What is the pressure of 0.25 moles of an ideal gas in a 5-litre container at 227°C? (R = 8.31 J mol⁻¹ K⁻¹)
**RMS speed** v_rms = √(3 R T/M) = √(3 k_B T/m) where M molar mass (kg/mol), m molecular mass (kg), k_B=1.38×10⁻²/³ J/K, R=8.314 J/mol·K, T absolute temperature (K). Proportional to √T and 1/√M, lighter gases faster at same T, e.g., H₂ faster than O₂, temperature increase raises v_rms as √T. PV = μ R T, P = (μ R T)/(V).T = 227 + 273 = 500 K, V = 5 × 10⁻³ m³.P = (0.25 × 8.31 × 500)/(5 × 10⁻³) = 2.0775 × 10⁵ Pa ≈ 2.08 atm (1 atm ≈ 10⁵ Pa). Substituting values gives 2.08 atm, which matches expected kinetic theory result,
Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > RMS Speed and Temperature Dependence