Equation of state for ideal gas from Helmholtz free-energy

Question

Starting from the definition of Helmholtz free energy: $$F:=U-TS$$ (where $U$ is the internal energy , $T$ temperature and $S$ entropy) we derive in few steps the following relation: $$F=-T\int \frac{U}{T^2}\mathrm d T+ \text{constant} \tag{1}$$

Now, we know also that Maxwell relations holds so at $T=\text{constant}$ we have: $$P=-\frac{\partial F}{\partial V} \tag{2}$$

In ideal gas the internal energy have the following form: $$U=\frac{3}{2} NT \tag{3}$$

If i substitute $(3)$ in $(1)$ and put the result in $(2)$ i should find the classical equation of state for ideal gas: $$ PV=NT \tag{4}$$ ...but from calculation i don't find this. Where is the error in my steps? It could be in the value of the constant ?

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Yes, wrong word. Anyway we can say something about this function a posteriori :

$$ p = - \frac{\partial F}{\partial V} = -C'(V) T. $$

Using (4) in this equation we obtain $$ C(V) = N * ln(V)+ constant$$

Answer 1

In 1) there is additive "constant" of integration. The integration is only over $T$, the terms may depend also on volume $V$ which can be arbitrary. Therefore the "constant" in that integration over $T$ can be actually a function of $V$:

$$ F(T,V) = -T\int \frac{U}{T^2}dT + C(V)T. $$

Since the first term, for an ideal gas, does not depend on volume, the only part relevant for calculating pressure from $F$ is the second term:

$$ p = - \frac{\partial F}{\partial V} = -C'(V) T. $$

The conclusion is, we cannot infer the familiar equation of state of ideal gas $p = nc_V T / V$ just from knowing $U = nc_V T$. The above result suggests large class of functions of $C(V)$ is consistent with $U=nc_VT$. But we did find at least that pressure must be proportional to temperature $T$.

In Callen there is a rationale for this - the equation $U=nc_VT$ is not the fundamental form, that is, $U$ is not expressed as function of its natural parameters $S,V$. If it was, we should be able to derive the equation of state from it.