Understanding thermodynamics of a stretched rubber band

[zenterix]

Homework Statement

The thermodynamics of a rubber band are given by the relation

$$dU=TdS+fdL+\mu dN$$

The force is defined in such a way that it is positive when the rubber band is stretched.

Relevant Equations

If we pull the rubber band in thermal isolation (ie no heat exchange), how does the temperature of the rubber band change?

You might find it useful to develop an expression for the heat capacity at constant length, $C_L$. Note that $C_L>0$.

Consider the function $U=U(T,L,N)$.

$$dU=\left (\frac{\partial U}{\partial T}\right )_{L,N} dT+\left (\frac{\partial U}{\partial L}\right )_{T,N} dL+\left (\frac{\partial U}{\partial N}\right )_{T,L} dN$$

and define

$$C_L\equiv\left (\frac{\partial U}{\partial T}\right )_{L,N}$$

By the 1st law we have

$$dU=dQ+dW=fdL>0$$

since $dQ=0$ by assumption, $dW=fdL$, and for stretching $f>0$ and $dL>0$.

I got this far.

Here is what the solution manual says

For the sake of thermal stability of materials, however, we know that heat capacities have to be larger than zero. The only possibility for $C_L=(\partial U/\partial T)_{L,N}$ to satisfy this condition is when $dT>0$ as well. That is, the temperature has to increase.

I do not understand this conclusion.

"Thermal stability" is lost on me, and has not been talked about in my course.

The result seems to be a purely mathematical one.

$$\left (\frac{\partial T}{\partial U}\right )_{L,N}=\frac{1}{C_L}$$

Sure, we have a positive change in $U$. But the heat capacity above is at constant length and we are changing the length. Why can we conclude from the above expression that $dT>0$ based on $dU>0$ when $L$ is changing?

 

[DrDu]

I think something is missing here which allows to estimate partial S/partial L|_T.

 

[DrDu]

Specifically, in which respect is a rubber band special as compared with any other solid? Also stretching an ordinary solid requires work and solids C_L is positive.