## Energetics of Nucleation

**Energetics of Nucleation**

I shall use the eqn numbers of my book. That way if you ask questions I can back to my book on nucleation which I wrote about a decade ago but never published (hope to some day, for now I call it “the book that noone will ever read”). As I have said in previous blogs. I first rewrote nucleation theory and then realized that it all means that thermodynamics needs a rewrite, hence I have spent the last decade (or so) rewriting thermodynamics. I know what a strange path I walk.

Herein we will discuss the work/energy required for various nucleation processes

** Bubble Nucleation**

Let subscript “*l*” represent the initial state that being the liquid or aqueous state, while subscript “*b*”, represent the second state that being the vaporous or gaseous bubble. The work required to nucleate a bubble (W_{b}), becomes:

W_{b}=A_{b}@+X(u_{b}-u_{l})+P_{b}V_{b}-P_{l}V_{l} 4.1.1

Where: *A _{b}* = bubble’s surface area, @= the surface tension of the tensile surface

*P _{b}* = the bubble’s internal pressure,

*V*= the bubble’s total volume. P

_{b}*= the surrounding liquid’s pressure,*

_{l}*V*= the volume of the cluster before bubble nucleation (i.e. The volume in the liquid or aqueous state). u= bonding energy per molecule.

_{l }In differential form we could rewrite eqn 4.1.1 as:

W_{b}=@dA+Xdu+d(PV) 4.1.1 (a)

The mean molecular volume in the gaseous state is generally three orders of magnitude greater than that in the liquid state, therefore: *V _{b}>>>V_{l}*. . Due to the curved tensile layer the pressure inside of the bubble (

*P*) is greater than the pressure of the surrounding liquid (

_{b}*P*). Based upon these two postulates, we can surmise the following:

_{l}P_{b}V_{b}>>>P_{l}V_{l } 4.1.2

Therefore, eqn 4.1.1 can be approximated by:

W_{b}=A_{b}@+X(u_{b}-u_{l})+P_{b}V_{b} 4.1.3

Breaking eqn 4.1.3 down we have the

- 1) Energy required to form a tensile layer: A
_{b}@ - 2) Energy required to break the liquid’s bonds , i.e. cohesive forces: X(u
_{b}-u_{l}) - 3) Work to form a bubble: P
_{b}V_{b}

The following illustrates that the work/energy required for bubble nucleation is always positive and is an increasing function of radius

**Traditional Nucleation theory**

Let us take a step back. The traditional consideration for the work required to form a bubble was based upon Williard Gibbs paper: *Equilibrium of Heterogeneous Substances* ( J.W. Gibbs, “The Scientific Papers”, Vol 1, Dover Publ. Inc. NY, 1961):

Based upon Gibb’s paper, throughout the 20^{th} century the work required to form a bubble was considered as being:

W_{b}=A_{b}@+VdP 3.7.3

It should be apparent to anyone who does not teach either nucleation theory or thermodynamics that eqn 3.7.3 is a ridiculous equation for bubble nucleation because it only considers the energy required to form the tensile layer and the work associated with pressure increase. For those who are cynical, may note that it is funny that in eqn 3.7.3 the traditionalists actually associate an isometric pressure increase with work.

To make matters worse eqn 3.7.3 was also applied to both droplet and globule nucleation.

** Globule Nucleation**

Of course eqn 3.7.3 should only have been applied to globule nucleation i.e. a lava lamp. For a liquid globule forming inside of a liquid, the work required is :

W_{gl}=A_{gl}+V_{gl}(dP) 3.7.3

Where the subscript “gl” represents globule.

**Discussion**

Part of the reason that eqn 3.7.3 was wrongly equally applied to all types of nucleation is that we beheld such a strong belief in the mess known as entropy and the second law. Or if you prefer we wrongly insisted in starting our thermodynamic theory with the partial derivative:

TdS=dE+PdV 1)

And from 1) we then subtracted the whole derivatives either d(TS) or d(E+PV) to obtain the other parts. This perverse logic is also illogically used to calculate things like Gibbs free energy and Helmholtz energy. I repeat that thermodynamics is the only place that I know of where we start with a part (partial derivative: PdV) and then subtract the whole [d(TS) or d(E+PV) ] from it in order to obtain the other parts. The two science greats who are perhaps the most to blame for embedded such illogical logic into our souls are Gibbs and Maxwell. But still to simply blame them is to not look at oneself in the mirror and admit that we are all equally gullible in this matter.

