Determining pressure in Interstellar space

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SUMMARY

The pressure in interstellar space, characterized by a hydrogen atom density of approximately 1 atom/cm³ and a temperature of 3 K, is estimated to be 4.14 x 10-17 Pa using the ideal gas law (PV = nRT). The calculation involves converting the number density to moles and applying the appropriate constants, including the Boltzmann constant. The root mean square velocity (Vrms) of hydrogen atoms is calculated to be approximately 273.53 m/s, which is consistent with the expected low velocities of particles in such a sparse environment.

PREREQUISITES
  • Understanding of the ideal gas law (PV = nRT)
  • Familiarity with kinetic theory of gases
  • Knowledge of Boltzmann's constant and its application
  • Basic unit conversions between atoms, moles, and kilograms
NEXT STEPS
  • Explore the implications of low-density gas behavior in astrophysics
  • Learn about the kinetic theory of gases and its applications in different environments
  • Investigate the calculation of root mean square velocity (Vrms) in various gas conditions
  • Study the effects of temperature on gas pressure and density in astrophysical contexts
USEFUL FOR

Astronomers, physicists, and students studying astrophysics or thermodynamics, particularly those interested in the properties of gases in low-density environments like interstellar space.

pdonovan
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Homework Statement


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.


Homework Equations


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

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?

Thank you for any help/advice!
 
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Different approach, I get the same number:

<br /> \frac{10^6 \, \text{atom}} {\text{m}^3} \cdot<br /> \frac{1 \, \text{mole}} {6.0221415 \times 10^{23} \, \text{atom}} \cdot<br /> 8.3144621 \, \text{J}/\text{K}/\text{mole} \cdot<br /> 3 \, \mbox{K} \approx 4.142 \times 10^{-17} \, \text{pascal}<br />
 
The only formula you need is PV = nRT
 
That is exactly one of the formulae that pdonovan, Spinnor.
 
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.
 
Your value for m is correct (assuming units of kilograms; always carry units around). Your math is wrong somewhere. Show your work.
 
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.
 
Yeah, your answer is too large. About 2 m/s too large. sqrt(74819.28) is about 273.53.

As a sanity check, you can always compute sqrt((3 * boltzmann's constant * 3 kelvin) / (1 amu)). You will get the same answer.
 
Then which values are wrong in the Vrms = sqrt(3KbT / m) formula? I have T = 3 and if m = 1, then Vrms = 1.11 x 10 ^ -11 which is wrong. And if m =1660x10^-27g it is too big (around 8.6).
 

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