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Ratio of kinetic energies 3 gm of $H_{2}$ and 4 gm of $O_{2}$ at $27^{0} C$ is

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1:12

3:4

12:1

1:3

answer is C.

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Detailed Solution

$3 gms H_{2}; {KE}_{1} = \frac{3}{2} nRT$

$4 gm O_{2}; {KE}_{2} = \frac{3}{2} nRT$ same temperature

$Rate of \frac{{KE}_{1}}{{KE}_{2}} = \frac{\frac{3}{2}}{\frac{4}{32}} = \frac{3}{2} \times \frac{32}{4} = 12 : 1$

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