Past exam of the physics B course of the University of Cambridge / 2014 / Paper 2 / Section C / C11 / Solution (film calculations)

Reading the approximately straight trend in the supplied graph gives $\gamma (330\mathrm{K})\simeq 22\mathrm{mN}{\mathrm{m}}^{-1}$ and ${\gamma }^{\prime }\simeq -0.13\mathrm{mN}{\mathrm{m}}^{-1}{\mathrm{K}}^{-1}$. These are graph-read estimates, not more precise material constants. For $\Delta A=10^{-4}{\mathrm{m}}^2$, reversible isothermal stretching givesBoth are positive: stretching does mechanical work on the film and requires heat input to prevent cooling. Its internal energy increases by about $6.5\mu \mathrm{J}$.

Write $\gamma =\alpha -\beta T$, with $\beta =-{\gamma }^{\prime }\simeq 1.3\times 10^{-4}\mathrm{J}{\mathrm{m}}^{-2}{\mathrm{K}}^{-1}$. Constant $C_A$ then makes the reversible adiabatic conditionwhere $T_{\ast }$ is an arbitrary fixed reference temperature; absorbing it into the constant gives the requested logarithm convention. HenceHere $C_A=0.070\mathrm{J}{\mathrm{K}}^{-1}$. Since $\gamma -T{\gamma }^{\prime }=\alpha$, integration gives $\Delta U=C_A(T_f-T_i)+\alpha \Delta A$. To first order this isThere is no heat exchange on this reversible adiabat. Cooling and increasing internal energy are compatible: positive area work more than offsets the decrease of the temperature-dependent energy term.