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Question

(C) The average translational kinetic energy of an ideal gas molecule is given by the formula:$E = \frac{3}{2} k_B T$where $k_B$ is the Boltzmann cons

Vedclass · Accepted Answer

$0.048$

Vedclass · Answer

(C) The average translational kinetic energy of an ideal 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.From this formula,it is clear that the translational kinetic energy depends only on the temperature $T$ of the gas.It is independent of the nature of the gas (i.e.,independent of molar mass).Since the temperature $T$ is the same for both ${O_2}$ and ${N_2}$ molecules,their average translational kinetic energies will be equal.Therefore,$E_{N_2} = E_{O_2} = 0.048 \; eV$.

The average translational kinetic energy of ${O_2}$ (molar mass $32$) molecules at a particular temperature is $0.048 \; eV$. The translational kinetic energy of ${N_2}$ (molar mass $28$) molecules in $eV$ at the same temperature is:

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