At T K, the ratio of kinetic energies of 4 g | vedclass

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(D) The kinetic energy $(KE)$ of $n$ moles of an ideal gas is given by $KE = n 	imes \frac{3}{2} RT$.For $4$ $g$ of $H_2$, the number of moles $n_{H_

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$8: 1$

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(D) The kinetic energy $(KE)$ of $n$ moles of an ideal gas is given by $KE = n 	imes \frac{3}{2} RT$.For $4$ $g$ of $H_2$, the number of moles $n_{H_2} = \frac{4 \ g}{2 \ g/mol} = 2$ $mol$.So, $KE_{H_2} = 2 	imes \frac{3}{2} RT = 3RT$.For $8$ $g$ of $O_2$, the number of moles $n_{O_2} = \frac{8 \ g}{32 \ g/mol} = 0.25$ $mol$ or $\frac{1}{4}$ $mol$.So, $KE_{O_2} = \frac{1}{4} 	imes \frac{3}{2} RT = \frac{3}{8} RT$.The ratio $KE_{H_2} : KE_{O_2} = 3RT : \frac{3}{8} RT = 1 : \frac{1}{8} = 8 : 1$.

At $T$ $K$, the ratio of kinetic energies of $4$ $g$ of $H_{2(g)}$ and $8$ $g$ of $O_{2(g)}$ is

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