Discussion
Let f(T, l) be the tension in a rubber band of length l, T being the absolute temperature. Let F denote Helmholtz free energy and S the entropy. Then :
Answer: B
The correct pair is \(\left.\partial F/\partial l\right|_{T} = f\) and \(\left.\partial F/\partial T\right|_{l} = -S\).
For a rubber band, the work done on it in stretching by dl is \(f\,dl\), playing the role that \(-p\,dV\) plays for a gas. So \(dU = T\,dS + f\,dl\), and the Helmholtz free energy \(F = U - TS\) gives
\[ dF = -S\,dT + f\,dl \]
Reading off the partial derivatives: at constant T, \(\partial F/\partial l = f\); at constant l, \(\partial F/\partial T = -S\). The first option swaps the roles of S and f, the third multiplies them by l, and the fourth sets the tension derivative to zero; none matches the differential.
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