12
$\begingroup$

As I know, the gases are insulators, because their particles are mainly electrically neutral, and thus there is no movable charges in them.

But maybe it shouldn't be always so, for example if a gas has a significant proportion of charged particles, I think the electrical conductivity isn't impossible. If at least one of the electrons of its particles bounds very weakly, maybe even at room conditions is it possible to have enough low resistance to be considered as a conductor.

Am I right? Does a such gas already exist? Or there is some mechanism which prohibits this idea?

Extensions:

  1. Maybe a such gas could be considered as plasma, but I think from this aspect we could see the plasma as gas as well.

  2. I am thinking at around 1 atm pressure and 300 K temperature, in the gas, not around the box around it. Electrostatic discharge is also problematic, because in most gases it can happen only far from the room conditions and isn't an equilibrium state.

  3. If there is none, I am looking for the gas which is the possible nearest to the these 3 conditions (300 K, 1 atm, conductive).

$\endgroup$
4
$\begingroup$

The gas inside discharge neon lamp is not really hot. Also, think of St. Elmo's fires - these definitely do appear at ambient pressure and are "cold", i.e., normally they would not burn anything. But there is a catch: here we are looking at strongly non-equilibrium situations, caused by strong electric fields.

As for the gases which would behave like that in normal conditions, I haven't heard of such. Some compounds turn into plasma easier than others, but still that requires many hundreds °C. Among the easiest are complex halogenides like $\ce{KBF}_4$ or $\ce{KAlCl}_4$. At 1000K they may be not vaporized yet, but have some vapor pressure. Also, it's not like the vapors consist entirely of ions - no, far from that, the ions are still a minority, but that's enough to create some significant conductivity.

$\endgroup$
  • $\begingroup$ Mercury boils already at 357$^{\circ}\mathrm{C}$! Maybe some metal mixture do even lower! Is it conductive? (Btw, it is not about mercury vapor lamps, I am thinking here on the ohmic resistance of 1atm mercury vapor in equillibric state, without any discharge effect.) $\endgroup$ – peterh Sep 17 '15 at 22:47
  • $\begingroup$ No; why would it? The mercury vapor is atomic. $\endgroup$ – Ivan Neretin Sep 17 '15 at 22:51
  • $\begingroup$ Because mercury is a conductor, although not a really good one. I think, if the fluid mercury has enough moveable charges, maybe as a vapor should it have as well. $\endgroup$ – peterh Sep 17 '15 at 22:53
  • 2
    $\begingroup$ No, it should not. Liquid mercury has some free electrons due to the fact that it is a metal, with metallic bond and all that comes with it. Gaseous mercury does not have metallic bond, or indeed any bond whatsoever; surely it should not have any free charges. $\endgroup$ – Ivan Neretin Sep 17 '15 at 23:00
  • $\begingroup$ A plasma is very hot if we consider the distribution of the energies of the different atoms which are present. A plamsa might have very little heat in it compared with a normal gas of the same volume as in many plasmas the density of gas atoms is very low. $\endgroup$ – Nuclear Chemist May 19 '18 at 5:23
6
$\begingroup$

Air at STP does conduct a tiny bit due to ionization by cosmic rays; this might even provide a path for lightning leaders. "Alex V. Gurevich of the Lebedev Physical Institute [et al] suggest that... cosmic ray... might provide a conductive path that initiates lightning."

In addition, ionization-type smoke detectors use a little radioactive material (e.g. $\ce{^241Am}$) to increase conductivity, which can be detected with a high-impedance amplifier.

$\endgroup$
  • $\begingroup$ Although I was looking for equilibrium-state gas, thank you very much this answer, too! $\endgroup$ – peterh Sep 9 '15 at 12:45
  • $\begingroup$ +1 for radioactive ionisation. I expect that in a sufficiently high radiation flux air (or perhaps even easier, some other gasses) could become significantly ionised. Searching for papers on breakdown voltages and corona discharge inside nuclear reactors might give further clues. $\endgroup$ – KalleMP Sep 4 '17 at 12:54
6
$\begingroup$

It depends on what you are prepared to consider a gas and what you are prepared to consider room conditions.

The gas inside all discharge lamps (fluorescent lamps and neon lamps in shop signs, for example) conducts electricity and the lamps work under normal room conditions. However, the gases are often at low pressure and are, strictly speaking, plasmas not room temperature neutral gases. Some lamps, especially high intensity ones like the xenon discharge lamps in floodlights) exist with high pressure gas when working so low pressure isn't the key.

And air itself will conduct electricity given the right conditions. Lightning couldn't exist if it didn't. Nor could you get the (much safer) discharges from Tesla Coils or Van de Graaff generators. But, again, the effect relies on the creation of a plasma from the breakdown of the neutral molecules in the gas.

So the answer is "yes" but only if you allow plasmas created from the medium to count as a normal gas.

$\endgroup$
  • $\begingroup$ Ok, but although a lamps works obviously on room conditions, the filling gas therein doesn't: it has a high temperature and small pressure for that. The case is similar in lightning: it is only a short-term process and has also very high temperature. The filling gas mix in the halogen lamps doesn't have significantly lower excitation energy as any other, they are only used because the halogen-cycle improves the lifetime of the lamp. $\endgroup$ – peterh Sep 7 '15 at 18:49
  • $\begingroup$ @peterh I think that's halogen filament lamps where the halogen is used to regenerate the tungsten filament. Halogen discharge lamps use doping by metal halide salts to alter the colour of the gas discharge in the gas plasma (some transition metal halides have specific emission lines that can make white light when combined). $\endgroup$ – matt_black Sep 7 '15 at 18:53
1
$\begingroup$

We need to distingish between conduction mechanisms at very high electric field strengths such as that in St-Elmmo's fire and the conduction which occurs at low electric field strengths.

When the electric field is very high, the breakdown of the dielectric occurs in some sort of casscade. What will happen is a fast moving electron or other particle will cause ionization of molecules. The electric field will accelerate these charged particles and this will allow the process to occur again. In this way the density of charged particles will increase, this conversion of unionizised molecules / atoms into electrons and cations will increase the temperture of the gas / plasma.

Some of the ionized gas atoms will emit photons, these photons can sometimes cause ionization of other gas molecules / atoms. Or they will take energy away from the spark.

As more and more current flows through the spark the number of electrons and ions will become higher. For a DC spark to be stable in a given system then there will be a given current which must flow. In an AC system when the frequency is low (like 50 Hz) the spark may snuff out as the current drops to zero during the cycle. The cations and the electrons during this time may be able to recombine thus lowering the charge carrier density so far that the spark is snuffed out.

In a RF plasma the charge carriers have too little time to recombine during the zero current momments in the cycle. Here the plasma / gas remains conductive during the whole cycle.

Now with a low electric field the casscasde of reactions caused by the rapid acceleration of the electrons does not occur. Here the conduction of the gas is dictated by the radiation level (either gamma, beta, alpha, X rays or hard UV) which creates electrons and cations.

It is interesting to note that during the Fukushima accident that the radioactive release at one point modified the earth's electric field. You can read an academic paper about this here.

$\endgroup$

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.