By what factor will the rms speed of gas molecules increase if the temperature is increased from $20 ^\circ \textrm{C}$ to $160 ^\circ \textrm{C}$?

VIDEO TRANSCRIPT

This is Giancoli Answers with Mr. Dychko. We're going to divide the two rms speeds in order to see by what factor they change when the temperature changes. So, the rms speed of the first temperature is going to be square root of 3 times Boltzmann's constant times the first temperature, T1, over the mass of a single molecule. And then the rms speed in the second case is going to be the same formula with T2, it's the same molecule, so it's the same mass there. So, our V rms2 divided by V rms1 is V rms2 multiplied by the reciprocal of V rms1 instead of dividing by a fraction of multiplying by its reciprocal, so, that's way you see square root m over 3 K T1 here. And the m's cancel the square root of 3 K's cancel and we're left with square root T2 over T1. These temperatures have to be written in kelvin, as always, for these formulas. So. That's square root 160 degrees Celsius plus 273.15 divided by 20 degree Celsius plus 273.15. And that gives about 1.2 is the factor by which the rms speed changes when the temperature changes from 22 degrees Celsius to 160 degree Celsius.