 Solution

Interpret the stated transition as an upper critical miscibility temperature: cooling increases the effective $\chi$ through ${\chi }_c$. Above $T_c$, the homogeneous mixing free energy is convex throughout $0<\varphi <1$, with no two-phase common tangent. Below $T_c$, it acquires a region with $f^{\prime \prime }<0$, bordered by the spinodal. A lower straight line touches the curve at two binodal compositions ${\varphi }_{-}$ and ${\varphi }_{+}$; this common-tangent construction gives the equilibrium polymer-poor and polymer-rich phases.

The precise raw curve need not have two equal minima: subtracting an affine function of $\varphi$ changes reference chemical potentials without changing curvature or coexistence compositions. The free-energy panels above subtract the tangent at ${\varphi }_c$ from the upper-temperature curve and the common tangent from the lower-temperature curve. This makes the common tangent and the convex-versus-nonconvex distinction visible.

If the mean initial composition ${\varphi }_0$ lies inside the low-temperature spinodal region, an arbitrarily small separation into equal-volume regions with compositions ${\varphi }_0\pm \delta \varphi$ changes the mean free energy bySmall fluctuations therefore grow spontaneously: this is spinodal decomposition. Interfacial costs limit very short-wavelength fluctuations, and the polymer-poor and polymer-rich domains subsequently coarsen towards their binodal compositions.

If ${\varphi }_0$ lies just outside a spinodal branch but still inside the binodal, then $f^{\prime \prime }({\varphi }_0)>0$. Small fluctuations increase free energy, so the uniform mixture is metastable. Separation requires a sufficiently large nucleus to overcome the interfacial barrier, followed by nucleation and growth. There is no guarantee that this happens rapidly. If the initial composition is outside the binodal as well, it remains a stable uniform phase after cooling.

For a mean composition in the coexistence interval, conservation of polymer fixes the volume fraction $\alpha$ of the rich phase by the lever rule:The temperatures alone do not determine which mechanism occurs without the initial composition. Numerical free-energy curves additionally require $\chi (T)$; the plotted example uses $\chi (T)={\chi }_c{T}_c/T$ with absolute temperatures $T_c=298.15\mathrm{K}$, $T_1=303.15\mathrm{K}$ and $T_2=293.15\mathrm{K}$.