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I know their mathematical definitions and how these terms are interrelated (mathematically) but I fail to understand the physical meaning of none but one which is INTERNAL ENERGY .

It seems implausible to me that these are just mathematical terms that serve the purpose that

  1. If $T, V, N$ are known

we use $F=F(T, V, N) $ where $F$: Free Energy or Helmholtz Free Energy

  1. If $T, P, N$ are known

we use $G=G(T, P, N)$ where $G$: Gibbs Free Energy

  1. If $S, P, N$ are known

we use $H=H(S, P, N)$ where $H$: Enthalpy

and that's all. They have no physical significance?

What I know of $U$(Internal Energy) is that it is a measure of kinetic energy of system molecules and hence also the system temperature. The more the molecular K. E., more heat energy produced due to molecular collisions and hence more the temperature.

I am expecting similar physical explanations to other thermodynamic variables which I couldn't find even on other stack exchange threads!

Mechanic
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    Consider defining variable names. –  Oct 27 '18 at 12:53
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    The fundamental quantities are U and S. The others are just convenient parameters to work with in solving many different types of problems. – Chet Miller Oct 27 '18 at 12:54
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    Each part of this question has been asked previously on Physics SE (several times) for example https://physics.stackexchange.com/q/296741/ and https://physics.stackexchange.com/q/362075/ for the physical significance of enthalpy or here https://physics.stackexchange.com/q/149493/ for the Gibbs free energy – By Symmetry Oct 27 '18 at 13:58

2 Answers2

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Formally, those quantities -- Helmholtz free energy $F(T, V, N)$, Gibbs free energy $G(P, T, N)$, and enthalpy $H(P, V, N)$ -- are Legendre transforms of the internal energy $U(S, V, N)$. This serves to the purpose of treating a thermodynamic system considering different quantities as the independent variables, as you have mentioned. The physical significance of those thermodynamic potentials, then, has to do with the various sorts of constraints one can put on a system, and how these constraints influence the equilibrium state.

First, let's consider the one you take as the most physically meaningful, the internal energy. When a system is thermally insulated -- that is, cannot exchange heat with its surroundings, and so loosely speaking its entropy should stay constant -- we know the equilibrium state attained by the system is the state of minimum internal energy (that claim is equivalent to the most widespread one that says that in an isolated system, which can't exchange energy of any kind with its surroundings, the equilibrium state maximizes the entropy). So the internal energy really does seem to be what one expects it to be: it is the quantity that is extremized by a system which cannot exchange heat with its surroundings, just like in classical mechanics the potential energy is extremized in the equilibrium state. The claim that the internal energy is a measure of kinetic energy and thus temperature is oversimplified, and really applies only to the classical ideal gas, where you neglect all interactions between the constituents of the gas. In a more general description, essentially the internal energy is really the total energy (accounting for kinetic and potential energy) stored in the system.

Now, imagine that instead of leaving the system thermally insulated -- that is, keeping its entropy constant --, you put it in contact with a thermal bath that holds its temperature fixed. Now, neither internal energy nor entropy are held constant in the system; the system will then tend to reach a physical state in which it extremizes a quantity that brings together both the entropy and the internal energy, and this happens to be exactly the Helmholtz free energy.

The Helmholtz free energy may also be interpreted as a measure of the "useful" energy that can be extracted from a system at constant temperature -- the work you can extract from a system at constant temperature can only be less than or equal to (minus) the change in the Helmholtz free energy. You can then try and interpret the formula $F = U - TS$ as an expression of the total energy of the system ($U$), minus the "useless" energy stored in a disordered fashion contributing to the entropy ($TS$). The Helmholtz free energy plays a central role in the canonical ensemble, and also in Landau's theory of second order phase transitions; you may want to take a look at these topics too.

Imagine now that you put your system in contact with a pressure reservoir (the analogous of a thermal bath, if you want to keep the pressure in your system constant). Essentially what that means is that the system now can exchange mechanical work with its surroundings, and that has the effect of influencing how the systems reaches its equilibrium state. Grossly speaking, in that case what you want to extremize is the internal energy of the system, plus the amount of work that had to be done against the pressure reservoir in order to put the system into that configuration, and that is exactly what is accounted for in the enthalpy $H = U + PV$; if you want a quicker explanation, check the answer mentioned above What exactly is enthalpy? .

As for the Gibbs free energy, following what has been written above, it is specially useful when the system can exchange both heat and work with its surroundings, which is to be considered now as both a pressure and a temperature reservoir. Now, you are taking into account both of the effects described above: the total internal energy $U$, the "useless" energy lost to disorder of the system $-TS$, and the work done against the pressure reservoir $PV$. I'd really recommend you check the answers in Gibbs free energy intuition for a deeper description of that.

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U, F, G, and H are sometimes referred to as Thermodynamic Potentials. A nice (in my opinion) explanation of the physical significance of these properties in relation to entropy (S) and system work can be found on the Hyperphysics web site under "Thermodynamic Potentials".

Hope this helps.

Bob D
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