How Does Temperature Affect the Heat Capacity of a Rubber Band?

[llabesab16]

1. The problem: Suppose a rubber band obeys the equation of state: L = f (b/T), where L is length, f is tension, b is a constant, and T is temperature. For this rubber band, determine (δC<sub>L/δL)T, i.e. the change in the constant length heat capacity with length at constant temperature. You should be able to determine this using the following Maxwell Relation: (δf/δT)L = - (δS/δL)T, where S is the entropy



2. equations: dS = [(δS/δL)T]dL + [(δS/δT)L]dT
The constant length heat capacity, CL = T[(δS/δT)L]




3. Attempt: since L = f (b/T), TL = fb, and d(TL) = d(fb). b is constant, and hold L constant and use chain rule to get LdT = bdf. This implies that (δf/δT)L = L/b. Using the Maxwell relation, this implies that L/b = - (δS/δL)T. Also, using the equation of state, L/b = f/T, so - (δS/δL)T = f/T also. As mentioned above, dS = [(δS/δL)T]dL + [(δS/δT)L]dT. [(δS/δL)T] = - f/T, and [(δS/δT)L] is the constant length heat capacity divided by T, [CL]/T. Since dS is an exact differential, another Maxwell relation results from this: [(δ[[CL]/T]/δL)T] = -[(δ(f/T)/δT)L]. Using the chain rule, I got that (δCL/δL)T = f/T - 1

I hope this is clear enough. Let me know if I need to clarify anything. Second opinions would be greatly appreciated! Thanks</sub>

 

[Mapes]

Isn't

-\left(\frac{\partial(f/T)}{\partial T}\right)_L=-\left(\frac{\partial(L/b)}{\partial T}\right)_L

equal to zero? And since you have this as equal to

\left(\frac{\partial(C_L/T)}{\partial L}\right)_T=\frac{1}{T}\left(\frac{\partial C_L}{\partial L}\right)_T

would the change in the constant-length heat capacity with length at constant temperature also be zero?