Why are 4 different thermodynamical potential?

[kompabt]

What is the reason that there are 4 different thermodynamical potential?
How can I know, which of them should I use?

 

[pabloenigma]

the 4 functions are Legendre differential transforms of each other,and are functions of different sets of variables

Use will be according to the problem.
For eg, H=H(S,P),ie Enthalpy is a function Of pressure,and can be used in problems where pressure is a controlled variable/independent variable.

Simmilarly the Gibbs function,G=G(P,T) can be used for processes where temperature and pressure are independent variables

for a good summary,refer to the thermodynamics text by Zemansky and Dittman,or the Mathematical Methods text by Margenau and Murphy

 

[Zeppos10]

the 4 functions are Legendre differential transforms of each other,and are functions of different sets of variables

the Legendre transform in thermodynamics is Science Fiction. see
https://www.physicsforums.com/showthread.php?t=313535&referrerid=219693

Use will be according to the problem. For eg, H=H(S,P),ie Enthalpy is a function Of pressure,and can be used in problems where pressure is a controlled variable/independent variable. Simmilarly the Gibbs function,G=G(P,T) can be used for processes where temperature and pressure are independent variables.

Enthalpy H (note: look for H(S,p) written as an explicit function of S and p: nowhere to be found) is a function of the external pressure: this can be considered an independent variable, but is best considered as a (commonly constant) parameter. Similar for G where p and T are the parameters.
compare this to this discussion:
https://www.physicsforums.com/showthread.php?t=456648&referrerid=219693

for a good summary,refer to the thermodynamics text by Zemansky and Dittman,or the Mathematical Methods text by Margenau and Murphy

Zemansky & Dittman (1981) say nothing about a Legendre Transform.
Leave Margenau & Murphy to the mathematicians.

 

[espen180]

Please see the following Wikipedia articles to see how the Gibbs free energy and the Helmholtz free energy are obtained from the internal energy via Legendre transforms.
http://en.wikipedia.org/wiki/Gibbs_free_energy#Derivation
http://en.wikipedia.org/wiki/Helmholtz_free_energy#Definition
http://en.wikipedia.org/wiki/Legendre_transformation#Thermodynamics

 

[kanato]

Enthalpy H (note: look for H(S,p) written as an explicit function of S and p: nowhere to be found)

See for instance, Claude Garrod, Statistical Mechanics and Thermodynamics pp. 148 or F. Schawbl, Statistical Mechanics pp. 77. Both books write down H(S,p).

 

[pabloenigma]

the Legendre transform in thermodynamics is Science Fiction.

Sir,i will read your link later,and reply.But even Callen,and the Mechanics text of Herbert-Goldstein,affirms the use of Legendre transforms in definition of these functions.and even if there is reason enough to think,LT is not needed in definition of these potential functions,I gez its best not to confuse the person who asked this question,coz it does seem he is a beginner

Zemansky & Dittman (1981) say nothing about a Legendre Transform.
Leave Margenau & Murphy to the mathematicians.

And The 7th edition of Zemansky Dittman,which I own,does say about Legendre transforms.And as far as I know,Margenau and Murphy is not so much a text for mathematicians,its a Mathematical Methods text for Physicists.But then,I'm just a college Junior.

 

[pabloenigma]

See for instance, Claude Garrod, Statistical Mechanics and Thermodynamics pp. 148 or F. Schawbl, Statistical Mechanics pp. 77. Both books write down H(S,p).

Both Zemansky-Dittman(7th edition) and Herbert Goldstein write down H(S,P).
where dH is given by TdS + VdP

 

[Zeppos10]

Both Zemansky-Dittman(7th edition) and Herbert Goldstein write down H(S,P).
where dH is given by TdS + VdP

if dH=TdS+Vdp then it seems to me that H=H(T,S,V,p) ie a function of 4 variables: according to the prevailing theory only two variables are independent though.
Since H is commonly applied in conditions of constant pressure, it might be admissable to eliminated p from the list: now you are left with H(T,V,S).
Zemansky-Dittman (1981) gives the same equation but without reference to a Legendre transform.
the book of Goldstein (1950) is not about thermodynamics but about classical mechanics and I do not believe Goldstein can claim any authority in the field of thermodynamics.
He does not provide any proof that H or G are Legendre transforms either: its a suggestion at best.

 

[Zeppos10]

Please see the following Wikipedia articles to see how the Gibbs free energy and the Helmholtz free energy are obtained from the internal energy via Legendre transforms.
http://en.wikipedia.org/wiki/Gibbs_free_energy#Derivation
http://en.wikipedia.org/wiki/Helmholtz_free_energy#Definition
http://en.wikipedia.org/wiki/Legendre_transformation#Thermodynamics

the wiki about enthalpy does NOT refer to the Legendre transform at all.
The wiki on Gibbs free energy does in the derivation, but all textbooks published before 1960 and most textbooks published after 1960 (eg Smith & VanNess) do not need the Legendre transform but are still able to derive the same equations.

 

[Zeppos10]

What is the reason that there are 4 different thermodynamical potential?
How can I know, which of them should I use?

This seems to be a valid question, and not only for beginners.
By calling U,H,G and F thermodynamic potentials and/or calling them Legendre transforms it seems that they are more or less equivalent to each other. This is not the case.
Internal energy and enthalpy are forms of energy that show up in the energy balances of systems. However, the energy (U,H) of these systems is not equal to the capacity to do (usefull) work: this quantity is given by G, assuming a given T and p of the environment. (For systems at constant V, F is used, but this function is used very little.)
In chemical engineering only H and G are used, for all practical purposes.

 

[Mapes]

if dH=TdS+Vdp then it seems to me that H=H(T,S,V,p) ie a function of 4 variables: according to the prevailing theory only two variables are independent though.

T and V are equivalent to (\partial H/\partial S)_P and (\partial H/\partial P)_S, respectively. There are only two independent variables.

By calling U,H,G and F thermodynamic potentials and/or calling them Legendre transforms it seems that they are more or less equivalent to each other.

No one here has claimed that the thermodynamic potentials are "more or less equivalent to each other."

 

[pabloenigma]

if dH=TdS+Vdp then it seems to me that H=H(T,S,V,p)
Zemansky-Dittman (1981) gives the same equation but without reference to a Legendre transform.
.

Zemansky-Dittman(7th edition) makes an EXPLICIT reference to Legendre Transforms.

 

[Zeppos10]

Zemansky-Dittman(7th edition) makes an EXPLICIT reference to Legendre Transforms.

Zemansky died in 1981, so the 1981 edition is the last one he aproved of: others have tampered with is legacy, and if he knew he would have turned in his grave. The first part is a fact, the second part is an opinion ofcourse.

 

[pabloenigma]

Zemansky died in 1981, so the 1981 edition is the last one he aproved of:

yes,Mark Zemansky died.But Richard H Dittman is alive,and he updated this edition.He mentions the use of Legendre transforms among the valuable additions to this edition.

others have tampered with is legacy, and if he knew he would have turned in his grave.

Sorry,if I sound rude,but the thing is,it does seem,you understand Zemansky more than Dittman...