Q. The speed of sound through oxygen at T K is v m$s^{-1}$. As the temperature becomes 2T and oxygen gas dissociates into atomic oxygen, the speed of sound

Solution:

The rms velocity of sound in gas is
$\hspace35mm v_{rms((molecule)} = \sqrt{\frac{\gamma RT}{M}}$
$\hspace40mm = \sqrt{\frac{1.4 \times RT}{M}} \hspace30mm ....(i)$
when the oxygen dissociates into atomic oxygen, its molecular mass becomes atomic mass, so, M $= \frac{M}{2}$ and
$T = 2 T \, given \, and \, \gamma = 1.66$
$\hspace35mm v_{rms(atomic)} = \sqrt{\frac{1.66 \times R \times 2T}{M/2}}$
$\hspace50mm =\sqrt{\frac{1.66 \times R \times 2T \times 2}{M}} \hspace30mm ....(ii)$
From Eqs. (i) and (ii)
$\frac{v_{rms(atomic)}}{v_{rms(molecule)}} = \sqrt{\frac{1.66 \times R \times 2T \times 2}{M}} \times \sqrt{\frac{M}{1.4 \times RT}}$
$\hspace40mm = \sqrt{\frac{1.66 \times 4}{1.4}} = 2.18$
Hence, $v_{rms(atomic)} = 2.18 \, v_{rms(molecule)}$
$\hspace40mm = 2 v_{rms(molecular)}$

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