The other side of eqn 3.7.3 is that Gibbs 350 page paper (*Equilibrium of Heterogeneous Substances*) was shortened into a 50 page treatise. To make matters worse both were published in the same book ( J.W. Gibbs, “The Scientific Papers”, Vol 1, Dover Publ. Inc. NY, 1961). So when referenced human nature meant that we all read his treatise rather than his 350 page novel.

The above sounds innocent enough until you realize that Gibbs & others made a mistake in going from his 350 page novel to his 50 page treatise, that being he forgot that eqn 3.7.3 only applies to globule nucleation. You see in his 50 page treatise he applies it to all nucleation processes. In other words like a pack of lemmings we all adhered to the misconception that eqn 3.7.3 actually applies to all nucleation processes including bubble nucleation.

Again we can see how we humans are easily mislead. It also shows how lazy we are, as we only seemed to read Gibbs 50 page treatise rather than his 350 page novel. I certainly did so the first time opening his book. It was not until the third or fourth time around did I bother to read the whole novel, it was like reading war and peace written in differential equations.

Just so that you know in his350 page novel Gibbs ^{(pg 255)} states: Critical work “*is the work which would be required to form (by a reversible process) the heterogeneous globule in the interior of a very large mass having initially the uniform phase of the exterior mass. For this work is equal to the increment of energy of the system when the globule is formed without change in entropy or volume of the system*”

Sadly he does not write the above in his 50 page treatise. And what a mess this has all become.

** Back to Bubble nucleation**

So let us say that I wanted to calculate the energy required to nucleate a bubb

The process of calculating the work required to nucleate a water vapor bubble:

- 1) Calculate the pressure within the bubble using eqn 4.2.7: P
_{b}=1.64@/r_{b}+P_{atm}

- 2) Calculate how many molecules are inside of the bubble, using: X=P
_{b}V_{b}/kT_{b}

- 3) From steam tables, obtain the latent heat per molecule at the bubble’s internal pressure , and multiply it by the
*X*molecules inside of the bubble. Therefore, obtaining the total energy required as latent heat . - 4) Calculate how much energy is required to raise the
*X*liquid molecule’s temperature to its boiling point (*T*_{b}at*P*)._{b}

- 5) Calculate the work required to form the tensile layer: A
_{b}@

- 6) Calculate the total work required to nucleate the bubble based upon:

W_{b}=A_{b}@+X(l_{p})+Xc_{y}(T_{b}-T_{w}) 4.5.4

where:

l_{p} is the latent heat per molecule at the bubble’s pressure , c_{y} is the heat capacity per molecule

X is the number of molecules, T_{b} is bubble’s temperature = superheat if you prefer

T_{w} is the water’s normal boiling point at 1 atm pressure, A_{b} is bubble’s surface area

@ is the surface tension

I must also emphasize that I used eqn 4.2.7: P_{b}=1.64@/r_{b}+P_{atm} instead of Young laplace’s equation which is wrongly applied to bubbles. For my papers concerning this see my website http://newthermodynamics.com and click on my papers: See:

1) Energetics of Nucleation: 2004 Physics Essays

2) Pressure within a bubble revisited: 2014 Physics Essays

3) The Considerations of forces perpendicular to tensile layers in bubbles, droplets, capillary rise and depression: 2007 Nature and Science

Sorry but I cannot give out all my secrets- just kidding it is more a case of what belongs in blog versus a book. I can if anyone wants discuss what is wrong is the Young Laplace equation in another blog on another day but you should understand it by reading my paper’s 2) and 3) above. As for how I arrive at eqn 4.5.4 will also come out in my book (if it ever actually gets published before I die. If not ce la vie let the experts figure it out. lol)

Another note: What value for surface tension of water, should we use in our analysis? If we assume that the temperature of the tensile layer is closer to that of the surrounding water than that of the bubble’s internal vapors, then we would approximate the water’s surface tension by 0.07 N/m, at 20 ^{o}C (room temperature) for all bubbles

I gave you the basics of bubble stability in my previous blog and now the energy required to nucleate. Next blog should be about probability or nucleation processes themselves. I still have not figured out what I should write but it will be coming soon. Before we finish we may as well consider droplet nucleation.

