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This is a bizarre question, but here goes...

If all of the water in the oceans were boiled into steam by the newly forming molten earth, could the atmosphere retain the steam? In other words, how much "space" does the water of the oceans occupy? If converted into steam, would the "space" of the atmosphere be great enough to contain the expanded water vapors?

mvw
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David Caldarola
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    This isn't a mathematics question - probably Earth Science SE or Chemistry SE would be a better fit. – Zev Chonoles Jun 13 '15 at 02:22
  • Where else would it go? It would make the atmosphere rather thicker... – Robert Israel Jun 13 '15 at 02:37
  • This may be a global climate change problem... – DVD Jun 13 '15 at 02:53
  • Might also be well suited for world building stack exchange – Paddling Ghost Jun 13 '15 at 03:00
  • The issue isn't one of "space", but rather it is a matter of what the typical speed of the water molecules is relative to the Earth's escape velocity. Steam at around 373 K would not contain a sufficiently greater energy density than water vapor around the present-day Earth does. Considering that surface escape velocity for the Earth is 11.2 km/sec, the steam would pretty much remain part of the Earth's atmosphere. The atmosphere would just become a bit denser and extend a bit further above the surface. (continued) – colormegone Jun 13 '15 at 03:36
  • There would be a greater loss rate of molecules from the atmosphere, since the proportion of molecules which do reach escape velocity or faster would be larger. But the atmosphere would still retain just about all of the water vapor. What would make a difference is if the temperature of the accreting Earth from impact is much hotter than the boiling point of water (say, several hundred K or more), in which case the vapor would pretty largely escape. – colormegone Jun 13 '15 at 03:40
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    I believe physics is a better place for this, and earth science may be better yet. Physics can compute how much more dense the atmosphere would become, how much higher into space it would extend, and how much more gas would escape. This is a good question, but in the wrong place. – Ross Millikan Jun 13 '15 at 03:56
  • Thanks for all of the info. I have found pretty much what I was looking for. David. – David Caldarola Jun 16 '15 at 03:57

1 Answers1

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Using mostly geometry for estimation:

In other words, how much "space" does the water of the oceans occupy?

Earth radius is about $r = 6400 \mbox{ km}$. Water covers about $70\%$ of the earth surface, estimation: $A_w = 4 \pi r^2 = 360,302,978 \mbox{ km}^2$

Wikipedia lists $360,570,000 \mbox{ km}^2$.

Average ocean depth is $d = 3.8 \mbox{ km}$. So a crude estimate would be a water volume of about $V_w = A_w d = 1,369,151,317 \mbox{km}^3$.

I found an estimation of the water volume as $V_w = 1,386,000,000 \mbox{km}^3$.

If converted into steam, would the "space" of the atmosphere be great enough to contain the expanded water vapors?

A mol of water weighs about $18\mbox{ g}$ this is a $18 \mbox{ ml}$ volume. As a gas it would take about 22 litres (depending on temperature and pressure). That is roughly an increase in volume about $1200$ times.

How much should we extend the earth radius for this? $$ 1200 V_w = \frac{4}{3}\pi (r + dr)^3 - \frac{4}{3}\pi dr^3 = \frac{4}{3}\pi (r^3 + 3r^2 \, dr + O(dr^2)) \Rightarrow \\ dr \approx \frac{\frac{3}{4} \frac{1200 V_w}{\pi} - r^3}{3r^2} = 1059 \mbox{ km} $$ Wikipedia lists that about $75\%$ of the atmosphere is contained within $11 \mbox{ km}$. The space border is about $100 \mbox{ km}$ (Kármán line).

So this would need a much larger volume. In reality the temperature is low (shrinks the gas volume) as is the pressure (expands the gas volume). However I saw only compression factors in the range of 1 to 10 for low temperatures.

So my crude guess is that the needed volume to contain the atmosphere plus water as some form of steam would be vastly greater than what our present atmosphere contains.

mvw
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