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Water boils at positive temperatures when put into a vacuum. Is this the case with all liquids, e.g. mercury?

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Superfluid liquids technically do not boil with bubbles, but they evaporate nevertheless from the surface. When helium stops to boil, it is a clear sign of suprfluid transition. –  firtree Aug 24 at 11:44

2 Answers 2

I take your question as

Is there any substance with condensed (solid or liquid) equilibrium phase at zero pressure?

No, because of statistical physics.

Let's consider two things. (1) The potential energy of interaction between molecules. (2) The thermal energy distribution for molecules.

The potential energy of interaction can generally be of any forms, with attraction, repulsion, extremums, but it is always $0$ at the infinite distance $r\to 0$, which means two molecules become free when they are far enough. Any bound state does then have energy below $0$, and any state with energy above $0$ is unbound, such that molecules move away even if they were close at some instant.

The thermal energy distribution at equilibrium always has some kind of "tail" in high energies, in the form of $\exp(-E/k_B T)$. It is the common feature of Fermi-Dirac, Bose-Einstein and Maxwell-Boltzmann distributions, while all differences lie in low energies. That said, there is no energy limit above which the probability for a molecule would be $0$.

These two facts together say that any condensed phase at zero pressure and $T>0$ would lose molecules, never reaching equilibrium. Though the rate of this process could be extremely slow and experimentally irrelevant.

What makes a condensed phase stable at some non-zero pressure? Then there are always some external molecules incoming that compensate the loss by evaporated molecules.

Two other cases are worth mentioning. First is atoms and atomic nuclei, aren't they stable? The same reasoning applies to them, but their binding energy is rather high, and thus the probability of evaporating even one particle at room temperature is extremely low (exponent is a very quickly diminishing function). Though at higher temperatures they reach equilibrium in plasma and nucleonic plasma respectively. These plasmas can be made at arbitrary low pressure, then there would be no atoms or nuclei.

Second case is quarks in nucleon. Here the energy of interaction does not vanish at infinity, so nucleons are true bound systems (at finite temperature). Though that is not independent of temperature: at some very high temperature there exists a sea of free gluons and quark-antiquark pairs, and the energy of interaction changes, becoming non-binding at infinity, like it happens in quark-gluon plasma.

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The boiling point of liquids depends on temperature and pressure. If the pressure of the medium that the liquid presents increases, the boiling point of the liquid increases as well. Since perfect vacuum has no pressure, all liquids boil in a perfect vacuum. However, there is no such a thing as perfect vacuum(even space is %99.99 vacuum). If you are asking if all liquids boil in space, the answer is no. If there is not sufficient heat then some liquids won't boil in space. In addition, if the heat of vaporization and the vapor pressure of a liquid at a certain temperature is known, the boiling point can be calculated by using the Clausius–Clapeyron equation thus:

$T_B = \Bigg(\frac{1}{T_0}-\frac{\,R\,\ln(\frac{P}{P_0})}{\Delta H_{vap}}\Bigg)^{-1}$

where:
$T_B $= the boiling point at the pressure of interest (in K)

$R$ = the ideal gas constant, 8.314 J · K−1 · mol−1

$P$ = is the vapour pressure of the liquid at the pressure of interest, either atm or kPa depending on the standard pressure used

$P_0$ = is some pressure where the corresponding T_0 is known, (usually data available at 1atm or 100kPa)

$\Delta H_{vap}$ = the heat of vaporization of the liquid, J · mol−1 at P_0

$T_0$ = the boiling temperature, in K

$\ln$ = the natural logarithm

You can calculate the boiling point of the liquid by this and then you should convert it to heat since the vacuum has no molecules(thus temperature). If the heat presents in the vacuum is larger than the heat you've calculated than you can say the liquid boils in that medium.

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In the limit $P\to 0$ the formula gives $T_B\to 0$. The liquid itself does have some heat so it boils until the rest is frozen into ice, and then it continues to sublimate (evaporate being dry). –  firtree Aug 24 at 12:04

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