Question

The ratio of average molecular kinetics energy of $PC{l}_5$ to that of $O_2$ both at $300K$ is:

• A. $1:1$
• B. $7:2$
• C. $176:1$
• D. $2:7$

Answer 1

The correct option is A $1:1$

Given:Temperature$=300K$

To find:Ratio of average molecular kinetic energy of $PC{l}_5$ and$O_2$

Solution:

We know that, average molecular kinetic energy for a molecule$=\frac{3}{2}kT$ for a molecule.

(Where k is Boltzmann constant and T is temperature)

$\therefore$$\therefore$Average molecular kinetic energy of$PC{l}_5$$=\frac{3}{2}k\left(300\right)$

And that of $O_2$ is$=\frac{3}{2}k\left(300\right)$$=\frac{3}{2}k\left(300\right)$

$\therefore$$\therefore$Required ratio$=1:1$

Hence (a) is the correct answer.