# Expanding plasma

1. Oct 30, 2004

### rich_mmiv

Hi,

When air is heated to a point where it turns into a plasma, is there any equation which governs how much the gas expands in the explosion?

Thanks

2. Oct 30, 2004

### Tide

That's rather vague. Generally, you would not have a single equation and what you use depends on what you want to find out and the physical situation. A set of fluid equations, Maxwell's equations and thermodynamic relationships may all be needed.

3. Oct 30, 2004

### pervect

Staff Emeritus
You can treat a plasma pretty much as an ideal gas

As the nucear weapon FAQ suggests

NukeFaq

The law for adiabatic expansion and compression is given in the FAQ. Assuming the expansion is adiabiatic ignores the energy loss due to radiation. Presumably the expansion is happening quickly enough that the energy loss to radiation is small, even though the amount of radiation may be large. This assumption really needs to be checked. Assuming it's correct, we can write:

$$P * V^\gamma$$ = constant

where P is the pressure, V is the volume, and $$\gamma$$ is 5/3 for a monatomic gas

Combine this with the ideal gas law, and one should be able to find the solution to your question.

The ideal gas law is

P * V = N * R * T

P and V have been previously defined, N is the number of moles in the gas, R is the universal gas constant, and T is the temperature

There's a lot more info in the FAQ, it's worth giving a closer read if you are interested in this problem.

To discuss your particular problem in more detail, I am assuming we are heating up a some number of moles N of air originally at some temperature T to a new, much higher temperature T', dissasociating it in the process.

Assuming the air dissasociates, we will have some greater number of moles of the new gas N'. Since nitrogen is the main component of air, N2 changing to N1 will roughly double the number of moles of gas. This process will of course require a lot of energy. The small percentage of air that is triatomic, such as CO2 and water vapor will make this factor slightly larger than 2.

Originally we had P * V = N * R * T

Keeping the volume constant, the ratio of the new pressure to the old pressure (which would typically be atmospheric pressure) is

P'/P = (N'/N) * (T'/T)

where N'/N should be roughly 2 as I argued previously.

If the expansion then continues adiabatically until P'=P, the volume should increase by a factor of

$$(\,\frac{N'}{N}\frac{T'}{T}\,)^{\frac{1}{\gamma}}$$

Disclaimer - I think this analysis is correct, but I don't offer a guarantee, thermo isn't one of my strong points.