Gas Laws and Volume-Temperature Relations

Gas Laws and Volume-Temperature Relations

What is the average translational kinetic energy of a nitrogen molecule (N₂) at 300 K? (k_B = 1.38 × 10⁻²³ J K⁻¹)

**Ideal gas law** P V = n R T governs gas laws, at constant pressure V ∝ T, so temperature increase 300 K→600 K doubles volume 24→48 L. Charles' law quantitative prediction V₂ = V₁×(T₂/T₁), illustrating direct proportionality, absolute temperature must be in kelvin. Average translational KE per molecule = (3)/(2) k_B T.Substitute: (3)/(2) × 1.38 × 10⁻²/³ × 300 = 6.21 × 10⁻²¹ J . Substituting values gives 6.21 × 10⁻²¹ 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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Gas Laws and Volume-Temperature Relations

The mean free path of a gas is 1.2 × 10⁻⁷ m with a number density of 4.0 × 10²⁵ m⁻³. What is the molecular diameter?

**Charles' law** V₁/T₁ = V₂/T₂ at constant pressure, volume proportional to absolute temperature (K), Gay-Lussac P₁/T₁ = P₂/T₂ at constant volume, Boyle's law P₁V₁ = P₂V₂ at constant temperature, combined ideal gas law P V = n R T, R=8.314 J/mol·K. For V₁=24 L T₁=300 K T₂=600 K, V₂= V₁ T₂/T₁=48 L, volume doubles when T doubles at constant P. l = (1)/(√(2) n π d²), d² = (1)/(√(2) n π l).d² = (1)/(1.414 × 4.0 × 10²⁵) × 3.14 × 1.2 × 10⁻⁷ = (1)/(2.13 × 10⁻¹⁹) ≈ 4.69 × 10⁻²⁰.d = √(4.69 × 10⁻²⁰) ≈ 2.17 × 10⁻¹⁰ m. Substituting values gives 2.17

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Gas Laws and Volume-Temperature Relations

What is the temperature at which the rms speed of nitrogen molecules is 1032 m/s? (Molecular mass of N₂ = 28 u, k_B = 1.

**Gas laws** Boyle, Charles, Gay-Lussac are special cases of ideal gas equation, for constant pressure volume-temperature relation V ∝ T, for constant temperature pressure-volume inverse, for constant volume pressure-temperature direct, enabling calculation of new volume from temperature ratio. v_rms = √((3k_B T)/(m)), m = 28 × 10⁻³⁶.02 × 10²³ = 4.65 × 10⁻²⁶ kg.1032² = 3 × 1.38 × 10⁻²/³ × T4.65 × 10⁻²⁶, T = 1.065 × 10⁶ × 4.65 × 10⁻²⁶/⁴.14 × 10⁻²/³ ≈ 1196 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 =

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Gas Laws and Volume-Temperature Relations

A gas has a C_v of 17.1 J mol⁻¹ K⁻¹. What is its C_p? (R = 8.31 J mol⁻¹ K⁻¹)

**Ideal gas law** P V = n R T governs gas laws, at constant pressure V ∝ T, so temperature increase 300 K→600 K doubles volume 24→48 L. Charles' law quantitative prediction V₂ = V₁×(T₂/T₁), illustrating direct proportionality, absolute temperature must be in kelvin. C_p = C_v + R.C_p = 17.1 + 8.31 = 25.41 J mol⁻¹ K⁻¹ ≈ 25.4 J mol⁻¹ K⁻¹. Substituting values gives 25.4 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.

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Gas Laws and Volume-Temperature Relations

The rms speed of a gas is 550 m/s at 275 K. At what temperature will the rms speed be 1100 m/s?

**Charles' law** V₁/T₁ = V₂/T₂ at constant pressure, volume proportional to absolute temperature (K), Gay-Lussac P₁/T₁ = P₂/T₂ at constant volume, Boyle's law P₁V₁ = P₂V₂ at constant temperature, combined ideal gas law P V = n R T, R=8.314 J/mol·K. For V₁=24 L T₁=300 K T₂=600 K, V₂= V₁ T₂/T₁=48 L, volume doubles when T doubles at constant P. v_rms ∝ √(T), (v₂)/(v₁) = √((T₂)/(T₁)).(1100)/(550) = √((T₂)/(275)), 2 = √((T₂)/(275)).Square both sides: 4 = (T₂)/(275), T₂ = 1100 K. Substituting values gives 1100 K, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P

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Gas Laws and Volume-Temperature Relations

What is the collision frequency of a gas molecule with a mean free path of 1.2 × 10⁻⁷ m and average speed of 480 m/s?

**Gas laws** Boyle, Charles, Gay-Lussac are special cases of ideal gas equation, for constant pressure volume-temperature relation V ∝ T, for constant temperature pressure-volume inverse, for constant volume pressure-temperature direct, enabling calculation of new volume from temperature ratio. Collision frequency = ()/(l).(480)/(1.2 × 10⁻⁷) = 4.0 × 10⁹ s⁻¹. Substituting values gives 4.0 × 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.

