The ratio of the mean kinetic energy of hydro | vedclass

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Question

(B) The mean kinetic energy $(E)$ 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 t

Vedclass · Accepted Answer

$2:3$

Vedclass · Answer

(B) The mean kinetic energy $(E)$ 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.Since $k_B$ is a constant,the mean kinetic energy is directly proportional to the absolute temperature $(E \propto T)$.Given temperatures are $T_1 = 300 \, K$ and $T_2 = 450 \, K$.The ratio of the mean kinetic energies is $\frac{E_1}{E_2} = \frac{T_1}{T_2}$.Substituting the values,we get $\frac{E_1}{E_2} = \frac{300}{450} = \frac{2}{3}$.Therefore,the ratio is $2:3$.

The ratio of the mean kinetic energy of hydrogen and nitrogen at temperatures $300 \, K$ and $450 \, K$ respectively is:

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