RMS Speed and Temperature Dependence

RMS Speed and Temperature Dependence

The rms speed of a gas is 450 m/s at 225 K. At what temperature will the rms speed be 900 m/s?

**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 ∝ √(T), (v₂)/(v₁) = √((T₂)/(T₁)).(900)/(450) = √((T₂)/(225)), 2 = √((T₂)/(225)).Square both sides: 4 = (T₂)/(225), T₂ = 900 K. Substituting values gives 900 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.

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RMS Speed and Temperature Dependence

What is the molar specific heat at constant volume (C_v) for a diatomic gas with no vibrational modes? (R = 8.31 J mol⁻¹

**RMS speed** v_rms = √(3 R T/M) = √(3 k_B T/m) where M molar mass (kg/mol), m molecular mass (kg), k_B=1.38×10⁻²/³ J/K, R=8.314 J/mol·K, T absolute temperature (K). Proportional to √T and 1/√M, lighter gases faster at same T, e.g., H₂ faster than O₂, temperature increase raises v_rms as √T. For a diatomic gas (rigid rotator), 5 degrees of freedom (3 translational + 2 rotational).C_v = (5)/(2) R = (5)/(2) × 8.31 = 20.775 J mol⁻¹ K⁻¹ ≈ 20.8 J mol⁻¹ K⁻¹ . Substituting values gives 20.8 J mol⁻¹ K⁻¹, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π

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RMS Speed and Temperature Dependence

A gas mixture contains 12 g of helium and 28 g of nitrogen. What is the ratio of their partial pressures?

**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. P = (μ RT)/(V), P_HeP_N₂ = μ_Heμ_N₂.μ_He = (12)/(4) = 3 mol, μ_N₂ = (28)/(28) = 1 mol.Ratio = (3)/(1) = 3:1. Substituting values gives 3:1, 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.

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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.

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RMS Speed and Temperature Dependence

A solid has a molar specific heat capacity of 25 J mol⁻¹ K⁻¹ at room temperature. How many degrees of freedom does each

**RMS speed** v_rms = √(3 R T/M) = √(3 k_B T/m) where M molar mass (kg/mol), m molecular mass (kg), k_B=1.38×10⁻²/³ J/K, R=8.314 J/mol·K, T absolute temperature (K). Proportional to √T and 1/√M, lighter gases faster at same T, e.g., H₂ faster than O₂, temperature increase raises v_rms as √T. For a solid, C = f × (R)/(2) × N_A = f × (R)/(2) × 1 = f × (R)/(2).Given C = 25 J mol⁻¹ K⁻¹, R = 8.31 J mol⁻¹ K⁻¹.25 = f × (8.31)/(2), f = (25 × 2)/(8.31) ≈ 6. Substituting values gives 6, which matches expected kinetic theory result, confirming mean

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RMS Speed and Temperature Dependence

The mean free path of a gas molecule is 2 × 10⁻⁷ m at a certain pressure. If the pressure is doubled, what is the new me

**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. Mean free path l = (1)/(√(2) n π d²), where n ∝ P at constant T.If P doubles, n doubles, so l halves.New l = 2 × 10⁻⁷/2 = 1 × 10⁻⁷ m . Substituting values gives 1 × 10⁻⁷ m, 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.

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RMS Speed and Temperature Dependence

A vessel contains helium gas at 27°C with a pressure of 2 atm. If the temperature increases to 127°C at constant volume,

**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. For constant volume: (P₁)/(T₁) = (P₂)/(T₂).T₁ = 27 + 273 = 300 K, T₂ = 127 + 273 = 400 K, P₁ = 2 atm.P₂ = P₁ × (T₂)/(T₁) = 2 × (400)/(300) = 2.67 atm . Substituting values gives 2.67 atm, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal

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RMS Speed and Temperature Dependence

What is the pressure of 0.25 moles of an ideal gas in a 5-litre container at 227°C? (R = 8.31 J mol⁻¹ K⁻¹)

**RMS speed** v_rms = √(3 R T/M) = √(3 k_B T/m) where M molar mass (kg/mol), m molecular mass (kg), k_B=1.38×10⁻²/³ J/K, R=8.314 J/mol·K, T absolute temperature (K). Proportional to √T and 1/√M, lighter gases faster at same T, e.g., H₂ faster than O₂, temperature increase raises v_rms as √T. PV = μ R T, P = (μ R T)/(V).T = 227 + 273 = 500 K, V = 5 × 10⁻³ m³.P = (0.25 × 8.31 × 500)/(5 × 10⁻³) = 2.0775 × 10⁵ Pa ≈ 2.08 atm (1 atm ≈ 10⁵ Pa). Substituting values gives 2.08 atm, which matches expected kinetic theory result,

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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.

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RMS Speed and Temperature Dependence

A container holds 0.5 moles of an ideal gas at 2 atm pressure and 27°C. What is the volume of the gas? (R = 8.31 J mol⁻¹

**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. Ideal gas equation: PV = μ R T.T = 27 + 273 = 300 K, P = 2 × 1.01 × 10⁵ Pa = 2.02 × 10⁵ Pa, μ = 0.5 mol.V = (μ R T)/(P) = (0.5 × 8.31 × 300)/(2.02 × 10⁵) = 6.17 × 10⁻³ m³ = 6.17 litres . Substituting values gives 6.2 litres, which

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RMS Speed and Temperature Dependence

At what temperature is the rms speed of oxygen molecules 600 m/s? (Molecular mass of O₂ = 32 u, k_B = 1.38 × 10⁻²³ J K⁻¹

**RMS speed** v_rms = √(3 R T/M) = √(3 k_B T/m) where M molar mass (kg/mol), m molecular mass (kg), k_B=1.38×10⁻²/³ J/K, R=8.314 J/mol·K, T absolute temperature (K). Proportional to √T and 1/√M, lighter gases faster at same T, e.g., H₂ faster than O₂, temperature increase raises v_rms as √T. v_rms = √((3k_B T)/(m)), m = 32 × 10⁻³⁶.02 × 10²³ = 5.32 × 10⁻²⁶ kg.600² = 3 × 1.38 × 10⁻²/³ × T5.32 × 10⁻²⁶, T = 3.6 × 10⁵ × 5.32 × 10⁻²⁶/⁴.14 × 10⁻²/³ ≈ 463 K. Substituting values gives 463 K, which matches expected kinetic theory result, confirming mean free path

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RMS Speed and Temperature Dependence

A gas occupies 28.0 litres at STP. How many moles are present? (Molar volume at STP = 22.4 litres)

**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. Number of moles (μ) = VolumeMolar volume.μ = (28.0)/(22.4) = 1.25 mol. Substituting values gives 1.25 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.

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