$A$ gas mixture consists of molecules of $A, B$ and $C$ with masses $m_A > m_B > m_C$. Rank the three types of molecules in decreasing order of $(a)$ average $K.E.$,$(b)$ $rms$ speeds.

(N/A) In a gas mixture at thermal equilibrium,all molecules are at the same temperature. The average kinetic energy of a gas molecule is given by $K.E._{avg} = \frac{3}{2} k_B T$. Since $k_B$ is the Boltzmann constant and $T$ is the same for all molecules in the mixture,the average kinetic energy is the same for all: $(K.E.)_A = (K.E.)_B = (K.E.)_C$.
$(b)$ The $rms$ speed of a gas molecule is given by $v_{rms} = \sqrt{\frac{3 k_B T}{m}}$,where $m$ is the mass of the molecule. Since $v_{rms} \propto \frac{1}{\sqrt{m}}$,the molecule with the smallest mass will have the highest $rms$ speed. Given $m_A > m_B > m_C$,the decreasing order of $rms$ speeds is $(v_{rms})_C > (v_{rms})_B > (v_{rms})_A$.