Calculating RMS Speed of Nitrogen Molecules

AI Thread Summary
The discussion revolves around calculating the root mean square (RMS) speed of nitrogen molecules in a given volume and pressure. The user initially calculated the temperature using the ideal gas law, resulting in T = 223.9 K, and determined the mass of nitrogen molecules. However, they faced issues with their final RMS speed calculation, which yielded an incorrect value. Participants suggested checking unit conversions and the correct molar mass of nitrogen, emphasizing that volume should remain in cubic meters rather than converting to liters. Despite attempts to adjust calculations based on feedback, the user continued to struggle with the correct RMS speed, indicating a need for further clarification on molar mass and unit consistency.
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^{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!
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top