Ideal Gas Statistics

The Maxwell-Boltzmann speed distribution describes the range of molecular speeds in an ideal gas. It predicts the most probable, mean, and rms speeds, and explains transport properties like viscosity and thermal conductivity.

Key Concepts

Maxwell-Boltzmann distribution: f(v) ∝ v² e^{-mv²/2k_BT}
Most probable speed: v_p = √(2k_BT/m)
Mean speed: ⟨v⟩ = √(8k_BT/πm)
RMS speed: v_rms = √(3k_BT/m)
Pressure derivation: P = (1/3)(N/V)m⟨v²⟩

Key Equations

Maxwell-Boltzmann distribution

f(v) = 4\pi n\left(\frac{m}{2\pi k_BT}\right)^{3/2}v^2 e^{-mv^2/2k_BT}

Most probable speed

v_p = \sqrt{\frac{2k_BT}{m}}

Mean speed

\langle v \rangle = \sqrt{\frac{8k_BT}{\pi m}}

RMS speed

v_{\rm rms} = \sqrt{\frac{3k_BT}{m}}

Worked Example

Example Problem

Problem

Find v_rms for N₂ molecules (m = 4.65×10⁻²⁶ kg) at T = 300 K.

Solution

v_rms = √(3k_BT/m) = √(3×1.38×10⁻²³×300/4.65×10⁻²⁶) = √(2.67×10⁵) = 517 m/s.

Practice

Exercises

7 problems

1 of 7

Find v_rms for O₂ (m=5.32×10⁻²⁶ kg) at T=300 K in m/s.

Your answer

m/s

2 of 7

Find the most probable speed v_p for N₂ (m=4.65×10⁻²⁶ kg) at T=300 K in m/s.

Your answer

m/s

3 of 7

Find the mean speed ⟨v⟩ for He (m=6.65×10⁻²⁷ kg) at T=300 K in m/s.

Unlock Exercise 3