
#1
Sep1005, 04:52 PM

P: 14

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 10e27) = 4.648e26 kg Then I subsitituted them all back into the original equation V rms = [(3*1.38e23 J/K*223.9 K)/(4.648e26 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!! 



#2
Sep1005, 04:57 PM

P: 1,445

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?? 



#3
Sep1005, 04:58 PM

P: 14

that's why I used the conversion factor from atms to pascals (1.013e5). Is that the wrong conversion?




#4
Sep1005, 05:00 PM

P: 1,445

RMS Speed 



#5
Sep1005, 05:30 PM

P: 14

Then the molar mass I found by dividing (4.648e26 kg)/1600 mol and got 2.905e29 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! 



#6
Sep1105, 09:36 AM

P: 14

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)] = 4.65e23 g or 4.65e26 kg Then I subsitituted them all back into the original equation V rms = [tex]\sqrt{\frac{(3)(1.38e23 J/K)(223.9 K)}{(4.65e26 kg)}}[/tex] V rms = 446.4 m/s Am I going in the right direction? Thank you!!! 


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