Nitrogen gas is at \[{300^0}C\] temperature. Find the temperature (in K) at which the rms speed of a \[{H_2}\] molecule would be equal to the rms speed of a nitrogen molecule.
Answer
249.9k+ views
Hint: Before we proceed to solving the problem. Let us know about the root mean square velocity (speed). The root mean square velocity is the square root of the mean of squares of the velocity of individual gas molecules.
Formula Used:
The formula for the root mean square speed is,
\[{V_{rms}} = \sqrt {\dfrac{{3RT}}{M}} \]
Where, R is universal gas constant, T is temperature and M is molar mass of gas.
Complete step by step solution:
Consider a nitrogen gas at \[{300^0}C\]. We need to find the value of temperature (in K) at which the rms speed of an \[{H_2}\] molecule would be equal to the RMS speed of a nitrogen molecule. For that, we have the formula for finding rms speed of nitrogen as,
\[{\left( {{V_{rms}}} \right)_{{N_2}}} = \sqrt {\dfrac{{3RT}}{M}} \]
We need to convert the temperature into kelvin, that is
\[T = \left( {300 + 273} \right)K\]
\[ \Rightarrow T = 573K\] and the molar mass of nitrogen is, \[M = 28\]
Substitute the value of temperature and molar mass in above equation we get,
\[{\left( {{V_{rms}}} \right)_{{H_2}}} = \sqrt {\dfrac{{3R\left( {573} \right)}}{{28}}} \]
Now, the rms speed of hydrogen is,
\[{\left( {{V_{rms}}} \right)_{{H_2}}} = \sqrt {\dfrac{{3RT}}{M}} \]
The molar mass of hydrogen is, \[M = 2\]. When the rms speed of a \[{H_2}\] molecule is equal to the rms speed of a nitrogen molecule, then,
\[{\left( {{V_{rms}}} \right)_{{H_2}}} = {\left( {{V_{rms}}} \right)_{{N_2}}}\]
\[\Rightarrow \sqrt {\dfrac{{3RT}}{2}} = \sqrt {\dfrac{{3R\left( {573} \right)}}{{28}}} \]
\[\Rightarrow \dfrac{T}{2} = \dfrac{{573}}{{28}}\]
\[\Rightarrow T = \dfrac{{573 \times 2}}{{28}}\]
\[\therefore T = 40.92K\]
Therefore, the temperature at which the rms speed of a \[{H_2}\] molecule would be equal to the rms speed of a nitrogen molecule is 40.92K.
Note: The variable R is also used to denote the resistance. But in this case the R we are using is the universal gas constant and M is the molar mass. This universal gas constant value depends on the type of gas and remains the same every time.
Formula Used:
The formula for the root mean square speed is,
\[{V_{rms}} = \sqrt {\dfrac{{3RT}}{M}} \]
Where, R is universal gas constant, T is temperature and M is molar mass of gas.
Complete step by step solution:
Consider a nitrogen gas at \[{300^0}C\]. We need to find the value of temperature (in K) at which the rms speed of an \[{H_2}\] molecule would be equal to the RMS speed of a nitrogen molecule. For that, we have the formula for finding rms speed of nitrogen as,
\[{\left( {{V_{rms}}} \right)_{{N_2}}} = \sqrt {\dfrac{{3RT}}{M}} \]
We need to convert the temperature into kelvin, that is
\[T = \left( {300 + 273} \right)K\]
\[ \Rightarrow T = 573K\] and the molar mass of nitrogen is, \[M = 28\]
Substitute the value of temperature and molar mass in above equation we get,
\[{\left( {{V_{rms}}} \right)_{{H_2}}} = \sqrt {\dfrac{{3R\left( {573} \right)}}{{28}}} \]
Now, the rms speed of hydrogen is,
\[{\left( {{V_{rms}}} \right)_{{H_2}}} = \sqrt {\dfrac{{3RT}}{M}} \]
The molar mass of hydrogen is, \[M = 2\]. When the rms speed of a \[{H_2}\] molecule is equal to the rms speed of a nitrogen molecule, then,
\[{\left( {{V_{rms}}} \right)_{{H_2}}} = {\left( {{V_{rms}}} \right)_{{N_2}}}\]
\[\Rightarrow \sqrt {\dfrac{{3RT}}{2}} = \sqrt {\dfrac{{3R\left( {573} \right)}}{{28}}} \]
\[\Rightarrow \dfrac{T}{2} = \dfrac{{573}}{{28}}\]
\[\Rightarrow T = \dfrac{{573 \times 2}}{{28}}\]
\[\therefore T = 40.92K\]
Therefore, the temperature at which the rms speed of a \[{H_2}\] molecule would be equal to the rms speed of a nitrogen molecule is 40.92K.
Note: The variable R is also used to denote the resistance. But in this case the R we are using is the universal gas constant and M is the molar mass. This universal gas constant value depends on the type of gas and remains the same every time.
Recently Updated Pages
JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Isoelectronic Definition in Chemistry: Meaning, Examples & Trends

Ionisation Energy and Ionisation Potential Explained

Iodoform Reactions - Important Concepts and Tips for JEE

Introduction to Dimensions: Understanding the Basics

Instantaneous Velocity Explained: Formula, Examples & Graphs

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

Hybridisation in Chemistry – Concept, Types & Applications

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

Understanding the Electric Field of a Uniformly Charged Ring

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

Derivation of Equation of Trajectory Explained for Students

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

JEE Advanced 2026 - Exam Date (Released), Syllabus, Registration, Eligibility, Preparation, and More

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

Important Questions For Class 11 Physics Chapter 1 Units and Measurement - 2025-26

NCERT Solutions For Class 11 Physics Chapter 1 Units And Measurements - 2025-26

CBSE Notes Class 11 Physics Chapter 4 - Laws of Motion - 2025-26

