Question 27.4: What would. be the size of a nucleus in equilibrium with a s......
What would. be the size of a nucleus in equilibrium with a supersaturation of 0.029, under the conditions of Example 27.3?
This is solved from the Kelvin equation. Here α = 1 + 0.029 = 1.029. Substitution of the values for V_{ M }, \sigma_a, v, R, \text { and } T from Example 27.3 into Eq. (27.9) gives
\ln \alpha=\frac{4 V_M \sigma}{ν R T L} (27.9)
\ln 1.029=\frac{4 \times 37.51 \times 2.5}{2 \times 8.3143 \times 10^7 \times 300 L}From this,
L=2.63 \times 10^{-7} \ cm \quad \text { or } \quad 2.63 \ nmNote that this is consistent with the size of a nucleus containing a few hundred particles about 0.3 nm in diameter. If σ were 80 ergs/cm², α would be 2.5 for a 2.63-nmparticle, and the supersaturation required would be 150 percent-dearly an impossible value for a soluble salt like KCl.
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