Answer
The correct answer is:
(c) is $\sqrt{2}$ times the original speed
Work Step by Step
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
