Hello all,
I am confused about the following fact. Does minimization of free energy always mean that entropy is maximized? are they always complimentary to each other or in certain conditions?
I am doubtful because I came across a text which said for isothermal systems ( i.e. temperature doesn't change) these are complimentary, whereas for case of non-isothermal systems, minimization of free energy does not always mean maximization of entropy. May someone help.
Thanks
By my comparison, exceptions occur when ions are formed in solution. The standard Gibbs energy of formation Zn2+ (aq)=- 147.06 KJ / mol.The change in the standard entropy is - 112.1 J / mole K. Formation of order as a result of hydration of ions.
What follows is somewhat simplified.
Consider closed system. Write first law of thermodynamics as
dU/dt = Q - p dV/dt
where t is time, U is internal energy, Q is heat transfer rate, p is pressure at the moving boundary, V is volume.
Write second law as
dS/dt - Q/T = Sgen >= 0
where S is entropy, T is thermodynamic temperature of the boundary (assumed homogeneous along the boundary) and Sgen is entropy generation, which is non-negative, and vanishes in equilibrium.
For adiabatic process at constant volume Q=0, so that first law gives dU/dt = 0 and 2nd law gives dS/dt >= 0. This states that entropy grows (>0) until the final equilibrium state is reached (=0). Hence, entropy grows to a maximum.
Now eliminate Q between both laws, to get
dU/dt - T dS/dt + p dV/dt = - T Sgen
The most commonly used free energies, the Gibbs and Helmholtz free energies, are defined only for isothermal processes (isothermal isobaric for Gibbs free energy, and isothermal isochoric for Helmholtz free energy). The temperature can vary in intermediate states, but at the very least the initial and final temperatures must be equal. In this case the Gibbs or Helmholtz free energy associated with a process has a simple physical interpretation as the maximum work W-yield the process can yield if spontaneous, or the minimum work W-cost required to enable it if nonspontaneous. If a spontaneous process is allowed to occur without doing work, then the entropy of the Universe increases by W-yield/T under isothermal conditions. Under nonisothermal conditions (not even initial and final states at the same temperature), the Helmholtz and Gibbs functions are still state functions, depending only on the current state of a system and not on its history. But they no longer have a simple physical interpretation in terms of work.
@Henning
Thanks for your kind reply.
I would like to clarify the following in order to ensure that I understood the things correctly.
When you write "
dU/dt - T dS/dt + p dV/dt = - T Sgen
what actually has to be minimized or maximized depends on the boundary conditions for which there are many possibilities, constant T at the wall or adiabatic, prescribed pressure or constant volume, or even constant entropy (which is essentially impossible to control). As long as the boundary conditions are such that the system can reach a proper equilibrium state (e.g., all walls should have the same temperature), no matter what the boundary conditions are, the equilibrium states are of same type: for simple substances: homogeneous temperature, homogeneous chemical potential, and zero velocity. For reacting mixtures you also have the law of mass action.
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