Estimating Air Molecule Motion at STP

Click For Summary

Homework Help Overview

The original poster attempts to estimate the time it takes for an air molecule to move into the region occupied by another molecule at standard temperature and pressure (STP), focusing on the average separation and mean speed of nitrogen molecules.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of molecular separation and average velocity using kinetic theory, questioning the validity of the initial assumptions and calculations.

Discussion Status

Some participants have provided guidance on correcting the mass used in calculations, leading to a revised velocity and time estimate. There is ongoing exploration of the reasonableness of these new values.

Contextual Notes

Participants are navigating the complexities of molecular mass and separation, with references to external resources for validation of the distance between molecules in a gas.

mlostrac
Messages
83
Reaction score
0

Homework Statement



From the average separation between air molecules at STP, and their mean speed, estimate how long it would take one molecule to move into the region occupied by another. Assume that air consists mainly of nitrogen molecules.


Homework Equations


1/2mv^2 = 3/2KT


The Attempt at a Solution



Distance between molecules at STP:

(6.02 x 10^23) x D^3 = 22.4 x 10^-3 m^3
D^3 = 3sqrt 22.4x10^-3 / 6.02 x 10^23
D = 3.34 x 10^-9 m

Since mass of N2 = 28 g/ mol, I used this to find the average velocity of the molecules:

1/2mv^2 = 3/2KT
v(rms) = sqrt(3KT/m)
= sqrt (3K (273) / .028 kg
= 6.35 x 10^-10 m/s

I then just plugged my two answers from above into the velocity equation,
velocity = distance/time
time = D/V
time = 3.34 x 10^-9 / 6.25 x 10^-10 = 5.3 seconds

Does what I did make sense? The velocity looks awfully small, so I'm not sure. Any help appreciated!
 
Physics news on Phys.org
The "m" is the mass of each molecule. You've noted that the mass for each mole is 0.028 kg. You need to divide this by 6.02E23 to get the mass per molecule.
 
Ah ha! Good eye, thank you for that.

My new velocity is 493 m/s, and my new time is therefore 6.77 x 10^-12 seconds.

Do these numbers make sense?
 
I think so. At first I was thinking your distance was way too small (it's on the order of the atomic width). But I can't see anything wrong with logic. So I googled "average separation of atoms in a gas" and got the following link:

http://www.practicalphysics.org/go/Guidance_34.html;jsessionid=alZLdQlAHb1?topic_id=4&guidance_id=1

and one from here:

https://www.physicsforums.com/showthread.php?t=43728

Both confirm your number.
 
Last edited by a moderator:
Thanks very much for all your help
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 16 ·
Replies
16
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
8K
  • · Replies 4 ·
Replies
4
Views
16K
  • · Replies 7 ·
Replies
7
Views
7K
  • · Replies 8 ·
Replies
8
Views
8K
  • · Replies 9 ·
Replies
9
Views
4K
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K