Question:
The R.M.S. speed of oxygen molecules at 27 °C is $v$. At 927 °C the rms speed will be:
The R.M.S. speed of oxygen molecules at 27 °C is $v$. At 927 °C the rms speed will be:
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If the absolute temperature (in Kelvin) becomes 4 times, the speed doubles because of the square root relationship.
Updated On: May 5, 2026
- $v$
- $v/2$
- $2v$
- $4v$
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The Correct Option is C
Solution and Explanation
Step 1: Concept
The Root Mean Square (RMS) speed is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$, so $v \propto \sqrt{T}$.
Step 2: Meaning
Initial Temperature $T_1 = 27 + 273 = 300$ K. Final Temperature $T_2 = 927 + 273 = 1200$ K.
Step 3: Analysis
$\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{1200}{300}} = \sqrt{4} = 2$.
Step 4: Conclusion
Therefore, $v_2 = 2v$.
Final Answer: (C)
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