Discussion
The thermodynamic identity for the ideal rubber band is \(dU = Tds + \tau dL\) where T is the temperature, \(\tau\) is the tension, U is the internal energy of the rubber band, S is its entropy, and L is its length. Which of the following Maxwell's relation is correct for this rubber band system is ....
Answer: A
The correct relation is \(-(\partial S/\partial L)_{T} = (\partial\tau/\partial T)_{L}\).
Start from the Helmholtz free energy \(F = U - TS\). Using \(dU = TdS + \tau dL\):
\[ dF = -S\,dT + \tau\,dL \]
so \(S = -(\partial F/\partial T)_{L}\) and \(\tau = (\partial F/\partial L)_{T}\). Equating the mixed second derivatives of \(F\) gives the keyed Maxwell relation. It is the rubber-band analogue of \((\partial S/\partial V)_{T} = (\partial p/\partial T)_{V}\), with the sign changed because the tension term enters \(dU\) as \(+\tau dL\) where the gas has \(-p\,dV\).
The option without the minus sign misses exactly that sign. The other two pair variables wrongly: from \(dU\) itself the correct relation is \((\partial T/\partial L)_{S} = (\partial\tau/\partial S)_{L}\), with no minus sign, and the Gibbs-type function \(U - TS - \tau L\) gives \((\partial S/\partial\tau)_{T} = (\partial L/\partial T)_{\tau}\), again without the minus sign of the last option.
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