Question

A certain organ pipe, open at both ends, produces a fundamental frequency of 280Hz in air. If the pipe is filled with helium at the same temperature, what fundamental frequency f_He will it produce? Take the molar mass of air to be 28.8g/mol and the molar mass of helium to be 4.00g/mol . I have already tried 877Hz and it is incorrect!

Answer 1

gamma(air) = 1.4, gamma(He) = 5/3
Now, c = sqrt(gamma*pressure/p)

where p = density, c = sound velocity

Also, the frequency f is proportional to the sound velocity c, and pressure is the same in both cases,
Hence F(He) = F(air) * sqrt[(gamma(He)/gamma(air)) * (p(air)/p(He))]
= 287 * sqrt[(5/3)/1.4 * 28.8/4] = 840.25 Hz

Answer 2

The fundamental frequency in an organ pipe is proportional to the speed of sound.

For a pipe open at both ends,like in this problem, it is given by [1]:
f = v/4L
(v speed of sound in the pipe, L the length of the pipe)

Thus you'll find for the ratio of the frequencies in the same pipe:
f_He/f_Air = v_He/v_Air

Speed of sound in an ideal gas is given by [2]:
v = v(?·R·T/M)
(? heat capacity ratio, R universal gas constant, T absolute Temperature, M molar mass)

At the same temperature you get for the ratio of speed of sounds:
v_He/v_Air = v[ ?_He·M_Air/ (?_Air·M_He) ]

Hence:
f_He/f_Air = v_He/v_Air = v[ ?_He·M_Air/ (?_Air·M_He) ]
or
f_He = f_Air · v[ ?_He·M_Air/ (?_Air·M_He) ]

Heat capacity ratio for a monatomic ideal gas like helium is [3]
?_He = 5/3
Air consist mainly of diatomic oxygen and nitrogen. So it has
?_Air = 7/5

Therefore:
f_He = 280-Hz · v[ (5/3)·28.8g/mol / ((7/5)·4.0g/mol) ] = 819.75 Hz

Answer 3