Two gases A and B are at absolute temperature | vedclass

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(C) The average kinetic energy $K$ of a gas molecule is given by the formula $K = \frac{3}{2} k_B T$,where $k_B$ is the Boltzmann constant and $T$ is

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$7: 6$

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(C) The average kinetic energy $K$ of a gas molecule is given by the formula $K = \frac{3}{2} k_B T$,where $k_B$ is the Boltzmann constant and $T$ is the absolute temperature.Since $K \propto T$,the ratio of the average kinetic energy of gas $B$ to gas $A$ is given by $\frac{K_B}{K_A} = \frac{T_B}{T_A}$.Given $T_A = 360 \ K$ and $T_B = 420 \ K$.Substituting these values,we get $\frac{K_B}{K_A} = \frac{420}{360} = \frac{7}{6}$.Thus,the ratio is $7: 6$.

Two gases $A$ and $B$ are at absolute temperatures $360 \ K$ and $420 \ K$ respectively. The ratio of average kinetic energy of the molecules of gas $B$ to that of gas $A$ is

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