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If the atmosphere were cooled to liquid, say the planet's average atmospheric temperature were reduced to 70K, how deep (above sea level - frozen oceans of water) would the resulting ocean be?

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I'm not sure it would be entirely liquefiable as the atmosphere gets thinner and thinner, but if we ignore that, then it's a simple matter of calculating density and ocean cover.

78% Nitrogen, 0.807 g/mL Source: Liquid Nitrogen

21% Oxygen, 1.141 g/mL Source: Liquid Oxygen

Ignoring the 1% of trace gases, average density of liquefied air, 0.87 g/mL, or 87% the density of water.

Mass of air, 1 ATM is equivalent to 33.9 feet of water. Source: link

and the Earth is 71% covered by oceans. From there it's just math. 33.9*(1/.87)*(1/.71) = 54.88 feet. That's quite a bit less than the oceans would rise by freezing solid (which might not happen given volcanism and warmth coming from the earth, but if even half of the average depth of 14,000 feet of oceans froze, a 9% increase in volume, we'd be looking at some 600 feet of sea level rise by freezing.

Colder temperatures might cause the liquid nitrogen/oxygen to contract a bit further but that's a ballpark answer - 54 feet, 10 1/2 inches. Some of the air might dissolve into the oceans prior to the oceans turning to ice, cause colder water can hold more gas, so the actual answer might be a little bit less than that.

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  • $\begingroup$ @Fred: But being liquid, they'd then flow downhill to the now-frozen oceans. $\endgroup$ – jamesqf Jun 25 '15 at 6:13
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For the atmosphere to be so cold it turns to liquid, extreme Snowball Earth conditions would exist. Any atmospheric gases that would turn into liquids would form a layer on the frozen surface of the Earth, that includes any surface rock that may be exposed and frozen water from oceans, lakes, etc.

This source, gives the amount of nitrogen and oxygen in the atmosphere is $78.084\%$ and $20.946\%$ and the total mass of the atmosphere as ${5.148} 10^{18} kg$ respectively. From this,

The mass of nitrogen in the atmosphere is $4.0198$ x $10^{18}$ $kg$

The mass of oxygen in the atmosphere is $1.0783$ x $10^{18}$ $kg$

The density of liquid nitrogen is $804$ $kg/m^3$

The density of liquid oxygen is $1140$ $kg/m^3$

Density is mass divided by volume, $\rho$ $=$ $m/v$

Using this the volume of liquid gas can be found.

The volume of liquid nitrogen, from the atmosphere would be

$4.0198$ x $10^{18}$/$804$ $=$ $5$ x $10^{15}$ $m^3$ $=$ $5$ x $10^{6}$ $km^3$

The volume of liquid oxygen, from the atmosphere would be

$1.0783$ x $10^{18}$/$1140$ $=$ $9.459$ x $10^{14}$ $m^3$ $=$ $9.459$ x $10^{5}$ $km^3$

These liquids would lie of the surface of the frozen Earth. The surface area of the Earth is $510,072,000$ $km^2$

Thus, the height above the surface of the Earth that liquid oxygen would rise to is

$9.459$ x $10^5$ / $510,072,000$ $=$ $0.001854$ $km$ $=$ $1.854$ $m$

and the height to which liquid nitrogen would rise, would be,

$5$ x $10^6$ / $510,072,000$ $=$ $0.009803$ $km$ $=$ $9.803$ $m$

Because liquid oxygen would form first due to its higher boiling point, the combined height due to both nitrogen and oxygen becoming liquids would be

$1.85$ + $9.80$ = $11.65$ m

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