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Consider a phase diagram of Oxygen Nitrogen.

Now my question is : if a take a box containing air (80 % N2 - 20 % O2) at standard pressure and I start cooling it from room temperature to -196°C (BP of N2) what will be the composition of the liquid I obtain at -196°C ? What I understand is that at the dew point of air (-192°C) the composition of the mixture will be almost 50-50. Bt what will happen to this composition if I cool the box to -193°C, then -194°C... to -196°C? How will the composition of the liquid phase change ? and what will it be at -196°C ?

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  • $\begingroup$ The link provided (hxxp://liquidair.org.uk/full-report/report-chapter-nine) is now linking to a compromised web page. Do not click! $\endgroup$
    – Mark Beadles
    Commented May 2, 2019 at 14:21

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You need to do a dew point calculation for the two component mixture of oxygen and nitrogen. Dew point calculations are described here: https://en.wikibooks.org/wiki/Introduction_to_Chemical_Engineering_Processes/Vapor-Liquid_equilibrium#Dew_Point. In addition, if you want higher purity of either oxygen or nitrogen than you obtain from the dew point calculation, you would need to run the mixture to a distillation column and distill it.

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