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## The Derivation of Kelvin's Equation

### Kelvin's Derivation

Consider a tube of radius r in a jar from which the air has been evacuated and water is allowed to evaporate until an equilibrium is reached. The tube stands in water and water has risen to a height h in the tube.

Let σ be the surface tension for the water and pw and ρw (Note that ρ is the Greek letter rho) are the pressure and density of the water whereas pv and ρv are the pressure and density of the water vapor.

The volume of the column of water is πr2h and its weight is πr2h(ρwv)g, where g is the acceleration due to gravity. This is the weight of the water less the weight of the water vapor it displaces. Since the density of the vapor is negligible compared to the density of the liquid the density difference in the equation can be replaced in the above formula by simple the density of the liquid. This weight must be balanced by the force created by surface tension in the meniscus in the tube; i.e., 2πrσ. Thus,

#### πr2hρwg = 2πrσ which reduces to rhρwg = 2σ or h = 2σ/(rρwg)

On the other hand, the difference of the vapor pressure at height h compared to the vapor pressure in the water at the base of the tube is approximately the negative of the weight of the vapor over a height h; i.e.,

#### Δp = pw - pv = ρvhgand thush = Δp/(ρvg)

Equating these two expressions for h gives:

#### Δp/(ρvg) = 2σ/(grρw)) orΔp = 2σρv/(rρw) which is a version of Kelvin's equation.

By the ideal gas law ρv can be replaced by pv/RvT, where Rv and T are the gas constant for water vapor and the temperature, respectively.

With the above replacements the previously presented version of Kelvin's equation becomes:

#### Δp/pv = 2σ/(rρwRvT)

This is still not quite the standard version of the Kelvin Equation.

#### Instead of Δp = gρvh the correct formfrom the hydrostatic equationshould be: pv/pw = exp[-gh/RvT] and thus h = ln(pw/pv)RvT/g

Now when the two expressions for h are equated the result is:

#### ln(pw/pv)RvT/g = 2σ/(grρw) orln(pv/pw) = -2σ/(rρwRvT)

or, equivalently

###### pv/pw = exp[-2σ/(rρwRvT)]

This is the standard form of the Kelvin Equation.

## Derivation Utilizing Gibbs Free Energy

Gibbs free energy is a thermodynamic quantity for a substance that does not change as the substance changes phase under conditions of equilibrium between the phases.

The formal definition of Gibbs free energy G for a parcel of gas of volume V, pressure p and temperature T is:

#### G = U +pV -TΦ

where U is the internal energy of the parcel and Φ is its entropy. It is usually convenient to express the extensive thermodynamic variables per unit mass; i.e.,

#### g = u + pα -Tφ

In differential form this is:

#### dg = du + pdα + αdp - Tdφ - φdT but sincedu + pdα = Tdφ the differential form reduces to dg = αdp - φdT

Suppose liquid water and water vapor are in equilibrium at ps and T. If the vapor pressure p is above the level of ps it is said to be supersaturated. The excess Gibbs free energy above its level at ps can be evaluating the following integral:

#### ΔG = ∫ppsαdp'

For an ideal gas α = RwT/p. Thus the integral reduces to:

#### ΔG = ∫ppsRwT(dp'/p') = RwT.ln(p/ps)

This is the latent heat from the supersaturation of the vapor that can support the surface tension energy of the droplet.

The Gibbs free energy for a droplet of radius r, which has an area of 4πr2 and volume of (4/3)πr3, is

#### G = 4πr2σ - (4/3)πr3RwT.ln(p/ps)

For convenience let S represent the saturation ratio of p/ps.

The relationship between Gibbs free energy and the radius of the droplet for various saturation ratios is shown below.

Under conditions of constant temperature and pressure a change will occur spontaneously if and only if there is a decrease in Gibbs free energy. Thus the radius of the droplet will change if its Gibbs free energy is above its minimum. The only equilibrium radius that is possible is the one that minimizes the Gibbs free energy. This could be a radius of zero or a radius indefinitely large.

The value of r* that maximizes the Gibbs free energy is the relevant quantity here. Its significance is that any droplet with a radius above that level will spontaneously grow in size and any droplet with a radius of less than that value will spontaneously diminish in size until it disappears. The previous diagram indicates that the higher the saturation ratio the smaller is the critical size for a droplet to grow.

The value of r* that maximizes G may be found by finding the value of r that makes dG/dr equal to zero; i.e.,

#### dG/dr = 8πrσ - 4πr2RwT.ln(S) = 0 Dividing by 4πr gives rRwT.ln(S) = 2σ so r* = 2σ/(RwT.ln(S))

Solving for S gives

###### S = exp[2σ/(rRwT)] orp/ps = exp[2σ/(rRwT)]

which is the Kelvin Equation.