# Why is the the differential of Helmoltz free energy dT dependent?

My question is simple : we use Hemloltz free energy "A" to study equilibrium of system under transformation at T and N constants.

We have A=E-T.S, but dA=dE-TdS-SdT

Why is the differential dT dependent as we construct this function to study equilibrium for system where the temperature is imposed ?

 : More detailed question :

In fact what I understood is that for any system where the external work is only done by pressure forces we have :

$$dU+PdV-TdS < 0$$ (it comes from first principle and using the fact that $dS=\frac{\delta Q}{T} + \delta S^c$ with $\delta S^c > 0$).

And we define a thermodynamic potential $\Phi$ as $d \Phi = dU+PdV-TdS$. The equilibrium will then be done if $d \Phi =0$

It is not possible to define a general $\Phi$, but if we work in (T,V,N) ($dT=dV=dN=0$), we have : $d \Phi = dU+PdV-TdS = d(U-T S)$, so $\Phi = A = U-T S$.

Then to construct F we assumed $dT=0$, so why do we consider after that $dT$ is not equal to 0 ? I don't get the logic.

• Nobody has an idea (I don't know how to up a post) ? – StarBucK Sep 23 '16 at 11:20
• We didn't assume dT=0 to "construct" F. We just defined F = U - TS. This definition certainly does not require fixed temperature. Now, it's true that the statement $dF \leq 0$ is true for systems coupled to a thermal bath of fixed temperature, but that doesn't mean that we can't still talk about the temperature dependence of $F$. – Dominic Else Sep 24 '16 at 14:00
• Yeah what I meant is to define F=U-TS we putted dT=0 in dU+PdV-TdS to find F=U-TS But indeed we can make T dependent it is just that if F is T dependent then the condition F=Fmin at equilibrium will not be true. Thank you – StarBucK Sep 24 '16 at 15:09

Because when you differ, you want to know the difference in energy between two states that feature different thermodynamic conditions. The A is energy given a particular N,V,T tuple. That means that A actually depend on that tuple and when you want to compute the difference in A between two different NVT tuples you have to define A as a function of NVT:
$$\mathrm dA = -\,p~\mathrm dV + S~\mathrm dT + X~\mathrm d\mu$$
• I think my question wasn't precised enough, I edited my main post to give more details on what I don't understand. To summarise : I can understand that we want to compute the difference in A between two different N V T systems but the logic of construction of A does'nt follow the same logic so I don't get it (to construct it we assume $dT=0$). – StarBucK Sep 22 '16 at 18:01