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The temperature at which $H_2$ has the same RMS speed (at 1 atm) as that of $O_2$ at NTP is

$(a)\;37K\qquad(b)\;17K\qquad(c)\;512K\qquad(d)\;27K$

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$U_{rms} H_2 = \sqrt{[\large\frac{3RT_1}{2}]}$
$U_{rms} O_2 = \sqrt{[\large\frac{3RT_2}{32}]}$
$[T_2 = 273K]$
$\therefore \large\frac{T_1}{2} = \large\frac{T_2}{32}$
$T_1 = \large\frac{2\times 273}{32} $
$\;\;\;\;\;=17K$
Hence answer is (b)
answered Mar 9, 2014 by sharmaaparna1
 

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