Chapter 14, Problem 88A

Interpretation Introduction

Interpretation: The molar mass of an unknown gas should be calculated if a certain number of moles of this gas passes to a tiny hole in $75sec$ whereas the same moles of the oxygen gas pass to the same hole in $30sec$ .

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 88A

The molar mass of the unknown gas is ${\text{M}}_{\text{X}}=200u$ .

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 = $\text{32 u}$

Rate of effusion of unknown gas = $\text{n moles / 75 sec}$

Rate of effusion of oxygen gas = $\text{n moles / 30 sec}$

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

  $\begin{array}{l}\frac{\text{Rate of effusion of gas X}}{{\text{Rate of effusion of gas O}}_{\text{2}}}\text{=}\sqrt{\frac{{\text{M}}_{\text{O}}{}_{\text{2}}}{{\text{M}}_{\text{X}}}}\\ \frac{\text{n moles /75 sec }}{\text{n moles /30 sec}}\text{=}\sqrt{\frac{\text{32 u}}{{\text{M}}_{\text{X}}}}\\ \frac{\text{30 sec }}{\text{75 sec}}\text{=}\sqrt{\frac{\text{32 u}}{{\text{M}}_{\text{X}}}}\\ 0.4\sec \text{=}\sqrt{\frac{\text{32 u}}{{\text{M}}_{\text{X}}}}\\ \text{0}\text{.16=}\frac{\text{32 u}}{{\text{M}}_{\text{X}}}\\ {\text{M}}_{\text{X}}=\frac{\text{32 u}}{\text{0}\text{.16}}\\ {\text{M}}_{\text{X}}=200u\end{array}$

Conclusion

As the time of effusion increases for the same moles, the rate of effusion decreases.