What is the root mean square speed of a gas molecule at absolute temperature $T$ with molar mass $M$?

Solution

1. The kinetic energy of an ideal gas is given by $KE=\frac{3}{2}RT$ per mole.
2. Equating this to the translational kinetic energy formula, we have $\frac{1}{2}M{v}_{rms}^2=\frac{3}{2}RT$.
3. Solving for the root mean square speed, we isolate $v_{rms}$ to get $v_{rms}=\sqrt{\frac{3RT}{M}}$.
4. Hence the answer is (A).