# Positively charged gases

Theoretically, if you could could have a positively charged gas say O$$^+$$ enclosed (by a sufficiently inert material) around a conductor which is grounded, what would the reaction be at the conductor if the conductor was made of;

1) Tungsten, 2) Stainless Steel, 3) Copper.

Would the gas gain electrons from the conductor and form O$$_2$$, react with the conductor to form a coating, react with each other, react with the container or something else?

• So the $\rm O^+$ is in an inert container, and this container is in grounded tungsten (steel, etc)? – peterh - Reinstate Monica Dec 10 '19 at 17:10
• The O$^+$ is in an inert container with a grounded metal rod protruding from the bottom of the container in contact with the O$^+$. There is no way the O$^+$ can escape from the container as it is essentially sealed from the outside environment. All it can do is interact with itself, the metal grounded rod or the container walls. – ED9909 Dec 10 '19 at 17:20
• Colleagues, please refrain from close votes. This is a genuine question which naturally arises in some students' minds every now and then. Sending them away will not help them realize why this question should not be asked. – Ivan Neretin Dec 10 '19 at 19:45
• Could you please elaborate how you got the idea for this question? It`s very, well, unusual, to say the least. ;) – Karl Dec 10 '19 at 21:19

You totally can't have a gas (or any other state of matter) made of positive ions alone.

That's not what you asked, though. The question was: what if we theoretically could have it. Well, then in a pretty short instant you will stop having it. The said instant would be rather violent, too(*).

To put things into perspective, imagine an indestructible spherical vessel about as big as to contain 1 mole of gas at normal conditions, which implies the radius of 17.5 cm. Now let's magically put 1 mole of $$\ce{O+}$$ inside. That would give it an electric charge of about 96500 coulombs, which is quite a lot. The electric field outside the sphere would be $$|\vec E\,|={1\over4\pi\varepsilon_0}\cdot{q\over r^2}=2.836\cdot10^{16}\;\rm V/m$$

Do you know how big is this? Wrong! it is bigger than you think. Mere $$3\cdot10^6\;\rm V/m$$ would cause a breakdown in dry air. Good solid insulators can handle more, but I think we may safely assume that $$10^{10}$$ is enough to break anything that is made of atoms. Mind you, we have six orders of magnitude on top of that(†). The conductor will be instantly vaporized and thrown out with immense force, enough to start many nuclear reactions and create a bunch of weird elementary particles. What metal it was made of would be the least of your concerns.

So it goes.

(*) Reminds me of a certain what-if episode, but they used a bigger sphere.
(†) Thought we would break the Schwinger limit, but we didn't.

• Thanks for the answer. If there was only a limited number of positive ions where the electric field strength wouldn't break down the whole system, what would happen? – ED9909 Dec 11 '19 at 10:28
• Then they will simply rush to the grounded conductor, neutralize, and form a layer of metal oxide. Some might recombine into O2 and bounce away. – Ivan Neretin Dec 11 '19 at 10:35
• It is similar to a calculation I sometimes get undergrads to do, consider a tennis ball sized lump of NaCl with a 1 % excess of sodium ions. How much energy would you have in the tennis ball sized lump of "special salt" ? It is a lot of energy, if I recall rightly. – Nuclear Chemist Dec 29 '19 at 20:07