 Solution

Over the plotted range, approximate the surface tension by $\gamma (T)={\gamma }_0-\beta T$ with constant positive $\beta =-{\gamma }^{\prime }$, and take $C_A=70\mathrm{mJ}{\mathrm{K}}^{-1}=0.070\mathrm{J}{\mathrm{K}}^{-1}$. A reversible adiabat hasIntegrating givesThe arbitrary fixed reference temperature $T_{\ast }$ makes the logarithm dimensionless; changing it only shifts the integration constant. This is the dimensionally precise version of writing $C_A\ln T+\beta A$ as constant. Each term has units of entropy. Between two states,If the graph had substantial curvature, the entropy differential would still hold, but this particular constant-$\beta$ exponential would no longer follow.