Question

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 to the second harmonic of pipe A is equal
to the frequency of the third harmonic of the fundamental
mode in pipe B, determine the value of $M_A/M_B.$
(b) Now the open end of the pipe B is 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 1

(a) Frequency of second harmonic in pipe A
= frequency of third harmonic in pipe B
$\therefore 2\left(\frac{v_A}{2l_A}\right)=3\left(\frac{v_B}{4l_B}\right)$
or $\frac{v_A}{v_B}=\frac{3}{4}(as{l}_A=l_B)$
or $\frac{\sqrt{\frac{{\gamma }_{A}R{T}_A}{M_A}}}{\sqrt{\frac{{\gamma }_{B}R{T}_B}{M_B}}}=\frac{3}{4}$
or $\sqrt{\frac{{\gamma }_A}{{\gamma }_B}}\sqrt{\frac{M_B}{M_A}}=\frac{3}{4}(as{T}_A=T_B)$
$\therefore \frac{M_A}{M_B}=\frac{{\gamma }_A}{{\gamma }_B}\left(\frac{16}{9}\right)$
$=\left(\frac{5/3}{7/5}\right)\left(\frac{16}{9}\right)({\gamma }_A=\frac{5}{3}and{\gamma }_B=\frac{7}{5})$
or $\frac{M_A}{M_B}=\left(\frac{25}{21}\right)\left(\frac{16}{9}\right)=\frac{400}{189}$
(b) Ratio of fundamental frequency in pipe A and in pipe B is
$\frac{f_A}{f_B}=\frac{v_A/2l_A}{v_B/2l_B}=\frac{v_A}{v_B}(as{l}_A=l_B)$
$=\frac{\sqrt{\frac{{\gamma }_{A}R{T}_A}{M_A}}}{\sqrt{\frac{{\gamma }_{B}R{T}_B}{M_B}}}=\sqrt{\frac{{\gamma }_A}{{\gamma }_B}.\frac{M_B}{M_A}}(as{T}_A=T_B)$
Substituting $\frac{M_B}{M_A}=\frac{189}{400}$ from part (a), we get
$\frac{f_A}{f_B}=\sqrt{\frac{25}{21}\times \frac{189}{400}}=\frac{3}{4}$