Question:
Which of the following relations is correct for root means square speed of a gas at temperature T? (m = mass of molecule, K = Boltzmann constant)
Which of the following relations is correct for root means square speed of a gas at temperature T? (m = mass of molecule, K = Boltzmann constant)
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Recall the expression for root mean square speed in terms of Boltzmann's constant.
Updated On: Nov 13, 2026
- \(C_{rms}=\sqrt{\frac{3KT}{m}}\)
- \(C_{rms}=\sqrt{\frac{8KT}{\pi m}}\)
- \(C_{rms}=\sqrt{\frac{2KT}{m}}\)
- \(C_{rms}=\sqrt{\frac{KT}{m}}\)
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept
From kinetic theory, the mean kinetic energy per molecule is \(\tfrac{1}{2}mc_{rms}^2=\tfrac{3}{2}KT\).
Step 2: Solve
\(c_{rms}=\sqrt{\dfrac{3KT}{m}}\).
Step 3: Compare the options
The expressions in B, C and D are the mean speed, the most probable speed and an incorrect form, not the rms speed.
Final Answer:
The rms speed is \(\sqrt{3KT/m}\). \[ \boxed{\text{(A) }C_{rms}=\sqrt{\dfrac{3KT}{m}}} \]
From kinetic theory, the mean kinetic energy per molecule is \(\tfrac{1}{2}mc_{rms}^2=\tfrac{3}{2}KT\).
Step 2: Solve
\(c_{rms}=\sqrt{\dfrac{3KT}{m}}\).
Step 3: Compare the options
The expressions in B, C and D are the mean speed, the most probable speed and an incorrect form, not the rms speed.
Final Answer:
The rms speed is \(\sqrt{3KT/m}\). \[ \boxed{\text{(A) }C_{rms}=\sqrt{\dfrac{3KT}{m}}} \]
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