChemistry 2 жыл бұрын
You're quite welcome
@dragonfly9786
@dragonfly9786 3 жыл бұрын
sir, gibbs free energy and likewise in helmholtz free energy, why are they not called simply gibbs energy and helmholtz energy becasue dimensionally they ARE energy. what is so 'free' here?
@PhysicalChemistry
@PhysicalChemistry 3 жыл бұрын
The more modern terminology is indeed Gibbs energy and Helmholtz energy. The "free energy" terms are older, and slightly out of fashion, but still very common. The idea is that the "free" energy is the energy that is free (available) to us to do useful work. If some of the energy change must be spent on decreasing the entropy, the Gibbs or Helmholtz energies tell us how much energy will be left over after paying this entropy tax. It's not free as in "free beer", but free as in "free on Friday night": available or not already committed.
@hyperduality2838
@hyperduality2838 2 жыл бұрын
@@PhysicalChemistry Making predictions is syntropic process -- teleological. Increasing the number of dimensions or states is an entropic process -- co-homology. Decreasing the number of dimensions or states is a syntropic process -- homology. Homology is dual to co-homology. Syntropy is dual to entropy -- the 4th law of thermodynamics. From a converging, convex or syntropic perspective everything divergent, concave or entropic -- the 2nd law of thermodynamics. Time for a large beer.
@GALADIUM
@GALADIUM 3 жыл бұрын
Sir, how does the - ∆(TS) = -T∆S - S∆T ?
@PhysicalChemistry
@PhysicalChemistry 3 жыл бұрын
Via the product rule. When you're taking a derivative, the product rule tells you that d/dx(f·g) = (df/dx)·g + f·(dg/dx). It works the same way when you're taking a differential (as in the video) or a finite change Δ (as in your question)
@GALADIUM
@GALADIUM 3 жыл бұрын
I understand Sir. Thank you so much
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