Two narrow cylindrical pipes `A and B` have the same length. Pipe `A` is open at both ends and is filled with a monoatomic gas of molar mass `M_(A)`. Pipe `B` is open at one end and closed at the other end, and is filled with a diatomic gas of molar mass `M_(B)`. Both gases are at the same temperature.
(a) If the frequency of the second harmonic of the fundamental mode in pipe `A` is equal to the frequency of the third harmonic of the fundamental mode in pipe `B`, determine the value of `M_(B)//M_(B)`.
(b) Now the open end of pipe `B` is also closed (so that the pipe is closed at both ends). Find the ratio of the fundamental frequency in pipe `A` to that in pipe `B`.

Answer

(i) `2.116`
(ii) `(3)/(4)`

Explanation

(i) Frequency of second harmonic in pipe
A = frequency of third harmonic in pipe B
`:. 2((V_(A))/(2l_(A)))=3((V_(B))/(4l_(B)))`
`rArr (V_(A))/(V_(B))=(3)/(4) ("as" l_(A)=l_(B))rArr (sqrt((gamma_(A)RT_(A))/(M_(A))))/(sqrt((gamma_(B)RT_(B))/(M_(B))))=(3)/(4)`
`rArr sqrt((gamma_(A))/(gamma_(B)))sqrt((M_(B))/(M_(A)))=(3)/(4)("as" T_(A)=T_(B))`
`:. (M_(A))/(M_(B))=(gamma_(A))/(gamma_(B))((16)/(9))=((5//3)/(7//5))((16)/(9))`
`(gamma_(A)=(5)/(3) and gamma_(B)=(7)/(5))`
`rArr(M_(A))/(M_(B))=((25)/(21))((16)/(9))=(400)/(189)`
(ii) Ratio of fundamental frequency in pipe A and in pipe B is :
`(f_(A))/(f_(B))=(V_(A)//2l_(A))/(V_(B)//2l_(B))=(V_(a))/V_(B)("as" l_(A)=l_(B))`
`= (sqrt((gamma_(A)RT_(A))/(M_(A))))/(sqrt((gamma_(B)RT_(B))/(M_(B))))=sqrt((gamma_(A))/(gamma_(B)).(M_(B))/(M_(A)))("as" T_(A)=T_(B))`
Substituting `(M_(B))/(M_(A))=(189)/(400)` from part (i), we get
`(f_(A))/(f_(B))=sqrt((25)/(21)xx(189)/(400))=(3)/(4)`.