Chapter 14, Problem 87A

Interpretation Introduction

Interpretation: The molar mass of the gas should be calculated if it effuses four times as fast as oxygen gas.

Concept Introduction: According to Graham’s law of effusion, the 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:

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

Answer to Problem 87A

The molar mass of the unknown gas is ${\text{M}}_{\text{X}}=1u$ .

Explanation of Solution

Gases can diffuse and effuse due to their high kinetic energy. The rate of effusion of gases is directly proportional to their temperature of it and also depends on the molar mass of the gases.

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

Molar mass of oxygen gas = $\text{16 u}$

Thus the molar mass of the unknown gas X must be:

  $\begin{array}{l}\frac{\text{Rate of effusion of gas X}}{{\text{Rate of effusion of gas O}}_{\text{2}}}\text{=}\sqrt{\frac{{\text{M}}_{\text{O}}{}_{\text{2}}}{{\text{M}}_{\text{X}}}}\\ \frac{\text{4}}{\text{1}}\text{=}\sqrt{\frac{\text{16 u}}{{\text{M}}_{\text{X}}}}\\ {\text{M}}_{\text{X}}=1\end{array}$

Conclusion

As the molar mass of the gas increases, the rate of effusion decreases.