The speed of the sound in oxygen $\mathrm{O}_2$ at a certain temperature is $460 \mathrm{~ms}^{-1}$. The…

The speed of the sound in oxygen $\mathrm{O}_2$ at a certain temperature is $460 \mathrm{~ms}^{-1}$. The speed of the sound in helium (He) at the same temperature will be (assume both the gases to be ideal)
  1. $330 \mathrm{~ms}^{-1}$
  2. $1420 \mathrm{~ms}^{-1}$
  3. $500 \mathrm{~ms}^{-1}$
  4. $650 \mathrm{~ms}^{-1}$

Solution

For Oxygen, Molar mass, $M_1=32$ Heat capacity ratio at constant pressure to that at constant volume, $ \gamma_1=C_p / C_V=7 / 5 $ [for diatomic gas] Speed of sound, $v_1=460 \mathrm{~m} / \mathrm{s}$ For Helium, Molar mass, $M_2=4$ Heat capacity ratio $\left(C_p / C_V\right)$, $ \gamma_2=\frac{5}{3} \quad \text { [for monoatomic gas] } $ Let speed of sound $=v_2$ Using the expression of speed of sound in an ideal gas at certain temperature $(T)$ is $ \begin{array}{rlrl} & v=\sqrt{\frac{\gamma R T}{M}} \\ \therefore & v_1 & =\sqrt{\frac{\gamma_1 R T}{M_1}}...(i) \\ \text { and } & v_2 & =\sqrt{\frac{\gamma_2 R T}{M_2}}...(ii) \end{array} $ Dividing Eq. (i) by Eq. (ii), we get $ \frac{v_1}{v_2}=\sqrt{\frac{\gamma_1 R T}{M_1}} \times \sqrt{\frac{M_2}{\gamma_2 R T}}=\sqrt{\frac{\gamma_1}{\gamma_2} \times \frac{M_2}{M_1}} $ Substituting the values, we get $ \frac{460}{v_2}=\sqrt{\frac{7}{5} \times \frac{4}{32} \times \frac{3}{5}} \Rightarrow v_2=1420 \mathrm{~m} / \mathrm{s} $ Hence, speed of sound in helium gas at the same temperature is $1420 \mathrm{~m} / \mathrm{s}$

Asked in: AP EAMCET 2021 (24 Aug Shift 1)

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