Chapter 13: Kinetic Theory

• KE = ⅔k_BT
• PV = ⅔Nk_BT = nRT
• v_rms = √(3RT/M)
• ∆U = nC_v∆T

Q1. Find the average kinetic energy of a gas molecule at 300K.

Solution

KE = ⅔k_BT = ⅔ × 1.38 × 10⁻⁺ × 300
= 6.21 × 10⁻⁺ J ≈ 3.88 × 10⁻₈ eV

Q2. Find v_rms for oxygen at 27°C. (M = 32 g/mol, R = 8.314)

Solution

v_rms = √(3RT/M) = √(3 × 8.314 × 300 / 0.032)
= √(233812.5) ≈ 483.6 m/s