#specific heat ratio

Collision Frequency and Mean Free Path Variation

The ratio of specific heats (gamma) for a triatomic gas with no vibrational modes is:

**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. Triatomic gas: 3 translational + 3 rotational = 6 degrees of freedom.C_v = 3R, C_p = C_v + R = 4R.γ = (C_p)/(C_v) = (4R)/(3R) = (4)/(3) = 1.33. Substituting values gives 1.33, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P V =

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 gas has a C_v of 24.93 J mol⁻¹ K⁻¹. What is the ratio of specific heats (gamma)? (R = 8.31 J mol⁻¹ K⁻¹)

**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_p = C_v + R = 24.93 + 8.31 = 33.24 J mol⁻¹ K⁻¹.γ = (C_p)/(C_v) = (33.24)/(24.93) ≈ 1.33. Substituting values gives 1.33, 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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Molecular Mass Density and Ideal Gas Equation

A gas has a C_p of 20.8 J mol⁻¹ K⁻¹. What is the ratio of specific heats (gamma)? (R = 8.31 J mol⁻¹ K⁻¹)

**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. C_v = C_p - R = 20.8 - 8.31 = 12.49 J mol⁻¹ K⁻¹ ≈ 12.5.γ = (C_p)/(C_v) = (20.8)/(12.5) ≈ 1.66 ≈ 1.67. Substituting values gives 1.67, 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

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Degrees of Freedom and Molar Specific Heat

A gas has a C_v of 20.8 J mol⁻¹ K⁻¹. What is the ratio of specific heats (gamma)? (R = 8.31 J mol⁻¹ K⁻¹)

**Molar specific heat** from equipartition C_v = f/2 R, C_p = f/2 R + R, γ = C_p/C_v =1+2/f, for f=3 γ=1.67, f=5 γ=1.4, f=6 γ=1.33, explaining specific heat variation with molecular structure, degrees of freedom determine heat capacity. C_p = C_v + R = 20.8 + 8.31 = 29.11 J mol⁻¹ K⁻¹.γ = (C_p)/(C_v) = (29.11)/(20.8) ≈ 1.40. Substituting values gives 1.40, 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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