Thermodynamics: Free Energy Confusion

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Homework Statement

Hi all, I'm having quite a big problem trying to understand the concept of enthalpy and free energy. I feel that a good way to sort this out is to write out what I understand about these things and have people correct me. Thanks in advance for any assistance.

I'm learning thermodynamics from Schroeder's text.

Enthalpy, $H = U + PV$

At constant pressure, Enthalpy is the total energy required to create a system with internal energy $U$out of nothing. This total energy is the sum of the system's internal energy and the compression work needed to create space for the system.

Enthalpy change, $\Delta H = \Delta U + \Delta PV = \Delta Q + \Delta W_{other}$ where $\Delta U = \Delta Q + \Delta W_{other}$

At constant pressure, $\Delta H$ is the amount of work needed to increase the internal energy of a system while doing more compression work to create even more space for it.
It also says that I can only retrieve energy in the form of heat and non-compression work by drawing from a system's enthalpy.

Gibbs' Free Energy $G = U - TS + PV$

At constant pressure and temperature, $G$ is the total energy required to create a system with internal energy $U$ out of nothing. This total energy is the sum of the system's internal energy and the compression work needed to create space for the system. However, it also takes into account the amount of heat $TS_{final}$we can draw from the environment while building the system.

Gibbs' Free Energy Change$\Delta G = \Delta U - T\Delta S + P\Delta V = \Delta W_{other} + \Delta Q - T\Delta S$

At constant pressure and temperature, $\Delta G$ represents the amount of energy needed to increase a system's internal energy while doing compression work on its surroundings. However, we can offset the environment's contribution of heat by deducting $T\Delta S$. It also shows that we can draw $\Delta W_{other}$ at most (quasistatic change), by decreasing a system's $G$ since $\Delta Q \leqslant T\Delta S$.

Question: Why is $\Delta Q \leqslant T\Delta S$ the case?

Helmholtz Free Energy $F = U - TS$

At constant temperature, this represents the amount of energy needed to create a system with internal energy $U$, while offsetting environmental heat contributions $TS_{final}$.

Question: Why does this not include compression work $PV$? While I understand that pressure is not taken to be constant, do we not need to do work to create space for the system?

Helmholtz Free Energy Change $\Delta F = \Delta U - T \Delta S = \Delta Q + \Delta W_{all} - T\Delta S$

This tells us the amount of work needed to increase a system's internal energy while taking into account environmental contributions of heat $T\Delta S$. It also tells us that should we draw from a system's $F$, we can obtain energy in the form of $\Delta W_{all}$ at most (quasistatic changes). $\Delta Q \leqslant T\Delta S$

Question: Again, not too sure why $\Delta Q \leqslant T\Delta S$ and why no compression-work term is included.

Please feel free to add on points that are important and of course, correct any misconceptions.

Thanks very much in advance!

Homework Equations

The Attempt at a Solution

 

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Would anyone be so kind as to assist? Thank you!