Problem Statement
Find the fraction of N₂ molecules at $T=300\ \text{K}$ with speed exceeding Earth’s escape velocity $v_{esc}=11.2\ \text{km/s}$.
Given Information
- See problem statement for all given quantities.
Physical Concepts & Formulas
The Maxwell-Boltzmann distribution describes the statistical distribution of molecular speeds in an ideal gas at thermal equilibrium. It arises from the maximisation of entropy subject to fixed total energy and particle number. Three characteristic speeds are defined: most probable $v_p$, mean $\langle v \rangle$, and root-mean-square $v_{\text{rms}}$.
- $v_p = \sqrt{2RT/M}$ — most probable speed
- $\langle v \rangle = \sqrt{8RT/(\pi M)}$ — mean speed
- $v_{\text{rms}} = \sqrt{3RT/M}$ — root-mean-square speed
- $f(v) = 4\pi n\left(\frac{M}{2\pi RT}\right)^{3/2} v^2 e^{-Mv^2/2RT}$ — distribution function
Step-by-Step Solution
Step 1 — Identify given quantities and set up the problem: For $v_{esc} \gg v_p$, the Maxwell tail approximation:
Step 2 — Apply the relevant physical law or equation: $$F(v>v_{esc}) \approx \frac{4}{\sqrt{\pi}}\left(\frac{m v_{esc}^2}{2k_BT}\right)e^{-mv_{esc}^2/(2k_BT)}$$
Step 3 — Solve algebraically for the unknown: Compute $x = \frac{m v_{esc}^2}{2k_BT} = \frac{M v_{esc}^2}{2RT} = \frac{0.028\times(1.12\times10^4)^2}{2\times8.314\times300} = \frac{0.028\times1.254\times10^8}{4988} = \frac{3.51\times10^6}{4988} \approx 704$.
Step 4 — Substitute numerical values with units: $$F \approx \frac{4}{\sqrt{\pi}}\times704\times e^{-704} \approx 1590\times e^{-704} \approx 10^{-304}$$
Step 5 — Compute and check the result: Essentially zero. This explains why Earth’s atmosphere retains nitrogen — almost no N₂ molecules have sufficient speed to escape.
Worked Calculation
$$F(v>v_{esc}) \approx \frac{4}{\sqrt{\pi}}\left(\frac{m v_{esc}^2}{2k_BT}\right)e^{-mv_{esc}^2/(2k_BT)}$$
$$F \approx \frac{4}{\sqrt{\pi}}\times704\times e^{-704} \approx 1590\times e^{-704} \approx 10^{-304}$$
$$\text{Numerical result} = \text{given expression substituted with values}$$
For $v_{esc} \gg v_p$, the Maxwell tail approximation:
$$F(v>v_{esc}) \approx \frac{4}{\sqrt{\pi}}\left(\frac{m v_{esc}^2}{2k_BT}\right)e^{-mv_{esc}^2/(2k_BT)}$$
Compute $x = \frac{m v_{esc}^2}{2k_BT} = \frac{M v_{esc}^2}{2RT} = \frac{0.028\times(1.12\times10^4)^2}{2\times8.314\times300} = \frac{0.028\times1.254\times10^8}{4988} = \frac{3.51\times10^6}{4988} \approx 704$.
$$F \approx \frac{4}{\sqrt{\pi}}\times704\times e^{-704} \approx 1590\times e^{-704} \approx 10^{-304}$$
Essentially zero. This explains why Earth’s atmosphere retains nitrogen — almost no N₂ molecules have sufficient speed to escape.
Answer
$$\boxed{F \approx \frac{4}{\sqrt{\pi}}\times704\times e^{-704} \approx 1590\times e^{-704} \approx 10^{-304}}$$
Physical Interpretation
Maxwell’s thermodynamic relations connect seemingly unrelated thermodynamic derivatives, allowing quantities that are hard to measure directly (like entropy changes at constant pressure) to be computed from measurable ones.
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