Question

The speed of sound through oxygen gas at $T\,{\text{K}}$ is $v\,{\text{m}} \cdot {{\text{s}}^{ - 1}}$. At the temperature becomes $2T$ and oxygen gas dissociated into atomic oxygen, the speed of sound is:
A. Remains the same
B. Becomes $2v$
C. Becomes $\sqrt 2 v$
D. None of these

Answer 1

Use the equation for the speed of the sound through a gas at temperature. This equation gives the relation between the adiabatic constant, gas constant, temperature of gas and molecular mass of gas.
Formula used:
The speed $v$ of sound through a gas at temperature ${T_K}$ is
$v = \sqrt {\dfrac{{\gamma R{T_K}}}{M}}$ …… (1)
Here, $\gamma$ is the adiabatic index, $R$ is the gas constant and $M$ is the molecular mass of gas.

Complete step by step answer:
The speed of sound through oxygen gas at $T\,{\text{K}}$ is $v\,{\text{m}} \cdot {{\text{s}}^{ - 1}}$.
The adiabatic constant $\gamma$ for diatomic gas and monatomic gas are $\dfrac{7}{5}$ and $\dfrac{5}{3}$ respectively.
Rewrite equation (1) for the speed of sound through oxygen gas.
$v = \sqrt {\dfrac{{\dfrac{7}{5}RT}}{{{M_{{O_2}}}}}}$
Here, ${M_{{O_2}}}$ is the mass of an oxygen atom.
At the temperature becomes $2T$ and oxygen gas dissociates into atomic oxygen.
Rewrite equation (1) for the speed $v'$of sound through monatomic oxygen.
$v' = \sqrt {\dfrac{{\dfrac{5}{3}R\left( {2T} \right)}}{{{M_O}}}}$
$\Rightarrow v' = \sqrt {\dfrac{{\dfrac{{10}}{3}RT}}{{{M_O}}}}$
Here, ${M_O}$ is the mass of monatomic oxygen.
The mass ${M_{{O_2}}}$ of oxygen molecules is twice as that of mass ${M_O}$ of monatomic oxygen.
${M_{{O_2}}} = 2{M_O}$
Divide the velocity $v'$ of sound through monoatomic oxygen by the velocity $v$ of sound through oxygen molecules.
$\dfrac{{v'}}{v} = \dfrac{{\sqrt {\dfrac{{\dfrac{{10}}{3}RT}}{{{M_O}}}} }}{{\sqrt {\dfrac{{\dfrac{7}{5}RT}}{{{M_{{O_2}}}}}} }}$
$\Rightarrow \dfrac{{v'}}{v} = \sqrt {\dfrac{{10RT}}{{3{M_O}}} \times \dfrac{{5{M_{{O_2}}}}}{{7RT}}}$
$\Rightarrow \dfrac{{v'}}{v} = \sqrt {\dfrac{{10}}{{3{M_O}}} \times \dfrac{{5{M_{{O_2}}}}}{7}}$
Substitute $2{M_O}$ for ${M_{{O_2}}}$ in the above equation.
$\Rightarrow \dfrac{{v'}}{v} = \sqrt {\dfrac{{10}}{{3{M_O}}} \times \dfrac{{5\left( {2{M_O}} \right)}}{7}}$
$\Rightarrow \dfrac{{v'}}{v} = \sqrt {\dfrac{{100}}{{21}}}$
$\Rightarrow v' = \dfrac{{10}}{{\sqrt {21} }}v$
Therefore, the speed of sound through monoatomic oxygen is $\dfrac{{10}}{{\sqrt {21} }}v$.

So, the correct answer is “Option D”.

Note:
Since the oxygen molecule contains two oxygen atoms, the mass of the oxygen molecule is twice as that of the mass of monatomic oxygen.
The adiabatic constant is different for monoatomic and diatomic gas.