
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. \[16:1\]
Answer
218.7k+ views
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.
Complete step by step answer:
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 }} $$ = \sqrt {\dfrac{{\gamma RT}}{M}} $
Where $v$= speed of sound
$\gamma $= Adiabatic constant for gas
$P$= Pressure of gas
$\rho $= density of gas
R= Universal gas constant
T= Temperature of gas
M= Molecular weight of gas
The given parameters in above question are
Hydrogen$ \Rightarrow {M_{{H_2}}} = 2$
Oxygen$ \Rightarrow {M_{{O_2}}} = 32$
Temperature is the same for both gases as per given condition. And both gases have the same atomicity so the value of adiabatic constant will also be the same.
So from the above mentioned formula we can say that $\gamma RT$ will have constant value for both gases and the value of speed of sound will depend only on the molecular mass of gas.
$v \propto \sqrt {\dfrac{1}{M}} $
After comparing both gases, we get
$\dfrac{{{v_{{H_2}}}}}{{{v_{{O_2}}}}} = \sqrt {\dfrac{{{M_{{O_2}}}}}{{{M_{{H_2}}}}}} $
$\dfrac{{{v_{{H_2}}}}}{{{v_{{O_2}}}}} = \sqrt {\dfrac{{32}}{2}} = \dfrac{4}{1}$
So the ratio of speed of sound in hydrogen and oxygen gas will be 4 : 1.
The correct answer is option (B).
Note: Here it is important to note that we have taken the same value of $\gamma$ for both the gases as hydrogen and oxygen exist as diatomic molecules and hence they will have the same degrees of freedom and hence the same $\gamma$.
Complete step by step answer:
According to kinetic theory of gases the speed of sound in a gas medium depends on many factors like atomicity of gas, pressure of gas and molecular weight of gas molecules. This relation of speed of sound is given by
$v = \sqrt {\dfrac{{\gamma P}}{\rho }} $$ = \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 $\gamma$.
Recently Updated Pages
Two discs which are rotating about their respective class 11 physics JEE_Main

A ladder rests against a frictionless vertical wall class 11 physics JEE_Main

Two simple pendulums of lengths 1 m and 16 m respectively class 11 physics JEE_Main

The slopes of isothermal and adiabatic curves are related class 11 physics JEE_Main

A trolly falling freely on an inclined plane as shown class 11 physics JEE_Main

The masses M1 and M2M2 M1 are released from rest Using class 11 physics JEE_Main

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

Derivation of Equation of Trajectory Explained for Students

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Angle of Deviation in a Prism

Understanding Collisions: Types and Examples for Students

Understanding Atomic Structure for Beginners

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Units And Measurements Class 11 Physics Chapter 1 CBSE Notes - 2025-26

NCERT Solutions For Class 11 Physics Chapter 8 Mechanical Properties Of Solids

Motion in a Straight Line Class 11 Physics Chapter 2 CBSE Notes - 2025-26

NCERT Solutions for Class 11 Physics Chapter 7 Gravitation 2025-26

How to Convert a Galvanometer into an Ammeter or Voltmeter

