Chapter 12.1, Problem 3PP

Interpretation Introduction

Interpretation:

The rate of effusion for an unknown gas, having a molar mass twice the gas that effuses at a rate of 3.6 mol/min needs to be determined.

Concept introduction:

According to the Graham’s law of effusion, rate of effusion of any gas is always inversely proportional to the square root of the mass of gaseous particles. The mathematical expression for this law is:

$\frac{\text{Rate of effusion of gas 1}}{\text{Rate of effusion of gas 2}}\text{=}\sqrt{\frac{\text{M2}}{\text{M1}}}$

In terms of time for gases; this can be written as:

$\frac{\text{Effusion time of gas 1}}{\text{Effusion time of gas 2}}\text{=}\sqrt{\frac{\text{M1}}{\text{M2}}}$

Answer to Problem 3PP

$\text{Rate of effusion of gas 1}=2.55\text{ mol/}\min$

Explanation of Solution

Given information :

Rate of effusion for gas-2 = 3.6 mol/min

Molar mass of gas-1 = 2 x molar mass of gas-1

$\frac{\text{Rate of effusion of gas 1}}{\text{Rate of effusion of gas 2}}\text{=}\sqrt{\frac{\text{M2}}{\text{M1}}}$

Putting the values in the equation:

$\begin{array}{l}\frac{\text{Rate of effusion of gas 1}}{\text{Rate of effusion of gas 2}}\text{=}\sqrt{\frac{\text{M2}}{\text{M1}}}\\ \frac{\text{Rate of effusion of gas 1}}{\text{3}\text{.6 mol/min}}\text{=}\sqrt{\frac{\text{M2}}{\text{2}\times \text{M2}}}\\ \frac{\text{Rate of effusion of gas 1}}{\text{3}\text{.6 mol/min}}\text{= }0.707\\ \text{Rate of effusion of gas 1}=0.707\times \text{3}\text{.6 mol/}\min \\ \text{Rate of effusion of gas 1}=2.55\text{ mol/}\min \end{array}$

Conclusion

$\text{Rate of effusion of gas 1}=2.55\text{ mol/}\min$