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 ?

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 $$ C(V) = N * ln(V)+ constant$$

  • $\begingroup$ 1) is a general relation? $\endgroup$ Apr 26, 2020 at 12:54
  • $\begingroup$ I would say yes. From Maxwell relations we have "S= -∂F/∂T" for V=const. Insert in the definition of F, put F under the derivative wrt T and obtain : -U/T^2 = ∂/∂T ( F/T). Then simply integrate. No other assumption in this derivation $\endgroup$
    – maru0032
    Apr 26, 2020 at 13:00

1 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.

  • $\begingroup$ Yes, i suppose that the trick was in the constant. But if i start from P=NT/V we have that -dF/dV = NT/V. Integrati h we find F=...+C. We cant deduce the value of the constant from this relation and the previuos? $\endgroup$
    – maru0032
    Apr 26, 2020 at 23:44
  • $\begingroup$ Both integrations introduce one unknown function: the first integration introduced function $C(V)$, the second integration introduced another function $D(T)$. There is no constant in numerical sense, these are functions. $\endgroup$ Apr 26, 2020 at 23:51
  • $\begingroup$ But a posteriori we can deduce the form of this function ,right?(see what i’ve added) $\endgroup$
    – maru0032
    Apr 27, 2020 at 12:59
  • $\begingroup$ Yes, that seems possible. $\endgroup$ Apr 27, 2020 at 19:45

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