Problem Statement
Estimate the mean free path of nitrogen molecules at 300 K and 1 atm pressure. (Diameter of N₂ molecule $d = 3.7 \times 10^{-10}$ m, $P = 1.01 \times 10^5$ Pa)
Given Information
- All quantities, constants, and constraints stated in the problem above
- Physical constants used as needed (see Concepts section)
Physical Concepts & Formulas
This problem draws on fundamental physical principles. The key is to identify which conservation law or field equation governs the system, then apply it systematically. Dimensional analysis can always be used to verify that the final answer has the correct units. Working from first principles — rather than memorising formulas — builds deeper understanding and allows tackling novel problems.
- Identify the relevant physical law (Newton’s laws, conservation of energy/momentum, Maxwell’s equations, etc.)
- State the mathematical form of that law as it applies here
- Check dimensions at every step: both sides of an equation must have the same units
Step-by-Step Solution
Problem Statement
Estimate the mean free path of nitrogen molecules at 300 K and 1 atm pressure. (Diameter of N₂ molecule $d = 3.7 \times 10^{-10}$ m, $P = 1.01 \times 10^5$ Pa)
Given Information
- $d = 3.7 \times 10^{-10}$ m
- $P = 1.01 \times 10^5$ Pa
- $T = 300$ K
- $k_B = 1.38 \times 10^{-23}$ J/K
Physical Concepts & Formulas
The mean free path is the average distance a molecule travels between successive collisions:
$$\lambda = \frac{1}{\sqrt{2}\pi d^2 n}$$
where $n = P/(k_B T)$ is the number density.
Step-by-Step Solution
Step 1: Find number density $n$.
$$n = \frac{P}{k_BT} = \frac{1.01 \times 10^5}{1.38 \times 10^{-23} \times 300} = \frac{1.01 \times 10^5}{4.14 \times 10^{-21}} = 2.44 \times 10^{25} \text{ m}^{-3}$$
Step 2: Compute $\lambda$.
$$\lambda = \frac{1}{\sqrt{2}\pi (3.7 \times 10^{-10})^2 \times 2.44 \times 10^{25}}$$
$$= \frac{1}{\sqrt{2}\pi \times 1.369 \times 10^{-19} \times 2.44 \times 10^{25}}$$
$$= \frac{1}{\sqrt{2}\pi \times 3.34 \times 10^{6}}$$
$$= \frac{1}{1.4142 \times 3.14159 \times 3.34 \times 10^6}$$
$$= \frac{1}{1.485 \times 10^7} \approx 6.7 \times 10^{-8} \text{ m}$$
Step 3: Convert.
$$\lambda \approx 67 \text{ nm}$$
Worked Calculation
$$n = \frac{P}{k_BT} = 2.44 \times 10^{25} \text{ m}^{-3}$$
$$\lambda = \frac{1}{\sqrt{2}\pi d^2 n} \approx \frac{1}{\sqrt{2}\pi (3.7\times10^{-10})^2 (2.44\times10^{25})} \approx 6.7 \times 10^{-8} \text{ m}$$
Answer
$\lambda \approx 67$ nm
Physical Interpretation
At atmospheric pressure, N₂ molecules travel only about 67 nm between collisions — about 180 times their own diameter. This short mean free path explains why gases diffuse slowly despite their high molecular speeds (~500 m/s). The frequent collisions create a “random walk” pattern that is far less efficient than straight-line travel.
Worked Calculation
Substituting all given numerical values with their units into the derived formula:
$$\text{Numerical result} = \text{given expression substituted with values}$$
Answer
$$\boxed{\lambda = \frac{1}{\sqrt{2}\pi d^2 n} \approx \frac{1}{\sqrt{2}\pi (3.7\times10^{-10})^2 (2.44\times10^{25})} \approx 6.7 \times 10^{-8} \text{ m}}$$
Physical Interpretation
The answer should be checked for dimensional consistency and physical reasonableness: is the magnitude in the expected range for this type of problem? Does the answer change in the correct direction when parameters are varied (e.g., increasing mass should increase momentum, increasing distance should decrease field strength)? These sanity checks are as important as the calculation itself.
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