What is the pressure exerted by 0.1 mole of an ideal gas in a 2-litre container at 127°C? (R = 8.31 J mol⁻¹ K⁻¹)
**Gas mixtures** ideal gas law applies to each component, P_total = Σ n_i R T/V, partial pressure P_i = n_i R T/V, mole tion X_i = n_i/n_total, P_i = X_i P_total, enabling calculation of individual pressures from composition, important for kinetic theory and chemistry. PV = μ R T, P = (μ R T)/(V).T = 127 + 273 = 400 K, V = 2 × 10⁻³ m³.P = (0.1 × 8.31 × 400)/(2 × 10⁻³) = 1.663 × 10⁵ Pa ≈ 1.66 atm (1 atm ≈ 10⁵ Pa). Substituting values gives 1.66 atm, which matches expected kinetic theory result, confirming mean free path λ =
Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Partial Pressures and Gas Mixtures