Estimating Pressure, RMS Speed of Hydrogen in Space

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Homework Help Overview

The discussion revolves around estimating the pressure and root mean square (RMS) speed of hydrogen atoms in a low-density environment in space, specifically at a temperature of 3 K. The original poster presents a series of calculations and equations related to the ideal gas behavior of hydrogen atoms, including number density and thermal energy considerations.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the calculation of pressure using the ideal gas law and question the correctness of the RMS speed formula for atoms versus molecules. There are discussions about converting units and relating the number of atoms to the volume they occupy. Some participants suggest using average kinetic energy to find relationships between energy and the number of atoms.

Discussion Status

There is ongoing dialogue with participants providing feedback on calculations and suggesting alternative approaches. Some participants express confusion about the application of chemistry concepts to the problem, while others clarify the relationships between pressure, energy, and volume in the context of ideal gases.

Contextual Notes

Participants note potential issues with the original poster's understanding of number density in atoms per cubic centimeter and the implications for calculations. There is also mention of the original poster's unfamiliarity with chemistry concepts, which may affect their approach to the problem.

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Homework Statement


in space far from stars is filled with a very low density of hydrogen atoms, the number density is about 1atom/cm^3 and T is about 3degreeK

a. estimate the pressure in space, answer in Pa and atm.
b. what is the rms speed of the atoms
c. what is the edge length L of an LxLxL cube of gas with 1J of thermal energy
k=1.38x106-23
1atm=101.3kPa
atomic mass of hydrogen=1.67x10^-27kg

Homework Equations


N/V=n
vrms=sqrt(3kT/m)
N/V*M=m
PV=NkT


The Attempt at a Solution


a. P=(N/V)kT=(1/1x10-6)*1.38x10^-23*3=4.14x10^-17Pa=4x10^-22atm?
b.m=(N/V)M=(1/1x10^-6)1.67x10^-27=1.67x10^-22kg
so vrms=sqrt(9*1.38x10^-23/1.67x10^-22)=.86m/s?

and c. i have no idea, please someone help, this micro macro stuff is stupid... I am not sure about the vrms formula is correct, since in the book it says its for molecules, but i can't find anything on atoms... I am pretty sure i have done something wrong for the number density, as i don't know, coz its in atoms/cm^3... and I am just stuck please some help me...
 
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ok well i think m is just atomic mass so vrms=sqrt(9*1.38x106-23/1.67x10^-27)=273m/s i think
please can anyone tell me if i am doing this right?
 
Hi fredrick08,

What is the average kinetic energy of one atom? And so how many atoms would you need to give 1 J of energy? Once you have that, how is the number of atoms related to the volume they occupy?
 
How about just converting to moles per volume? A gas is 22.4 liters per mole at STP. Just use the gas laws from chemistry.
 
Hi sirzerp,

Perhaps I'm not understanding your train of thought; how will converting to moles and calculating volumes at STP help find the volume containing 1 J of energy at 3K?
 
(a) looks good to me

(b)
vrms=sqrt(9*1.38x106-23/1.67x10^-27)

Right idea, but one of the numbers is incorrect here.
 
alphysicist said:
Hi sirzerp,

Perhaps I'm not understanding your train of thought; how will converting to moles and calculating volumes at STP help find the volume containing 1 J of energy at 3K?

We know the number of moles per volume and temp. Just treat it as an ideal gas right?

Once you know pressure, is not pressure a direct energy unit?
 
Last edited:
Hi Redbelly98,

Redbelly98 said:
(a) looks good to me

(b)


Right idea, but one of the numbers is incorrect here.

There is a 6 where a ^ should be; but I think other than that the numbers are correct. Are you seeing something that I am overlooking?
 
sirzerp said:
We know the number of moles per volume and temp. Just treat it as an ideal gas right?

Once you know pressure, is not pressure a direct energy unit?

Yes, so you could use

<br /> P =\frac{2}{3}\frac{U}{V}<br />

and solve for V; but I don't think you need to convert to moles or calculate things at STP.



-------------------

My original suggestion also gave the volume in two step from the given data. The number of molecules giving an energy U is:

<br /> U=\frac{3}{2} N k T<br />

and since the density is (1 atom/cm^3) then number for N is the volume (in cm^3).
 
  • #10
alphysicist said:
There is a 6 where a ^ should be; but I think other than that the numbers are correct. Are you seeing something that I am overlooking?

Oops, I was looking at the first factor of 9, thinking it should be a 3. But he had already multiplied 3 with the T=3K.

Sorry! :redface:

(b) looks good to me now!
 
  • #11
ok ty heaps, srry i was away, just got up lol, ok it makes sense now, srry for also confusing anyone, unfortunately I've never done chemistry before, so i don't really know the chem formulas just the ones in my physics book...
 

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