Two gases A and B are at absolute temperature | vedclass

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(A) The average kinetic energy $(E_{avg})$ of a gas molecule is given by the formula $E_{avg} = \frac{3}{2} k_B T$,where $k_B$ is the Boltzmann consta

Vedclass · Accepted Answer

$6: 5$

Vedclass · Answer

(A) The average kinetic energy $(E_{avg})$ of a gas molecule is given by the formula $E_{avg} = \frac{3}{2} k_B T$,where $k_B$ is the Boltzmann constant and $T$ is the absolute temperature.Since $k_B$ is a constant,the average kinetic energy is directly proportional to the absolute temperature $(E_{avg} \propto T)$.Given temperatures are $T_A = 350 \ K$ and $T_B = 420 \ K$.The ratio of the average kinetic energy of gas $B$ to that of gas $A$ is $\frac{E_B}{E_A} = \frac{T_B}{T_A}$.Substituting the values,we get $\frac{E_B}{E_A} = \frac{420}{350} = \frac{42}{35} = \frac{6}{5}$.Thus,the ratio is $6: 5$.

Two gases $A$ and $B$ are at absolute temperatures $350 \ K$ 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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