Question:
The R.M.S. speed of oxygen molecules at \(27^\circ\text{C}\) is \(v\). At \(927^\circ\text{C}\), its R.M.S. speed will be
The R.M.S. speed of oxygen molecules at \(27^\circ\text{C}\) is \(v\). At \(927^\circ\text{C}\), its R.M.S. speed will be
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R.M.S. speed depends on absolute temperature as \(v_{\text{rms}}\propto \sqrt{T}\).
Updated On: May 7, 2026
- \(v\)
- \(\frac{v}{2}\)
- \(2v\)
- \(4v\)
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The Correct Option is C
Solution and Explanation
Concept:
R.M.S. speed of gas molecules is proportional to the square root of absolute temperature:
\[
v_{\text{rms}}\propto \sqrt{T}
\]
Step 1: Convert temperatures into kelvin. Initial temperature: \[ T_1=27^\circ\text{C}=27+273=300\ \text{K} \] Final temperature: \[ T_2=927^\circ\text{C}=927+273=1200\ \text{K} \]
Step 2: Use proportionality: \[ \frac{v_2}{v_1}=\sqrt{\frac{T_2}{T_1}} \] \[ \frac{v_2}{v}=\sqrt{\frac{1200}{300}} \] \[ \frac{v_2}{v}=\sqrt{4} \] \[ \frac{v_2}{v}=2 \] \[ v_2=2v \] Therefore, \[ \boxed{2v} \]
Step 1: Convert temperatures into kelvin. Initial temperature: \[ T_1=27^\circ\text{C}=27+273=300\ \text{K} \] Final temperature: \[ T_2=927^\circ\text{C}=927+273=1200\ \text{K} \]
Step 2: Use proportionality: \[ \frac{v_2}{v_1}=\sqrt{\frac{T_2}{T_1}} \] \[ \frac{v_2}{v}=\sqrt{\frac{1200}{300}} \] \[ \frac{v_2}{v}=\sqrt{4} \] \[ \frac{v_2}{v}=2 \] \[ v_2=2v \] Therefore, \[ \boxed{2v} \]
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