Rubber Physics (Part 8)

We can determine a part of the free energy when we integrate the force $f$ over $l$ : $\begin{aligned} F(T, l) &= - \int_{l_0}^l f dl + F_0(T) \\ &= \frac{b}{2} T (l - l_0)^2 + F_0(T) \end{aligned}$ Note, that we have a constant of integration, $F_0(T)$ , that still depends on $T$ . Nevertheless, we can get an expression for the entropy $S(T,l) = - \left( \frac{\partial F}{\partial T} \right)_l = - \frac{b}{2} (l - l_0)^2 + F_0'(T)$ We still don't know anything about $F_0'(T)$ . But we know, that the entropy is related to the heat capacity via $\begin{aligned} & & T \left( \frac{\partial S}{\partial T} \right)_l &= C_l = \text{const} \\ \Rightarrow & & dS &= \frac{C_l}{T} dT \quad \text{for } l = \text{const}\\ \Rightarrow & & S(T,l) &= \int_{T_0}^{T} \frac{C_l}{T} dT + S_0(l)\\ & & &= C_l \ln \frac{T}{T_0} + S_0(l) \end{aligned}$ Here, we have constant of integration, $S_0(l)$ , that only depends on $l$ . But if we compare both expressions for the entropy we can derive the full two-dimensional function $S(T, l) = - \frac{b}{2} (l - l_0)^2 + C_l \ln \frac{T}{T_0} + S_{00}$ with a constant $S_{00} = S_0(l_0)$ . In case of the adiabatic process, we have $S(T,l) = S_{00} = \text{const}$ . Therefore, we can solve this equation for $T$ : $\begin{aligned} & & \Delta S &= - \frac{b}{2} (l - l_0)^2 + C_l \ln \frac{T}{T_0} = 0 \\ \Rightarrow & & \ln \frac{T}{T_0} &= \frac{b}{2 C_l} (l - l_0)^2 \\ \Rightarrow & & T &= T_0 \exp \left(\frac{b}{2 C_l} (l - l_0)^2 \right) \end{aligned}$ For $l = 3 l_0 = 0.39 \,\text{cm}$ we get an temperature increase of $\Delta T = T - T_0 = \left(\exp \left(\frac{(0.26)^2}{2} \right) - 1\right) \cdot 293 \,\text{K} \approx 10 \,\text{K}$