What's the Gibbs-Duhem equation for a rubber band? Consider a rubber band of length $L$ held at tension $f$. For displacements between equilibrium states
$dE=TdS+fdL+\mu dn$
$\mu$ is the chemical potential of a rubber band and n is the mass or mole number
I star using Euler's Theorem:
$f(x_1,\dotso, x_n)=\sum^n_{i=1} \left(\frac{\partial f}{\partial x_i}\right)_{x_j}x_i$
where $f(x_1,\dotso, x_n)=E(S,\vec{X})$ then
$E(S,\vec{X})=\left(\frac{\partial E}{\partial S}\right)_{\vec{X}}S+\left(\frac{\partial E}{\partial \vec{X}}\right)_S\vec{X}$
where:
$\left(\frac{\partial E}{\partial S}\right)_{\vec{X}}=T$
$\left(\frac{\partial E}{\partial \vec{X}}\right)_S=\vec{f}$
then
$E(S,\vec{X})=TS+\vec{f}\vec{X}$
here's my question, what's next?, I want to use de Legendre transform in $\vec{f}\vec{X}$  in such a way that I get $\vec{f}\vec{X}=fdL+\mu dn$
 A: The relevant Gibbs-Duhem relation arises from subtracting thermodynamic differential relation from the Euler equation. Let's consider a 1-dimensional rubber-band for simplicity (so no vector signs). Also, there's no change of mass or particle number so I'll ignore the chemical potential $\mu$.
$$dE = TdS+fdL\tag{1}$$
$$E(S,L)=TS+fL\implies dE=TdS+SdT+fdL+Ldf \tag{2}$$
Subtracting these two gives
$$\boxed{SdT+Ldf=0}$$
If we now regard independent variables to be $L$ and $T$ (as it might be in an experimental setting):
$$\begin{align}
S(L,T)dT+Ldf(L,T)&=S(L,T)dT+L\left[\frac{\partial f}{\partial L}dL +\frac{\partial f}{\partial T}dT \right]\\
&=\left(S+L\frac{\partial f}{\partial T}\right)dT+\left(L\frac{\partial f}{\partial L}\right)dL \\
&=0
\end{align}$$
Now if you want to calculate how much hotter $dT$ a rubber band will get after you stretch it by $dL$.
$$\begin{align}
dT=-\frac{\frac{\partial f}{\partial L}}{s+\frac{\partial f}{\partial T}} dL
\end{align}$$
Where $s\equiv S/L$ is the entropy per unit length.
