= Solution
Over the plotted range, approximate the <surface tension> by $\gamma(T)=\gamma_0-\beta T$ with constant positive $\beta=-\gamma'$, and take $C_A=70\,\mathrm{mJ\,K^{-1}}=0.070\,\mathrm{J\,K^{-1}}$. A reversible adiabat has
$$
0=dS=\frac{C_A}{T}\,dT+\beta\,dA.
$$
Integrating gives
$$
\boxed{C_A\ln\left(\frac{T}{T_*}\right)+\beta A=\text{constant},
\qquad \beta\simeq1.3\times10^{-4}\,\mathrm{J\,m^{-2}K^{-1}}.}
$$
The arbitrary fixed reference <temperature> $T_*$ 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,
$$
\boxed{T_f=T_i\exp\left(-\frac{\beta\Delta A}{C_A}\right).}
$$
If the graph had substantial <curvature>, the <entropy> differential would still hold, but this particular constant-$\beta$ exponential would no longer follow.
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