At $T(\mathrm{~K})$, the ratio of kinetic energies of $4 \mathrm{~g}$ of $\mathrm{H}_2(g)$ and $8…
At $T(\mathrm{~K})$, the ratio of kinetic energies of $4 \mathrm{~g}$ of $\mathrm{H}_2(g)$ and $8 \mathrm{~g}$ of $\mathrm{O}_2(g)$ is
$1: 4$
$4: 1$
$2: 1$
$8: 1$
Solution
$\mathrm{KE}=\frac{3}{2} R T$ for 1 mole of the gas.
$\because 4 \mathrm{~g}$ of $\mathrm{H}_2$ gas has 2 moles of $\mathrm{H}_2$
$\therefore$ It has 2 times KE as compared to 1 mole of gas
But $8 \mathrm{~g}$ of $\mathrm{O}_2$ gas has $1 / 4$ moles of $\mathrm{O}_2$;
$\therefore$ It has one fourth part of the KE as compared to 1 mole of gas.
Hence, ratio of $\mathrm{KE}$ of $\mathrm{H}_2$ and $\mathrm{O}_2$
$
\mathrm{KE}_{\mathrm{H}_2}: \mathrm{KE}_{\mathrm{O}_2}=2: \frac{1}{4}=8: 1
$