I have $N$ ${\rm H}^+$ gas molecules in a sealed, electrically insulated container. What equation can I use to accurately calculate the pressure of the gas?

It seems to me that $PV = nRT$ will not apply to this scenario as the electrical forces between the molecules are significant, especially as the density of the gas increases.

Additional Thoughts

It is a complicated problem because there are a lot of interactions between energies and forces, lots of questions that need to be considered, for example:

  • Is the pressure at every point in the container the same, or does the shape of the container affect the pressure as the molecules attempt to be distant from each other?
  • Does the repulsive force of the gas cause the molecules to stay more or less stationary?
  • How much energy is stored in the electric fields between the molecules?

Why this is interesting

I am trying to determine how effective this device would be as a capacitor/battery. How does this device compare to Li-Ion batteries in storage capacity? Could this be practically used as a piezoelectric or linear actuator? See: https://i.sstatic.net/DBTWn.jpg, https://i.sstatic.net/48kr1.jpg

Any insight would be appreciated!

  • 3
    $\begingroup$ This isn't an ionised gas. An ionised gas has electrons too. $\endgroup$
    – ProfRob
    Aug 28, 2015 at 4:59
  • $\begingroup$ @RobJeffries. Ok. what should I call it? Also can you share a link with a definition of an ionized gas. $\endgroup$
    – Jakobovski
    Aug 28, 2015 at 15:46
  • $\begingroup$ An ionised gas is one where its electrons and nuclei have been separated. The electrons are still present in the gas. $\endgroup$
    – ProfRob
    Aug 28, 2015 at 21:51
  • $\begingroup$ Just a few notes: First, in the ideal gas you essentially linearly superpose the gas ingredients, so you can solve every gas on its own and then see add the results for all gas components. However, once you introduce interaction between gas particles, this is not possible any more and the whole set of components has to be solved together. As a result, a "superposition" of the ionized $H^+$ and electrons will have a very different behaviour than a formal mathematical sum of the pressures/energies when treated as single components. So make sure you're not wasting your time on something useless. $\endgroup$
    – Void
    Aug 30, 2015 at 10:38
  • $\begingroup$ Second, the Coulomb force is a long-range force, so this might be a really difficult mathematical problem to solve. The first correction you can obtain is by computing the change in potential energy with the configuration slightly varying in volume (i.e. for a slightly smaller charge density). This will give you the pressure exerted even purely "unthermodynamically", and it will always depend on the container shape. The second step is the virial expansion but the validity of the expansion will be limited only to very dilute $H^+$ gases. $\endgroup$
    – Void
    Aug 30, 2015 at 10:47

2 Answers 2


Pressure inside the container: zero
Effective pressure on the container: proportional to$\frac{N^2}{r^4}$
(where $r$ indicates the size of the container)

The situation: There are $N$ protons inside an otherwise empty container.
Assumptions: The container does not interact with the protons except for completely stopping them from moving outside (essentially, a potential wall). The container is perfectly insulated, not allowing outside influence.

The charges in vacuum will act as in a conducting material, since there is nothing bonding them to their immediate vicinity. Thus, the protons will have migrated to the edge of the container. (See Gaussian surfaces.) This image is quite different from normal gas molecules. This is because gaseous groups of atoms simply move about and bump into each other. A group of charged particles, however, will "want" to be as far away from each other as possible; minimising the force between them.

What, then, is left over of our concept of "pressure"? When wanting to know the pressure of something in a container, it is usually to know the pressure on the container walls, or wanting to know what pressure a gas feels when added into the container. However, these protons do not interact with matter as gas usually does in the macroscopic world. I don't know my quantum mechanics (or chemistry?) well enough to consider these kinds of interactions, which is why we will only consider a "potential wall", and not a material one. (Perhaps Steve B's answer brings some light on the topic?)

Pressure of a gas within the container

As with all Gaussian surfaces, the electric field created by these photons is zero on every point inside the container. This means that aside from the chemical effects of the gas molecules directly bumping into them (which may or may not be significant, I don't know), the protons have no effect on any gas molecules inside of the container. This may be interpreted as the protons having no contribution to the pressure inside the container.

Pressure on the container itself

First of all, the pressure experienced by the container is equal to the pressure experienced by the protons on the edge of the container. This is true simply because of Newton's third law, and the fact that the protons are not moving perpendicularly to the surface of the container. We denote this value as $p$.

Second of all, the pressure on the protons is equal to the energy density caused by the electrical field produced by the protons themselves. This is as true for charged particles as it is for gaseous ones[1].

Luckily, the energy density $U_E$ of an electrical field $E$ in vacuum is easily found[2]: $$p = U_E = \frac{1}{2}\varepsilon_0 E^2$$

Thus, if the electric field on the surface of the container can be found (which depends only on the number of protons and the shape/size of the container), the pressure can be calculated for any area on the surface of the container.

