A gas is at $20 ^\circ \textrm{C}$. To what temperature must it be raised to triple the rms speed of its molecules?

VIDEO TRANSCRIPT

This is Giancoli Answers with Mr. Dychko. The rms speed at the first temperature is going to be square root of 3 times Boltzmann's constant times the first temperature, T1, divided by the mass of a single molecule. And then the rms speed of the second temperature is going to be square root of 3 K T2 over m. And the m doesn't need a subscript because it's the same molecule, it's just a different temperature. So, we put a subscript on the T only and we know the V rms2, there's gonna be 3 times the rms speed of the first temperature. And so we substitute for V rms2 that's 3 K T2 over m square rooted. And that equals 3 times V rms1, 3 K T1 over m. And the square, well... Whole bunch of things cancel on both sides, you can multiply both sides by square root m over 3 K, and then that cancels away everything except for this square root 2 or T2 equals 3 times square root T1. And then square both sides. And you get T2 is 9 times T1. So, 9 times the first temperature written in kelvin, that's 20 degree Celsius plus 273.15. And that gives 2638.135 kelvin. And then we subtract away 273.15. And that gives 236.5 degrees Celsius.