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When you are diluting air in the system and/or properly evacuating it, you must consider water temperature. Methane is completely dissolved in water at 42 degrees Fahrenheit (5.5 C) but can be completely released (as a gas) at 58 degrees F (14.5 C). (source)

This would seem to imply that the enthalpy of solution for methane in water is negative (i.e. the process is exothermic). But, intuitively, it seems that you would have to input energy to force non-polar and polar to mingle.

Can anyone explain why this might be the case?

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    $\begingroup$ Did you check on how other gases are dissolved in water? $\endgroup$
    – tobias47n9e
    Sep 8, 2012 at 6:33
  • $\begingroup$ In that class? Yeah I looked for other specific linear alkanes, like ethane, and alkanes in general. No dice. $\endgroup$
    – readyready15728
    Sep 8, 2012 at 9:48
  • $\begingroup$ Spießbürger asked for "other" gases, not alkanes. Starting from propane or Butane things change a lot: this is about "permanent" gases only. Do what Spießbürger wrote. You will learn something very important. $\endgroup$
    – Georg
    Sep 8, 2012 at 14:36
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    $\begingroup$ "Spießbürger asked for 'other' gases, not alkanes." Other gases, which may not share important properties with methane. I have an idea of how carbon dioxide dissolves in water, for example. I still haven't got a clue about methane, which is unlike CO2. I was hoping for an answer, not an unhelpful, cryptic comment. $\endgroup$
    – readyready15728
    Sep 9, 2012 at 13:58
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    $\begingroup$ "Dissolution increases in liquid with decrease in temperature, as the" ... what? $\endgroup$
    – readyready15728
    Sep 9, 2012 at 14:09

2 Answers 2

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The enthalpy of solution going from the ideal gas state to the solution at infinite dilution is exothermic. However, more importantly, the Gibbs energy of solution is positive meaning that dissolution of methane into water is not a favored process. Most of this comes from the relatively large negative entropy change associated with the dissolution at room temperature. This positive Gibbs energy is manifested in the very small solubilities observed and gives rise to the whole concept of something being "hydrophobic." The fact that any methane dissolves in water is due to equilibrium effects based on a finite positive Gibbs energy. Since methane is dissolved to a greater extent at lower temperatures must mean that the Gibbs energy is less positive at the lower temperatures.

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  • $\begingroup$ Wilhelm et al. "Low Pressure Solubility of Gases in Liquid Water" Chemical Reviews. 1977. p.233 $\endgroup$
    – Jason Waldrop
    Sep 11, 2012 at 20:30
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Never mind, I think I found the answer.

In the gas phase, attractions between solute particles are negligible, while solvent-solute attractions may reduce the energy of solute particles, resulting in an exothermic process that favors lower solvent temperatures.

For the record.

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