# Spherical cloud of molecular nitrogen problem

1. Jan 26, 2006

### stunner5000pt

How big would a spherical cloud of molecular nitrogen at a uniformed pressure of 1 atm (1x10^5 Pascal) and a temperature of 300 K have to be for it to collapse under the mutual gravitational attraction of its molecules? i.e., What is the critical radius for Jean's collapse of a nitrogen cloud initially at this temperature and pressure? Compare your answer with the radius of the Earth and comment. Repeat for a cloud of molecular hydrogen at a pressure of 1 matm and 300 K and compare this with the radius of the Sun.

jean's criterion is
$$\frac{GM^2}{R_{C}} \geq \frac{3}{2} NkT$$ where N is the number of molecules
$$R_{C} \geq \frac{2}{3} \frac{GM^2}{NkT}$$
now over to PV = NkT [/tex]
im assuming this gas will form a sphere... so the volume is 4/3 pi r^3
$$\frac{GM^2}{R_{C}} \geq \frac{3}{2}PV = 2 \pi P R_{C}^3$$
$$R_{C}^4 \leq \frac{GM^2}{2 \pi P}$$ (1)
the probem is the MAss of this cloud
PV = n RT where n is the number of moles
$$PV = \frac{MRT}{M_{m}}$$ where Mm is the molar mass
substituting this into 1
$$R_{C}^4 \leq \frac{8GM_{m}^2}{9RT} \pi R_{C}^6$$
so
$$R_{C}^2 \leq \frac{9RT}{8GM_{m} \pi}$$
is this fine?

2. Jan 27, 2006

### stunner5000pt

3. Jan 27, 2006

### Staff: Mentor

The problem is outside of my area of expertise, and probably Space Tiger would have an answer.

However, it would seem the question in quotes is somewhat contradictory.

Any cloud with a mass that would cause it to collapse, would not have a uniform pressure of 1 atm (105 Pa), but rather a pressure gradient.

I am guessing that the answer is something like - To collapse, the attractive force due to gravity at the outer radius must be such that the molecular velocity (mean or most probable?) must be less than the escape velocity at that radius.

4. Jan 27, 2006

### SpaceTiger

Staff Emeritus
Alright, you went in several circles here and made a mistake near the end (your answer is only in terms of temperature), so I suggest starting over and looking at what you want your answer to be expressed in terms of. You're given a temperature and pressure, so what you want is the radius of the cloud in terms of its temperature and pressure.

You start with the Jeans condition, which has the following variables: R_C, M, T, and N. To answer the question, you want it to contain: R_C, T, and P. This means you'll want to express and M and N in terms of T and P. You're most of the way there with:

$$PV=NkT$$

and

$$PV=\frac{M}{M_m}RT$$

Given all these things, can you give me expressions for M and N in terms of T and P? Once you have that, it's just plug and chug.

5. Jan 28, 2006

### stunner5000pt

ok so we know $$R_{C} = \frac{2}{3} \frac{GM^2}{NkT}$$
$$N = \frac{PV}{kT}$$
$$M = \frac{PVM_{m}}{RT}$$
$$R_{C} \geq \frac{2}{3} \frac{GM^2}{PV}$$
$$R_{C} \geq \frac{2}{3} \frac{GPVM_{m}^2}{R^2 T^2}$$
and the volume is that a sphere $$V = \frac{4}{3} \pi R_{C}^3$$
\$$R_{C} \geq \frac{2}{3} \frac{GPM_{m}^2}{R^2 T^2} \frac{4}{3} \pi R_{C}^3$$
$$\frac{1}{R_{C}^2} \geq \frac{8}{9} \frac{GM_{m}^2 P \pi}{R^2 T^2}$$
invert and invert the signs as well
$$R_{C}^2 \leq \frac{9R^2 T^2}{8GM_{m}^2 P \pi}$$
good?

if you dont mind could you look at this thread of mine
the prof said the the problem is easier than it seems... but is it really so easy??

Last edited: Jan 28, 2006
6. Jan 29, 2006

### stunner5000pt

so is the 'corrected' derivation seen in post #3 correct? It does involve both pressure and temperature... is this correct?

7. Jan 31, 2006

### stunner5000pt

so is this correct? I just wanna finish off with this question!

8. Feb 1, 2006

### SpaceTiger

Staff Emeritus
I don't see any problems with your algebra. Plug in numbers and see what you get.

9. Feb 5, 2006

### stunner5000pt

i get an answer that is 2.1 * 10^4 m. My prof says that the answer should be between 10 and 100 times the earth's radius which i 6.38x10^6 m

10. Feb 6, 2006

### SpaceTiger

Staff Emeritus
I get something closer to what your teacher gets. Can you show explicitly the numbers and units of what you're plugging in?

11. Feb 6, 2006

### stunner5000pt

a possbile problem of mine could haev been that i was not using kg/mol for the molar mass
after tht correction i got 2.1 x 10^7 m

r = 8.314 J/mol K
T= 300 K
G = 6.67 x 10 ^ -11 N m^2/kg^2
P = 10^5 Pa
Mm = 0.02802 kg/mol (molecular nitrogen)

12. Feb 6, 2006

### SpaceTiger

Staff Emeritus
Ok, that's what I get too. It's still a bit smaller than expected, but there are numerical factors that we've neglected. For example, I think the proper Jeans condition is:

$$3NkT < \frac{GM^2}{R_c}$$

which should increase the answer by a factor of root 2. I don't see anything else wrong, but it's possible I'm missing something.