Chapter 12, Problem 42A

Interpretation Introduction

Interpretation:

The molar mass of gas with respect to effusion of $He$ needs to be calculated.

Concept introduction:

Rate of diffusion is inversely proportional to molar mass of gas.

Answer to Problem 42A

$36gm/mole$ is molar mass of gas.

Explanation of Solution

Rate of diffusion is explained by given Graham’s law,

$\frac{R_1}{R_2}=\sqrt{\frac{M_2}{M_2}}$ where $M_1,M_2$ are masses of gas.

For $He$ & gas, $\frac{R_{He}}{R_{gas}}=\sqrt{\frac{M_{gas}}{M_{He}}}$

According to values, $R_{He}=3R_G$

$\frac{3R_G}{R_G}=\sqrt{\frac{M_G}{4}}\Rightarrow \frac{3}{4}=\sqrt{\frac{M_G}{4}}\Rightarrow \frac{9}{1}=\frac{M_G}{4}$

$M_G=36gm/mole$

Diffusion and effusion are nature of gas which depends upon temperature and pressure of the gas.

Conclusion

Rate of diffusion is given by Graham’s law.