1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Helmholtz free energy problem

  1. Nov 21, 2015 #1
    1. The problem statement, all variables and given/known data

    Capture.PNG
    2. Relevant equations
    Maxwell relations

    3. The attempt at a solution
    I have an attempt at a solution, but I am not sure if I can replace the integral of dT in the helmholtz equation by the T I found using the internal energy. Does this make sense? Thanks

    new doc 17_1.jpg
     
  2. jcsd
  3. Nov 22, 2015 #2
    The problem statement asks you to express F as a function only of T and V. This is not the form of your final answer.

    Chet
     
  4. Nov 24, 2015 #3
    I still haven't figured out how to get rid of the entropy terms in the expression of Helmholtz free energy. Can I simply replace the S in the equation I found by heat capacity times temperature? (given dS=cdT). I don't know how to proceed after, any advice is appreciated. Am I on the correct path or did I start solving the problem wrong from the very beginning? Thanks!
     
  5. Nov 24, 2015 #4

    TSny

    User Avatar
    Homework Helper
    Gold Member

    In your integrations you treated S as constant which is not a valid assumption.

    You have an expression for T in terms of S and V. Try using that along with the definition of F in tems of U, S, and T.
     
  6. Nov 25, 2015 #5
    If I rewrite the expression for T and isolate S i get S=(27T^3V)/(64A). I can then replace the integral of -SdT to find F by the integral of -(27T^3V)/(64A)dT which equals -(27T^4V)/(256A).
    It seems to make sense since I get rid of the S terms in the first integral. However I don't see how I can get rid of the S term in the expression of pressure P, so I can't integrate PdV.

    I also considered using the definition F=U-TS but there again I get stuck with the S term in the expression of U. And if I use dF=dU-TdS-SdT and integrate I still have to do something with dU. I know I am probably missing some simple relation but I really can't think of anything that would rid me of S...
     
  7. Nov 25, 2015 #6
    Why are you integrating at all? You solved for S and got:
    $$S=\frac{27}{64}\frac{VT^3}{A^3}$$
    Why don't you just substitute that into the equation for U to get the internal energy in terms of T and V? Then why don't you just multiply S by T to get TS in terms of T and V? Then, F = U - TS.

    Chet
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Helmholtz free energy problem
  1. Helmholtz Free Energy (Replies: 0)

  2. Helmholtz free energy (Replies: 10)

Loading...