The rms speed of a gas is 550 m/s at 275 K. At what temperature will the rms speed be 1100 m/s?
**Charles' law** V₁/T₁ = V₂/T₂ at constant pressure, volume proportional to absolute temperature (K), Gay-Lussac P₁/T₁ = P₂/T₂ at constant volume, Boyle's law P₁V₁ = P₂V₂ at constant temperature, combined ideal gas law P V = n R T, R=8.314 J/mol·K. For V₁=24 L T₁=300 K T₂=600 K, V₂= V₁ T₂/T₁=48 L, volume doubles when T doubles at constant P. v_rms ∝ √(T), (v₂)/(v₁) = √((T₂)/(T₁)).(1100)/(550) = √((T₂)/(275)), 2 = √((T₂)/(275)).Square both sides: 4 = (T₂)/(275), T₂ = 1100 K. Substituting values gives 1100 K, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P
Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Gas Laws and Volume-Temperature Relations