## College Physics (4th Edition)

In an ideal gas, the average kinetic energy of the molecules is $\overline{KE} = \frac{3}{2}kT$ Then: $\frac{1}{2}m\overline{v^2} = \frac{3}{2}kT$ $v_{rms} = \sqrt{\frac{3kT}{m}}$ The rms speed is the speed of a molecule with the average kinetic energy $\overline{KE}$ The correct answer is: (b) speed of a molecule with the average kinetic energy