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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?
$(v_{rms})_1 < (v_{rms})_2 < (v_{rms})_3$ and $(\bar K)_1 = (\bar K)_2 = (\bar K)_3$
$(v_{rms})_1 = (v_{rms})_2 = (v_{rms})_3$ and $(\bar K)_1 = (\bar K)_2 > (\bar K)_3$
$(v_{rms})_1 > (v_{rms})_2 > (v_{rms})_3$ and $(\bar K)_1 < (\bar K)_2 > (\bar K)_3$
$(v_{rms})_1 > (v_{rms})_2 > (v_{rms})_3$ and $(\bar K)_1 < (\bar K)_2 < (\bar K)_3$
Solution
${v_{rms}} \propto \frac{1}{{\sqrt M }} \Rightarrow {\left( {{v_{rms}}} \right)_1} < {\left( {{v_{rms}}} \right)_2} < {\left( {{v_{rms}}} \right)_3}$
also in mixture temperature of each gas will be same, hence kinetic energy also remains same.
