If the number of molecules of {H_2} is double | vedclass

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(C) The average kinetic energy of a gas molecule is given by the formula $E = \frac{3}{2} k_B T$,where $k_B$ is the Boltzmann constant and $T$ is the

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$2:1$

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(C) The average kinetic energy of a gas molecule is given by the formula $E = \frac{3}{2} k_B T$,where $k_B$ is the Boltzmann constant and $T$ is the absolute temperature.The total kinetic energy of $N$ molecules is given by $K.E. = N 	imes \frac{3}{2} k_B T$.For hydrogen $(H_2)$: $E_{H_2} = N_{H_2} 	imes \frac{3}{2} k_B T$.For oxygen $(O_2)$: $E_{O_2} = N_{O_2} 	imes \frac{3}{2} k_B T$.Given that the number of molecules of $H_2$ is double that of $O_2$,we have $N_{H_2} = 2 N_{O_2}$.Therefore,the ratio of the total kinetic energy is:$\frac{E_{H_2}}{E_{O_2}} = \frac{N_{H_2} 	imes \frac{3}{2} k_B T}{N_{O_2} 	imes \frac{3}{2} k_B T} = \frac{N_{H_2}}{N_{O_2}} = \frac{2 N_{O_2}}{N_{O_2}} = \frac{2}{1}$.Thus,the ratio is $2:1$.

If the number of molecules of ${H_2}$ is double that of ${O_2}$,then the ratio of the kinetic energy of hydrogen to that of oxygen at $300 \, K$ is:

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