Chapter 12.1, Problem 2PP

Interpretation Introduction

Interpretation:

The ratio of diffusion rates for carbon monoxide and carbon dioxide 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; we can write as:

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

Answer to Problem 2PP

$\frac{\text{Rate of effusion of CO}}{{\text{Rate of effusion of gas CO}}_{\text{2}}}\text{= }1.25$

Explanation of Solution

Carbon monoxide and carbon dioxide gases have different rate of diffusion.

The ratio of rate of effusion of two gases can be calculated as follows:

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

Molar mass of CO = 28 g/mol

Molar mass of CO2 = 44 g/mol

Plug the values in the equation:

$\begin{array}{l}\frac{\text{Rate of effusion of CO}}{{\text{Rate of effusion of gas CO}}_{\text{2}}}\text{=}\sqrt{\frac{{\text{M}}_{{\text{CO}}_{\text{2}}}}{{\text{M}}_{\text{CO}}}}\\ \frac{\text{Rate of effusion of CO}}{{\text{Rate of effusion of gas CO}}_{\text{2}}}\text{=}\sqrt{\frac{\text{44 g/mol }}{\text{28 g/mol }}}=1.25\end{array}$

Conclusion

$\frac{\text{Rate of effusion of CO}}{{\text{Rate of effusion of gas CO}}_{\text{2}}}\text{= }1.25$