Question

At 300 K, the ratio of the average molecular kinetic energy of $U{F}_6$ to that of $H_2$ is:

• A. 2 : 7
• B. 7 : 2
• C. 6 : 2
• D. 1 : 1

Answer 1

The correct option is D 1 : 1

Average molecular K.E. $=\frac{3}{2}KT$

It means kinetic energy depends on $T$ only.

$\because$ Temperature of $H_2$ and $U{F}_6$ both is same.

$\therefore$ Ratio of average molecular K.E. $=1:1$.

Hence, the correct answer is option $\text{D}$.