
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
232.8k+ 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 Main 2023 April 6 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 6 Shift 2 Question Paper with Answer Key

JEE Main 2023 (January 31 Evening Shift) Question Paper with Solutions [PDF]

JEE Main 2023 January 30 Shift 2 Question Paper with Answer Key

JEE Main 2023 January 25 Shift 1 Question Paper with Answer Key

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Key

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

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

Waves Class 11 Physics Chapter 14 CBSE Notes - 2025-26

Mechanical Properties of Fluids Class 11 Physics Chapter 9 CBSE Notes - 2025-26

Thermodynamics Class 11 Physics Chapter 11 CBSE Notes - 2025-26

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

