= Solution
In a <regular solution>, ideal combinatorial <entropy> competes with a nonzero unlike-contact <energy>. For equal molecular <volumes> and <mole fraction> $\phi$ of species A, write mixing <free energy> per site as
$$
\frac{f(\phi)}{k_BT}
=\phi\ln\phi+(1-\phi)\ln(1-\phi)+\chi\phi(1-\phi),
$$
where
$$
\chi=\frac{z}{k_BT}\left[\epsilon_{AB}-\frac{\epsilon_{AA}+\epsilon_{BB}}2\right].
$$
Here $z$ is the <coordination number> and $\epsilon_{ij}$ a pair-contact <energy> counted consistently in the lattice model. A positive $\chi$ penalizes unlike contacts; a negative value favors them. The logarithmic terms lower <free energy> on mixing. The sign of the <enthalpy of mixing> alone therefore does not determine miscibility.
At fixed <temperature> and composition, a homogeneous phase is globally stable only if no division into two compositions lowers its <free energy>. The <common-tangent construction> selects coexisting compositions $\phi_a,\phi_b$, equivalent to equality of both species' <chemical potentials>. Local stability to infinitesimal fluctuations instead requires
$$
\frac{f''}{k_BT}=\frac1\phi+\frac1{1-\phi}-2\chi>0.
$$
Zero <curvature> gives the <spinodal>, while the coexistence boundary is the <binodal>. Between them a homogeneous state can be metastable and require <nucleation>; inside the <spinodal> its small composition fluctuations grow. The symmetric <regular solution> has
$$
\boxed{\phi_c=\tfrac12,\qquad\chi_c=2.}
$$
Below this contact penalty, the <free energy> is convex at every composition. Above it a miscibility gap is possible around the central composition. If $\chi=A/T$ with $A>0$, heating promotes mixing, but specific interactions or temperature-dependent noncombinatorial terms can produce other trends.
For two <polymers>, let $N_A,N_B$ count equal-volume segments per chain, and now let $\phi$ be the segment <volume fraction> of A. Connecting segments into chains reduces their independent translational <entropy>. The <binary Flory-Huggins polymer blend> has
$$
\frac{f}{k_BT}=\frac{\phi}{N_A}\ln\phi
+\frac{1-\phi}{N_B}\ln(1-\phi)+\chi\phi(1-\phi).
$$
The energetic contact term remains extensive in the number of segments, while the favorable <entropy> per segment is reduced by $1/N_A$ and $1/N_B$. This is the key <long-chain limit of polymer-blend miscibility>: a small positive incompatibility can overcome the mixing <entropy>. The lattice treatment is a controlled model for this argument, rather than a claim that every real blend has composition-independent $\chi$.<Thermodynamics of High Polymer Solutions by Paul J. Flory>[1942, pp. 51–55: chain-connected lattice configurations and combinatorial mixing entropy; later sections add the energetic contribution]{ref}
The local stability condition and critical interaction parameter become
$$
\frac1{N_A\phi}+\frac1{N_B(1-\phi)}-2\chi>0,\qquad
\boxed{\chi_c=\frac12\left(N_A^{-1/2}+N_B^{-1/2}\right)^2.}
$$
For equal chain <lengths>, $\chi_c=2/N$. Similar segment chemistry and matched cohesive interactions can reduce the unlike-contact penalty; favorable specific interactions, such as complementary <hydrogen bonding>, can make mixing favorable even for long chains. Lower <molar mass> increases mixing <entropy>. <temperature> can help when it lowers the effective $\chi$, but heating is not universally sufficient. A suitable common <solvent> can facilitate processing or ternary mixing, and a block-copolymer compatibilizer can stabilize fine domains; neither guarantees that the final solvent-free binary material is one <equilibrium> molecular phase.
An oil–water <emulsion> illustrates <emulsion stability versus molecular miscibility>. A <surfactant> adsorbs with its water-affine head toward water and its oil-affine tail toward oil, lowering <interfacial tension> and the <work> $\gamma\,dA$ needed to make additional interface. Breaking a fixed oil <volume> $V_o$ into spherical droplets of radius $R$ gives total <area> $3V_o/R$, so creating small droplets still requires substantial <area> and usually mechanical <work>.
Adsorbed <surfactants> can oppose <droplet coalescence> through charged double-layer repulsion, hydrated-head interactions or <steric repulsion> from sufficiently extended groups. Adequate coverage, <adsorption> time, ionic conditions and <film drainage> affect the lifetime. A lower <interfacial tension> does not itself remove the bulk miscibility gap, and a protected <emulsion> is commonly kinetically stabilized. A thermodynamically stable <microemulsion> is a distinct situation requiring an appropriate composition and free-energy balance. The two liquids must not be called molecularly miscible simply because droplets disperse visibly.
For the capillary, balance upward contact-line <force> against hydrostatic column weight, neglecting meniscus <volume> and gas <density>. For inner radius $r$,
$$
2\pi r\gamma\cos\theta=\rho g\pi r^2h,
\qquad h=\frac{2\gamma\cos\theta}{\rho gr}.
$$
The <contact angle> is measured through the liquid. <complete wetting> means $\theta_2=0$, so the <contact-angle change of capillary rise> gives
$$
\boxed{h_2=\frac{0.050}{\cos35^\circ}=0.06104\,\mathrm m\simeq6.10\,\mathrm{cm}.}
$$
This follows from the stipulated <wetting> change, holding radius, bulk liquid <surface tension> and <density> fixed. One should not replace that premise with a general claim that grease always improves glass <wetting>; if contamination also changes the liquid <surface tension>, the original height and new angle alone would not determine the new height.
For two wide vertical plates separated by gap $d$, use unit width. There are two upward contact-line contributions and liquid cross-sectional <area> $d$, giving the <capillary rise between parallel plates>:
$$
2\gamma\cos\theta=\rho gdh_{\rm plates},
\qquad\boxed{h_{\rm plates}=\frac{2\gamma\cos\theta}{\rho gd}.}
$$
Only one principal meniscus <curvature> contributes between wide plates, whereas a circular tube has two. For the same <fluid> and angle,
$$
\boxed{\frac{h_{\rm plates}}{h_{\rm tube}}=\frac rd.}
$$
If the plate gap equals the tube diameter, $d=2r$, the rise is half as large: the initial and completely <wetting> heights would be $2.50$ cm and $3.05$ cm. If instead $d=r$, the heights are equal. In either geometry, changing only the angle from $35^\circ$ to zero multiplies the height by $1/\cos35^\circ$. Since the question specifies no plate spacing, the general gap-dependent relation is the unambiguous result; the half-height answer requires the equal-gap-to-diameter convention.
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