Pipes, resonating frequencies and yeah some gases

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
10 replies · 2K views
mooncrater
Messages
215
Reaction score
18

Homework Statement


A closed organ pipe resonates in its fundamental mode at a frequency of ##200Hz## in ##O_2## at a certain temperature. If the pipe contains 2 moles of ##O_2## and 3 moles of ##O_3## are now added to it, then what will be the fundamental frequency of same pipe at same temperature?
[Given answer is ##172. 7Hz##]

Homework Equations


The relevant equation according to me is:
##v =√(\gamma P/\rho) ## (in a gas speed of a wave)

The Attempt at a Solution


What I did is:
Velocity of a wave in a gas=
$$v=√(\gamma P/\rho)$$
So using ##v=\nu\lambda##
We can say that
##\nu_1/\nu_2=√(\rho_2/\rho_1)##
And we know that
##\rho_1=4×16/V ## where V is the volume of pipe
And ##\rho_2=3×3×16+2×2×16/V=13×16/V##
Therefore my frequency is coming out to be
##110 Hz##
Which is wrong.. I know the whole of this seems to be wrong from the start... so how to do this?
 
on Phys.org
mooncrater said:

Homework Statement


A closed organ pipe resonates in its fundamental mode at a frequency of ##200Hz## in ##O_2## at a certain temperature. If the pipe contains 2 moles of ##O_2## and 3 moles of ##O_3## are now added to it, then what will be the fundamental frequency of same pipe at same temperature?
[Given answer is ##172. 7Hz##]

Homework Equations


The relevant equation according to me is:
##v =√(\gamma P/\rho) ## (in a gas speed of a wave)

The Attempt at a Solution


What I did is:
Velocity of a wave in a gas=
$$v=√(\gamma P/\rho)$$
So using ##v=\nu\lambda##
We can say that
##\nu_1/\nu_2=√(\rho_2/\rho_1)##
And we know that
##\rho_1=4×16/V ## where V is the volume of pipe
And ##\rho_2=3×3×16+2×2×16/V=13×16/V##
Therefore my frequency is coming out to be
##110 Hz##
Which is wrong.. I know the whole of this seems to be wrong from the start... so how to do this?
First what is the gamma for diatomic gas?
 
  • Like
Likes   Reactions: mooncrater
Okay, ##\gamma## Will also change for ##O_2## and ##O_3##.
For a diatomic gas it is 7/5 and for a triatomic gas it's 4/3.
 
Last edited:
Thanks I got it now:smile:
 
BvU said:
Care to enlighten us with your workings ?
Yup... why not...
I just put the values of ##\gamma ##
too, instead of cancelling them while comparing the resonant frequencies in the two cases.
 
mooncrater said:
Yup... why not...
I just put the values of ##\gamma ##
too, instead of cancelling them while comparing the resonant frequencies in the two cases.
But γ would not be 4/3 when you mix both of them.
What would be equivalent γ then?
 
Hmmm.. ##\gamma## can be calculated by comparing the internal energies of the gases in the two cases. :
## n_1C_v1ΔT+n_2C_v2ΔT=(n_1+n_2)C_{net}Δ T##
Cancelling ##ΔT##
##2×(7/5)R+3×(4/3)R=5×R/(\gamma-1)##
Cancelling ##R##
##\gamma=59/34##
But it's giving the answer 123. 51 Hz
 
mooncrater said:
Hmmm.. ##\gamma## can be calculated by comparing the internal energies of the gases in the two cases. :
## n_1C_v1ΔT+n_2C_v2ΔT=(n_1+n_2)C_{net}Δ T##
Cancelling ##ΔT##
##2×(7/5)R+3×(4/3)R=5×R/(\gamma-1)##
Cancelling ##R##
##\gamma=59/34##
But it's giving the answer 123. 51 Hz
What about P?