The ratio of average speed of an oxygen molecule to the rms speed of a nitrogen molecule at the same temperature is :

Question

The ratio of average speed of an oxygen molecule to the rms speed of a nitrogen molecule at the same temperature is : (A) ((3 π /7))1 / 2 (B) ((7/3 π ))1 / 2 (C) ((3/7 π ))√2 (D) ((7 π /3))1 / 2

Tardigrade Admin · Accepted Answer

( barV/Vrms)=√(8 × Mv 2/3 (π M)O2)=√(8 × 28/3 π × 32)=((7/3 π ))v2

Q. The ratio of average speed of an oxygen molecule to the rms speed of a nitrogen molecule at the same temperature is :

$(\frac{3 π}{7})^{1/2}$

$(\frac{7}{3 π})^{1/2}$

$(\frac{3}{7 π})^{2}$

$(\frac{7 π}{3})^{1/2}$

$\frac{V ˉ}{V _{r m s}} = \frac{8 \times M _{v 2}}{3 ( π M ) _{O_{2}}} = \frac{8 \times 28}{3 π \times 32} = (\frac{7}{3 π})^{v_{2}}$

Q. The ratio of average speed of an oxygen molecule to the rms speed of a nitrogen molecule at the same temperature is :

(73π​)1/2

(3π7​)1/2

(7π3​)2​

(37π​)1/2

Vrms​Vˉ​=3(πM)O2​​8×Mv2​​​=3π×328×28​​=(3π7​)v2​

$(\frac{3 π}{7})^{1/2}$

$(\frac{7}{3 π})^{1/2}$

$(\frac{3}{7 π})^{2}$

$(\frac{7 π}{3})^{1/2}$

$\frac{V ˉ}{V _{r m s}} = \frac{8 \times M _{v 2}}{3 ( π M ) _{O_{2}}} = \frac{8 \times 28}{3 π \times 32} = (\frac{7}{3 π})^{v_{2}}$