Molecular Mass Density and Ideal Gas Equation

Molecular Mass Density and Ideal Gas Equation

At what temperature is the rms speed of nitrogen molecules 600 m/s? (Molecular mass of N₂ = 28 u, k_B = 1.38 × 10⁻²³ J K

**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. v_rms = √((3k_B T)/(m)), m = 28 × 10⁻³⁶.02 × 10²³ = 4.65 × 10⁻²⁶ kg.600² = 3 × 1.38 × 10⁻²/³ × T4.65 × 10⁻²⁶, T = 3.6 × 10⁵ × 4.65 × 10⁻²⁶/⁴.14 × 10⁻²/³ ≈ 404 K. Substituting values gives 404 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

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Molecular Mass Density and Ideal Gas Equation

A gas has a C_p of 20.8 J mol⁻¹ K⁻¹. What is the ratio of specific heats (gamma)? (R = 8.31 J mol⁻¹ K⁻¹)

**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. C_v = C_p - R = 20.8 - 8.31 = 12.49 J mol⁻¹ K⁻¹ ≈ 12.5.γ = (C_p)/(C_v) = (20.8)/(12.5) ≈ 1.66 ≈ 1.67. Substituting values gives 1.67, 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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Molecular Mass Density and Ideal Gas Equation

A gas has a density of 1.2 kg m⁻³ at 2 atm and 400 K. What is its molecular mass? (R = 8.31 J mol⁻¹ K⁻¹, 1 atm = 1.01 ×

**Ideal gas equation** P V = n R T = (m/M) R T, density ρ = m/V = P M/(R T), molecular mass M (kg/mol), P pressure (Pa), T temperature (K). At given P,T density proportional to M, heavier gases denser, e.g., at 1.5 atm 300 K V=24 L n= P V/(R T)=1.5×1.013×10⁵×0.024/(8.314×300)≈1.46 mol. P = (ρ R T)/(M), M = (ρ R T)/(P).P = 2 × 1.01 × 10⁵ = 2.02 × 10⁵ Pa.M = (1.2 × 8.31 × 400)/(2.02 × 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

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Molecular Mass Density and Ideal Gas Equation

What is the average translational kinetic energy of an oxygen molecule at 600 K? (k_B = 1.38 × 10⁻²³ J K⁻¹)

**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. Average translational KE = (3)/(2) k_B T.(3)/(2) × 1.38 × 10⁻²/³ × 600 = 1.242 × 10⁻²⁰ J. Substituting values gives 1.242 × 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.

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Molecular Mass Density and Ideal Gas Equation

A gas at 1 atm and 273 K has a volume of 15 litres. If the temperature increases to 819 K at constant pressure, what is

**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. Charles’ law: (V₁)/(T₁) = (V₂)/(T₂).V₁ = 15 litres, T₁ = 273 K, T₂ = 819 K.V₂ = V₁ × (T₂)/(T₁) = 15 × (819)/(273) = 45 litres. Substituting values gives 45 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 and v_rms = √(3 R T/M) relations.

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Molecular Mass Density and Ideal Gas Equation

The mean free path of a gas is 1 × 10⁻⁷ m with a molecular diameter of 2 × 10⁻¹⁰ m. What is the number density of the ga

**Ideal gas equation** P V = n R T = (m/M) R T, density ρ = m/V = P M/(R T), molecular mass M (kg/mol), P pressure (Pa), T temperature (K). At given P,T density proportional to M, heavier gases denser, e.g., at 1.5 atm 300 K V=24 L n= P V/(R T)=1.5×1.013×10⁵×0.024/(8.314×300)≈1.46 mol. l = (1)/(√(2) n π d²), n = (1)/(√(2) π d² l).n = (1)/(1.414 × 3.14 × (2 × 10⁻¹⁰))² × 1 × 10⁻⁷ = (1)/(1.77 × 10⁻²⁶) ≈ 5.65 × 10²⁵ m⁻³. Substituting values gives 5.65 × 10²⁵ m⁻³, which matches expected kinetic theory result, confirming mean free path λ

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Molecular Mass Density and Ideal Gas Equation

A mixture of 0.5 moles of helium and 1.5 moles of oxygen is at 350 K in a 25-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 = 0.5 + 1.5 = 2, V = 25 × 10⁻³ m³.P = (2 × 8.31 × 350)/(25 × 10⁻³) = 2.326 × 10⁵ Pa ≈ 2.33 atm. Substituting values gives 2.33 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

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Molecular Mass Density and Ideal Gas Equation

What is the pressure of 0.8 moles of an ideal gas in a 16-litre container at 427°C? (R = 8.31 J mol⁻¹ K⁻¹)

**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. PV = μ R T, P = (μ R T)/(V).T = 427 + 273 = 700 K, V = 16 × 10⁻³ m³.P = (0.8 × 8.31 × 700)/(16 × 10⁻³) = 2.90625 × 10⁵ Pa ≈ 2.91 atm (1 atm ≈ 10⁵ Pa). Substituting values gives 2.91 atm, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2

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Molecular Mass Density and Ideal Gas Equation

A gas at 400 K has a pressure of 1 atm. If the temperature is reduced to 200 K at constant volume, what is the new press

**Ideal gas equation** P V = n R T = (m/M) R T, density ρ = m/V = P M/(R T), molecular mass M (kg/mol), P pressure (Pa), T temperature (K). At given P,T density proportional to M, heavier gases denser, e.g., at 1.5 atm 300 K V=24 L n= P V/(R T)=1.5×1.013×10⁵×0.024/(8.314×300)≈1.46 mol. Constant volume: (P₁)/(T₁) = (P₂)/(T₂).P₁ = 1 atm, T₁ = 400 K, T₂ = 200 K.P₂ = P₁ × (T₂)/(T₁) = 1 × (200)/(400) = 0.5 atm . Substituting values gives 0.5 atm, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas

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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

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Molecular Mass Density and Ideal Gas Equation

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

**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_Hev_N₂ = √(m_N)₂m_He.v_Hev_N₂ = √((28)/(4)) = √(7) ≈ 2.645. Substituting values gives 2.65:1, 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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Molecular Mass Density and Ideal Gas Equation

A gas at 1.5 atm and 300 K has a volume of 18 litres. If the pressure decreases to 0.75 atm at constant temperature, wha

**Ideal gas equation** P V = n R T = (m/M) R T, density ρ = m/V = P M/(R T), molecular mass M (kg/mol), P pressure (Pa), T temperature (K). At given P,T density proportional to M, heavier gases denser, e.g., at 1.5 atm 300 K V=24 L n= P V/(R T)=1.5×1.013×10⁵×0.024/(8.314×300)≈1.46 mol. Boyle’s law: P₁ V₁ = P₂ V₂.P₁ = 1.5 atm, V₁ = 18 litres, P₂ = 0.75 atm.V₂ = (P₁ V₁)/(P₂) = (1.5 × 18)/(0.75) = 36 litres. Substituting values gives 36 litres, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas

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