Mean Free Path and Molecular Diameter

Mean Free Path and Molecular Diameter

A gas occupies 67.2 litres at STP. How many molecules are present? (N_A = 6.02 × 10²³ mol⁻¹, molar volume at STP = 22.4

**Mean free path** λ = 1/(√2 n π d²) is average distance molecule travels between collisions, n number density (m⁻³), d molecular diameter (m), π≈3.14. Inversely proportional to n and d², larger n or d reduces λ. Rearranged d² = 1/(√2 n π λ), so d = √(1/(√2 n π λ)), enabling diameter estimation from measured λ and n. Number of moles (μ) = VolumeMolar volume = (67.2)/(22.4) = 3.0 mol.Number of molecules = μ × N_A = 3.0 × 6.02 × 10²³ = 1.806 × 10²⁴. Substituting values gives 1.806 × 10²⁴, which matches expected kinetic theory result, confirming mean free path λ =

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Mean Free Path and Molecular Diameter

At STP, 22.4 litres of oxygen gas (O₂) is present. What is the mass of this gas? (Molecular mass of O₂ = 32 u, 1 u = 1 g

**Molecular diameter from mean free path** uses λ =1/(√2 n π d²), solving d = √(1/(√2 n π λ)). At higher pressure n ∝ P, λ ∝1/P, so doubling P halves λ, illustrating pressure dependence of collision distance. At STP (273 K, 1 atm), 1 mole of any ideal gas occupies 22.4 litres (molar volume).Given volume = 22.4 litres, so number of moles (μ) = (22.4)/(22.4) = 1 mol .Mass = μ × molecular mass = 1 × 32 = 32 g . Substituting values gives 32 g, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal

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Mean Free Path and Molecular Diameter

What is the collision frequency of a gas molecule with a mean free path of 3.0 × 10⁻⁷ m and average speed of 600 m/s?

**Kinetic theory mean free path** λ = 1/(√2 π d² n) quantifies collision frequency. With n =1.0×10²⁵ m⁻³, λ=9×10⁻⁷ m, d² =1/(1.414×10²⁵×3.14×9×10⁻⁷)=2.5×10⁻²⁰ m², d≈1.58×10⁻¹⁰ m, typical molecular size ~10⁻¹⁰ m, consistent with gas kinetic theory. Collision frequency = ()/(l).(600)/(3.0 × 10⁻⁷) = 2.0 × 10⁹ s⁻¹. Substituting values gives 2.0 × 10⁹ s⁻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.

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Mean Free Path and Molecular Diameter

How much heat is required to raise the temperature of 0.2 moles of a diatomic gas by 10 K at constant volume, with no vi

**Mean free path** λ = 1/(√2 n π d²) is average distance molecule travels between collisions, n number density (m⁻³), d molecular diameter (m), π≈3.14. Inversely proportional to n and d², larger n or d reduces λ. Rearranged d² = 1/(√2 n π λ), so d = √(1/(√2 n π λ)), enabling diameter estimation from measured λ and n. Diatomic gas: 5 degrees of freedom, C_v = (5)/(2) R.Q = μ C_v Δ T = 0.2 × (5)/(2) × 8.31 × 10 = 83.1 J. Substituting values gives 83.1 J, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π

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Mean Free Path and Molecular Diameter

What is the total internal energy of 1 mole of a triatomic gas at 300 K with no vibrational modes? (R = 8.31 J mol⁻¹ K⁻¹

**Molecular diameter from mean free path** uses λ =1/(√2 n π d²), solving d = √(1/(√2 n π λ)). At higher pressure n ∝ P, λ ∝1/P, so doubling P halves λ, illustrating pressure dependence of collision distance. Triatomic gas: 6 degrees of freedom (3 translational + 3 rotational).U = 3 R T = 3 × 8.31 × 300 = 7479 J ≈ 7.48 kJ. Substituting values gives 7.48 kJ, 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.

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Mean Free Path and Molecular Diameter

The rms speed of a gas is 500 m/s at 250 K. At what temperature will the rms speed be 1000 m/s?

**Kinetic theory mean free path** λ = 1/(√2 π d² n) quantifies collision frequency. With n =1.0×10²⁵ m⁻³, λ=9×10⁻⁷ m, d² =1/(1.414×10²⁵×3.14×9×10⁻⁷)=2.5×10⁻²⁰ m², d≈1.58×10⁻¹⁰ m, typical molecular size ~10⁻¹⁰ m, consistent with gas kinetic theory. v_rms ∝ √(T), (v₂)/(v₁) = √((T₂)/(T₁)).(1000)/(500) = √((T₂)/(250)), 2 = √((T₂)/(250)).Square both sides: 4 = (T₂)/(250), T₂ = 1000 K. Substituting values gives 1000 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 and v_rms = √(3 R T/M) relations.

