Chapter 14, Problem 72A

Interpretation Introduction

Interpretation: The gas that effuses faster at the same temperature between oxygen and atomic argon is to be determined.

Concept introduction: According to Graham's law, the gas diffusing rate, or effusing rate, at a given temperature is inversely related to the square root of the gas's molecular mass.

Answer to Problem 72A

The oxygen effuses $1.1173$ times faster than atomic argon.

Explanation of Solution

The molecular mass of oxygen gas is $32{\text{ g mol}}^{-1}$ .

The molecular mass of atomic argon is $\text{39}{\text{.95 g mol}}^{-1}$ .

Since oxygen has a smaller molecular mass than atomic argon, it effuses faster than atomic argon.

The formula to calculate the rate of effusion is as follows:

  $\frac{Rat{e}_{{\text{O}}_{\text{2}}}}{Rat{e}_{\text{Ar}}}=\sqrt{\frac{M_{\text{Ar}}}{M_{{\text{O}}_{\text{2}}}}}$

Where,

• $M_{\text{Ar}}$ is the molecular mass of atomic argon.
• $M_{{\text{O}}_{\text{2}}}$ is the molecular mass of oxygen.
• $Rat{e}_{{\text{O}}_{\text{2}}}$ is the rate of effusion of oxygen.
• $Rat{e}_{\text{Ar}}$ is the rate of effusion of atomic argon.

Substitute the values of $M_{{\text{O}}_{\text{2}}}$ and $M_{\text{Ar}}$ in the above equation as follows:

  $\begin{array}{c}\frac{Rat{e}_{{\text{O}}_{\text{2}}}}{Rat{e}_{\text{Ar}}}=\sqrt{\frac{M_{\text{Ar}}}{M_{{\text{O}}_{\text{2}}}}}\\ =\sqrt{\frac{\text{39}\text{.95 }\bcancel{{\text{g mol}}^{-1}}}{\text{32 }\bcancel{{\text{g mol}}^{-1}}}}\\ =\sqrt{1.2484}\\ =1.1173\end{array}$

Therefore, oxygen effuses $1.1173$ times faster than atomic argon.

Conclusion

The oxygen effuses $1.1173$ times faster than atomic argon.