Determining pressure in Interstellar space

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Homework Help Overview

The discussion revolves around estimating the pressure in interstellar space, characterized by a very low density of hydrogen atoms at a temperature of approximately 3 K. Participants are tasked with calculating pressure in pascals and atmospheres using relevant equations.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore various equations related to gas laws, including PV = nRT and the average translational kinetic energy. There is an attempt to calculate pressure using different methods, with some questioning the correctness of their values for number density and molecular mass.

Discussion Status

Some participants have provided calculations that yield similar pressure values, while others express uncertainty about their results and seek clarification on their assumptions and calculations. There is ongoing exploration of the root causes of discrepancies in calculated values for Vrms.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the information they can use or the methods they can apply. There is a focus on ensuring that units are consistent throughout the calculations.

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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