Chapter 6, Problem 6.108QE

Interpretation Introduction

Interpretation:

The molar mass of a gas when time taken by oxygen gas and this gas to effuse is $\text{5}\text{.2 min}$ and $\text{8}\text{.3 min}$ respectively 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.

Another relation that describes the rate of effusion in terms of the time required by a particular gas to effuse is given as follows:

  $\frac{t_{\text{2}}}{t_{\text{1}}}=\sqrt{\frac{M_{\text{2}}}{M_{\text{1}}}}$

Here,

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

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

$t_{\text{1}}$ denotes the time taken by the first gas to effuse.

$t_{\text{2}}$ denotes the time taken by second gas to effuse.

Therefore the gases that have greater rates of effusion will require lesser time for effusion out of a hole compared to a gas that has smaller rates of effusion.