kinetic energy of $O_2 >$ kinetic energy of $SO_2$
kinetic energy of $O_2 < $ kinetic energy of $SO_2$
kinetic energy of both are equal.
None of these
Hide Solution
Verified By Collegedunia
The Correct Option isB
Solution and Explanation
$K E=\frac{3}{2} R T$$K E \propto T$$\frac{K E_{O_{2}}}{K E_{S O_{2}}}=\frac{T_{O_{2}}}{T_{S O_{2}}}=\frac{273}{546}=\frac{1}{2}$$K E_{S O_{2}}=2 K E_{O_{2}}$$\therefore K E_{S O_{2}}>2 K E_{O_{2}}$