What is the solution for finding F in terms of T and V without integrating?

[wololo]

Homework Statement

Homework Equations

Maxwell relations

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

 

[Chestermiller]

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

 

[wololo]

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!

 

[TSny]

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.

 

[wololo]

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...

 

[Chestermiller]

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...

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