To solve this problem, it will be necessary to apply the concepts related to the fundamental resonance frequency in a closed organ pipe.
This is mathematically given as
[tex]f_n (2n+1)(\frac{v}{4L})[/tex]
For fundamental frequency n is 0, then,
[tex]f_0 = \frac{v}{4L}[/tex]
When,
v = Velocity of sound
L = Length,
Rearranging to find the velocity,
[tex]v = f_0 (4L)[/tex]
[tex]v = (80Hz)(4)(1.3m)[/tex]
[tex]v = 416m/s[/tex]
Therefore the speed of sound in this gas is 416m/s
