#nitrogen

Collision Frequency and Mean Free Path Variation

A mixture of 0.4 moles of helium and 0.6 moles of nitrogen is at 400 K in a 25-litre container. What is the total pressu

**Collision frequency** Z = √2 n π d² v_avg, n number density, d molecular diameter, v_avg average speed, proportional to n and v_avg, mean free path λ = v_avg/Z =1/(√2 n π d²), inversely proportional to n, so λ ∝1/P at constant T because n ∝ P, collision frequency increases with pressure, λ decreases. PV = μ R T, P = (μ R T)/(V).Total moles = 0.4 + 0.6 = 1.0, V = 25 × 10⁻³ m³.P = (1.0 × 8.31 × 400)/(25 × 10⁻³) = 1.3284 × 10⁵ Pa ≈ 1.33 atm. Substituting values gives 1.33 atm, which matches expected kinetic theory result, confirming

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Collision Frequency and Mean Free Path Variation

View Details

Collision Frequency and Mean Free Path Variation

A gas mixture has equal numbers of helium and nitrogen molecules at 400 K. What is the ratio of their rms speeds? (Atomi

**Collision frequency** Z = √2 n π d² v_avg, n number density, d molecular diameter, v_avg average speed, proportional to n and v_avg, mean free path λ = v_avg/Z =1/(√2 n π d²), inversely proportional to n, so λ ∝1/P at constant T because n ∝ P, collision frequency increases with pressure, λ decreases. v_rms ∝ (1)/(√(m)), v_Hev_N₂ = √(m_N)₂m_He.v_Hev_N₂ = √((28)/(4)) = √(7) ≈ 2.65. Substituting values gives 2.65:1, 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 > Collision Frequency and Mean Free Path Variation

View Details

Molecular Mass Density and Ideal Gas Equation

At what temperature is the rms speed of nitrogen molecules 600 m/s? (Molecular mass of N₂ = 28 u, k_B = 1.38 × 10⁻²³ J K

**Ideal gas equation** combines Boyle, Charles, Avogadro laws, P V = N k_B T, N number of molecules, k_B Boltzmann constant, for 1 mole N_A=6.022×10²³, R = N_A k_B, enabling calculation of volume from P,T,n. v_rms = √((3k_B T)/(m)), m = 28 × 10⁻³⁶.02 × 10²³ = 4.65 × 10⁻²⁶ kg.600² = 3 × 1.38 × 10⁻²/³ × T4.65 × 10⁻²⁶, T = 3.6 × 10⁵ × 4.65 × 10⁻²⁶/⁴.14 × 10⁻²/³ ≈ 404 K. Substituting values gives 404 K, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P V = n R T

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

View Details

Molecular Mass Density and Ideal Gas Equation

A mixture of 1 mole of argon and 0.5 moles of nitrogen is at 450 K in a 15-litre container. What is the total pressure?

**Ideal gas equation** combines Boyle, Charles, Avogadro laws, P V = N k_B T, N number of molecules, k_B Boltzmann constant, for 1 mole N_A=6.022×10²³, R = N_A k_B, enabling calculation of volume from P,T,n. PV = μ R T, P = (μ R T)/(V).Total moles = 1 + 0.5 = 1.5, V = 15 × 10⁻³ m³.P = (1.5 × 8.31 × 450)/(15 × 10⁻³) = 3.74 × 10⁵ Pa ≈ 3.74 atm. Substituting values gives 3.74 atm, 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

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

View Details

Molecular Mass Density and Ideal Gas Equation

A gas mixture has equal masses of helium and nitrogen at 300 K. What is the ratio of their rms speeds? (Atomic mass: He

**Molecular mass and density** relation ρ = P M/(R T) allows density calculation, ideal gas law also P = n k_B T where n number density, molecular mass determines mass per molecule, density increases with pressure and decreases with temperature, inverse T dependence. v_rms ∝ (1)/(√(m)), v_Hev_N₂ = √(m_N)₂m_He.v_Hev_N₂ = √((28)/(4)) = √(7) ≈ 2.645. Substituting values gives 2.65:1, 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 > Molecular Mass Density and Ideal Gas Equation

View Details

Molecular Mass Density and Ideal Gas Equation

The rms speed of nitrogen molecules is 516 m/s at 300 K. What will it be at 600 K? (Molecular mass of N₂ = 28 u)

**Ideal gas equation** combines Boyle, Charles, Avogadro laws, P V = N k_B T, N number of molecules, k_B Boltzmann constant, for 1 mole N_A=6.022×10²³, R = N_A k_B, enabling calculation of volume from P,T,n. v_rms ∝ √(T), (v₂)/(v₁) = √((T₂)/(T₁)).(v₂)/(516) = √((600)/(300)) = √(2) ≈ 1.414.v₂ = 516 × 1.414 ≈ 729 m/s . Substituting values gives 729 m/s, 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 > Molecular Mass Density and Ideal Gas Equation

View Details

RMS Speed and Temperature Dependence

A gas mixture contains 12 g of helium and 28 g of nitrogen. What is the ratio of their partial pressures?

**Maxwell-Boltzmann distribution** gives distribution of speeds, v_rms = √(3kT/m), most probable v_mp = √(2kT/m), average v_avg = √(8kT/πm), all ∝ √T, ratio v_rms:v_avg:v_mp =1.732:1.596:1.414, temperature raises all speeds proportionally. P = (μ RT)/(V), P_HeP_N₂ = μ_Heμ_N₂.μ_He = (12)/(4) = 3 mol, μ_N₂ = (28)/(28) = 1 mol.Ratio = (3)/(1) = 3:1. Substituting values gives 3:1, 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.