Q. At the same temperature calculate the ratio of average velocity of $SO_{2}\ to\ CH_{4}.$
A) 2 : 3
B) 3 : 4
C) 1 : 2
D) 1 : 6
Explanation
(c)
$U_{av} \propto \frac{1}{\sqrt{M}}\$at constant temperature
$\frac{U_{av}\left( {SO}_{2} \right)}{U_{av}\ \left( {CH}_{4} \right)} = \sqrt{\frac{M_{{CH}_{4}}}{M_{{SO}_{2}}}} = \sqrt{\frac{16}{64}} = \frac{1}{2}$
$U_{{SO}_{2}}\ :U_{{CH}_{4}} = 1\ :2$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.