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# The ratio of speed of sound in Hydrogen to that in oxygen at the same temperature is:A. $1:4$B. $4:1$C. $1:1$D. $16:1$

Last updated date: 20th Jun 2024
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Hint: The given question is from kinetic theory of gases and speed of sound in gaseous medium. As we know that velocity of sound depends on adiabatic constant, pressure of gas and molecular mass of gas in a specific relation. So we use this relation to solve this problem.

According to kinetic theory of gases the speed of sound in a gas medium depends on many factors like atomicity of gas, pressure of gas and molecular weight of gas molecules. This relation of speed of sound is given by
$v = \sqrt {\dfrac{{\gamma P}}{\rho }}$$= \sqrt {\dfrac{{\gamma RT}}{M}}$
Where $v$= speed of sound
$\gamma$= Adiabatic constant for gas
$P$= Pressure of gas
$\rho$= density of gas
R= Universal gas constant
T= Temperature of gas
M= Molecular weight of gas
The given parameters in above question are
Hydrogen$\Rightarrow {M_{{H_2}}} = 2$
Oxygen$\Rightarrow {M_{{O_2}}} = 32$
Temperature is the same for both gases as per given condition. And both gases have the same atomicity so the value of adiabatic constant will also be the same.
So from the above mentioned formula we can say that $\gamma RT$ will have constant value for both gases and the value of speed of sound will depend only on the molecular mass of gas.
$v \propto \sqrt {\dfrac{1}{M}}$
After comparing both gases, we get
$\dfrac{{{v_{{H_2}}}}}{{{v_{{O_2}}}}} = \sqrt {\dfrac{{{M_{{O_2}}}}}{{{M_{{H_2}}}}}}$
$\dfrac{{{v_{{H_2}}}}}{{{v_{{O_2}}}}} = \sqrt {\dfrac{{32}}{2}} = \dfrac{4}{1}$
So the ratio of speed of sound in hydrogen and oxygen gas will be 4 : 1.
The correct answer is option (B).

Note: Here it is important to note that we have taken the same value of $\gamma$ for both the gases as hydrogen and oxygen exist as diatomic molecules and hence they will have the same degrees of freedom and hence the same $\gamma$.