Checking Average Atoms in 1 m^3 of Atmosphere at STP

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Discussion Overview

The discussion revolves around the calculation of the average number of atoms in a 1 cubic meter volume of atmosphere at standard temperature and pressure (STP). Participants explore different approaches to arrive at the number of atoms, considering the composition of air and the implications of diatomic molecules.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant presents a calculation yielding 88.14 moles of atoms based on the average atomic mass of atmospheric components.
  • Another participant points out that the initial calculation is incorrect due to the omission of the diatomic nature of nitrogen and oxygen, suggesting that this affects the number of atoms counted.
  • Some participants propose using Avogadro's principle and the ideal gas law (PV=nRT) as a simpler method to determine the number of moles without needing the specific composition of the gas mixture.
  • There is acknowledgment that while the approach is valid, the need to account for diatomic molecules means the final count of atoms would need adjustment.

Areas of Agreement / Disagreement

Participants generally disagree on the correctness of the initial calculation, with some supporting the need to consider diatomic molecules while others suggest alternative methods. The discussion remains unresolved regarding the exact number of atoms.

Contextual Notes

Participants note the importance of recognizing the diatomic nature of nitrogen and oxygen in the calculations, which affects the total atom count. There is also a mention of the ideal gas law as a potentially simpler approach, but it is not fully explored.

Who May Find This Useful

This discussion may be useful for individuals interested in gas laws, atmospheric chemistry, and calculations involving moles and atomic composition in gases.

mesa
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So I need a check on my calculations for the average number of atoms in a 1 meter cube of atmosphere at STP.

I came up with 88.14 mols, here is how I did it:

First I calculated the average amu using percentage in air by
Nitorgen 78.08% @ 14.007 amu
Oxygen 20.95% @ 15.999 amu
Argon .93% @ 39.948 amu

I took these numbers and multiplied them by their relative percentages to get average amu for atmosphere:
10.94 + 3.35 + .37 = 14.67 amu average

We know the average density of air at STP is 1.293x10^3g/m^3
so I divided by average amu and got 88.14 mols/m^3

Is this right?
 
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No, that's not correct - you forgot nitrogen and oxygen are present as diatomics. But approach per se is OK.

Note that there is much simpler way. Avogadro's principle says that no matter what the gas identity is, number of molecules/atoms in a given volume is always the same (as long as gas can be treated as ideal). That means you don't need to know the mixture composition, just use PV=nRT to calculate number of moles of gas in 1 cubic meter.
 


Borek said:
No, that's not correct - you forgot nitrogen and oxygen are present as diatomics. But approach per se is OK.

I'm aware that oxygen and nitrogen exist as diatomic molecules in our atmosphere, I happen to need the number of atoms as stated in the question and title of the post but I could see how that could be missed Borek :)

Borek said:
Note that there is much simpler way. Avogadro's principle says that no matter what the gas identity is, number of molecules/atoms in a given volume is always the same (as long as gas can be treated as ideal). That means you don't need to know the mixture composition, just use PV=nRT to calculate number of moles of gas in 1 cubic meter.

That's a good approach, however I would need to recalculate for number of atoms since (as you know) the IGL counts our diatomic molecules and Ar the same so the result would be slightly less than double of n.

Either way a fairly simple check and good suggestion.
 
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mesa said:
I happen to need the number of atoms as stated in the question and title

Yep, missed that, my mistake.

 
Last edited by a moderator:


Borek said:


Ha ha! Since when were chemists fans of Wilder?
Pretty risque stuff for 59' but that's how he was :)
 
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