Consider two ideal gases,A and B,at the same temperature.The rms speed of the molecules of gas A is twice that of gas B.How does the molecular mass of A compare to that of B?

Question

Quizplus · Accepted Answer

The rms speed of gas molecules is inversely proportional to the square root of their molecular mass. If the speed of A is twice that of B, then the mass of A must be one fourth that of B, according to the formula $v_{rms} = \sqrt{\frac{3kT}{m}}$, where $v_{rms}$ is the root mean square speed, $k$ is the Boltzmann constant, $T$ is the temperature, and $m$ is the molecular mass.

Consider Two Ideal Gases,A and B,at the Same Temperature