Ideal Gas- molecules applying pressure

Click For Summary

Homework Help Overview

The problem involves calculating the pressure exerted by nitrogen molecules striking a wall, focusing on the dynamics of elastic collisions and the relationship between force, area, and pressure. The subject area includes concepts from kinetic theory and gas laws.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to calculate the pressure using the number of molecules, their speed, and the area of the wall, while expressing uncertainty about the chemistry involved. Some participants suggest focusing on momentum rather than energy. Others question the assumptions made regarding the nature of collisions and the area conversion.

Discussion Status

Participants are actively engaging with the problem, with some providing hints and corrections regarding the approach to calculating force and pressure. There is a recognition of errors in the original calculations, and the discussion is evolving as the original poster revises their approach based on feedback.

Contextual Notes

There are indications of confusion regarding the definitions and assumptions related to elastic collisions and the conversion of area units. The original poster expresses a lack of confidence in their calculations, particularly concerning the force and final pressure values.

latitude
Messages
54
Reaction score
0

Homework Statement


In a period of 1 s, 5 x 10^23 nitrogen molecules strike a wall with an area of 8 cm^2. Assume the molecules move with a speed of 300 m/s and strike the wall head-on in elastic collisions. What is the pressure exerted on the wall? (The mass of one N2 molecule is 4.68 x 10^-26 kg.)


Homework Equations


P = 2/3(N/V)(1/2mo(v^2)avg = F/A = [mo/d(N)(v^2)]/A


The Attempt at a Solution


I found the moles of N2 (I think this is how you do it, I was never strong in chemistry and haven't taken it in two years...)
n = N/Na = 5 x 10^23/6.022x10^23 = 0.83 moles
I believe n = N/V as well, so...
P = 2/3(n)(1/2mo(v^2)avg
I'm thinking I need to find the average speed... since I have vx, I can do the force one

F = (4.68 x 10^-26)/d(5 x 10^23)(300^2)
So I need to find 'd'...

I'm stuck, I'm not sure whether to try and find 'd' (length of the wall the molecules are hitting), or try and find v.avg (I'm thinking I'd somehow do that with the time interval and the vx value, but I'm not sure how...)

Thanks in advance.
 
Physics news on Phys.org
You are off on a chemistry tangent; this is not a chemistry problem. You are given the average speed of the molecules hitting the wall, the number of elastic collisions with the wall during some time period, and the area of the wall. You aren't using this information!

Hint: Try looking at it from a perspective of momentum rather than energy.
 
Ahh!

So...
(F)deltaT = deltap = movx
F(1) = (4.68 x 10^-26 x 5 x 10^23)(300 m/s)
F = 7.02 N

P = F/A
P = 7.02/0.08 m^2
= 87.75 Pa??

Is this right?? I feel like I'm on a better track now, thanks so much! I'm just afraid that the Force is not correct, my final answer seems shady...
 
Two things are wrong:
(1) Your force calculation assumes perfectly inelastic collisions. The collisions are elastic.
(2) 8 cm2 is not 0.08 m2.
 
So...
instead, my force equation should be:
F(deltat) = -2mvx?
Which makes my answer -14.04, except I read in the textbook that the velocity component of an elastic collision is reversed because the mass of the wall is so much greater than the mass of the molecule. I don't quite understand the theory behind that?, but it makes my force positive and my final answer 17550 Pa
Thank you for all of your help!
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
1K
Replies
10
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
16
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
5K