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At what temperature is the root mean square (rms) speed of oxygen molecules ($O_2$) equal to that of helium molecules ($He$) at $27^\circ\text{C}$? (Given molar mass of $O_2 = 32\text{ g/mol}$, $He = 4\text{ g/mol}$)}

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Since oxygen is 8 times heavier than helium ($32 / 4 = 8$), its absolute temperature must be exactly 8 times higher to maintain the same rms speed: $8 \times 300\text{ K} = 2400\text{ K}$.

Updated On: May 31, 2026

$2400\text{ K}$
$2127\text{ K}$
$2427^\circ\text{C}$
$2127^\circ\text{C}$

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The Correct Option is A

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At what temperature is the root mean square (rms) speed of oxygen molecules ($O_2$) equal to that of helium molecules ($He$) at $27^\circ\text{C}$? (Given molar mass of $O_2 = 32\text{ g/mol}$, $He = 4\text{ g/mol}$)}

Show Hint

The Correct Option is A

Solution and Explanation

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