Solving Interstellar Gas Pressure: A Daunting Task

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Homework Help Overview

The discussion revolves around calculating the pressure of interstellar gas, specifically focusing on a scenario involving a typical region of interstellar space with a specified number of atoms and temperature. The subject area includes concepts from thermodynamics and the ideal gas law.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of the ideal gas law (PV=nRT) and the conversion of atoms to moles using Avogadro's number. Questions arise regarding unit consistency and the correct use of the gas constant.

Discussion Status

There is an ongoing exploration of the calculations involved, with some participants providing guidance on unit conversions and significant figures. Multiple interpretations of the problem setup and calculations are being examined, but no consensus has been reached regarding the final answer.

Contextual Notes

Participants note the importance of maintaining consistent units throughout the calculations and express concerns about significant figures affecting the results. There is also a mention of the appropriateness of posting in the homework section of the forum.

chineseman1
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I have been trying to do this questions for about 30 minutes right now and its a one dot questions. I think I am converting the atoms wrong. Could someone help me?

A typical region of interstellar space may contain 10^6 atoms per cubic meter (primarily hydrogen) at a temp of 100 K. What is the pressure of the gas?

>Okay, what I have done was use PV=nrT, but I am having trouble converting it. Do I take the atoms and divide by avogadros?

Thank You.
 
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It depends on what value of r you are using. Your units need to match.
 
I am using 8.31.
 
For future reference you should post homework problems in the homework section-where this thread will probably be moved to in a minute anyway.

Anyway, you have 106 atoms in 1 m3. 106 atoms is:
10^6 atoms\times\frac{1mol}{6.02\times 10^{23} atoms}\approx1.66mol
in one m3. Convert m3 to L by multiplying by 1,000 then you can use the ideal gas law.
 
chineseman1 said:
I am using 8.31.
Always keep in mid what units you are using. R=8.31\frac{J}{K\cdot mol} so you better not convert to liters after all because 1J=1\frac{kg\cdot m^2}{s^2}, so the unit of volume is the m3
 
Okay. So I was doing it right!
But when I do use the Ideal Gas Law I get 1.37E-19, while the answer in my book is 10^-15. So I really don't know what I am doing wrong here. I as doing P= nRT/V, so I just plugged in all my variables and still not getting right answer.
 
chineseman1 said:
Okay. So I was doing it right!
But when I do use the Ideal Gas Law I get

1.37E-19,

while the answer in my book is 10^-15. So I really don't know what I am doing wrong here. I as doing P= nRT/V, so I just plugged in all my variables and still not getting right answer.

Take another look at your bookkeeping on exponents.

And, review comments on significant figures.
 
chineseman1 said:
Okay. So I was doing it right!
But when I do use the Ideal Gas Law I get 1.37E-19, while the answer in my book is 10^-15. So I really don't know what I am doing wrong here. I as doing P= nRT/V, so I just plugged in all my variables and still not getting right answer.
What units is your answer in?
What are the units of the answer in the book?
Are you sure you didn't convert your volume to liters? You should keep it as 1m3
 
I got in! It was sig figs, the killer of me. Thanks to all you guy physics gurus!
 

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