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(C) The average translational kinetic energy of a gas molecule is given by the formula: $K_{avg} = \frac{3}{2} k_B T$,where $k_B$ is the Boltzmann con

Vedclass · Accepted Answer

$0.048$

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(C) The average translational kinetic energy of a gas molecule is given by the formula: $K_{avg} = \frac{3}{2} k_B T$,where $k_B$ is the Boltzmann constant and $T$ is the absolute temperature.Since the average translational kinetic energy depends only on the temperature $T$ and not on the nature or mass of the gas molecules,it is the same for all ideal gases at the same temperature.Given that the average translational kinetic energy of $O_2$ molecules is $0.048 \ eV$ at temperature $T$,the average translational kinetic energy of $N_2$ molecules at the same temperature $T$ will also be $0.048 \ eV$.

The average translational kinetic energy of $O_2$ molecules at a particular temperature is $0.048 \ eV$. The average translational kinetic energy of an equal number of $N_2$ molecules at the same temperature is (in $eV$):

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