# Thermodynamic process when nebula is heated

The basic thermodynamics problem is stated as follows.

The nebula contains a very tenuous gas of a given number density (atoms per volume) that is being heated to a given temperature. What is the gas pressure?

1. What are the basic assumptions that should be taken in solving this problem? There is no sealed container, obviously, but if nebulæ were to allowed to expand indefinitely then it would be an isobaric process, would it not? Seen how that would not allow us to determine the pressure (?), then some sort of constraint is to be placed. If we are to assume that it is, in fact, an isochoric process, would it be a simple matter of finding molar mass from given number density and plugging it in the formula of Ideal Gas Law ($p = nRT / V$) assuming the volume of 1 metre cubed?

2. Given the numbers ($1 × 10^{8}$ atoms per m$^3$, 7500K) what should be realistic order of magnitude for the answer (in $Pa$ or $atm$) for the purposes of assessment as many certainly would not have intuitive grasp of your average nebular pressures?

In general, for questions like this (advanced question from introductory chapter) is it detrimental to overthink the problem, that is: do I look for more difficulty than I should?

• It turned out that my assumption of isochoric process was correct, but it is still not intuitively obvious to me. It was more of a logical conclusion based on combination of what was given, and what was to be found, rather than a result of firm understanding of theory behind the processes. – theUg Dec 11 '12 at 1:29
• It is worth reading about the Virial theorem in connection with this. The system definitely has pressure, but if it is large enough for self-gravitation to be significant it is not a uniform. – dmckee --- ex-moderator kitten Mar 24 '13 at 13:00

2) An interesting look at how interstellar densities are actually measured in practice can be found in this article by Jenkins and Tripp. Figure 7 shows the distribution of pressures of atomic hydrogen in their study. Taking the peak to be $$\frac{P}{k} = 3\times10^3~\mathrm{K}/\mathrm{cm}^3,$$ with Boltzmann's constant $k = 1.4\times10^{-23}~\mathrm{J}/\mathrm{K}$, this corresponds to $$P = 4\times10^{-20}~\mathrm{J}/\mathrm{cm}^3 = 4\times10^{-14}~\mathrm{Pa}.$$ This can vary by a few orders of magnitude (the numbers you give are for slightly denser region), but it gives you a sense of how thin the stuff out there is.