# Homework Help: Determining pressure in Interstellar space

1. Sep 18, 2011

### pdonovan

1. The problem statement, all variables and given/known data
Interstellar space, far from any stars, is filled with a very low density of hydrogen atoms (H, not H2). The number density is about 1atom / cm3 and the temperature is about 3 K. Estimate the pressure in interstellar space. Give your answer in Pa and in atm.

2. Relevant equations
Average translational kinetic energy per molecule: Eavg = 1.5KbT = .5mv2
p = F/A = (1/3)(N/V)mVrms2
Eth = 1.5nRT
pV = nRT

3. The attempt at a solution
V = 10-6 m3
T = 3K
m = 1u = 1.66x10-27
n = 1.66x10-24 (not sure if this is right?)

pV = nRT --> p = 4.14x10-17
Where am I going wrong, or is this correct?

2. Sep 18, 2011

### D H

Staff Emeritus
Different approach, I get the same number:

$$\frac{10^6 \, \text{atom}} {\text{m}^3} \cdot \frac{1 \, \text{mole}} {6.0221415 \times 10^{23} \, \text{atom}} \cdot 8.3144621 \, \text{J}/\text{K}/\text{mole} \cdot 3 \, \mbox{K} \approx 4.142 \times 10^{-17} \, \text{pascal}$$

3. Sep 18, 2011

### Spinnor

The only formula you need is PV = nRT

4. Sep 18, 2011

### D H

Staff Emeritus
That is exactly one of the formulae that pdonovan, Spinnor.

5. Sep 18, 2011

### pdonovan

Thank you very much, that was the correct answer.

Now, how would I got about finding Vrms?

I know p = (1/3)(N/V)mVrms2
So, p = 4.14x10-17
N = 1
V = 10-6
m = 1u = 1.66x10-27
And found Vrms = .86m/s, but this is incorrect. I think my m or v might be incorrect.

6. Sep 18, 2011

### D H

Staff Emeritus
Your value for m is correct (assuming units of kilograms; always carry units around). Your math is wrong somewhere. Show your work.

7. Sep 18, 2011

### pdonovan

Now I have...

4.14x10-17pa = (1/3)(1/10-6)mVrms2

1.242x10-22 = mVrms2

74819.28 = Vrms2
Vrms = 275.64m/s

So something is definitely wrong, because the atom should be moving very slowly.

8. Sep 18, 2011

### D H

Staff Emeritus