Regarding root mean square velocity

  • Thread starter fireflies
  • Start date
  • #1
210
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I am stuck with this question: why can't we use the equation

c(rms)=√(3RT/M) for problems when no of moles is not 1?

Because I think the mole will be deduced as

c(rms)=√(3nRT/nM) =√(3RT/M)

For instance I found this problem, given atmospheric pressure is 10^5 Pa, molecular mass of
Nitrogen is 28, and at STP the density is 1.25 kg/m^3. What is the c(rms) at 100°C?

If the problem is solved with c(rms)=√(3P/rho) the answer is 576.35 which is the actual answer, and when taken the above equation the answer is 18.22

Then what's wrong with my concept?
 

Answers and Replies

  • #2
3,757
424
The formula is OK. Posiibly something wrong in the way you use it. Impossible to say if you don't show your actual calculation.
The number of moles has nothing to do with that formula. The average speed is an intensive parameter, it does not depend on the volume of gas taken into consideration.
 
  • #3
Khashishi
Science Advisor
2,815
493
Can you see that, for an ideal gas, RT/M = P/rho, using the ideal gas equation and the definition of density?
 
  • #4
210
12
Can you see that, for an ideal gas, RT/M = P/rho, using the ideal gas equation and the definition of density?
 
  • #5
210
12
Sorry, I had been trying to upload the calculation but due to some problems it was not uploading. So, here I'm typing the calculations:

Calculation 1 (using the first equation)

Crms=√(3RT/M) = √(3*8.314*373/28) = 18.228 (approx)

Calculation 2

we know PV=nRT,
or, PV= (m/M)RT
or, P=(m/V)*R*T/M
∴ P=ρRT/M

i.e. P/(ρ*T)=constant

so in this case we get P/(ρT)=P'/(ρ'T')
or, P'/ρ' = (PT')/(ρT)

Now, crms = √(3p'/ρ') = √{(3*10^5*373)/(1.25*273)} = 572.63(approx)

The later one is the solution and anser given. But it doesn't match with the answer of the previous one. Why?
 
  • #6
3,757
424
Your units in the first calculation are not consistent.
If you use R=8.31 this is in J/mol*K. Then you have to take the molar mass per mol too. It is 28g/mol which is .028 kg/mol. You need to use kilograms to be consistent with the Joules in the unit for R. So you are off by a factor of sqrt(1000) or about 32. Which will give you the same value as the second calculation (around 500 m/s).
 
  • Like
Likes fireflies
  • #7
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12
Yes, now the answer comes. Thanks for helping
 
  • #8
3,757
424
Yes, now the answer comes. Thanks for helping
You are welcome.:smile:
 

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