Chapter 9, Problem 9.89SP

Interpretation Introduction

Interpretation:

The temperature at which both the xenon and bromine molecules have same average speed has to be calculated.

Concept Introduction:

Root mean square value:

${\text{μ}}_{\text{rms}}\text{=}\sqrt{\frac{\text{3RT}}{{\text{M}}_{\text{m}}}}$

Where,

$\begin{array}{l}{\text{μ}}_{\text{rms}}\text{=}\text{root}\text{mean}\text{square}\text{value}\\ \text{R}\text{= ideal}\text{gas}\text{constant = 8}\text{.314}\frac{\text{joule}}{\text{mole}\text{.Kelvin}}\\ \text{K}\text{=}\text{absolute}\text{temperature}\text{in}\text{kelvin}\\ \text{M}\text{=}\text{molar}\text{mass}\text{of}\text{the}\text{molecule}\text{in}\text{kilograms}\end{array}$