
#1
Oct2812, 04:15 PM

P: 196

hello
This is a section from Callen, Herbert B  Thermodynamics and an Introduction to Thermostatistics "Any equilibrium state can be characterized either as a state of maximum entropy for given energy or as a state of minimum energy for given entropy. But these two criteria nevertheless suggest two different ways of attaining equilibrium. Let us consider a piston originally fixed at some point in a closed cylinder. We are interested in bringing the system to equilibrium without the constraint on the position of the piston. We can simply remove the constraint and allow the equilibrium to establish itself spontaneously. Here the entropy increases and the energy is maintained constant by the closure condition. This process is suggested by the maximum entropy principle. Alternatively, we can permit the piston to move very slowly, reversibly doing work on an external agent until it has moved to the position that equalises the pressure on the two sides. During this process energy is withdrawn from the system and entropy remains a constant (the process is reversible and no heat flows).This is the process suggested by the minimum energy principle. Independent of whether the equilibrium is brought about by either of these two processes, or by any other process, the final equilibrium state in each case satisfies both extrema conditions." According to the two processes described in the text, the first process refers to this till the equilibrium state is reached: (∂S/∂V)_{T} and the second process refers to (∂P/∂T)_{V} And according to the Maxwell's relations (the Helmholtz free energy minimized ) these two are equal. But according to the last statement : both of these satisfy the extremal condition, but need not be necessarily equal. So please tell me what is to be added in the theory so that the equivalence seems visible for a process. (i am not asking for a mathematical proof, the one based on the definition of exact differentials) Or someone please help me visualize that these two processes are actually same (also in what respect as per the Maxwell's relations?) Thank you for any help (I was confused in this section hence had to put it here, but i had no intention of copyright violation in any way) 



#2
Oct2812, 04:51 PM

P: 5,462

I have never heard of the text you refer to, but is the author really suggesting that both the value of entropy and energy can be specified at the same time?




#3
Oct2912, 11:29 AM

P: 196

Well, as far as i have got it, only one can describe the process sufficiently at a time.
So they probably can't be satisfied at the same time. the book is http://www.amazon.com/Thermodynamics.../dp/0471862568.... and if it is not wrong i can provide a link to pdf and the page i am referring to. 



#4
Oct3012, 07:09 AM

P: 51

Helmholtz free energy related query.
The text you are referring to is the ultimate book of thermodynamics,in my opinion,and also the most frequently cited source in research literature.
However,I dont really get where you bring in the Helmoltz function.The author says,that minimizing the energy holding the entropy constant gets you to the equilibrium state,the same equilibrium state that you would have reached if you had constrained the energy to a constant value(by isolating etc) and maximizing the entropy. Callen himself has mentioned the logical similarity of this thing with the isoperimetric problem in geometry.The Circle(analogous to equilibrium state) can be characterized either as the 2 dimensional figure of maximum area for a given perimeter,or of a minimum perimeter for a given area..Ref section 51 However I dont know where the Helmholtz function comes into this paragraph,and could you please throw more light on the same 



#5
Oct3012, 08:59 AM

P: 196

For the second process he said change in pressure is allowed so i jumped on to (∂P/∂T)_{V}. (I think i used the Maxwell's relation somewhere in my brain trying to fit them here, as i was trying to see these equations being equivalent for two processes, i guess i jumped on conclusions.....) So can you now help me with two processes (of the Maxwell's relations) and prove them to be equivalent in non mathematical way.. 



#6
Oct3012, 09:02 AM

P: 196

Well thanks for making me see that i used the very relations I wanted to see proved!




#7
Oct3012, 12:37 PM

P: 5,462

That book is frighteningly expensive.
I hope it actually makes better statements about equilibrium. Is a system for which entropy is a maximum or the internal energy is a minimum, but the volume is not constant in equilibrium? The answer to this has significant implications for the piston in a cylinder described in post#1 If the piston is a diathermal piston then the condition of maximum entropy corresponds to equilibrium. But if the piston is an adiabatic one then this is not the case, since dS_{total} = dS_{left} + dS_{right} = 0 for any position of the piston when the constraint is removed and the piston is free to move under the influence of the substances in the chambers. Where left and right refer to the chambers either side of the piston and the total refers to the whole system. 



#8
Oct3012, 11:40 PM

P: 51

Eitherway,even if the derivatives were right,there is no reason that they are equal.It just says,that both processes lead to the same final state,theres no reason that two random derivatives of the changing variables should be equal. 



#9
Oct3112, 12:10 AM

P: 51

Now,if I understand your point correctly,you are talking about a movable adiabatic wall. This a subtle from the point of view of Entropy maximum formalism,but can be done nonetheless.Writing dS=(1/T)dU+PdV,it can be shown that maximizing Entropy leads to the Mechanical equilibrium condition P1=P2. 



#10
Oct3112, 06:35 AM

P: 5,462

Pabloenigma
Whilst I agree with you that the equilibrium condition is equalty of pressures, it is very easy to glibly state "It can be shown that................." So I ask for your proof of your above statement. I think you cannot prove it using entropy maximization since dS = 0. Let us consider a sealed pipe containing a frictionless, adiabtic piston which divides the pipe into to chambers A and B and inviscid gasses in each chamber. Now let the volume of chamber A at equilibrium be V_{a} and consider a small change of state from equilibrium, to V_{a} + dV_{a} [tex]d{S_a} = \frac{{d{U_{}}}}{{{T_a}}} + \frac{{{P_a}}}{{{T_a}}}d{V_a} = 0[/tex] since [tex]d{U_a} =  {P_a}d{V_a}[/tex] Similarly dS_{b} = 0. So you cannot use the maximum entropy condition to specify equilibrium. 



