#STP

Partial Pressures and Gas Mixtures

A gas occupies 16.8 litres at STP. How many moles are present? (Molar volume at STP = 22.4 litres)

**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. Number of moles (μ) = VolumeMolar volume.μ = (16.8)/(22.4) = 0.75 mol. Substituting values gives 0.75 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 > Partial Pressures and Gas Mixtures

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Collision Frequency and Mean Free Path Variation

A gas occupies 44.8 litres at STP. How many moles are present? (Molar volume at STP = 22.4 litres)

**Collision frequency** Z = n σ v_rel, σ = π d² cross-section, v_rel = √2 v_avg, so Z ∝ n, for air at STP n≈2.5×10²⁵ m⁻³ d≈3×10⁻¹⁰ m λ≈68 nm, collision frequency ~10⁹ s⁻¹, illustrating frequent collisions at atmospheric pressure. Number of moles (μ) = VolumeMolar volume.μ = (44.8)/(22.4) = 2.0 mol. Substituting values gives 2.0 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 > Collision Frequency and Mean Free Path Variation

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Molecular Mass Density and Ideal Gas Equation

A gas occupies 89.6 litres at STP. How many moles are present? (Molar volume at STP = 22.4 litres)

**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. Number of moles (μ) = VolumeMolar volume.μ = (89.6)/(22.4) = 4.0 mol. Substituting values gives 4.0 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 > Molecular Mass Density and Ideal Gas Equation

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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.

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

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

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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

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Gas Laws and Volume-Temperature Relations

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Gas Laws and Volume-Temperature Relations

A gas occupies 5.6 litres at STP. How many moles of the gas are present? (Molar volume at STP = 22.4 litres)

**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. Number of moles (μ) = VolumeMolar volume.μ = (5.6)/(22.4) = 0.25 mol. Substituting values gives 0.25 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.

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Gas Laws and Volume-Temperature Relations

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Gas Laws and Volume-Temperature Relations

A gas occupies 56.0 litres at STP. How many moles are present? (Molar volume at STP = 22.4 litres)

**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. Number of moles (μ) = VolumeMolar volume.μ = (56.0)/(22.4) = 2.5 mol. Substituting values gives 2.5 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 > Gas Laws and Volume-Temperature Relations

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Gas Laws and Volume-Temperature Relations

A gas occupies a volume of 11.2 litres at STP. How many molecules are present in this gas? (N_A = 6.02 × 10²³ mol⁻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. Molar volume at STP = 22.4 litres/mol.Number of moles (μ) = (11.2)/(22.4) = 0.5 mol .Number of molecules = μ × N_A = 0.5 × 6.02 × 10²³ = 3.01 × 10²³ . Substituting values gives 3.01 × 10²³, 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 > Gas Laws and Volume-Temperature Relations

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Mean Free Path and Molecular Diameter

A gas occupies 67.2 litres at STP. How many molecules are present? (N_A = 6.02 × 10²³ mol⁻¹, molar volume at STP = 22.4

**Mean free path** λ = 1/(√2 n π d²) is average distance molecule travels between collisions, n number density (m⁻³), d molecular diameter (m), π≈3.14. Inversely proportional to n and d², larger n or d reduces λ. Rearranged d² = 1/(√2 n π λ), so d = √(1/(√2 n π λ)), enabling diameter estimation from measured λ and n. Number of moles (μ) = VolumeMolar volume = (67.2)/(22.4) = 3.0 mol.Number of molecules = μ × N_A = 3.0 × 6.02 × 10²³ = 1.806 × 10²⁴. Substituting values gives 1.806 × 10²⁴, which matches expected kinetic theory result, confirming mean free path λ =

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Mean Free Path and Molecular Diameter

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Mean Free Path and Molecular Diameter

At STP, 22.4 litres of oxygen gas (O₂) is present. What is the mass of this gas? (Molecular mass of O₂ = 32 u, 1 u = 1 g

**Molecular diameter from mean free path** uses λ =1/(√2 n π d²), solving d = √(1/(√2 n π λ)). At higher pressure n ∝ P, λ ∝1/P, so doubling P halves λ, illustrating pressure dependence of collision distance. At STP (273 K, 1 atm), 1 mole of any ideal gas occupies 22.4 litres (molar volume).Given volume = 22.4 litres, so number of moles (μ) = (22.4)/(22.4) = 1 mol .Mass = μ × molecular mass = 1 × 32 = 32 g . Substituting values gives 32 g, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Mean Free Path and Molecular Diameter

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Electrolytic Cells and Electrolysis and Faraday's Laws

In electrolysis of molten CaCl₂, 0.4 g of Ca (atomic mass 40 g/mol) is deposited. What volume of Cl₂ gas (STP) is produc

Cathode: Ca²⁺ + 2e⁻ → Ca , Moles = (0.4/40) = 0.01 mol , Charge = 0.01 × 2 × 96500 = 1930 C . Anode: 2Cl⁻ → Cl₂ + 2e⁻ , Moles Cl₂ = 0.01 mol , Volume = 0.01 × 22.4 = 0.224 L .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrolytic Cells and Electrolysis and Faraday's Laws

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