The relation between rms velocity, v rms and the most probable velocity, v mp of a gas is

Question

The relation between rms velocity, v rms and the most probable velocity, v mp of a gas is (A) v rms =v mp  (B) v rms =√(3/2) v mp  (C) v rms =√(2/3) v mp  (D) v rms =(2/3) v mp

Tardigrade Admin · Accepted Answer

v rms =√(3 k T/m), v mp =√(2 k T/m) (v rms /v mp )=√(3/2)

Q. The relation between $r m s$ velocity, $v_{r m s}$ and the most probable velocity, $v_{m p}$ of a gas is

$v_{r m s} = v_{m p}$

$v_{r m s} = \frac{3}{2} v_{m p}$

$v_{r m s} = \frac{2}{3} v_{m p}$

$v_{r m s} = \frac{2}{3} v_{m p}$

$v_{r m s} = \frac{3 k T}{m},$
$v_{m p} = \frac{2 k T}{m}$
$\frac{v _{r m s}}{v _{m p}} = \frac{3}{2}$

Q. The relation between rms velocity, vrms​ and the most probable velocity, vmp​ of a gas is

vrms​=vmp​

vrms​=23​​vmp​

vrms​=32​​vmp​

vrms​=32​vmp​

vrms​=m3kT​​, vmp​=m2kT​​ vmp​vrms​​=23​​

Q. The relation between $r m s$ velocity, $v_{r m s}$ and the most probable velocity, $v_{m p}$ of a gas is

$v_{r m s} = v_{m p}$

$v_{r m s} = \frac{3}{2} v_{m p}$

$v_{r m s} = \frac{2}{3} v_{m p}$

$v_{r m s} = \frac{2}{3} v_{m p}$

$v_{r m s} = \frac{3 k T}{m},$
$v_{m p} = \frac{2 k T}{m}$
$\frac{v _{r m s}}{v _{m p}} = \frac{3}{2}$