Edit to “Solution”

Valter edited
revision #20151 parent #20150 ← older
@@ -46,7 +46,10 @@Solution
$$D \approx \frac{1}{3} \cdot 6.0 \cdot 10^{-8} \cdot 587 \approx 1.17 \cdot 10^{-5} \text{ m}^2/\text{s}$$
Converting to square millimeters, we obtain $D \approx 12 \text{ mm}^2/\text{s}$.
+<b>Theory Reference:</b>
+For a detailed derivation of the mean free path utilized in this solution, you can refer to <i>Physics</i> by Halliday, Resnick, and Krane (HRK), Volume 1, Chapter 22, Section 22-3 ("The mean free path"), page 502. It thoroughly describes the effective cross-section concept ($\sigma = \pi d^2$) and the concentration relation from the ideal gas law ($1/n = kT/P$) that form the basis of these molecular-kinetic equations.
+
#### Answer
a. $W = W_1 - W_2 \approx \alpha \lambda \sqrt{kT/m}$
b. $D \approx 12 \text{ mm}^2/\text{s}$