How much energy would be released if 2L of (relatively pure) water compressed to 1L at room temperature was suddenly decompressed to atmospheric pressure? Even a first-order approximation would be helpful.

As far as I can tell, said water would end up sitting on the border between Ice-VII and Ice-X, as according to here ice-VII has a density of ~1.50g/cc and ice-X has a density of ~2.51 g/cc.

I am aware that increasing the density of water by 2x would require an rather high pressure (to put it mildly). It's not impossible, though - based on the chart here it's "only" about 7*10^10 Pa - and we've achieved higher sustained pressures in a laboratory setting before.

I'm not sure how to calculate this, though, as I have little data on either the density of the various phases as a function of pressure, or the energy of each phase change.

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    $\begingroup$ How much energy does it take to form the Ice-VII? Remember that work=force*distance. For a first approximation, assume the density of the phases doesn't depend on the pressure. $\endgroup$ – Peter Shor Jul 25 '16 at 1:57
  • $\begingroup$ There are "issues" with this question. The device that could compress water to the degree indicated, and the device that could hold the resultant pressure, probably does NOT exist for a volume of 2L. If you try this with a diamond-anvil cell, you might get useful results (for 0.1 ml of volume, or less). $\endgroup$ – David White Jul 25 '16 at 2:13
  • $\begingroup$ @DavidWhite - Consider it in the same vein as this question, then, if you wish. $\endgroup$ – TLW Jul 25 '16 at 3:33
  • $\begingroup$ @PeterShor - it actually goes from water to ice-VI to ice-VII to partially-to-ice-X, as far as I can tell. Quick estimate works out to something like 21MJ. $\endgroup$ – TLW Jul 25 '16 at 3:45
  • $\begingroup$ I've never worked with pressures as high as you are suggesting, but do know that at least with SCUBA - sized pressures I would rather prefer having a hydraulic line rupture than a gas line. While gas ruptures tend to release lots of energy, hydraulic ruptures are mild in comparison. Doesn't $\int{PdV}$ apply, and isn't dV small? $\endgroup$ – docscience Jul 25 '16 at 3:47

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