Internal Energy of Ideal Gases

Internal Energy of Ideal Gases

A gas has a C_v of 21.0 J mol⁻¹ K⁻¹. What is its C_p? (R = 8.31 J mol⁻¹ K⁻¹)

**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. C_p = C_v + R.C_p = 21.0 + 8.31 = 29.31 J mol⁻¹ K⁻¹ ≈ 29.3 J mol⁻¹ K⁻¹. Substituting values gives 29.3 J mol⁻¹ K⁻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.

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Internal Energy of Ideal Gases

View Details

Internal Energy of Ideal Gases

How much heat is required to raise the temperature of 0.5 moles of neon by 25 K at constant volume? (R = 8.31 J mol⁻¹ K⁻

**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. For monatomic gas, C_v = (3)/(2) R.Q = μ C_v Δ T = 0.5 × (3)/(2) × 8.31 × 25 = 155.8125 J ≈ 155.8 J. Substituting values gives 155.8 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

View Details

Internal Energy of Ideal Gases

What is the time between collisions for a gas molecule with a mean free path of 4.5 × 10⁻⁷ m and average speed of 450 m/

**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. tau = (l)/() = 4.5 × 10⁻⁷/4⁵⁰ = 1.0 × 10⁻⁹ s. Substituting values gives 1.0 × 10⁻⁹ 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.

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Internal Energy of Ideal Gases

View Details

Internal Energy of Ideal Gases

What is the average translational kinetic energy of a neon atom at 800 K? (k_B = 1.38 × 10⁻²³ J K⁻¹)

**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. Average translational KE = (3)/(2) k_B T.(3)/(2) × 1.38 × 10⁻²/³ × 800 = 1.656 × 10⁻²⁰ J. Substituting values gives 1.656 × 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 > Internal Energy of Ideal Gases

View Details

Internal Energy of Ideal Gases

A vessel contains 1 mole of a diatomic gas at 300 K. What is its total internal energy if vibrational modes are not exci

**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. For diatomic gas (rigid rotator): U = (5)/(2) μ R T.U = (5)/(2) × 1 × 8.31 × 300 = 6232.5 J ≈ 6.23 kJ . Substituting values gives 6.23 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 > Internal Energy of Ideal Gases

View Details

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

View Details

Internal Energy of Ideal Gases

What is the collision frequency of a gas molecule if its mean free path is 2 × 10⁻⁷ m and average speed is 500 m/s?

**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. Collision frequency = ()/(l).(500)/(2 × 10⁻⁷) = 2.5 × 10⁹ s⁻¹. Substituting values gives 2.5 × 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.

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Internal Energy of Ideal Gases

View Details

Internal Energy of Ideal Gases

A gas at 1 atm and 273 K has a density of 1.43 kg m⁻³. What is its molecular mass? (R = 8.31 J mol⁻¹ K⁻¹, 1 atm = 1.01 ×

**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. P = (ρ R T)/(M), M = (ρ R T)/(P).M = (1.43 × 8.31 × 273)/(1.01 × 10⁵) = 0.0321 kg/mol ≈ 32.1 g/mol. Substituting values gives 32 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 and v_rms = √(3 R T/M) relations.

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Internal Energy of Ideal Gases

View Details

Internal Energy of Ideal Gases

A gas occupies 33.6 litres at STP. How many molecules are present? (N_A = 6.02 × 10²³ mol⁻¹, molar volume at STP = 22.4

**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. Number of moles (μ) = VolumeMolar volume = (33.6)/(22.4) = 1.5 mol.Number of molecules = μ × N_A = 1.5 × 6.02 × 10²³ = 9.03 × 10²³. Substituting values gives 9.03 × 10²³, 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 > Internal Energy of Ideal Gases

View Details

Internal Energy of Ideal Gases

A gas mixture has equal masses of neon and nitrogen at 300 K. What is the ratio of their rms speeds? (Atomic mass: Ne =

**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. v_rms ∝ (1)/(√(m)), v_Nev_N₂ = √(m_N)₂m_Ne.v_Nev_N₂ = √((28)/(20.2)) ≈ √(1.386) ≈ 1.18. Substituting values gives 1.18:1, 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.

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Internal Energy of Ideal Gases

View Details

Internal Energy of Ideal Gases

A monatomic gas has a molar specific heat at constant pressure of 20.8 J mol⁻¹ K⁻¹. What is the value of C_v?

**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. For an ideal gas: C_p - C_v = R, where R = 8.31 J mol⁻¹ K⁻¹.C_v = C_p - R = 20.8 - 8.31 = 12.49 J mol⁻¹ K⁻¹ ≈ 12.5 J mol⁻¹ K⁻¹ . Substituting values gives 12.5 J mol⁻¹ K⁻¹, 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 > Internal Energy of Ideal Gases

View Details

Internal Energy of Ideal Gases

A gas at 2.5 atm and 500 K has a volume of 12 litres. If the pressure increases to 5 atm at constant temperature, what i

**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. Boyle’s law: P₁ V₁ = P₂ V₂.P₁ = 2.5 atm, V₁ = 12 litres, P₂ = 5 atm.V₂ = (P₁ V₁)/(P₂) = (2.5 × 12)/(5) = 6 litres. Substituting values gives 6 litres, 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 > Internal Energy of Ideal Gases

View Details