Chapter 5, Problem 5.95QP

Interpretation Introduction

Interpretation:

The ratio of rates of effusion of ${\text{N}}_{\text{2}}$ and ${\text{O}}_{\text{2}}$ under the same conditions has to be calculated

Concept Introduction:

Root-mean-square (rms):

The root-mean-square molecular speed (u) is given by the below formula.

$\begin{array}{l}u\text{=}\sqrt{\frac{3\text{RT}}{{\text{M}}_{\text{m}}}}\\ ={\left(\frac{3\text{RT}}{{\text{M}}_{\text{m}}}\right)}^{\frac{1}{2}}\end{array}$

Where,

R is the molar gas constant

T is the absolute temperature

${\text{M}}_{\text{m}}$ is the molar mass of the gas

Graham’s law of effusion:

$\text{Rate of effusion of molecules }\propto \frac{\text{1}}{\sqrt{{\text{M}}_{\text{m}}}}$

Where,

${\text{M}}_{\text{m}}$ is the molar mass of the gas