Question

A closed organ pipe resonates in its fundamental mode at a frequency of 200 Hz with $O_2$ in the pipe at a certain temperature. If the pipe now contains 2 moles of $O_2$ and 3 moles of ozone, than what will be the fundamental frequency of same pipe at same temperature ?

• A. 268.23 Hz
• B. 175.4 Hz
• C. 149.45 Hz
• D. None of these

Answer 1

The correct option is B 175.4 Hz

Let $\nu$ and $f$ represent the fundamental frequency and degree of freedom respectively.

As $f_{O_2}=5$ and $f_{O_3}=6$

For $O_2$ only $\gamma =\frac{2+5}{5}=1.4$

Thus equivalent degree of freedom $f_{eq}=\frac{f_{O_2}\times n_{O_2}+f_{O_3}\times n_{O_3}}{n_{total}}$

$f_{eq}=\frac{5\times 2+6\times 3}{2+3}=5.6$

${\gamma }_{eq}=\frac{2+f_{eq}}{f_{eq}}=\frac{2+5.6}{5.6}=1.36$

Equivalent molecular mass $M_{eq}=\frac{M_{O_2}\times n_{O_2}+M_{O_3}\times n_{O_3}}{n_{total}}$

$M_{eq}=\frac{32\times 2+48\times 3}{2+3}=41.6gm/mol$

As $v=\nu \lambda$ and $v=\sqrt{\frac{\gamma RT}{M}}$

$⟹\frac{{\nu }^{\prime }}{\nu }=\frac{v^{\prime }}{v}=\sqrt{\frac{{\gamma }_{eq}}{M_{eq}}\times \frac{M}{\gamma }}$ (as $\lambda$ is constant)

$\frac{{\nu }^{\prime }}{200}=\sqrt{\frac{1.36}{41.6}\times \frac{32}{1.4}}$

$⟹{\nu }^{\prime }\approx 172.9Hz$