Thermodynamic potential question,help!

  • #1
i have studied my lecture notes back to front and looked high and low for a clue how to do this question but i tried in a) constructing an expression for u and came up with DU= +TDS -PDV - FDL BAUT I COULD BE MILES OUT. could anyone help.

thanks

here it goes:


A piece of rubber of length L is subject to work by a hydrostatic pressure P and also
by a tensional force F.

a)Construct the expression for dU.
[Hint: the usual expression will have an extra term added to it corresponding
to the increment of work performed by F

b)Generate (i.e. invent new) thermodynamic potentials that have as proper vari-
ables (S; V; F); (S; P; F) and (T; P; F).
[Note: the potential R has (S; V; F) as proper variables if dR = xdS+ydV +zdF
where x; y; z are to be found.]
 

Answers and Replies

  • #2
you can't think of three thermo equations with variables (S; V; F); (S; P; F) and (T; P; F)?
 
  • #3
for example could it be DR =TDS + PDV + LDF FOR (SVF), AND FOR (SPF) DN = TDS + VDP + LDF ? I am just not sure what question means thats all, can anyone help?
 
  • #4
i suppose they want you to derive new (invent new?) potentials, but I would go ahead and use internal energy change and say
dU = dQ - dW
dQ = TdS
dW = PdV +ldf
now dU = TdS - (PdV + LdF)

stick to something basic and use the four potentials (internal, hemholtz, gibbs, and enthalpy) as a basis.
 

Related Threads on Thermodynamic potential question,help!

Replies
2
Views
425
  • Last Post
Replies
0
Views
2K
Replies
1
Views
1K
  • Last Post
Replies
0
Views
1K
  • Last Post
Replies
11
Views
5K
  • Last Post
Replies
2
Views
645
  • Last Post
Replies
6
Views
4K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
2
Views
1K
Top