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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.

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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.

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  • $\begingroup$ 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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