Physics of ISM: Pressure, Volumes, Masses

[sharrington3]

Homework Statement

Answer the following questions. Pleas show all of your work and your line of thinking and state your assumptions.

There are several phases of the ISM. Consider the following three phases with different typical number densities $n$ and temperatures $T$.

• The cold neutral medium (CNM) with $n \simeq 10 cm^{-3}$ and $T \simeq 100 K$;
• The warm ionized medium (WIM) with $n \simeq 0.1 cm^{-3}$ and $T \simeq 10^4 K$;
• The hot ionized medium (HIM) with $n \simeq 10^{-3} cm^{-3}$ and $T \simeq 10^6 K$;

(a) Compute the pressure of the three phases assuming the ideal gas law, and show that they are all equal. Describe the significance of this in a few sentences.

(b) If the gas in a galaxy were equally distributed between the phases so that 1/3 of the mass were in each of the three phases above, what would be the relative volumes filled by the three phases?

(c) For the Milky Way, half of the mass is thought to be in the warm phase, and it also resides in half of the volume of the galaxy. What are the relative volumes and masses of the other two phases?

Homework Equations

Ideal gas law
$PV = NRT$
$n = \frac{N}{V}$
$V_{total} = V_{CNM} + V_{WIM} + V_{HIM}$

The Attempt at a Solution

Part (a) was very simple to do. I found the pressure in each to be $8314 \frac{J}{cm^3 \cdot mol}$. For whatever reason, I cannot figure out part b, and I feel that c will easily follow. I've tried assuming the mass of the Milky Way, dividing it into thirds, and then using the molar mass of each phase of the ISM to find the relative volume of each phase. The problem is, the entire mass of the Milky Way is not in gaseous form for one, and another thing is that the CNM is mostly neutral hydrogen and the WIM is mostly singly ionized hydrogen. I can easily calculate the number of atoms of each necessary to make up their respective 1/3 of the mass of the galaxy, but the rest of the mass coming from the HIM is made up of everything else, so I can't calculate the number of particles in the HIM to find the amount of volume it occupies.

 

[mfb]

(a) Your pressure has a strange unit (especially with the mol). Why don't you calculate it in Pa?

(b) If you have 1kg of cold, 1kg of warm and 1kg of hot medium, what is the total volume they will occupy? Can you scale this, if necessary?
(c) can be solved similar to (b), right.

 

[sharrington3]

I converted my answer for pressure. I apologize, I simply forgot to do that before. I got that the pressure for each, assuming each follows the ideal gas law, would be about $1.38 \times 10^{-14} Pa$, which is right around the vacuum pressure out in space. For part (b), I made the assumption that number of molecules in each phase was the same, and I used that to calculate their relative volumes. I got a total volume that added up to 0.9999, so, about 99.99% of the total volume of a galaxy with such a distribution was represented (assuming that the volume of all three added to 1, where 1 is the normalized volume of whatever galaxy you're working with). I'm trying to solve these using only the information provided in the assignment proper.

I must be missing something important, or I'm an idiot, because I'm still having trouble with (c).

 

[mfb]

I don't understand your description of (b). What does the 0.9999 mean and how did you get it?

What did you try for (c)?

 

[paranormal]

For part (b), we can use the ideal gas law to find the number of moles of gas in each phase, and then use the number density to find the volume occupied by that many moles. Let's assume that the total mass of the Milky Way is 1 kg (this is just a rough estimate for simplicity). Then, we can calculate the number of moles of gas in each phase:

CNM: $n_{CNM} = \frac{1/3 \cdot M_{MW}}{m_{H} \cdot 0.1}$ (since 1/3 of the mass is in the CNM and the rest is made up of hydrogen atoms)
WIM: $n_{WIM} = \frac{1/3 \cdot M_{MW}}{m_{H} \cdot 1}$ (since 1/3 of the mass is in the WIM and the atoms are singly ionized)
HIM: $n_{HIM} = \frac{1/3 \cdot M_{MW}}{m_{H} \cdot 2}$ (since 1/3 of the mass is in the HIM and the atoms are doubly ionized)

Using the ideal gas law, we can find the volume occupied by each phase:

CNM: $V_{CNM} = \frac{n_{CNM}RT}{P} = \frac{1/3 \cdot M_{MW}}{m_{H} \cdot 0.1} \cdot \frac{8314 \cdot 100}{8314} = \frac{1}{3} \cdot \frac{M_{MW}}{m_{H}} = \frac{1}{3} \cdot \frac{1 kg}{1.67 \cdot 10^{-27} kg} = \frac{1}{3} \cdot 6 \cdot 10^{26} m^3 = 2 \cdot 10^{26} m^3$

WIM: ##V_{WIM} = \frac{n_{WIM}RT}{P} = \frac{1/3 \cdot M_{MW}}{m_{H} \cdot 1} \cdot \frac{8314 \cdot 10^4}{8314} = \frac{1}{3} \cdot \frac{M_{MW}}{m_{H}} \cdot