JEE MAIN - Physics (2022 - 25th June Morning Shift - No. 7)
The relation between root mean square speed (vrms) and most probable sped (vp) for the molar mass M of oxygen gas molecule at the temperature of 300 K will be :
$${v_{rms}} = \sqrt {{2 \over 3}} {v_p}$$
$${v_{rms}} = \sqrt {{3 \over 2}} {v_p}$$
$${v_{rms}} = {v_p}$$
$${v_{rms}} = \sqrt {{1 \over 3}} {v_p}$$
توضيح
$${v_{rms}} = \sqrt {{{3RT} \over M}} $$
$${v_p} = \sqrt {{{2RT} \over M}} $$
$$ \Rightarrow {v_{rms}} = \sqrt {{3 \over 2}} {v_p}$$