For a stretched rubber band, it is observed experimentally that the tension $f$ is proportional to the temperature $T$ if the length $L$ is held constant. Prove that:

(a) the internal energy $U$ is a function of temperature only;

So we have the first law:

$$ dU=T dS+fdL=dQ+dW $$

The work done on the elastic band is $0$ as it is held at a fixed length. $$ \implies dU=dQ=dT/C $$

where $C$ is heat capacity. By integrating we show that $U=T/C+A$ where $A$ is an arbitrary constant of integration.

However I feel my method implies that $TdS=dQ$ as both the length and work are held constant. However I don't think heating up an elastic band is reversible so this cannot be true, which would imply work is done.


You need to first express dS in terms of dT and dL, and then to apply one of the Maxwell equations for the partial derivative of S with respect to L at constant T. This leads to the equation $$dU=C_vdT+\left[f-T\left(\frac{\partial f}{\partial T}\right)_L\right]dL$$If f is proportional to T at constant L, then f must be of the functional form $f=T\alpha(L)$. If we substitute this into the equation for dU, we obtain:$$dU=C_vdT+(\alpha T-\alpha T)=C_vdT$$ So,$$\left(\frac{\partial U}{\partial T}\right)_L=C_v$$and $$\left(\frac{\partial U}{\partial L}\right)_T=0$$The second partial derivative of U with resect to partials of L and T must mathematically be independent of the order of differentiation. So, $$\frac{\partial^2 U}{\partial T \partial L}=\frac{\partial^2 U}{\partial L \partial T}=\left(\frac{\partial C_v}{\partial L}\right)_T=0$$ Therefore, $C_v$ for this material must be a function only of T. And, of course, the same must be true for U.

  • $\begingroup$ Could you please add a few intermediate steps leading up to the equation you have shown? I tried to work it out on my own but failed. What is given is that $\frac{\partial f}{\partial T}|_L=\alpha$ in which $\alpha$ is a constant. But $f$ in the second parenthesis on the RHS of your equation is a function of both $T$ and $L$ and it is not clear why $U$ should be a function of $T$ alone. $\endgroup$
    – Deep
    Aug 10 '17 at 5:36
  • $\begingroup$ @Deep. I hope my fleshing out of the analysis helps. I will be pleased to answer any additional questions. $\endgroup$ Aug 10 '17 at 12:19
  • $\begingroup$ +1 Nice. Why do you think a function of the form $f=\alpha T+g(L)$, where $\alpha$ is constant and $g$ is some function not applicable here? I ask because then we would have $dU=C_vdT+g(L)dL$. $\endgroup$
    – Deep
    Aug 11 '17 at 6:46
  • $\begingroup$ @Deep The problem statement says that f is proportional to T, not just linear in T. $\endgroup$ Aug 11 '17 at 12:18
  • $\begingroup$ I guess you are right. $\endgroup$
    – Deep
    Aug 11 '17 at 12:31

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