Example: Spherical container

The electric field on the surface of a conducting sphere is also well known[3]: $$E = \frac{Ne}{4\pi\varepsilon_0r^2}$$
(where $e$ is the elementary charge; the charge of a proton, and $r$ is the radius of the sphere)
This results in a pressure across the surface of the sphere of: $$p_{sphere} = \frac{N^2e^2}{32\pi^2\varepsilon_0r^4}$$

The pressure experienced by differently shaped containers will not be so easily found; the pressure would not even be constant across the surface, as opposed to regular gaseous pressure. However, pressure on differently shaped containers will still follow the same proportionality to the number of protons $N$ and the size of the container $r$.

As per rmhleo's comment, it is true that temperature will likely cause a distribution of protons with more at the edges and less in the centre. However, this effect should be minimal. Gas can extend space because the particles are neutral. Once a ranged force is present between these particles, however, the particles will clump up. An example of this would be the Earth's atmosphere (see that thin blue line?). Since the electromagnetic force is much, much stronger than gravity, this effect is observable even at scales much smaller than planetary.

To get a rough idea, the interior of the container is to the Gaussian surface as outer space is to Earth's atmosphere.

  • $\begingroup$ Adding references here, since I don't have enough reputation to add multiple links: [1] hyperphysics.phy-astr.gsu.edu/hbase/press.html#ed [2] hyperphysics.phy-astr.gsu.edu/hbase/electric/engfie.html [3] hyperphysics.phy-astr.gsu.edu/hbase/electric/elesph.html#c2 $\endgroup$
    – Gyoshi
    Sep 12, 2015 at 22:26
  • 1
    $\begingroup$ This answer seems right to me. However there is an important difference with a conducting sphere: the particles will not rest in the edge, because the kinetic energies do not dissipate into the medium. Rather, the picture should be protons bouncing in the frontier mostly accumulated, but the charge distribution inside the container should be higher near the edge and decreasing towards the center of the sphere. $\endgroup$
    – rmhleo
    Sep 13, 2015 at 0:18
  • $\begingroup$ @rmhleo You're right, kinetic energy of the particles (i.e. temperature) should cause some deviation from a perfect sphere of charge. However, this is true for all Gaussian surfaces. Particles may be bumped towards the interior of the container, but as more particles "deform" the Gaussian surface, these particles should dissuade further movement towards the interior. This does mean that temperature causes non zero electrical fields, but since it is said that E=0 on the interior of Gaussian surfaces, it must mean that this effect is minimal for any realistic temperatures. $\endgroup$
    – Gyoshi
    Sep 13, 2015 at 8:29
  • $\begingroup$ I this is the case probably for non isolated bodies, but I think that charges inside an isolated container cannot get rid of their kinetic energy, so they will be spread in the interior as much as the Coulomb overall potential allow them. $\endgroup$
    – rmhleo
    Sep 17, 2015 at 9:32
  • $\begingroup$ If the outside of the container was a conductor and inside a perfect insulator. And electrons could freely flow onto the conductor. Would there still be a pressure? Would the electrons on the conductor cancel out the force of the protons? $\endgroup$
    – Jakobovski
    Oct 2, 2015 at 17:46

A large number of positive ions without any negative charges cancelling out the charge is extremely difficult to achieve in practice, which is why no one ever talks about it. You certainly can't use PV=nRT.

An important consideration you left out is where are the negative charges? I understand that they are not in the container, but they are somewhere, perhaps right outside the container, getting as close as possible. In other words, if you draw the electric field lines emanating from the positive ions, they have to end at some negative charge somewhere. That's the negative charge I'm referring to.

Once you draw in the negative charges, you realize that you have a kind of capacitor. And you can (to some extent) use normal capacitor logic to think about what will happen. For example, if there are regions of the container where the negative and positive charge can approach very close together (for example, where the wall is unusually thin), then you expect the ions to pile up with a higher density at that location, and the electrons to pile up on the other side.

You still have entropy, so the ions will not be stationary. You need to pull out your electrochemistry textbook for the formulas describing how the ions trade off entropy versus the desire to get to low potential. See Debye length for an equation in that category, although I'm not sure it's exactly the one you want.

Continuing in your electrochemistry textbook, you also need to keep in mind various other considerations. For example, if there is a strong enough electric field across the container walls, you should expect chemical reactions to happen -- most simply, the ions might rip electrons off of the atoms in the container wall.

Since you have an ionic "solute" but no solvent and no counter-ions, this is not conventional electrochemistry. But it's similar physics. If you get really good at conventional electrochemistry derivations, then maybe you can alter the formulas to fit your situation.

  • $\begingroup$ 1. Yes it is hard to achieve, but probably possible. See NASA's ION engine, and spectrometry. 2. Yes this is a capacitor. In fact that is why I am in interested in this. If this is feasible it might be a very useful battery/capacitor. Also depending on the container it can act as a piezoelectric or linear actuator. $\endgroup$
    – Jakobovski
    Sep 13, 2015 at 7:26
  • $\begingroup$ @Jakobovski Or those novelty glass globes filled with fluorescent plasma. Not completely ionized, of course, but you can see that locally reducing capacitance causes the gas to pile up. $\endgroup$ Sep 13, 2015 at 7:44
  • $\begingroup$ @BlackbodyBlacklight - The "novelty glass globes filled with fluorescent plasma" is not an example. Even if it were completely ionized, it would be a mixture of positive and negative ions. $\endgroup$ Sep 13, 2015 at 12:36

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