Question 6.9: The surface tension modifies the melting point of particles....

The chemical equilibrium between an infinitely large solid particle and the liquid is given by,

μ_s (T_∞, p_0) = μ_\ell (T_∞, p_0) .

For a given radius r, this chemical equilibrium relation becomes,

\mu _s \Bigl(T_m ( r) , p_s(r)\Bigr) = \mu _\ell \Bigl(T_m ( r) , p_\ell (r)\Bigr) .

Thus, for a radius r + dr, this relation becomes,

\mu _s \Bigl(T_m ( r+dr) , p_s(r+dr)\Bigr) = \mu _\ell \Bigl(T_m ( r+dr) , p_\ell (r+dr)\Bigr) .

When expanding the pressure and the melting temperature to first-order in terms of the radius r, this result is recast as,

\mu _s \biggl(T_m (r)+ \frac{dT_m}{dr}dr , p_s (r) +\frac{dp_s}{dr}dr \biggr) = \mu _\ell \biggl(T_m (r)+ \frac{dT_m}{dr}dr , p_\ell (r) +\frac{dp_\ell}{dr}dr \biggr) .

Furthermore, when expanding the chemical potential to first-order in terms of the melting temperature and the pressure, we obtain,

\mu _s \Bigl(T_m (r) ,p_s (r)\Bigr) + \frac{\partial \mu _s}{\partial T_m} \frac{dT_m}{dr} dr + \frac{\partial \mu _s }{\partial p_s } \frac{dp_s }{dr} dr= \mu _\ell \Bigl(T_m (r) ,p_\ell (r)\Bigr) + \frac{\partial \mu _\ell }{\partial T_m} \frac{dT_m}{dr} dr + \frac{\partial \mu _\ell }{\partial p_\ell } \frac{dp_\ell }{dr} dr .

The expression for the Laplace pressure implies that,

\frac{dp_s}{dr} = – \frac{2\gamma _s}{r^2}           and      \frac{dp_\ell}{dr} = – \frac{2\gamma _\ell}{r^2}

Thus, the chemical equilibrium condition is reduced to,

\biggl(\frac{\partial \mu _s}{\partial T_m} – \frac{\partial \mu _\ell }{\partial T_m} \biggr) dT_m =2 \biggl(\gamma_s\frac{\partial \mu _s }{\partial p_s }-\gamma _\ell \frac{\partial \mu _\ell }{\partial p_\ell } \biggr) \frac{dr}{r^2} .

The Gibbs free energy differential reads,

dG = −S dT + V dp + μ  dN

which implies that,

\frac{\partial \mu }{\partial T} = \frac{\partial}{\partial T} \Bigl(\frac{\partial G}{\partial N} \Bigr) = \frac{\partial}{\partial N} \Bigl(\frac{\partial G}{\partial T} \Bigr) = – \frac{\partial S }{\partial  N} = – s .

\frac{\partial \mu }{\partial p} = \frac{\partial}{\partial p} \Bigl(\frac{\partial G}{\partial N} \Bigr) = \frac{\partial}{\partial N} \Bigl(\frac{\partial G}{\partial p} \Bigr) =  \frac{\partial V }{\partial  N} = ν .

The differential equation becomes,

(s_\ell − s_s) dT_m = 2 (γ_s ν_s − γ_\ell ν_\ell) \frac{dr }{r^2} .

The integration of this equation from temperature T_∞ and radius r = ∞ to temperature T_m (r) and radius r yields,

(s_\ell − s_s) \Bigl(T_m (r)- T_\infty \Bigr) = -\frac{2}{r} (γ_s ν_s − γ_\ell ν_\ell) .

Since \ell_{s\ell} = T_∞ (s_\ell − s_s),

T_∞ − T_m (r) = \frac{2 T_∞ }{\ell_{s\ell} r} (γ_s ν_s − γ_\ell ν_\ell) .

Thus, if γ_\ell ν_\ell < γ_s ν_s, then T_m (r) < T_∞, which is the case for certain metals.