#argon

Partial Pressures and Gas Mixtures

A gas mixture contains 10 g of neon and 40 g of argon. What is the ratio of their partial pressures? (Atomic mass: Ne =

**Gas mixtures** ideal gas law applies to each component, P_total = Σ n_i R T/V, partial pressure P_i = n_i R T/V, mole tion X_i = n_i/n_total, P_i = X_i P_total, enabling calculation of individual pressures from composition, important for kinetic theory and chemistry. P = (μ RT)/(V), P_NeP_Ar = μ_Neμ_Ar.μ_Ne = (10)/(20.2) ≈ 0.495 mol, μ_Ar = (40)/(39.9) ≈ 1.0025 mol.Ratio = (0.495)/(1.0025) ≈ 0.494 ≈ 1:2. Substituting values gives 1:2, 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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Partial Pressures and Gas Mixtures

How much heat is required to raise the temperature of 1 mole of argon gas by 20 K at constant volume? (R = 8.31 J mol⁻¹

**Dalton's law of partial pressures** total pressure P_total = Σ P_i, P_i = X_i P_total, X_i mole tion, each gas exerts pressure as if alone, ideal gas mixture P_i V = n_i R T, partial pressure proportional to mole tion, e.g., air 79% N₂ 21% O₂ P_N₂=0.79 atm P_O₂=0.21 atm at 1 atm total. For monatomic gas, C_v = (3)/(2) R.Heat Q = μ C_v Δ T = 1 × (3)/(2) × 8.31 × 20 = 249.3 J ≈ 249 J . Substituting values gives 249 J, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas

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

What is the average translational kinetic energy of an argon atom at 900 K? (k_B = 1.38 × 10⁻²³ J 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. Average translational KE = (3)/(2) k_B T.(3)/(2) × 1.38 × 10⁻²/³ × 900 = 1.863 × 10⁻²⁰ J. Substituting values gives 1.863 × 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

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 mixture of 1 mole of argon and 0.5 moles of nitrogen is at 450 K in a 15-litre container. What is the total pressure?

**Ideal gas equation** combines Boyle, Charles, Avogadro laws, P V = N k_B T, N number of molecules, k_B Boltzmann constant, for 1 mole N_A=6.022×10²³, R = N_A k_B, enabling calculation of volume from P,T,n. PV = μ R T, P = (μ R T)/(V).Total moles = 1 + 0.5 = 1.5, V = 15 × 10⁻³ m³.P = (1.5 × 8.31 × 450)/(15 × 10⁻³) = 3.74 × 10⁵ Pa ≈ 3.74 atm. Substituting values gives 3.74 atm, 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

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Molecular Mass Density and Ideal Gas Equation

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Molecular Mass Density and Ideal Gas Equation

A gas mixture has equal numbers of oxygen and argon molecules at 300 K. What is the ratio of their rms speeds? (Molecula

**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. v_rms ∝ (1)/(√(m)), v_O₂v_Ar = √(m_Ar)m_O₂.v_O₂v_Ar = √((39.9)/(32)) ≈ √(1.247) ≈ 1.117. Substituting values gives 1.12:1, 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 > Molecular Mass Density and Ideal Gas Equation

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RMS Speed and Temperature Dependence

The rms speed of argon molecules is 430 m/s at 300 K. What is the rms speed of nitrogen molecules at the same temperatur

**Temperature dependence of RMS speed** v_rms ∝ √T, doubling T increases v_rms by √2≈1.414, e.g., at 300 K v_rms for O₂ ≈483 m/s, at 600 K ≈683 m/s, illustrating kinetic theory relation between temperature and molecular motion, average kinetic energy ½ m v_rms² =3/2 k_B T. v_rms ∝ (1)/(√(m)), v_N₂v_Ar = √(m_Ar)m_N₂.v_N₂430 = √((39.9)/(28)) ≈ √(1.425) ≈ 1.193.v_N₂ = 430 × 1.193 ≈ 513 m/s. Substituting values gives 513 m/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 > RMS Speed and Temperature Dependence

