Question

If the total number of $H_2$ molecules is double of the $O_2$ molecules then the ratio of total kinetic energies pf $H_2$ to that of $O_2$ at 300 K is :-

• A. 1 : 1
• B. 1 : 2
• C. 2 : 1
• D. 1 : 3

Answer 1

The correct option is C 2 : 1

Kinetic energy for a single molecule is given as $E=\frac{3}{2}kT$

where $k$ is Boltzman's constant so for $2n$ molecules of Hydrogen the kinetic energy will be $E_{H_2}=2n\left(1.5kT\right)$

And that for $n$ molecules of $O_2$ will be $E_{O_2}=n\left(1.5kT\right)$

ratio of energi$es$ will be $E_{H_2}:E_{O_2}=2:1$

Option C is correct.