Skip to content
Login Register

#translational kinetic energy

2 public questions tagged with this topic.

Molecular Mass Density and Ideal Gas Equation

What is the temperature at which the average translational kinetic energy of a gas molecule is 1.035 × 10⁻²⁰ J? (k_B = 1

**Ideal gas equation** P V = n R T = (m/M) R T, density ρ = m/V = P M/(R T), molecular mass M (kg/mol), P pressure (Pa), T temperature (K). At given P,T density proportional to M, heavier gases denser, e.g., at 1.5 atm 300 K V=24 L n= P V/(R T)=1.5×1.013×10⁵×0.024/(8.314×300)≈1.46 mol. Average translational KE = (3)/(2) k_B T.1.035 × 10⁻²⁰ = (3)/(2) × 1.38 × 10⁻²/³ × T.T = 1.035 × 10⁻²⁰ × 23 × 1.38 × 10⁻²/³ = 500 K . Substituting values gives 500 K, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Molecular Mass Density and Ideal Gas Equation

View Details
Gas Laws and Volume-Temperature Relations

What is the average translational kinetic energy of a nitrogen molecule (N₂) at 300 K? (k_B = 1.38 × 10⁻²³ J K⁻¹)

**Ideal gas law** P V = n R T governs gas laws, at constant pressure V ∝ T, so temperature increase 300 K→600 K doubles volume 24→48 L. Charles' law quantitative prediction V₂ = V₁×(T₂/T₁), illustrating direct proportionality, absolute temperature must be in kelvin. Average translational KE per molecule = (3)/(2) k_B T.Substitute: (3)/(2) × 1.38 × 10⁻²/³ × 300 = 6.21 × 10⁻²¹ J . Substituting values gives 6.21 × 10⁻²¹ J, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P V = n R T and v_rms = √(3 R T/M) relations.

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Gas Laws and Volume-Temperature Relations

View Details