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RMS Speed and Temperature Dependence

The rms speed of nitrogen molecules is 516 m/s at 300 K. What is the rms speed of argon molecules at the same temperatur

**Maxwell-Boltzmann distribution** gives distribution of speeds, v_rms = √(3kT/m), most probable v_mp = √(2kT/m), average v_avg = √(8kT/πm), all ∝ √T, ratio v_rms:v_avg:v_mp =1.732:1.596:1.414, temperature raises all speeds proportionally. v_rms ∝ (1)/(√(m)), v_Arv_N₂ = √(m_N)₂m_Ar.v_Ar516 = √((28)/(39.9)) ≈ √(0.7017) ≈ 0.8375.v_Ar = 516 × 0.8375 ≈ 432 m/s. Substituting values gives 432 m/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 > RMS Speed and Temperature Dependence

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RMS Speed and Temperature Dependence

A mixture of 1 mole of neon and 2 moles of argon is at 500 K in a 30-litre container. What is the total pressure? (R = 8

**Temperature dependence of RMS speed** v_rms ∝ √T, doubling T increases v_rms by √2≈1.414, e.g., at 300 K v_rms for O₂ ≈483 m/s, at 600 K ≈683 m/s, illustrating kinetic theory relation between temperature and molecular motion, average kinetic energy ½ m v_rms² =3/2 k_B T. PV = μ R T, P = (μ R T)/(V).Total moles = 1 + 2 = 3, V = 30 × 10⁻³ m³.P = (3 × 8.31 × 500)/(30 × 10⁻³) = 4.155 × 10⁵ Pa ≈ 4.16 atm. Substituting values gives 4.16 atm, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π

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

The rms speed of helium molecules is 1370 m/s at 300 K. What is the rms speed of argon molecules at the same temperature

**Maxwell-Boltzmann distribution** gives distribution of speeds, v_rms = √(3kT/m), most probable v_mp = √(2kT/m), average v_avg = √(8kT/πm), all ∝ √T, ratio v_rms:v_avg:v_mp =1.732:1.596:1.414, temperature raises all speeds proportionally. v_rms ∝ (1)/(√(m)), v_Arv_He = √(m_He)m_Ar.v_Ar1370 = √((4)/(39.9)) ≈ √(0.1) ≈ 0.316.v_Ar = 1370 × 0.316 ≈ 433 m/s. Substituting values gives 433 m/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 > RMS Speed and Temperature Dependence

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Internal Energy of Ideal Gases

At what temperature is the rms speed of argon molecules 860 m/s? (Atomic mass of Ar = 39.9 u, k_B = 1.38 × 10⁻²³ J K⁻¹)

**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. v_rms = √((3k_B T)/(m)), m = 39.9 × 10⁻³⁶.02 × 10²³ = 6.63 × 10⁻²⁶ kg.860² = 3 × 1.38 × 10⁻²/³ × T6.63 × 10⁻²⁶, T = 7.396 × 10⁵ × 6.63 × 10⁻²⁶/⁴.14 × 10⁻²/³ ≈ 1185 K. Substituting values gives 1200 K, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2

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

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Internal Energy of Ideal Gases

A mixture of 0.3 moles of nitrogen and 0.7 moles of argon is at 350 K in a 20-litre container. What is the total pressur

**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. PV = μ R T, P = (μ R T)/(V).Total moles = 0.3 + 0.7 = 1.0, V = 20 × 10⁻³ m³.P = (1.0 × 8.31 × 350)/(20 × 10⁻³) = 1.45575 × 10⁵ Pa ≈ 1.46 atm. Substituting values gives 1.46 atm, which matches expected kinetic theory result, confirming mean free path λ

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

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Internal Energy of Ideal Gases

A gas mixture has equal volumes of helium and argon at the same temperature and pressure. What is the ratio of their rms

**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. v_rms ∝ (1)/(√(m)), v_Hev_Ar = √(m_Ar)m_He = √((39.9)/(4)) ≈ √(10) ≈ 3.16. Substituting values gives 3.16:1, 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

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