Calculating RMS Speed of Nitrogen Molecules

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SUMMARY

The discussion focuses on calculating the root mean square (RMS) speed of nitrogen molecules using the equation Vrms = (3KT/m)^(1/2). The user initially calculated the temperature (T) using the ideal gas law PV=nRT, resulting in T = 223.9 K. The mass of nitrogen (N2) was determined to be 4.648e-26 kg. After substituting these values into the RMS speed equation, the user obtained Vrms = 446.5 m/s, which was incorrect. Further discussion revealed unit conversion issues and the need for accurate molar mass calculations, leading to a revised RMS speed of 564,826 m/s, which still appeared incorrect.

PREREQUISITES
  • Understanding of the ideal gas law (PV=nRT)
  • Knowledge of root mean square speed calculation
  • Familiarity with unit conversions, particularly between atmospheres and pascals
  • Basic concepts of molecular mass and molar mass calculations
NEXT STEPS
  • Review the ideal gas law and its applications in thermodynamics
  • Learn about the derivation and application of the RMS speed formula
  • Study unit conversion techniques, especially for pressure and volume
  • Investigate the calculation of molar mass and its significance in gas laws
USEFUL FOR

Students in chemistry or physics, educators teaching gas laws, and anyone involved in thermodynamic calculations will benefit from this discussion.

Kali8972
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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!
 
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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^{3} 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 = \sqrt{\frac{3KT}{m}}



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

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

(noting that the conversion:1.013e5 \frac{\frac{N}{m^{2}}}{atm} 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^{2}) = [(28)/(6.02e23)] = 4.65e-23 g or 4.65e-26 kg



Then I subsitituted them all back into the original equation

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



Am I going in the right direction?
Thank you!
 

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