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

Question

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

Answer 1

$U_{rms} H_2 = \sqrt{[\large\frac{3RT_1}{2}]}$

$U_{rms} O_2 = \sqrt{[\large\frac{3RT_2}{32}]}$

$[T_2 = 273K]$

$\therefore \large\frac{T_1}{2} = \large\frac{T_2}{32}$

$T_1 = \large\frac{2\times 273}{32} $

$\;\;\;\;\;=17K$

Hence answer is (b)