Gibbs Free Energy: Work in H-TS Explained

AI Thread Summary
Gibbs free energy (G) is defined as G = H - TS, where H represents enthalpy and TS is the temperature-entropy product. When creating a system from a volume of zero at constant pressure and temperature, the work required is represented by G, not just heat (Q). Enthalpy accounts for the total energy needed to establish the system, including work done against atmospheric pressure. The Gibbs free energy allows for some energy input from the environment as heat, which is essential for maintaining constant temperature during the process. Thus, G reflects the net energy needed to create the system while considering external contributions.
iScience
Messages
466
Reaction score
5
the gibbs free energy is defined to be H-TS. In my thermo book, it says that if i were to create a system out from volume V=0 at constant pressure and temperature, the work that i would need to provide is G=H-TS. But for constant pressure i thought the work terms in enthalpy canceled out and were zero and that all that was left was the heat (Q), i thought that's what made the enthalpy quantity useful. why then does the H-TS quantity have work in it? ie, i thought H-TS would just be Q-TS but apparently it's not.. why is this not true?
 
Physics news on Phys.org
You have
$$
G = H - TS = U + PV - TS
$$
Considering that you are creating a system from nothing, enthalpy is the total energy you will have to furnish to create the system and puch away the atmosphere to make place for it. With the Gibbs free energy, you can consider that you do not have to furnish all the energy, as some can come from the environment in the form of heat (assuming constant temperature). So, when creating a system from nothing, ##G## is what you need to supply to create the system at ##T=0## and puch away the atmosphere to make room for it, considering that the environment will take care of bringing it to the right temperature.
 
Thread 'Question about pressure of a liquid'
I am looking at pressure in liquids and I am testing my idea. The vertical tube is 100m, the contraption is filled with water. The vertical tube is very thin(maybe 1mm^2 cross section). The area of the base is ~100m^2. Will he top half be launched in the air if suddenly it cracked?- assuming its light enough. I want to test my idea that if I had a thin long ruber tube that I lifted up, then the pressure at "red lines" will be high and that the $force = pressure * area$ would be massive...
I feel it should be solvable we just need to find a perfect pattern, and there will be a general pattern since the forces acting are based on a single function, so..... you can't actually say it is unsolvable right? Cause imaging 3 bodies actually existed somwhere in this universe then nature isn't gonna wait till we predict it! And yea I have checked in many places that tiny changes cause large changes so it becomes chaos........ but still I just can't accept that it is impossible to solve...
Back
Top