Internal energy as a function of V and T in the first law

[RedCliff]

In thermodynamics, the internal energy (U) is the function of the volume (V) and temperature (T), U = U (T).
Therefore, according to the mathematics, dU = π dV + Cv dT, where π is internal pressure and Cv is the constant volume heat capacity.
Meanwhile, according to the 1st law, dU = δ w + δq, when under reversible process, dU = - p dV + C dT.
If we compare both, dU = π dV + Cv dT = - p dV + C dT. Since both V and T are free variables, we can get,
π = -p and Cv = C.
Obvious this is not right result. Based on Maxwell equation:
π = T (∂p/∂T)_V -p
Which part is wrong in my derivation?

Thanks in advance!

 

[mitochan]

Hi.
1st law of thermodynamics is
$$dU=TdS-pdV$$
changing independent parameter pair from (S,V) to (T,S)
$$dU=TdS-pdV=T(\frac{\partial S}{\partial T})_V dT + T(\frac{\partial S}{\partial V})_T dV - pdV$$

 

[RedCliff]

Hello,
Thank you for your replying! I fully agree with you on this point. If we express dU with all state functions, that is, dU = TdS - pdV, yes, your derivation is correct.

But my question is: what part is wrong in my above derivation as I use dU = -pdV + CdT?

However, you post might provide me a clue to the answer to my question:
I guess it is because C is not a state function but a path function. I remember I once read a paragraph saying the derivation/integration related to path functions are different from those related to state functions. In science, we mostly deal with state functions. Therefore, from dU = π dV + Cv dT = -pdV + C dT, we cannot conclude that
π = -p and Cv = C.

 

[mitochan]

Hi.
Referring to the eauation of my post, the equation you used is correct in case
$$T(\frac{\partial S}{\partial V})_T=0$$
I do not think this stands. We should have double energy or entropy for double volume or two similar systems regarded as united one.You may be interested in Helmholtz free energy F=U-TS.
Introducing F, the first law of thermodynanmics is
$$dF=dU-TdS-SdT=-pdV-SdT$$
Is the RHS the form you expect?