## College Physics (4th Edition)

The correct answer is: (c) is $\sqrt{2}$ times the original speed
In an ideal gas, the average kinetic energy of the molecules is $\overline{KE} = \frac{3}{2}kT$ Then: $\frac{1}{2}m\overline{v^2} = \frac{3}{2}kT$ $v_{rms} = \sqrt{\frac{3kT}{m}}$ We can write an expression for the original rms speed: $v_{rms} = \sqrt{\frac{3kT}{m}}$ We can find the new rms speed: $v_{rms}' = \sqrt{\frac{3k(2T)}{m}}$ $v_{rms}' = \sqrt{2}\times \sqrt{\frac{3kT}{m}}$ $v_{rms}' = \sqrt{2}\times v_{rms}$ The correct answer is: (c) is $\sqrt{2}$ times the original speed