I've been looking on the net for a reference which shows a phase diagram for elements at least from $\ce{H}$ up to $\ce{Ne}$. Specifically, I'm looking which of these elements can be solid or liquid in a vacuum.

I haven't really found anything, except some anecdotical comments here and there that hydrogen and helium evaporate in a vacuum at any temperature.

Any suggestions as to where these data might be published?

  • $\begingroup$ For the most part, I would expect any of the elements that usually exist as diatomic molecules such as hydrogen, fluorine, oxygen, nitrogen. Would fly apart extremely fast since nothing holds them together. While a substance like carbon graphite or diamond is all connected, and the vacuum probably would not break the covalent bonds. $\endgroup$ – Leonardo Sep 13 '12 at 7:33

This may be what suffices you are looking for :

This may get you started:

It is a very good question, which I came across as well. You could just google-image search phase diagram + ELEMENT to copy and past the most appealing charts into a table in your editor of choice. If the intend is personal use, this might be the fastest route.

wolfram alpha should allow you to look up the state of the element at a given pressure as follows:

sodium phase at 0 bar

Helium itself is a particularly interesting subject.

I hope this helps for a start...

  • 1
    $\begingroup$ +1 for link to Wolfram alpha... at least I now have a partial phase diagram for ethanol (who knew it would be so rare?!) $\endgroup$ – AJP Oct 11 '15 at 10:18

If you want to know the temperature at atmospheric pressure below which an element is a solid, just look up the melting point of that element.

As for evaporation of solids, ALL materials evaporate. That is, all materials exhibit a positive non-zero vapor pressure. That vapor pressure will vary with temperature. When the vapor pressure hits 1 atm, you are at the boiling point.

For phase diagrams of the elements simply Google for them. One promising source is here


from thermodynamical point of view at absolute zero pressure each and every compound will evaporate, given enough time. So, it is better to stick with finite (i.e. small, but non-zero) pressure.

Then again, phase diagrams usually has 'triple point' (at which solid, liquid and gas coexists) with three areas - lower-right one, where gas is stable, left one, where solid is stable, and upper, where liquid is stable, with barrier between liquid and gas fading at some point. This means, that liquid is doubtful to exist in vacuum.

Given that, you can hope for slowly evaporating solids, which are ones with crystal structure hold by metallic or covalent bond. Evaporation speed should drop with boiling point raise, which drops in the row of $\ce{C}$ (3915 K), $\ce{B}$ (3927 K), $\ce{Be}$ (2725 K), $\ce{Li}$ (1342 K). $\ce{Be}$ is sometimes used as construction material in satellite construction, so it should work fine, but is very toxic. Carbon should work fine as well, as to my knowledge was used in thermonuclear fusion program (where both vacuum and hight temperature stability are required) , but is hard to work with. $\ce{Li}$ and $\ce{B}$ are out of consideration except VERY special application, as $\ce{Li}$ is highly corrosive and flammable and $\ce{B}$ is almost as hard to work with as diamond.


All elements exist in as a solid in at least some pressure and temperature conditions, i.e. in some region of their phase diagram. In fact, all elements present a solid form at zero pressure (or atmospheric pressure) if you cool them enough, i.e. when they reach 0 K. All… except helium, which requires some pressure to solidify.

  • $\begingroup$ sounds promising, but i would like to know at what temperature does this happen for, say, hydrogen, at vacuum pressure $\endgroup$ – lurscher Sep 13 '12 at 14:26
  • $\begingroup$ @lurscher in your question, does “vacuum pressure” mean “zero pressure”? $\endgroup$ – F'x Sep 13 '12 at 14:52
  • $\begingroup$ yeah, vacuum pressure means zero pressure $\endgroup$ – lurscher Sep 13 '12 at 14:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.