Deriving pV = nRT from molecular collisions

problems.proof.stepOf

problems.proof.prevStep problems.proof.nextStep

\tfrac12 m\overline{v^2} = \tfrac32 k_BT \ \Rightarrow\ pV = \tfrac13 N \cdot 3k_BT = Nk_BT = nRT

problems.proof.analysis

The definition of absolute temperature (from kinetic theory) is that the average translational kinetic energy per molecule equals $\tfrac32 k_BT$ . Substituting into $pV=\tfrac13Nm\overline{v^2}=\tfrac23N(\tfrac12m\overline{v^2})$ from the previous step recovers exactly $pV=Nk_BT=nRT$ (with $n=N/N_A$ , $R=N_Ak_B$ ).

problems.proof.pitfall. This is not a mathematically rigorous 'proof' — it relies on the ideal-gas model (molecules as points, elastic collisions, no interaction between molecules except during collisions). For real gases at high pressure, correction terms are needed (e.g. the Van der Waals equation).