The average translational kinetic energy of n | vedclass

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Question

(D) The average translational 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

Vedclass · Accepted Answer

$0.084$

Vedclass · Answer

(D) The average translational 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.From this relation,it is clear that the translational kinetic energy depends only on the temperature $T$ and is independent of the molar mass of the gas.Given for nitrogen $(N_2)$: $E_1 = 0.042 \ eV$ at temperature $T_1 = T$.For oxygen $(O_2)$: The temperature is doubled,so $T_2 = 2T$.Since $E \propto T$,we have:$\frac{E_2}{E_1} = \frac{T_2}{T_1} = \frac{2T}{T} = 2$Therefore,$E_2 = 2 	imes E_1 = 2 	imes 0.042 \ eV = 0.084 \ eV$.

The average translational kinetic energy of nitrogen (molar mass $28$) molecules at a particular temperature is $0.042 \ eV$. The translational kinetic energy of oxygen molecules (molar mass $32$) in $eV$ at double the temperature is:

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