
If ${V_H},\,{V_N}\,and\,{V_O}$ denotes the root mean square velocities of molecules of hydrogen, nitrogen and oxygen respectively at a given temperature then:
(A) ${V_H} > {V_N} > {V_O}$
(B) ${V_H} = {V_N} = {V_O}$
(C) ${V_O} > {V_H} > {V_N}$
(D) ${V_N} > {V_O} > {V_H}$
Answer
163.5k+ views
Hint: Let start with finding the relation between the root mean square velocity and the masses of the molecules of Hydrogen, Nitrogen and oxygen. Then on the basis of the relation between their masses get the relation between the root mean square velocity of the molecules given.
Formula Used:
${V_{rms}}$ is given by:
${V_{rms}} = \sqrt {\dfrac{{3RT}}{M}} $
Where, R is constant
T is temperature
And M is the molecular mass of the particle.
Complete answer:
First start with the given information in the question:
Let ${V_{rms}}$ be the root mean square velocity.
Now, the root mean square velocity of Hydrogen be ${V_H}$ ,
the root mean square velocity of Nitrogen be ${V_N}$
and the root mean square velocity of Oxygen be ${V_O}$
Now, we know that ${V_{rms}}$ is given by:
${V_{rms}} = \sqrt {\dfrac{{3RT}}{M}} $
So, from the above equation we can find that root mean square velocity is inversely proportional to the mass of the molecule.
${V_{rms}} \propto \dfrac{1}{\surd{M}}$
Now, we know that;
${M_H} = $ , ${M_N} = 28$ and ${M_0} = 32$
Means, ${M_H} < {M_N} < {M_O}$
Therefore, ${V_H} > {V_N} > {V_O}$
Hence the correct answer is Option(A).
Note: Here we have used the formula of root mean square velocity in terms of the universal gas constant, temperature of the gas and the mass of the gas molecule. As all the other quantities were constant hence we get the required relation for the root mean square velocity and the mass of the gas.
Formula Used:
${V_{rms}}$ is given by:
${V_{rms}} = \sqrt {\dfrac{{3RT}}{M}} $
Where, R is constant
T is temperature
And M is the molecular mass of the particle.
Complete answer:
First start with the given information in the question:
Let ${V_{rms}}$ be the root mean square velocity.
Now, the root mean square velocity of Hydrogen be ${V_H}$ ,
the root mean square velocity of Nitrogen be ${V_N}$
and the root mean square velocity of Oxygen be ${V_O}$
Now, we know that ${V_{rms}}$ is given by:
${V_{rms}} = \sqrt {\dfrac{{3RT}}{M}} $
So, from the above equation we can find that root mean square velocity is inversely proportional to the mass of the molecule.
${V_{rms}} \propto \dfrac{1}{\surd{M}}$
Now, we know that;
${M_H} = $ , ${M_N} = 28$ and ${M_0} = 32$
Means, ${M_H} < {M_N} < {M_O}$
Therefore, ${V_H} > {V_N} > {V_O}$
Hence the correct answer is Option(A).
Note: Here we have used the formula of root mean square velocity in terms of the universal gas constant, temperature of the gas and the mass of the gas molecule. As all the other quantities were constant hence we get the required relation for the root mean square velocity and the mass of the gas.
Recently Updated Pages
Uniform Acceleration - Definition, Equation, Examples, and FAQs

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Degree of Dissociation and Its Formula With Solved Example for JEE

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

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

Motion in a Straight Line Class 11 Notes: CBSE Physics Chapter 2

NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line

Important Questions for CBSE Class 11 Physics Chapter 1 - Units and Measurement
