By Kent W. Mayhew

*Boiling here on Earth*

Boiling here on Earth requires the energy to break the liquid water molecule's bonds (*dU*)
plus the irreversible work done onto the atmosphere [(*PdV*)atm ] A.K.A. Lost work. This is expressed by the equation defining
the latent heat of vaporization (*L*vap), here on Earth’s surface, that being:

*L*vap = *dU*+(*PdV*)atm (1)

__Boiling__ in Space

Consider the boiling of water in outer space. There is no requirement of work therefore in order to boil water in outer space you would only need the energy required to break the water molecules cohesive bonds. Therefore for boiling in outer space we expect that:

*L*vap (in outer space) = *dC* (2)

Eqn (2) was written with energy associated the water molecule's cohesive forces as *dC* instead of the bonding energy (dU)
in eqn (1). The reason being is that there may be some pressure functionality to liquid water’s bonding energy, which is a bit more
obvious if one thinks in terms of cohesive forces. If one considers pressure as part of the cohesive forces binding the
liquid molecules together, then as the pressure decreases the energy required for boiling would decrease. This is most likely the
reality!

If there is no pressure functionality associated with the water’s bonding energy then:

*dC*= *dU * (3)

In
which case:

*L*vap (in outer space) = *dU* (4)

No matter if our reality is defined by eqn (2) or (4) the fact remains that boiling in outer space will require significantly less energy than boiling water on Earth's surface, therefore boiling in outer space will occur at a significantly lower temperature.

*Boiling
in the Space lab*

Next consider boiling water in the space lab orbiting Earth. There is no gravity thus vaporization should
not involve work being passed onto the surrounding atmosphere as occurs here on Earth i.e. no lost work as defined by:

(*PdV*)atm = 0 (5)

Since there is no lost work as defined by (5) , then boiling in the space lab should require less energy than here on Earth.

Investigating further: The space lab is in an isometric (closed rigid) system hence there could be work associated with any isometric
pressure increase (or potential work), as defined VdP , i.e. an isometric pressure increase inside of the space lab. As a potential
to do work, it is not necessarily lost work. Even so let us play the game and say our first inclination might be to write:

*L*vap = *dU*+*V*lab*dP* (6)

How accurate is eqn (6)? First of all, eqn (6) assumes that there is no pressure dependence in the water molecule’s bonding energy. How accurate is that? Probably not an exact description but it maybe suffice for a rough approximation.

Furthermore,
eqn (6) assumes that the space lab’s volume is constant, which is a valid assumption!

Boiling in the space lab involves the mass transfer of water molecules into the isometric lab, hence the lab’s pressure should increase i.e. the mean molecular volume in the gaseous state must decrease. Therefore boiling in the space lab might need to consider any dependence of pressure with molecular volume. And if the gas inside of the space lab is ideal (PV=NkT) then the work might become a logarithmicfunction such as:

W=(NkT)In(Pf/Pi) (7)

Eqn (7) allows us to realize that as the pressure increases then the greater the work would be for another liquid molecule to change states from a liquid to vapour. However it is lacking in the fact that there should be something to account for the change in number of vaporous molecules i.e dN. If you have some insight into this plesae let me know, as I am far from perfect - call me human

Anyhow substituting 7) into 6) gives:

*L*vap = *dU*+* *(NkT)In(Pf/Pi) (8)

Again the above logarithmic functions are based upon the ideal gas
law would be an isothermal function hence ignores that this may also result in heat being added to the space lab’s interior, due in
part to increased molecular friction (inelastic collisions) i.e. the natural *P*-*T* relationship.

Certainly the understanding of
boiling in a rigid body in outer space with complete precision may become rather complex. However: Imagine the far-fetched possibility
that the space lab had some elaborate engineering that manages to keep the space lab isobaric. Since this engineering marvel keeps
everything in the lab isobaric then the latent heat of vaporization would equal negative of latent heat of condensation. Hence:

*L*vap(spacelab) = *dU* (9)

The realization is
that eliminating gravity simply means that no work was actually required. Of course this is not exactly true in the case of
the space lab because the pressure increases due to boiling and it eradication by the space lab’s devices may not necessarily be instantaneous
and continuous.

No matter we can now think of it this way, weight of the atmosphere on Earth means there is a downward
force that expanding system’s must overcome, and this force is non-existent in space.

Therefore boiling in the space lab is nothing
more than an increase to both the thermal energy and the number of gaseous molecules in the space lab. If in its interior pressure
and/or temperature were allowed to increase then the ability of the gas within the lab to do work will increase. Even so no work is
actually done.

There has been research done in zero gravity boiling by Herman Merte^{6} and others. They have found that due the lack
of buoyancy and convection in weightless environments, the boiling tends to involve large often singular bubbles, as the bubble tends
to stay near the heater, rather than form the cascade of smaller bubbles as normally witnessed here on Earth. It should be stated
that such experiments were performed under the premises of traditional thought, rather than the understanding presented in my book.

This website is copyright of Kent W. Mayhew who in 2018 resides in Ottawa Ontario Canada

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