If the number of hydrogen molecules is twice | vedclass

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(C) The average kinetic energy of a gas molecule depends only on the absolute temperature $T$ and is given by $K_{avg} = \frac{3}{2} k_B T$.Since both

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$2 : 1$

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(C) The average kinetic energy of a gas molecule depends only on the absolute temperature $T$ and is given by $K_{avg} = \frac{3}{2} k_B T$.Since both gases are at the same temperature,the average kinetic energy per molecule is the same for both hydrogen and oxygen.Let $N_H$ be the number of hydrogen molecules and $N_O$ be the number of oxygen molecules. Given $N_H = 2N_O$.The total kinetic energy $E$ is given by $E = N 	imes K_{avg}$.Therefore,the ratio of total kinetic energies is $\frac{E_H}{E_O} = \frac{N_H 	imes K_{avg}}{N_O 	imes K_{avg}} = \frac{N_H}{N_O}$.Substituting $N_H = 2N_O$,we get $\frac{E_H}{E_O} = \frac{2N_O}{N_O} = \frac{2}{1}$.Thus,the ratio is $2 : 1$.

If the number of hydrogen molecules is twice the number of oxygen molecules at the same temperature,what is the ratio of their total kinetic energies? (Hydrogen : Oxygen)

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