Calculating RMS Speed of Nitrogen Molecules

Click For Summary

Homework Help Overview

The discussion revolves around calculating the root mean square (RMS) speed of nitrogen molecules in a specified volume and pressure, using the ideal gas law and kinetic theory equations. The original poster presents their calculations and seeks assistance in identifying potential errors in their approach.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to calculate the RMS speed using the equation V rms = (3KT/m)^(1/2) and seeks to determine the temperature (T) using the ideal gas law. They express uncertainty about their calculations and results.
  • Participants question the unit conversions used in the calculations, particularly regarding the volume and the conversion from atm to pascals.
  • There are inquiries about the molar mass of nitrogen and how it relates to the number of moles in the context of the problem.
  • Some participants suggest checking unit consistency and the correctness of the molar mass calculation.

Discussion Status

The discussion is ongoing, with participants providing suggestions and raising questions about unit conversions and calculations. The original poster has made adjustments based on feedback but continues to express confusion about the results. There is no explicit consensus yet on the correct approach or solution.

Contextual Notes

Participants note that the volume should be considered in liters rather than cubic meters, and there is a discussion about the implications of this on temperature calculations. The original poster's calculations of temperature and molar mass are under scrutiny, with suggestions for re-evaluation.

Kali8972
Messages
14
Reaction score
0
I have been having difficulties with this problem and I was wondering if I could get some help with it.

Q. What is the rms speed of nitrogen molecules contained in a 7.0 m3 volume at 4.20 atm if the total amount of nitrogen is 1600 mol?

I figured I would have to use this equation to solve it:

V rms = (3KT/m)^(1/2)

But first I would need to find T, so I found it through this equation:

PV=nRT
(4.20 atm)(1.013e5 N/m^2/atm)(7.0m3) = (1600 mol)(8.315 J/mol K)(T)
and T = 223.9 K

I also needed to find m

m (N2) = (28)(1.66 x 10e-27) = 4.648e-26 kg

Then I subsitituted them all back into the original equation

V rms = [(3*1.38e-23 J/K*223.9 K)/(4.648e-26 kg)]^(1/2)
V rms = 446.5 m/s

But that isn’t the right answer. I don’t know what I am doing wrong, but I was wondering if anyone could help point me in the right direction, it would be greatly appreciated.

Thank you!
 
Physics news on Phys.org
for the PV = nRT part, see if all your units cancel out. Keep in mind that the R value you have used will only cancel out pascals.

also how do you fin d out mass from the number of moles
number of moles = mass / molar mass
whati s the molar mass of N2??
 
that's why I used the conversion factor from atms to pascals (1.013e5). Is that the wrong conversion?
 
Kali8972 said:
that's why I used the conversion factor from atms to pascals (1.013e5). Is that the wrong conversion?

i didnt see you had already done that. OK now what should the units of volume be? Volume is always given in litres. not cubic metres. However 1L = 1 dm^3 (decimeter)
 
stunner5000pt said:
for the PV = nRT part, see if all your units cancel out. Keep in mind that the R value you have used will only cancel out pascals.

also how do you fin d out mass from the number of moles
number of moles = mass / molar mass
whati s the molar mass of N2??

Ok I took all your suggestions and I found the new Volume to be 7000 L, which would make the new T about 223859 Kelvin.

Then the molar mass I found by dividing (4.648e-26 kg)/1600 mol and got 2.905e-29

I subsitituted all those in and got 564826 m/s and that still seems far off.

I think I messed up the molar mass, any ideas on what I am doing wrong?

Thanks so much for all your help! I really appreciate it! :smile:
 
I'm still having troubles with this problem, any help would be appreciated, since some of the last suggestions didn't work out too well.


Q. What is the rms speed of nitrogen molecules contained in a 7.0 m[tex]^{3}[/tex] volume at 4.20 atm if the total amount of nitrogen is 1600 mol?


I figured I would have to use this equation to solve it:

V rms = [tex]\sqrt{\frac{3KT}{m}}[/tex]



But first I would need to find T, so I found it through this equation:

PV=nRT
(4.20 atm)(1.013e5 [tex]\frac{\frac{N}{m^{2}}}{atm}[/tex])([tex]7 m^{3}[/tex]) = (1600 mol)(8.315 [tex]\frac{J}{mol K}[/tex])(T)
and T = 223.9 K

(noting that the conversion:1.013e5 [tex]\frac{\frac{N}{m^{2}}}{atm}[/tex] to change atms to Pascals was used and the volume should NOT be converted to L because it makes the temperature to large and it doesn't cancel with my conversion factor )



I also needed to find m

m (N[tex]^{2}[/tex]) = [(28)/(6.02e23)] = 4.65e-23 g or 4.65e-26 kg



Then I subsitituted them all back into the original equation

V rms = [tex]\sqrt{\frac{(3)(1.38e-23 J/K)(223.9 K)}{(4.65e-26 kg)}}[/tex]
V rms = 446.4 m/s



Am I going in the right direction?
Thank you!
 

Similar threads

  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 21 ·
Replies
21
Views
3K
Replies
15
Views
3K
  • · Replies 6 ·
Replies
6
Views
5K
  • · Replies 3 ·
Replies
3
Views
16K
  • · Replies 2 ·
Replies
2
Views
6K
  • · Replies 1 ·
Replies
1
Views
5K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K