
If the root means square velocity is \[{\left( {{\rm{3}}{{\rm{0}}^{\rm{2}}}{\rm{R}}} \right)^{\frac{{\rm{1}}}{{\rm{2}}}}}\] at \[{\rm{27^\circ C}}\] then calculate the molar mass of gas in kilograms.
A. 1
B. 2
C. 4
D. 0.001
Answer
232.8k+ views
Hint: Root mean square velocity or r.m.s velocity is interpreted as the square root of the mean of the squares of the distinct velocities.
It is signified by \[{{\rm{V}}_{{\rm{rms}}}}\].
Formula Used:
\[{{\rm{V}}_{{\rm{rms}}}}{\rm{ = }}\sqrt {\frac{{{\rm{3RT}}}}{{\rm{M}}}} \] where
R = Universal gas constant
T = Temperature
M = Gram molecular mass of the gas molecule
Complete Step by Step Solution:
A gas comprises molecules present at large distances.
The molecules are independent to change positions in the area in which it is enclosed.
These molecules have continual motion in varied directions at varied velocities.
They keep on striking each other and the walls of the container which is defined as a molecular collision.
The expression for r.m.s velocity is given by the expression:
Given
\[{{\rm{V}}_{{\rm{rms}}}}{\rm{ = }}\sqrt {\frac{{{\rm{3RT}}}}{{\rm{M}}}} \]
\[{{\rm{V}}_{{\rm{rms}}}}{\rm{ = }}{\left( {{\rm{3}}{{\rm{0}}^{\rm{2}}}{\rm{R}}} \right)^{\frac{{\rm{1}}}{{\rm{2}}}}}\]
For calculation of rms velocity, the temperature must be in Kelvin.
So, temperature=27°C=(27+273)=300K
\[ \Rightarrow \sqrt {{{\left( {{\rm{30}}} \right)}^{\rm{2}}}{\rm{R}}} {\rm{ = }}\sqrt {\frac{{{\rm{3R}}\left( {{\rm{300K}}} \right)}}{{\rm{M}}}} \]
Squaring both sides we get,
\[ \Rightarrow {\left( {{\rm{30}}} \right)^{\rm{2}}}{\rm{ = }}\frac{{{\rm{3}}\left( {{\rm{300K}}} \right)}}{{\rm{M}}}\]
\[ \Rightarrow {\rm{M = 3}}\left( {{\rm{300}}} \right){\rm{30 \times 30g}}\]
So, M=1g=0.001kg
Hence, the molecular mass of the gas is 0.001kg.
So, option D is correct.
Note: r.m.s velocity and the square root of temperature have a direct relationship. Root mean square velocity and the square root of molecular mass are inversely related. While calculating the numerical value of r.m.s velocity, the temperature is to be taken in kelvin and molar mass is to be taken in kg always to get the value of r.m.s velocity in m/s.
It is signified by \[{{\rm{V}}_{{\rm{rms}}}}\].
Formula Used:
\[{{\rm{V}}_{{\rm{rms}}}}{\rm{ = }}\sqrt {\frac{{{\rm{3RT}}}}{{\rm{M}}}} \] where
R = Universal gas constant
T = Temperature
M = Gram molecular mass of the gas molecule
Complete Step by Step Solution:
A gas comprises molecules present at large distances.
The molecules are independent to change positions in the area in which it is enclosed.
These molecules have continual motion in varied directions at varied velocities.
They keep on striking each other and the walls of the container which is defined as a molecular collision.
The expression for r.m.s velocity is given by the expression:
Given
\[{{\rm{V}}_{{\rm{rms}}}}{\rm{ = }}\sqrt {\frac{{{\rm{3RT}}}}{{\rm{M}}}} \]
\[{{\rm{V}}_{{\rm{rms}}}}{\rm{ = }}{\left( {{\rm{3}}{{\rm{0}}^{\rm{2}}}{\rm{R}}} \right)^{\frac{{\rm{1}}}{{\rm{2}}}}}\]
For calculation of rms velocity, the temperature must be in Kelvin.
So, temperature=27°C=(27+273)=300K
\[ \Rightarrow \sqrt {{{\left( {{\rm{30}}} \right)}^{\rm{2}}}{\rm{R}}} {\rm{ = }}\sqrt {\frac{{{\rm{3R}}\left( {{\rm{300K}}} \right)}}{{\rm{M}}}} \]
Squaring both sides we get,
\[ \Rightarrow {\left( {{\rm{30}}} \right)^{\rm{2}}}{\rm{ = }}\frac{{{\rm{3}}\left( {{\rm{300K}}} \right)}}{{\rm{M}}}\]
\[ \Rightarrow {\rm{M = 3}}\left( {{\rm{300}}} \right){\rm{30 \times 30g}}\]
So, M=1g=0.001kg
Hence, the molecular mass of the gas is 0.001kg.
So, option D is correct.
Note: r.m.s velocity and the square root of temperature have a direct relationship. Root mean square velocity and the square root of molecular mass are inversely related. While calculating the numerical value of r.m.s velocity, the temperature is to be taken in kelvin and molar mass is to be taken in kg always to get the value of r.m.s velocity in m/s.
Recently Updated Pages
Hess Law of Constant Heat Summation: Definition, Formula & Applications

Disproportionation Reaction: Definition, Example & JEE Guide

JEE General Topics in Chemistry Important Concepts and Tips

JEE Extractive Metallurgy Important Concepts and Tips for Exam Preparation

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

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

Trending doubts
JEE Main 2026: Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

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

Hydrocarbons Class 11 Chemistry Chapter 9 CBSE Notes - 2025-26

Thermodynamics Class 11 Chemistry Chapter 5 CBSE Notes - 2025-26

Equilibrium Class 11 Chemistry Chapter 6 CBSE Notes - 2025-26

Organic Chemistry Some Basic Principles And Techniques Class 11 Chemistry Chapter 8 CBSE Notes - 2025-26

NCERT Solutions For Class 11 Chemistry Chapter 7 Redox Reactions (2025-26)

