Question:
The RMS velocity of a gas at temperature \( T \) is \( v \). If temperature becomes \( 4T \), new RMS velocity is:
The RMS velocity of a gas at temperature \( T \) is \( v \). If temperature becomes \( 4T \), new RMS velocity is:
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When temperature increases by a factor of \( n \), the RMS velocity increases by a factor of \( \sqrt{n} \). Since the temperature quadrupled (\( 4 \times \)), the velocity doubled (\( \sqrt{4} = 2 \)).
Updated On: Jun 3, 2026
- \( v \)
- \( 2v \)
- \( 4v \)
- \( v/2 \)
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The Correct Option is B
Solution and Explanation
Concept:
The root mean square (RMS) velocity of gas molecules is a measure of the speed of particles in a gas, which is directly linked to the kinetic energy and absolute temperature of the gas.
• Formula: \( v_{rms} = \sqrt{\frac{3RT}{M}} \)
• \( R \): Universal gas constant.
• \( T \): Absolute temperature (in Kelvin).
• \( M \): Molar mass of the gas.
• Proportionality: \( v_{rms} \propto \sqrt{T} \).
Step 1: Establishing the relationship between velocity and temperature.
From the formula, we see that for a given gas (constant \( M \)), the RMS velocity is directly proportional to the square root of its absolute temperature: \[ v \propto \sqrt{T} \quad \Rightarrow \quad \frac{v_1}{v_2} = \sqrt{\frac{T_1}{T_2}} \]
Step 2: Substituting the given values.
Initial velocity (\( v_1 \)) = \( v \)
Initial temperature (\( T_1 \)) = \( T \)
Final temperature (\( T_2 \)) = \( 4T \)
Let the new RMS velocity be \( v_2 \). \[ \frac{v}{v_2} = \sqrt{\frac{T}{4T}} \]
Step 3: Solving for the new velocity.
\[ \frac{v}{v_2} = \sqrt{\frac{1}{4}} = \frac{1}{2} \] Cross-multiplying gives: \[ v_2 = 2v \]
• Formula: \( v_{rms} = \sqrt{\frac{3RT}{M}} \)
• \( R \): Universal gas constant.
• \( T \): Absolute temperature (in Kelvin).
• \( M \): Molar mass of the gas.
• Proportionality: \( v_{rms} \propto \sqrt{T} \).
Step 1: Establishing the relationship between velocity and temperature.
From the formula, we see that for a given gas (constant \( M \)), the RMS velocity is directly proportional to the square root of its absolute temperature: \[ v \propto \sqrt{T} \quad \Rightarrow \quad \frac{v_1}{v_2} = \sqrt{\frac{T_1}{T_2}} \]
Step 2: Substituting the given values.
Initial velocity (\( v_1 \)) = \( v \)
Initial temperature (\( T_1 \)) = \( T \)
Final temperature (\( T_2 \)) = \( 4T \)
Let the new RMS velocity be \( v_2 \). \[ \frac{v}{v_2} = \sqrt{\frac{T}{4T}} \]
Step 3: Solving for the new velocity.
\[ \frac{v}{v_2} = \sqrt{\frac{1}{4}} = \frac{1}{2} \] Cross-multiplying gives: \[ v_2 = 2v \]
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