Degrees of Freedom and Molar Specific Heat

Degrees of Freedom and Molar Specific Heat

A gas at 1.5 atm and 300 K has a density of 1.2 kg m⁻³. What is its molecular mass? (R = 8.31 J mol⁻¹ K⁻¹, 1 atm = 1.01

**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. P = (ρ R T)/(M), M = (ρ R T)/(P).P = 1.5 × 1.01 × 10⁵ = 1.515 × 10⁵ Pa.M = (1.2 × 8.31 × 300)/(1.515 × 10⁵) = 0.01975 kg/mol ≈ 19.75 g/mol ≈ 20 g/mol. Substituting values gives 20 g/mol, 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 > Degrees of Freedom and Molar Specific Heat

View Details

Degrees of Freedom and Molar Specific Heat

What is the average translational kinetic energy of a hydrogen molecule at 700 K? (k_B = 1.38 × 10⁻²³ J 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. Average translational KE = (3)/(2) k_B T.(3)/(2) × 1.38 × 10⁻²/³ × 700 = 1.449 × 10⁻²⁰ J. Substituting values gives 1.449 × 10⁻²⁰ 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 > Degrees of Freedom and Molar Specific Heat

View Details

Degrees of Freedom and Molar Specific Heat

The mean free path of air molecules at STP is 2.9 × 10⁻⁷ m. If temperature doubles at constant pressure, what is the new

**Degrees of freedom** f counts independent motions, monatomic 3 translational, diatomic 3 translational +2 rotational =5 at room T, vibrational adds at high T, molar specific heat at constant volume C_v = f/2 R, at constant pressure C_p = C_v + R, ratio γ = C_p/C_v = (f+2)/f, monatomic γ=5/3≈1.67, diatomic γ=7/5=1.4. l = (1)/(√(2) n π d²), n = (N)/(V), and PV = Nk_B T.At constant P, V ∝ T, so n ∝ (1)/(T). If T doubles, n halves, l doubles.New l = 2 × 2.9 × 10⁻⁷ = 5.8 × 10⁻⁷ m . Substituting values gives 5.8 × 10⁻⁷ m, which matches expected

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

View Details

Degrees of Freedom and Molar Specific Heat

At what temperature is the rms speed of oxygen molecules 964 m/s? (Molecular mass of O₂ = 32 u, k_B = 1.38 × 10⁻²³ J K⁻¹

**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. v_rms = √((3k_B T)/(m)), m = 32 × 10⁻³⁶.02 × 10²³ = 5.32 × 10⁻²⁶ kg.964² = 3 × 1.38 × 10⁻²/³ × T5.32 × 10⁻²⁶, T = 9.29 × 10⁵ × 5.32 × 10⁻²⁶/⁴.14 × 10⁻²/³ ≈ 1194 K. Substituting values gives 1200 K, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P V = n

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

View Details

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

View Details

Degrees of Freedom and Molar Specific Heat

A gas at 4 atm and 500 K has a volume of 20 litres. If the pressure increases to 8 atm at constant temperature, what is

**Degrees of freedom** f counts independent motions, monatomic 3 translational, diatomic 3 translational +2 rotational =5 at room T, vibrational adds at high T, molar specific heat at constant volume C_v = f/2 R, at constant pressure C_p = C_v + R, ratio γ = C_p/C_v = (f+2)/f, monatomic γ=5/3≈1.67, diatomic γ=7/5=1.4. Boyle’s law: P₁ V₁ = P₂ V₂.P₁ = 4 atm, V₁ = 20 litres, P₂ = 8 atm.V₂ = (P₁ V₁)/(P₂) = (4 × 20)/(8) = 10 litres. Substituting values gives 10 litres, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P

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

View Details

Degrees of Freedom and Molar Specific Heat

The rms speed of a gas is 300 m/s at 150 K. At what temperature will the rms speed be 600 m/s?

**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. v_rms ∝ √(T), (v₂)/(v₁) = √((T₂)/(T₁)).(600)/(300) = √((T₂)/(150)), 2 = √((T₂)/(150)).Square both sides: 4 = (T₂)/(150), T₂ = 600 K. Substituting values gives 600 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.

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

View Details

Degrees of Freedom and Molar Specific Heat

How much heat is required to raise the temperature of 0.1 moles of a monatomic gas by 30 K at constant volume? (R = 8.31

**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. Monatomic gas: C_v = (3)/(2) R.Q = μ C_v Δ T = 0.1 × (3)/(2) × 8.31 × 30 = 37.395 J ≈ 37.4 J. Substituting values gives 37.4 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 > Degrees of Freedom and Molar Specific Heat

View Details

Degrees of Freedom and Molar Specific Heat

A gas at 2 atm and 400 K has a volume of 6 litres. If the pressure decreases to 1 atm at constant temperature, what is t

**Degrees of freedom** f counts independent motions, monatomic 3 translational, diatomic 3 translational +2 rotational =5 at room T, vibrational adds at high T, molar specific heat at constant volume C_v = f/2 R, at constant pressure C_p = C_v + R, ratio γ = C_p/C_v = (f+2)/f, monatomic γ=5/3≈1.67, diatomic γ=7/5=1.4. Boyle’s law: P₁ V₁ = P₂ V₂.P₁ = 2 atm, V₁ = 6 litres, P₂ = 1 atm.V₂ = (P₁ V₁)/(P₂) = (2 × 6)/(1) = 12 litres. Substituting values gives 12 litres, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P

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

View Details

Degrees of Freedom and Molar Specific Heat

A mixture of 2 moles of helium and 3 moles of nitrogen is at 400 K in a 50-litre container. What is the total pressure?

**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. PV = μ R T, P = (μ R T)/(V).Total moles = 2 + 3 = 5, V = 50 × 10⁻³ m³.P = (5 × 8.31 × 400)/(50 × 10⁻³) = 3.324 × 10⁵ Pa ≈ 3.32 atm. Substituting values gives 3.32 atm, 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 > Degrees of Freedom and Molar Specific Heat

View Details

Degrees of Freedom and Molar Specific Heat

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

View Details

Degrees of Freedom and Molar Specific Heat

What is the average translational kinetic energy of an argon atom at 500 K? (k_B = 1.38 × 10⁻²³ J K⁻¹)

**Degrees of freedom** f counts independent motions, monatomic 3 translational, diatomic 3 translational +2 rotational =5 at room T, vibrational adds at high T, molar specific heat at constant volume C_v = f/2 R, at constant pressure C_p = C_v + R, ratio γ = C_p/C_v = (f+2)/f, monatomic γ=5/3≈1.67, diatomic γ=7/5=1.4. Average translational KE = (3)/(2) k_B T.(3)/(2) × 1.38 × 10⁻²/³ × 500 = 1.035 × 10⁻²⁰ J. Substituting values gives 1.035 × 10⁻²⁰ 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 > Degrees of Freedom and Molar Specific Heat

View Details