#11
Oct3112, 06:41 AM

P: 5,462

A Dhingra
The thread carries the name Helmholtz free energy in the title and suggests some confusion embodied in the textbook concerning equilibrium. The following may be of use when reading the textbook. Starting with the following second law condition [tex]\begin{array}{l} \Delta S \ge \int {\frac{{\delta q}}{T}} \\ \delta q \le TdS \\ \end{array}[/tex] Thus [tex]dU  TdS + PdV \le 0[/tex] With the equality sign holding at equilibrium, From which can be derived a shedload of equilibrium conditions: Substituting H=U+PV [tex]dH  TdS  VdP \le 0[/tex] Thus at constant entropy and pressure [tex]{\left( {\Delta H} \right)_{SP}} \le 0[/tex] A system at constant entropy and pressure is in equilibrium when its enthalpy is at a minimum. Substituting A=UTS [tex]dA + SdT + PdV \le 0[/tex] Or [tex]{\left( {\Delta A} \right)_{TV}} \le 0[/tex] A system at constant T and V is in equilibrium when its Helmholtz free energy is a minimum Substituting G=U+PVTS [tex]dG + SdT  VdP \le 0[/tex] A system at constant T and P is in equilibrium when its Gibbs free energy is a minimum [tex]{\left( {\Delta G} \right)_{TP}} \le 0[/tex] Thus we have the following set of 5 alternative conditions with equality holding for equilibrium. They may not all be available as shown in post#10 [tex]\begin{array}{l} {\left( {\Delta S} \right)_{UV}} \ge 0 \\ {\left( {\Delta U} \right)_{SV}} \le 0 \\ {\left( {\Delta H} \right)_{SP}} \le 0 \\ {\left( {\Delta A} \right)_{TV}} \le 0 \\ {\left( {\Delta G} \right)_{TP}} \le 0 \\ \end{array}[/tex] Note you always need to specify two variables to define equilibrium. This was one of my original points. 



#12
Oct3112, 08:40 AM

P: 51

My point is,the theory says(as i quoted above post 9) the entropy is maximized over the manifold of constrained equilibrium states.That is,if we remove the constraint(that of the piston being fixed here),the piston reverts to the state which maximizes the entropy.and this reverting is ofcourse subject to the conditon that U1+U2=0,from where P1=P2 follows.The theory doesnt exclude that 



#13
Oct3112, 08:49 AM

P: 5,462

Pabloenigma, you now have it right. You need one of the other (energy) criteria I listed in post#11 to establish equilibrium.




#14
Oct3112, 08:55 AM

P: 51

http://en.wikipedia.org/wiki/Herbert_Callen I insist,because, I think, in the above discussion I havent really done justice to this book. In my country, a theoritical physicist Palas Pal(http://www.saha.ac.in/theory/palashbaran.pal/) once said that there are 2 kinds of books in thermodynamics :good and bad.If Callen is of the first kind,all the rest are of the second kind. 



#15
Nov112, 04:29 AM

P: 5,462

Unfortunately a script error crept into the collection of formulae at the end of my post#11 so the last four have the wrong direction to their inequalities.
The correct list is [tex]\begin{array}{l} {\left( {dS} \right)_{UV}} \ge 0 \\ {\left( {dU} \right)_{SV}} \le 0 \\ {\left( {dH} \right)_{SP}} \le 0 \\ {\left( {dA} \right)_{TV}} \le 0 \\ {\left( {dG} \right)_{SP}} \le 0 \\ \end{array}[/tex] 



#16
Nov112, 09:07 AM

P: 196

What would make such two random derivatives equal, the final states(which you said does not) or it is something to do with the equivalence principle of maximum entropy and minimum energy? How does this happen exactly, please tell me where to read about it (if in Callen) or anywhere else? Or please explain this through two processes....The same thing I was trying to do, making equations of the processes and then checking if they will be equal or not, but my derivatives turned out to be wrong,I guess. 



#17
Nov112, 12:12 PM

P: 196

In fact it might not be possible to form any such derivatives at the equilibrium because there won't be any change in teh states any further. Such derivatives can be used only during the process if the the equation of state exists. So the derivatives used by Maxwell have actually nothing to do with the equivalence principle of entropy maximum and energy minimum, right? 



#18
Nov212, 03:56 PM

P: 5,462

Since you did not respond to the bulk of my post #11 I'm not really quite sure where you are having difficulty. You asked several questions mixed together.
Do you fully appreciate the distinction between differences (Δx) , differentials, dx, partial differentials ∂x and small variations δx ? The distinction comes into play when we derive Maxwells relations. Have you seen the derivation? Are you aware that whilst each partial specifies one constant variable, two are in total specified as I said before? Can you apply say [tex]{\left( {\frac{{\partial S}}{{\partial V}}} \right)_T} = {\left( {\frac{{\partial P}}{{\partial T}}} \right)_V}[/tex] to the piston in cylinder example? I do not want to waste time writing out a lot of stuff if you are not going to read it, but we can go through it if you like. 


Register to reply 
Related Discussions  
Helmholtz free energy  Introductory Physics Homework  5  
Helmholtz free energy  Classical Physics  0  
Helmholtz free energy  Advanced Physics Homework  10  
Helmholtz Free Energy  Advanced Physics Homework  0  
Helmholtz free energy  Classical Physics  3 