The graph below shows that the work required for bubble nucleation is an increasing function of radius

**Droplet Nucleation**

I repeat that droplet nucleation was wrongly thought of in terms of eqn 3.7.3

Based upon a combination of intuition and our new understandings, we realize that: The work required to nucleate a droplet should be the summation of three work terms, that being the work required to form the tensile layer (Ad@), the work required as latent heat of condensation [-L] and the work required to increase the pressure within the droplet.

Where, *A _{d }*= surface area of the droplet, and @ = the droplet’s surface tension.

Ignoring the work associated with the pressure increase inside of the droplet, then the work required to nucleate a droplet, would be:

W_{d}=A_{d}@-L 6.1.3

Now the latent heat of condensation does not involve work, while the latent heat of vaporization does involve work I.e. the upward displacement of our atmosphere (W=PdV).

Accordingly for X molecules:

The latent heat of vaporization = X(u_{l}-u_{g}) +PdV = U_{l}-U*g* +PdV (3)

While

The latent heat of condensation= X(u_{g}-u_{l}) = U_{g}-U_{l} (4)

Where u_{g} is the bonding energy per molecule in gaseous state, u_{l} is the bonding energy per molecule in liquid state

U_{g} is the total bonding energy of X molecules in gaseous state and U_{l} is the total bonding energy of X molecules in liquid state

Based upon 3) and 4) we realize that the latent heat of vaporization is not equal and opposite to the latent heat of condensations as is wrongly traditionally taught.

In terms of* X* molecules nucleating into a droplet we can rewrite eqn 6.1.3 as:

W_{d}=A_{d}@-X(u_{l}-u_{g}) 6.1.3 (a)

Next comes a whole bunch of math in my book that no one will ever read (assuming that it gets published lol).

Now just look at eqn 6.1.3. We have positive energy associated with the tensile layer formation (A_{d}@), which is a function of surface area and negative energy associated with the energy of condensation [X(u_{l}-u_{g})], which is a function of volume. So if I were to plot this I would get the following:

Table 6.1.1

x = 1 |
r_{d}=v@/u_{l} |
x=4 |
r_{d}=4v@/u_{l} |

x = 2 |
r_{d}=2v@/u_{l} |
x=5 |
r_{d}=5v@/u_{l} |

x = 3 |
r_{d}=3v@/u_{l } |
x=6 |
r_{d}=6v@/u_{l} |

Plotting the work required for a droplet to nucleate, as a function of radius, gives Graph 6.1.1. Note: Graph 6.1.1 is designed to give the general shape and *x*-intercepts. It is not intended to represent the energy requirements for any droplet in particular.

Now in my previous blog I discussed that small droplets were not stable. The reason is given in graph 6.1.1, that being small droplets require positive energy to nucleate while large ones do not. Specifically:

Obviously a droplet, requires positive work for nucleation (endothermic), when:

0<r_{d}<6v@/u_{l} 6.1.25

A droplet, becomes negative work (exothermic) when:

r_{d}>6v@/u_{l} 6.1.26

** A discussion concerning droplets **

A droplet requires energy to nucleate up until it reaches a critical size at which point it becomes stable from an energy perspective. Does this not explain why droplets need something to nucleate against. Consider seeding clouds wherein salts, metal etc are used to cause rain. We now realize that in order for a droplet to nucleate it must extract energy until it gets big enough that it is a stable droplet. And no wonder why salts and metals work best, it is because they have relatively high thermal conductivity hence will release they energy the easiest.

Okay I could go on and on but you get the idea. Anyhow I trust that I must be boring some of you by now.

Cheers and all the best and again thanks for reading what I write.