Physics, asked by srinivasrti4295, 11 months ago

Thermodynamic potentials without unconstrained variables?

Answers

Answered by arbabali12
0
Principle of minimum energy in thermodynamics states

(∂U∂V)s=0(∂U∂V)s=0

at equilibrium U(S,V)U(S,V) with fixed entropy is at its extremum for the unconstrained variable V.V.

But what if VV is fixed as well? The situations like this occurred while I was dealing with the other thermodynamics potentials such as F(T,V)F(T,V) or H(S,p).H(S,p). F(T,V)F(T,V) is obviously 2-dimensional(for closed system), since SS and ppcan be calculated from partial derivatives of FF with respect to the independent variables TT and V.V. It seems to me that a fixed T,VT,V system does not allow any unconstrained variables, and evolve at all. How can I understand the change of free energy along constant TT and VV processes?

I have another question. Since thermodynamic state variables can be properly defined only at equilibrium according to my textbook, so do potentials which are functions of T,S,V,p,T,S,V,p, I thought. But while studying 3-d F−T,VF−T,V space, I found even at points apart from minimum-FF point, every thermodynamic variable can be properly defined. Do those points indicate equilibrium states for given thermodynamic variables? At first I thought the minimum-FF point indicates the unique equilibrium, but now I know this is nonsense since T=p=0T=p=0 at that point.

Last question: If thermodynamic potentials such as FF can be defined only at equilibrium, I think we need some FF-like quantity for non-equilibrium state to describe the state evolving from non-equilibrium to its equilibrium. Sometimes I find an obvious non-equilibrium state that should go to the corresponding obvious equilibrium state. For example, dividing a box into two parts, filling one part with ideal gas, and removing the wall(system connected to heat reservoir, and total box volume fixed). Quantities like pressure for the whole box is poorly defined since the box is half filled at first, but I could calculate free energy for the initial system just regarding the gas-filled system only, and this led to ΔF<0ΔF<0for the whole process(I know this somehow contradicts my first question). Is this calculation valid? If this is valid, I think it's possible to imagine one (although not physically relevant) additional dimension (related to non-equilibrium property of the system) can be added to the F−T,VF−T,Vspace, and FF along this new dimension (with other variables-T,VT,V-fixed) is minimized when the non-equilibrium property vanishes. In my opinion this idea resolves my first question, but I'm not sure whether this is a sound argument

Answered by sushmita
0
0 \div 0 \\ \: \: \: -


Thermodynamic potential. Athermodynamic potential (in fact, rather energy, than potential) is a scalar quantity used to represent the thermodynamic state of a system. The concept ofthermodynamic potentials was introduced by Pierre Duhem in 1886. Josiah Willard Gibbs in his papers used the term fundamental functions.
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