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(B) The average kinetic energy $(E)$ of a gas molecule is directly proportional to its absolute temperature $(T)$,given by the relation $E = \frac{3}{

Vedclass · Accepted Answer

$6:7$

Vedclass · Answer

(B) The average kinetic energy $(E)$ of a gas molecule is directly proportional to its absolute temperature $(T)$,given by the relation $E = \frac{3}{2} k_B T$,where $k_B$ is the Boltzmann constant.Since $E \propto T$,the ratio of the average kinetic energies of the molecules of two gases at temperatures $T_1$ and $T_2$ is given by $\frac{E_1}{E_2} = \frac{T_1}{T_2}$.Given $T_1 = 300 \ K$ and $T_2 = 350 \ K$.Therefore,$\frac{E_1}{E_2} = \frac{300}{350} = \frac{6}{7}$.

Two gases are at absolute temperatures $300 \ K$ and $350 \ K$ respectively. The ratio of the average kinetic energy of their molecules is:

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