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Gas Laws and Volume-Temperature Relations

A gas has a C_v of 27.0 J mol⁻¹ K⁻¹. What is its C_p? (R = 8.31 J mol⁻¹ K⁻¹)

**Ideal gas law** P V = n R T governs gas laws, at constant pressure V ∝ T, so temperature increase 300 K→600 K doubles volume 24→48 L. Charles' law quantitative prediction V₂ = V₁×(T₂/T₁), illustrating direct proportionality, absolute temperature must be in kelvin. C_p = C_v + R.C_p = 27.0 + 8.31 = 35.31 J mol⁻¹ K⁻¹ ≈ 35.3 J mol⁻¹ K⁻¹. Substituting values gives 35.3 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.

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Gas Laws and Volume-Temperature Relations

A gas has a volume of 22.4 litres at STP. How many moles are present if the temperature is raised to 546 K at constant p

**Charles' law** V₁/T₁ = V₂/T₂ at constant pressure, volume proportional to absolute temperature (K), Gay-Lussac P₁/T₁ = P₂/T₂ at constant volume, Boyle's law P₁V₁ = P₂V₂ at constant temperature, combined ideal gas law P V = n R T, R=8.314 J/mol·K. For V₁=24 L T₁=300 K T₂=600 K, V₂= V₁ T₂/T₁=48 L, volume doubles when T doubles at constant P. At STP, 22.4 litres = 1 mole.Charles’ law: (V₁)/(T₁) = (V₂)/(T₂), but moles remain constant at constant P.Initial μ = 1 mol, remains 1 mole as V adjusts with T. Substituting values gives 1.0 mol, which matches expected kinetic theory result, confirming mean free path

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Gas Laws and Volume-Temperature Relations

A gas mixture has equal masses of helium and oxygen. What is the ratio of their number of molecules?

**Gas laws** Boyle, Charles, Gay-Lussac are special cases of ideal gas equation, for constant pressure volume-temperature relation V ∝ T, for constant temperature pressure-volume inverse, for constant volume pressure-temperature direct, enabling calculation of new volume from temperature ratio. Let mass = m. N = (m)/(M) × N_A.For He (M = 4): N_He = (m)/(4) × N_A.For O₂ (M = 32): N_O₂ = (m)/(32) × N_A.Ratio N_HeN_O₂ = (m)/(4)(m)/(32) = (32)/(4) = 8:1. Substituting values gives 8:1, 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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Gas Laws and Volume-Temperature Relations

A gas has a molar specific heat at constant volume of 20.8 J mol⁻¹ K⁻¹. What is its C_p? (R = 8.31 J mol⁻¹ K⁻¹)

**Ideal gas law** P V = n R T governs gas laws, at constant pressure V ∝ T, so temperature increase 300 K→600 K doubles volume 24→48 L. Charles' law quantitative prediction V₂ = V₁×(T₂/T₁), illustrating direct proportionality, absolute temperature must be in kelvin. C_p = C_v + R.C_p = 20.8 + 8.31 = 29.11 J mol⁻¹ K⁻¹ ≈ 29.1 J mol⁻¹ K⁻¹. Substituting values gives 29.1 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.

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Gas Laws and Volume-Temperature Relations

A gas at 3 atm and 600 K has a density of 0.96 kg m⁻³. What is its molecular mass? (R = 8.31 J mol⁻¹ K⁻¹, 1 atm = 1.01 ×

**Charles' law** V₁/T₁ = V₂/T₂ at constant pressure, volume proportional to absolute temperature (K), Gay-Lussac P₁/T₁ = P₂/T₂ at constant volume, Boyle's law P₁V₁ = P₂V₂ at constant temperature, combined ideal gas law P V = n R T, R=8.314 J/mol·K. For V₁=24 L T₁=300 K T₂=600 K, V₂= V₁ T₂/T₁=48 L, volume doubles when T doubles at constant P. P = (ρ R T)/(M), M = (ρ R T)/(P).P = 3 × 1.01 × 10⁵ = 3.03 × 10⁵ Pa.M = (0.96 × 8.31 × 600)/(3.03 × 10⁵) = 0.0158 kg/mol ≈ 15.8 g/mol ≈ 16 g/mol. Substituting values gives 16 g/mol, which matches

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Gas Laws and Volume-Temperature Relations

What is the heat required to raise the temperature of 0.2 moles of a triatomic gas by 10 K at constant volume? (R = 8.31

**Gas laws** Boyle, Charles, Gay-Lussac are special cases of ideal gas equation, for constant pressure volume-temperature relation V ∝ T, for constant temperature pressure-volume inverse, for constant volume pressure-temperature direct, enabling calculation of new volume from temperature ratio. Triatomic gas: 6 degrees of freedom (3 translational + 3 rotational).C_v = 3R, Q = μ C_v Δ T = 0.2 × 3 × 8.31 × 10 = 49.86 J . Substituting values gives 49.86 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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