Question

Calculate the mean free path of a gas of randomly-moving hard spheres of 2 nanometer radius, when the density of spheres is 3.00 x 10¹⁹ per cubic meter.

• A. 1.17 × 10^-4 m
•  B. 9.76 × 10^-5 m
•  C. 4.70 × 10^-4 m
•  D. 1.88 × 10^-3 m
•  E. None of the above.

Answer 1

The mean free path or the average distance traveled between collisions can be calculated by using

$\lambda =v\cdot t_{mean}= \frac{V}{4\pi \sqrt{2}r²N}= \frac{1}{4\pi \sqrt{2}r² \rho}$

Then, we substitute the density of molecules $(\rho=N/V)$ and the molecule radius r (in meters) to find $\lambda$ :

$\lambda = \frac{1}{4\pi \sqrt{2}r² \rho}= \frac{1}{4\pi \sqrt{2}(2\times 10^{-9}m)² (3\times10^{19}\,molecules/m^{3})}=4.67\times10^{-4}\,m/molecule$

In conclusion, the correct answer is C. 4.70 × 10^-4 m

Reference:

• Young, H. D., Freedman, R. A., & Ford, A. L. (2006). Sears and Zemansky's university physics (Vol. 1). Pearson education.