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#triatomic gas

8 public questions tagged with this topic.

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

How much heat is required to raise the temperature of 0.3 moles of a triatomic gas by 15 K at constant volume, ignoring

**Internal energy of ideal gas** U = f/2 n R T depends only on temperature, f degrees of freedom, n moles, R=8.314 J/mol·K, for monatomic f=3 U=3/2 n R T, diatomic f=5 at moderate T U=5/2 n R T, independent of pressure or volume, only T matters for ideal gas. Triatomic gas: 6 degrees of freedom, C_v = 3R.Q = μ C_v Δ T = 0.3 × 3 × 8.31 × 15 = 112.185 J ≈ 112.2 J. Substituting values gives 112.2 J, 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 > Internal Energy of Ideal Gases

How much heat is required to raise the temperature of 0.15 moles of a triatomic gas by 25 K at constant volume, with no

**Equipartition theorem** energy ½ k_B T per degree of freedom per molecule, f degrees give U = f/2 k_B T per molecule, f/2 R T per mole, internal energy function of T only for ideal gas, no intermolecular potential. Triatomic gas: 6 degrees of freedom, C_v = 3 R.Q = μ C_v Δ T = 0.15 × 3 × 8.31 × 25 = 93.4875 J ≈ 93.5 J. Substituting values gives 93.5 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 > Internal Energy of Ideal Gases

What is the total internal energy of 0.3 moles of a triatomic gas at 300 K with no vibrational modes? (R = 8.31 J mol⁻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. Triatomic gas: 6 degrees of freedom, U = 3 μ R T.U = 3 × 0.3 × 8.31 × 300 = 2243.7 J ≈ 2.24 kJ. Substituting values gives 2.24 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.

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Degrees of Freedom and Molar Specific Heat

What is the heat required to raise the temperature of 0.2 moles of a triatomic gas by 10 K at constant volume? (R = 8.31

**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. Triatomic gas: 6 degrees of freedom (3 translational + 3 rotational).C_v = 3R, Q = μ C_v Δ T = 0.2 × 3 × 8.31 × 10 = 49.86 J . Substituting values gives 49.86 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

How much heat is required to raise the temperature of 0.4 moles of a triatomic gas by 15 K at constant volume, with no v

**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. Triatomic gas: 6 degrees of freedom, C_v = 3 R.Q = μ C_v Δ T = 0.4 × 3 × 8.31 × 15 = 149.58 J ≈ 149.6 J. Substituting values gives 149.6 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

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.

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

What is the total internal energy of 0.8 moles of a triatomic gas at 500 K with no vibrational modes? (R = 8.31 J mol⁻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, U = 3 μ R T.U = 3 × 0.8 × 8.31 × 500 = 9969.6 J ≈ 9.97 kJ. Substituting values gives 9.97 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.

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