Physical Chemistry of Foods

applicability of the Kelvin equation; these will be discussed in Sections 13.6
and 14.2.
Some examples are given in Table 10.4, to show the influence of some
variables on the results. It is seen that, especially for gases, a considerable
increase in solubility can occur for fairly large bubbles. For an air bubble of
1 mm diameter, the solubility ratio would be about 5, and such a bubble
would soon dissolve. It is indeed true that such very small gas bubbles are
rarely observed. Although Eq. (10.9) does not tell us anything about the rate
at which changes due to increased solubility occur, it may be clear that a
bubble that starts dissolving will generally do so at an ever faster rate,
because its radius decreases, whence the solubility excess increases during
the process.
The increase in solubility with decreasing radius has some important
consequences.

Nucleation. If a new phase should be formed because a dissolved

substance has become supersaturated, this may nevertheless not

occur. Formation of a new phase needs nucleation, i.e., the

spontaneous formation of regions of a few times molecular size in

diameter, say 2 nm. For sucrose, for instance, the data in Table 10.4

then predict a solubility ratio of 2.4, and nucleation would thus need

a supersaturation of that magnitude, otherwise the tiny sucrose

TABLE10.4 Increase in Solubility of Dispersed Substances Due to

Surface Curvature

Disperse phase Water Triglyc. oil Air Sucrose

Continuous phase triglyc. oil water water saturated soln.

r (mm) 0.5 0.5 50 0.01

g (mN)?m
^110 a 10 a 40 a 5

VD (m^3 ?mol
^1 )1.8? 10 
^5 7.7? 10 
^4 0.024 2.2? 10 
^4

x^0 (nm) 0.14 6 770 0.9

s(r)/s? 1.0003 1.012 1.016 1.09

r¼radius (of curvature).

g¼interfacial tension.

VD¼molar volume (disperse phase).

x^0 ¼characteristic length.

s(r)¼solubility of disperse phase if radius isr.

s?¼solubility if 1/r¼0.

asurfactant present.

Calculated according to Eq. (10.9) forT¼300 K.