The normal melting and boiling points of $\ce{O2}$ are $\pu{-218 ^\circ C}$ and $\pu{-183 ^\circ C}$, respectively. Its triple point is at $\pu{-219 ^\circ C}$ and $\pu{1.14 Torr}$, and it's critical point is $\pu{-119 ^\circ C}$ and $\pu{49.8 atm}$. Will $\ce{O2 (s)}$ float or sink on $\ce{O2 (l)}$?

I quickly sketched the phase diagram of oxygen:

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Pressure of the molecules will affect how compact the molecules are in solid and liquid, so I guess we can explain whether it will float or sink using the slope?

How does the slope tell us about the densities of $\ce{O2}$ solid and liquid ?

And in addition, how do I use intermolecular forces to explain this?

  • $\begingroup$ Floating or sinking is a matter of comparing the densities, or molar volumes of two phases; even though one can easily find the density of an ideal gas ($V_m = RT/p$), I'm unaware of how a phase diagram can be used for it. from what I remember, solid oxygen is heavier and will sink. $\endgroup$
    – andselisk
    Dec 12, 2017 at 6:35

1 Answer 1


Yes, we can do that.

In fact, it is only the sign of the slope that matters. Say, you stand right below the gas/liquid curve; you apply a little more pressure, and you step from gas to liquid. How does that correspond to Le Chatelier's principle? Well, just like that: the system responds to increased pressure by decreasing its own volume, hence the liquid must be denser that the gas (which we of course already knew from independent sources).

Guess what?

The same reasoning applies verbatim to the liquid/solid curve, since it is sloped in the same direction.

As to the intermolecular forces, I think the usual hand-waving along the lines of "less freedom $\to$ less motion $\to$ lower volume" will do, and even that is redundant. Having solid denser than liquid is normal. If anything, it is the opposite situation that would require an explanation.


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