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The majority of the Moon's mass is oxygen. With energy it could chemically be liberated from its silicon et cetera compounds and be turned into oxygen gas. Given its low gravity, lack of magnetosphere and vicinity to the Solar wind, could enough oxygen gas be produced on the Moon to keep up with the loss to space, so that a pure oxygen atmosphere breathable to humans could be maintained for some time (thousands or millions of years)? Inert gasses would be a problem, I suppose, because of the lack of nitrogen, carbon and nobel gasses, but oxygen is all that we need. Something like a pure oxygen atmosphere at a fifth, maybe a tenth, of Earth's atmospheric pressure is survivable. Aside from technological, economical and mental sanity reasons, is it also somehow physically impossible? Could an object like the Moon have a thick atmosphere even temporarily? Titan does.

How would one estimate the kind of energy required to produce enough oxygen gas, as a fraction or multiple of the Solar irradiance on the Moon?

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    $\begingroup$ looks like worldbuilding question. Here is aluminium production energy 54 MJ/kg - it is half Al half O2 by mass. For each square meter of surface you need 60t of oxigen, minimum for 1bar.(lets skip pyre oxigen atmosphere dangers) that means roughly you have rip about 60m deep of moon crust(probably more), all around of moon. Energy $3.24 TJ/m^2$ for surface of moon. $\endgroup$ – MolbOrg Jul 3 '16 at 13:23
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    $\begingroup$ Relevant (but not a duplicate): Could the Moon keep an atmosphere? $\endgroup$ – DarkDust Jul 3 '16 at 13:45
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The Moon is capable to have an Earth-like atmosphere. Although the escape velocity on the Moon is much smaller as on the Earth, it is still around 6 Mach.(*)

If the Moon had an atmosphere, it would lose it in around some tens of millions of years(*). This is why it had long gone if it ever had, but this would be enough for us, in our time scale, which is only some thousands of years on the most optimistic sense.

Nitrogene, it doesn't have the Moon, but it is not really needed. We don't use it for anything. What matters, is the partial pressure of the O2 in the atmosphere. In a 100% O2 atmosphere with around 0.2 bar pressure (20% or the air is O2) we could breathe without any major problem.

But, the Moon has a much weaker gravity, it is only around 1/6 or the Earth. Which requires to have around 6 times more of per-area mass of the atmosphere, to get the same pressure. Which means, around of 1.2 kg of O2 is needed over every $cm^2$ to get the required O2 pressure (on the Earth, it is around 1kg).

The Moons radius is also around 1/6 of the Earth, so its surface area is 1/36 Earth. The surface area of the Earth is $5.1*10^8 km^2$, so it is $14*10^6 km^2$, which is $1.4*10^7*10^{10}=1.4*10^{17} cm^2$. This requires around $1.6*10^{17}$ kg of $O_2$.

The lunar soil is mainly from the mineral named regolith, it is essentially a mix of different metal oxides. It melts around at $1200K$, which is essentially lava. On the Earth, a similar thing is coming from the vulcanos.

Well, most of its components have a much higher melting point (for example, $Al_2 O_3$ has many thousands of K), but not all of them. The molten salt mixes have normally a much smaller melting point, this is why this around 1200K would be enough.

It is bad electric conductor, but it is enough conductive to be electrolysable, and this around $1000K$ is not an unreachable high temperature. Although most of the electric energy will simply heat the lava, a significant part will produce $O_2$. The electric effectivity of different electrolysis processes are in the order of 10-60%, so we can now calculate with their geometric mean, which is around 25%.

The burning heat of the Al is around 22 $\frac{MJ}{kg}$, its molar mass is 27. Thus, burning of 1kg of Al gives 22MJ of energy, it is 37 mol, which consumes 37*1.5 (Al2O3!), so 55 mol of $O$. This 55 mol of oxigene is 888g.

Thus, to produce an Earth-like O2 atmosphere on the Moon, we would need to produce around $1.8*10^{17} kg$ of Al. This would require around $4*10^{18} MJ$ of energy. It is $4*10^{24} J$. Considering an effectivity of 20%, it is around $2*10^{25} J$.

The yearly Al production on the Earth is around 40million tons. To produce this mass of Al we would need 4.5millions of years - on the Earth.

According to this article, on the Earth were around 184 TWh solar energy produced in 2015. It is $6.62*10^{17} J$. Thus, to produce the required $2*10^{25} J$ of energy, we would need around 30million years.

But, from the other side: although these project doesn't seem feasible in our lifetime, maybe sometimes, in a better world, it would be possible to make automatized factories by robots who are capable to build the required number of PV cells and electrolysis chemical factories on the Moon. Actually, if we have enough robots, we can do anything, and robots can be also produced by robots.

(*) I am very happy to google some reference for ask.

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  • $\begingroup$ about 11 years worth of solar power collected over the whole surface of the Moon. Solar irradiation of 53.1 petawatt. Create lenses (solar sail like mirrors?) big enough... $\endgroup$ – SF. Jul 5 '16 at 21:11
  • $\begingroup$ @SF. True... I wish I would once work on such a project. :-( Btw, in our current world, we can't even terraform the Sahara. $\endgroup$ – peterh Jul 5 '16 at 21:46
  • $\begingroup$ and with current state of hibernation technology we'd overshoot by a couple centuries. ;) $\endgroup$ – SF. Jul 5 '16 at 22:07
  • $\begingroup$ @SF. Consider a network of tremendous photovoltaic plants, a network of water desalination plants and a network of palm farms. It could give work to a billion of people and also solve the oil problem of the world, on a zero-CO2 way. The only reason why it doesn't happen that nobody cares on it. $\endgroup$ – peterh Jul 5 '16 at 22:07
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    $\begingroup$ @DNA We doesn't get it from the athmosphere, we get it from the proteins we eat. And the plants get it from the nitrate salts in the earth (or, a few of them, by microbes who are really capable to get it from the air). $\endgroup$ – peterh Aug 12 '16 at 4:02

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