On the Size Dependence of Molar and Specific Properties of Independent Nano-phases and Those in Contact with Other Phases

Abstract

Nano-materials are materials with at least one nano-phase. A nano-phase is a phase with at least one of its dimensions below 100 nm. It is shown here that nano-phases have at least 1% of their atoms along their surface layer. The ratio of surface atoms is proportional to the specific surface area of the phase, defined as the ratio of its surface area to its volume. Each specific/molar property has its bulk value and its surface value for the given phase, being always different, as the energetic states of the atoms in the bulk and in the surface layer of a phase are different. The average specific/molar property of a nano-phase is modeled here as a linear combination of the bulk and surface values of the same property, scaled with the ratio of the surface atoms. That makes the performance of all nano-phases proportional to their specific surface area. As the characteristic size of the nano-phase is inversely proportional to its specific surface area, all specific/molar properties of nano-phases are inversely proportional to the characteristic size of the phase. This is applied to the size dependence of the molar Gibbs energy of the nano-phase, which appears to be in agreement with the thermodynamics of Gibbs. This agreement proves the general validity of the present model on the size dependence of the specific/molar properties of independent nano-phases. It is shown that the properties of nano-phases are different for independent nano-phases (surrounded only by their equilibrium vapor phase) and for nano-phases in multi-phase situations, such as a liquid nano-droplet in the sessile drop configuration.

Appendix: On the Specific Surface Area of Phases with Different Shapes

The specific surface area of a phase depends on its size and on its shape. First, let us consider the simplest 3D object, a sphere with radius r (m). The equations for its surface area and for its volume are: $A_{\text{sphere}} = 4 \times \pi \times r^{2}$, $V_{\text{sphere}} = 4 \times \pi \times r^{3} /3$. Substituting these two equations into Eq 1, the equation for the specific surface area for a sphere is obtained as:

$$ A_{{{\text{sp}},{\text{sphere}}}} = \frac{3}{r} $$

(Aa)

Now, let us consider a cube with side length a (m) with the known equations for its surface area and volume: $A_{\text{cube}} = 6 \times a^{2}$, V_cube = a³. Substituting these two equations into Eq 1, the equation for the specific surface area for a cube is obtained as:

$$ A_{{{\text{sp}},{\text{cube}}}} = \frac{6}{a} $$

(Ab)

As follows from Eq Aa-Ab, the specific surface area of any object is inversely proportional to its characteristic size. Thus, nano-phases have high specific surface areas, compared to micro-phases or macro-phases. This is the major special property of nano-phases that makes them so special compared to micro-phases and macro-phases.

Now, let us find that special value of the radius of a sphere (r_{sphere–cube}, m), which provides the same volume as a cube of side length a. To obtain this, the volumes of these two bodies should be identical: $4 \times \pi \times r_{{{\text{sphere}} - {\text{cube}}}}^{3} /3 = a^{3}$. Let us express from here r_{sphere–cube} and substitute it into Eq Aa:

$$ A_{{{\text{sp}},{\text{sphere}} - {\text{cube}}}} = \frac{4.836}{a} $$

(Ac)

Comparing Eq Ab-Ac, one can see that a sphere has a smaller specific surface area than a cube of the same volume by 19.3%. Let us declare that among all 3D bodies the sphere has the smallest specific surface area. This is because spheres have no corners.

Let us consider also a cylinder of radius r_cap. In the first approximation, let us consider only its wall and neglect its ends. Then: $A_{{{\text{cyl}} - {\text{wall}}}} = 2 \times r_{\text{cap}} \times \pi \times L$, $V_{\text{cyl}} = r_{\text{cap}}^{2} \times \pi \times L$, where L (m) is the length of the cylinder. Substituting these two equations into Eq 1, the equation for the specific surface area for a cylindrical wall is obtained as:

$$ A_{{{\text{sp}},{\text{cyl}} - {\text{wall}}}} = \frac{2}{{r_{\text{cap}} }} $$

(Ad)

Equation Ad is approximately valid for long cylinders (L ≫ r_cap), even if their ends are taken into account. For short cylinders, the ends of the cylinder should also be taken into account: $A_{\text{cyl}} = 2 \times r_{\text{cap}} \times \pi \times L + 2 \times r_{\text{cap}}^{2} \times \pi$. Substituting this equation and $V_{\text{cyl}} = r_{\text{cap}}^{2} \times \pi \times L$ into Eq 1, the equation for the specific surface area for a cylinder is obtained as:

$$ A_{{{\text{sp}},{\text{cyl}}}} = \frac{2}{{r_{\text{cap}} }} + \frac{2}{L} $$

(Ae)

Finally, let us consider the specific surface area of a thin film of a small thickness d (m) surrounded by two large parallel surface areas A (m²). Neglecting the negligible side surface areas compared to two large surface areas A, the surface area and the volume of the thin film are written as: $A_{\text{film}} = 2 \times A$, $V_{\text{film}} = A \times d$. Substituting these two equations into the definition of the specific surface area Eq 1, the specific surface area of a thin film is obtained as:

$$ A_{{{\text{sp}},{\text{film}}}} = \frac{2}{d} $$

(Af)

Note that the two large surface areas A usually correspond to different types of interfaces, such as film/substrate and film/gas.