Question

The ratio of average kinetic energy of $H_2$ and $C{H}_4$ at $27{}^{\circ }C$ will be:

• A. 1 : 1
• B. 1 : 8
• C. 8 : 1
• D. 4 : 1

Answer 1

The correct option is A 1 : 1

We know,
$Avg.KE=\frac{3}{2}kT$
Where $k$ = boltzmann constant.
Temperature is the same in both cases here.
Since the average kinetic energy depends only on temperature so both $H_2$ and $C{H}_4$ will have the same average kinetic energy and the required ratio will be $1:1$.