#degrees of freedom

Partial Pressures and Gas Mixtures

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 =

**Partial pressure concept** for mixture of non-reacting gases, total pressure sum of partials, each gas behaves independently, kinetic theory still holds with effective n = Σ n_i, mean free path depends on total n and cross-sections, mixture properties weighted by mole tions. 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 T and v_rms = √(3 R T/M) relations.

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Partial Pressures and Gas Mixtures

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Collision Frequency and Mean Free Path Variation

A solid has a molar specific heat capacity of 24.9 J mol⁻¹ K⁻¹. How many degrees of freedom per atom does it have? (R =

**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. C = f × (R)/(2), 24.9 = f × (8.31)/(2).f = (24.9 × 2)/(8.31) ≈ 5.99 ≈ 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 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

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Collision Frequency and Mean Free Path Variation

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 variation** λ ∝1/n ∝1/P at constant T, λ ∝ T/P, temperature increase increases λ because n decreases at constant P, but also v increases, overall λ ∝ T/P, for gas at 2 atm λ=4×10⁻⁷ m, at 4 atm λ=2×10⁻⁷ m halves when pressure doubles, as n doubles. 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 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

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RMS Speed and Temperature Dependence

A solid has a molar specific heat capacity of 25 J mol⁻¹ K⁻¹ at room temperature. How many degrees of freedom does each

**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. For a solid, C = f × (R)/(2) × N_A = f × (R)/(2) × 1 = f × (R)/(2).Given C = 25 J mol⁻¹ K⁻¹, R = 8.31 J mol⁻¹ K⁻¹.25 = f × (8.31)/(2), f = (25 × 2)/(8.31) ≈ 6. Substituting values gives 6, which matches expected kinetic theory result, confirming mean

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > RMS Speed and Temperature Dependence

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RMS Speed and Temperature Dependence

A solid has a molar specific heat capacity of 24.5 J mol⁻¹ K⁻¹. How many degrees of freedom per atom does it have? (R =

**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. C = f × (R)/(2), 24.5 = f × (8.31)/(2).f = (24.5 × 2)/(8.31) ≈ 5.9 ≈ 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 T and v_rms = √(3 R T/M) relations.

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > RMS Speed and Temperature Dependence

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Internal Energy of Ideal Gases

How many degrees of freedom does a diatomic molecule have if its vibrational mode is active?

**Ideal gas internal energy** proportional to temperature, U = (f/2) R T per mole, monatomic 3/2 R T, diatomic 5/2 R T, change ΔU = f/2 n R ΔT, for temperature increase internal energy rises, explaining why heating gas at constant volume raises U entirely as heat. Diatomic molecule: 3 translational + 2 rotational + 1 vibrational (2 modes: KE and PE).Total degrees of freedom = 3 + 2 + 2 = 7. Substituting values gives 7, 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 > Internal Energy of Ideal Gases

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Degrees of Freedom and Molar Specific Heat

What is the ratio of C_p to C_v for a gas with 3 translational and 2 rotational degrees of freedom?

**Specific heat relation** C_p - C_v = R for ideal gas per mole, Mayer's relation, due to work done at constant pressure, degrees of freedom include translational, rotational, vibrational, each quadratic term contributes ½ R to C_v. Degrees of freedom = 3 + 2 = 5.C_v = (5)/(2) R, C_p = C_v + R = (5)/(2) R + R = (7)/(2) R.γ = (C_p)/(C_v) = (7)/(2) R(5)/(2) R = (7)/(5) = 1.4. Substituting values gives 1.4, 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 > Degrees of Freedom and Molar Specific Heat

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Gas Laws and Volume-Temperature Relations

A solid has a molar specific heat capacity of 25.5 J mol⁻¹ K⁻¹. How many degrees of freedom per atom does it have? (R =

**Gas laws** Boyle, Charles, Gay-Lussac are special cases of ideal gas equation, for constant pressure volume-temperature relation V ∝ T, for constant temperature pressure-volume inverse, for constant volume pressure-temperature direct, enabling calculation of new volume from temperature ratio. C = f × (R)/(2), 25.5 = f × (8.31)/(2).f = (25.5 × 2)/(8.31) ≈ 6.14 ≈ 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 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

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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

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Mean Free Path and Molecular Diameter

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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

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Mean Free Path and Molecular Diameter

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Mean Free Path and Molecular Diameter

A solid has a molar specific heat capacity of 25.5 J mol⁻¹ K⁻¹. How many degrees of freedom per atom does it have? (R =

**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. C = f × (R)/(2), 25.5 = f × (8.31)/(2).f = (25.5 × 2)/(8.31) ≈ 6.14 ≈ 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 T and v_rms = √(3 R T/M) relations.

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Mean Free Path and Molecular Diameter

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