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(A) The $r.m.s.$ speed of a gas molecule is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$.Since the temperature $T$ is the same for all gases

Vedclass · Accepted Answer

$(v_{rms})_1 < (v_{rms})_2 < (v_{rms})_3$ and $(\bar K)_1 = (\bar K)_2 = (\bar K)_3$

Vedclass · Answer

(A) The $r.m.s.$ speed of a gas molecule is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$.Since the temperature $T$ is the same for all gases in the mixture,$v_{rms} \propto \frac{1}{\sqrt{M}}$.Given $m_1 > m_2 > m_3$,it follows that $(v_{rms})_1 The average kinetic energy of a gas molecule is given by $\bar K = \frac{3}{2} k_B T$.Since the temperature $T$ is the same for all gases in the mixture,the average kinetic energy $\bar K$ is the same for all types of molecules,i.e.,$(\bar K)_1 = (\bar K)_2 = (\bar K)_3$.

$A$ gas mixture consists of molecules of type $1, 2$ and $3$,with molar masses $m_1 > m_2 > m_3$. $v_{rms}$ and $\bar K$ are the $r.m.s.$ speed and average kinetic energy of the gases. Which of the following is true?

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