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RideAnS [48]
3 years ago
9

A certain organ pipe, open at both ends, produces a fundamental frequency of 272 HzHz in air. Part A If the pipe is filled with

helium at the same temperature, what fundamental frequency fHefHef_He will it produce? Take the molar mass of air to be 28.8 g/molg/mol and the molar mass of helium to be 4.00 g/molg/mol .
Chemistry
1 answer:
vladimir2022 [97]3 years ago
7 0

Answer:

fundamental frequency in helium = 729.8 Hz

Explanation:

Fundamental frequency of an ope tube/pipe = v/2L

where v is velocity of sound in air = 340 m/s; λ is wave length of wave = 2L ; L  is length of  the pipe

To find the length of the pipe,

frequency  = velocity of sound / 2L

272 = 340 / 2 L

L = 0.625 m

If the pipe is filled with helium at the same temperature, the velocity of sound will change as well as the frequency of note produced since velocity is directly proportional to frequency of sound.

Also, the velocity of sound is inversely proportional to  square root of molar mass of gas; v ∝ 1/√m

v₁/v₂ = √m₂/m₁

v₁ = velocity of sound in air, v₂ = velocity of sound in helium, m₁ = molar mass of air, m₂ = molar mass of helium

340 / v = √4 / 28.8

v₂ = 340 / 0. 3727

v₂  =  912.26  m /s  

fundamental frequency in helium  = v₂ / 2L

fundamental frequency in helium = 912.26 / (2 x 0.625)

fundamental frequency in helium = 729.8 Hz

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