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Mean Free Path and Molecular Diameter

A solid has a molar specific heat capacity of 24.4 J mol⁻¹ K⁻¹. How many degrees of freedom per atom does it have? (R =

**Mean free path** λ = 1/(√2 n π d²) is average distance molecule travels between collisions, n number density (m⁻³), d molecular diameter (m), π≈3.14. Inversely proportional to n and d², larger n or d reduces λ. Rearranged d² = 1/(√2 n π λ), so d = √(1/(√2 n π λ)), enabling diameter estimation from measured λ and n. C = f × (R)/(2), 24.4 = f × (8.31)/(2).f = (24.4 × 2)/(8.31) ≈ 5.87 ≈ 6. Substituting values gives 6, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P V = n R

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Mean Free Path and Molecular Diameter

A gas has a C_p of 35.4 J mol⁻¹ K⁻¹. What is the ratio of specific heats (gamma)? (R = 8.31 J mol⁻¹ K⁻¹)

**Molecular diameter from mean free path** uses λ =1/(√2 n π d²), solving d = √(1/(√2 n π λ)). At higher pressure n ∝ P, λ ∝1/P, so doubling P halves λ, illustrating pressure dependence of collision distance. C_v = C_p - R = 35.4 - 8.31 = 27.09 J mol⁻¹ K⁻¹.γ = (C_p)/(C_v) = (35.4)/(27.09) ≈ 1.31. Substituting values gives 1.31, 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.

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Mean Free Path and Molecular Diameter

The mean free path of a gas molecule is 2.0 × 10⁻⁶ m at 0.1 atm. What will it be at 0.4 atm if temperature remains const

**Kinetic theory mean free path** λ = 1/(√2 π d² n) quantifies collision frequency. With n =1.0×10²⁵ m⁻³, λ=9×10⁻⁷ m, d² =1/(1.414×10²⁵×3.14×9×10⁻⁷)=2.5×10⁻²⁰ m², d≈1.58×10⁻¹⁰ m, typical molecular size ~10⁻¹⁰ m, consistent with gas kinetic theory. l ∝ (1)/(n), n ∝ P. If P increases by 4 times (0.1 to 0.4), n increases 4 times, l reduces to (1)/(4).New l = 2.0 × 10⁻⁶/4 = 5.0 × 10⁻⁷ m. Substituting values gives 5.0 × 10⁻⁷ m, 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.

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Mean Free Path and Molecular Diameter

A solid has a molar specific heat capacity of 26.5 J mol⁻¹ K⁻¹. How many degrees of freedom per atom does it have? (R =

**Mean free path** λ = 1/(√2 n π d²) is average distance molecule travels between collisions, n number density (m⁻³), d molecular diameter (m), π≈3.14. Inversely proportional to n and d², larger n or d reduces λ. Rearranged d² = 1/(√2 n π λ), so d = √(1/(√2 n π λ)), enabling diameter estimation from measured λ and n. C = f × (R)/(2), 26.5 = f × (8.31)/(2).f = (26.5 × 2)/(8.31) ≈ 6.38 ≈ 6. Substituting values gives 6, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P V = n R

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Mean Free Path and Molecular Diameter

A gas has a C_v of 12.7 J mol⁻¹ K⁻¹. What is the ratio of specific heats (gamma)? (R = 8.31 J mol⁻¹ K⁻¹)

**Molecular diameter from mean free path** uses λ =1/(√2 n π d²), solving d = √(1/(√2 n π λ)). At higher pressure n ∝ P, λ ∝1/P, so doubling P halves λ, illustrating pressure dependence of collision distance. C_p = C_v + R = 12.7 + 8.31 = 21.01 J mol⁻¹ K⁻¹.γ = (C_p)/(C_v) = (21.01)/(12.7) ≈ 1.65. Substituting values gives 1.65, 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.

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Mean Free Path and Molecular Diameter

The rms speed of argon molecules is 430 m/s at 300 K. What is the rms speed of neon molecules at the same temperature? (

**Kinetic theory mean free path** λ = 1/(√2 π d² n) quantifies collision frequency. With n =1.0×10²⁵ m⁻³, λ=9×10⁻⁷ m, d² =1/(1.414×10²⁵×3.14×9×10⁻⁷)=2.5×10⁻²⁰ m², d≈1.58×10⁻¹⁰ m, typical molecular size ~10⁻¹⁰ m, consistent with gas kinetic theory. v_rms ∝ (1)/(√(m)), v_Nev_Ar = √(m_Ar)m_Ne.v_Ne430 = √((39.9)/(20.2)) ≈ √(1.975) ≈ 1.405.v_Ne = 430 × 1.405 ≈ 604 m/s. Substituting values gives 604 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.

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