Concentration of Molecules in the Air

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SUMMARY

The discussion focuses on calculating the concentration of molecules in the atmosphere at an altitude of 300 km, where the atmospheric pressure is approximately 10-8 mmHg and the temperature is 500 K. Participants utilized the Ideal Gas Law, specifically the equation n/V = P/RT, to derive the number of molecules per milliliter. The universal gas constant R was taken as 0.08206 atm·L/(K·mol), leading to a final calculation of approximately 192,764,220 molecules/mL. Participants emphasized the importance of significant figures in their calculations.

PREREQUISITES
  • Understanding of the Ideal Gas Law (PV = nRT)
  • Familiarity with Avogadro's number (NA = 6.022 x 1023 molecules/mol)
  • Basic knowledge of atmospheric pressure and temperature units
  • Ability to perform unit conversions (e.g., L to mL)
NEXT STEPS
  • Study the Ideal Gas Law applications in different atmospheric conditions
  • Learn about the significance of significant figures in scientific calculations
  • Explore the effects of altitude on atmospheric pressure and temperature
  • Investigate molecular concentration calculations in various gases
USEFUL FOR

Students in chemistry or physics, educators teaching gas laws, and professionals in atmospheric science or aerospace engineering will benefit from this discussion.

MysticDude
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Homework Statement


At a height of 300km above the Earth's surface, an astronaut finds that the atmospheric pressure is about 10-8 mmHg and the temperature 500 K. How many molecules are there per milliliter at this altitude?


Homework Equations





The Attempt at a Solution


I first thought that I had to calculate the amount of moles in area then multiply that by Avogadro's number. But then I realized that there is no amount of volume given here, only the height, as I was thinking about using the Ideal Gas Law. All I need is the volume, but I don't know how to get it. Was I even on the right track here?
 
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PV = nRT \implies V/n = RT/P \implies n/V = P/RT. To simplify your calculation, take the universal gas constant to be: R = 0.08206 \frac{\textrm{atm}\cdot\textrm{L}}{\textrm{K}\cdot\textrm{mol}}.
 
Coto said:
PV = nRT \implies V/n = RT/P \implies n/V = P/RT. To simplify your calculation, take the universal gas constant to be: R = 0.08206 \frac{\textrm{atm}\cdot\textrm{L}}{\textrm{K}\cdot\textrm{mol}}.

Yes, but I'm missing the volume. I only have the 300 km in the air for information.
 
They're looking for mols/mL ... so no, you don't need a volume, you need an n/V ... which only requires knowledge of P and T. (Liter is a measure of volume).
 
You are given volume. They ask for number of molecules per milliliter.
 
Okay, so by using Coto's equation, I get 3.207 x 10-13\frac{mol}{L}.
So to get \frac{molecules}{milliliter}, I would have to do this:
3.207 x 10^{-13}\frac{mol}{L} x \frac{1 L}{1000 mL} x 6.022x10^{23}\frac{molecules}{mol}...in order to get my answer, right?
 
Last edited:
Exactly... except N_A = 6.022 \cdot 10^{23}\frac{\textrm{moleculues}}{\textrm{mol}}. Not -23 :).
 
Coto said:
Exactly... except N_A = 6.022 \cdot 10^{23}\frac{\textrm{moleculues}}{\textrm{mol}}. Not -23 :).

I changed that a minute after I saw the negative. I guess the picture didn't update o.o

Anyway, I got something like 192764220 molecules/milliliter. Can you compare?
 
Close enough to what I get. Watch significant figures.
 
  • #10
Borek said:
Close enough to what I get. Watch significant figures.
I know that I should watch significant numbers. I just plugged in everything directly into my calculator. Thanks a bunch for checking!
 

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