Chapter 6, Problem 6.106QE

Interpretation Introduction

Interpretation:

The ratio of the rate of effusion of ${\text{N}}_{\text{2}}$ to that of ${\text{SO}}_{\text{2}}$ gas for a container that has the equimolar amount of the two gases has to be calculated.

Concept Introduction:

Graham’s law states that the rate of effusion of gases varies inversely to the square root of its molar mass. Mathematically it can be stated as follows:

  $\frac{{\text{Rate}}_{\text{1}}}{{\text{Rate}}_{\text{2}}}=\sqrt{\frac{M_{\text{2}}}{M_{\text{1}}}}$

Here,

$M_{\text{2}}$ denotes the molar mass of second gas.

$M_{\text{1}}$ denotes the molar mass of first gas.

${\text{Rate}}_{\text{1}}$ denotes the rate of effusion of first gas.

${\text{Rate}}_{\text{2}}$ denotes the rate of effusion of second gas.