#chemistry

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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Collision Frequency and Mean Free Path Variation

What is the volume of 0.3 moles of an ideal gas at 2.5 atm and 127°C? (R = 8.31 J mol⁻¹ K⁻¹)

**Mean free path variation** λ ∝1/n ∝1/P at constant T, λ ∝ T/P, temperature increase increases λ because n decreases at constant P, but also v increases, overall λ ∝ T/P, for gas at 2 atm λ=4×10⁻⁷ m, at 4 atm λ=2×10⁻⁷ m halves when pressure doubles, as n doubles. PV = μ R T, V = (μ R T)/(P).T = 127 + 273 = 400 K, P = 2.5 × 1.01 × 10⁵ = 2.525 × 10⁵ Pa.V = (0.3 × 8.31 × 400)/(2.525 × 10⁵) = 3.95 × 10⁻³ m³ ≈ 3.95 litres. Substituting values gives 3.95 litres, which matches expected kinetic theory

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Collision Frequency and Mean Free Path Variation

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RMS Speed and Temperature Dependence

What is the pressure of 0.4 moles of an ideal gas in a 4-litre container at 327°C? (R = 8.31 J mol⁻¹ K⁻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. PV = μ R T, P = (μ R T)/(V).T = 327 + 273 = 600 K, V = 4 × 10⁻³ m³.P = (0.4 × 8.31 × 600)/(4 × 10⁻³) = 4.986 × 10⁵ Pa ≈ 5.0 atm (1 atm ≈ 10⁵ Pa). Substituting values gives 5.0 atm, which matches expected kinetic theory result, confirming mean free path

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > RMS Speed and Temperature Dependence

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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

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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Mean Free Path and Molecular Diameter

What is the total internal energy of 1 mole of a triatomic gas at 300 K with no vibrational modes? (R = 8.31 J mol⁻¹ K⁻¹

**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. Triatomic gas: 6 degrees of freedom (3 translational + 3 rotational).U = 3 R T = 3 × 8.31 × 300 = 7479 J ≈ 7.48 kJ. Substituting values gives 7.48 kJ, 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 > Mean Free Path and Molecular Diameter

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Mean Free Path and Molecular Diameter

A solid has a molar specific heat capacity of 24.4 J mol⁻¹ K⁻¹. How many degrees of freedom per atom does it have? (R =

**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. C = f × (R)/(2), 24.4 = f × (8.31)/(2).f = (24.4 × 2)/(8.31) ≈ 5.87 ≈ 6. Substituting values gives 6, 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 > Mean Free Path and Molecular Diameter

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CHEMISTRY

What is the number of unpaired electrons in [Cr(Hâ‚‚O)6]^{3+ ?

Given: What is the number of unpaired electrons in [Cr(H₂O)6]^{3+ ? These values define the system as per NCERT data. Formula: Cr³⁺ ( d³ ) with H₂O in an octahedral field has 3 unpaired electrons ( t_{2g³ ), as it’s a d³ system.. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Substituting values like 1.2 × 10⁻⁵, 236 J kg⁻¹ K⁻¹, CH₃CH₂NH₂ etc. into the formula and simplifying step by step. Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom and Periodicity, Topic: Mole concept, atomic models and periodic trends.

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CHEMISTRY

How many optical isomers are possible for [Cr(Câ‚‚Oâ‚„)3]^{3- ?

[Cr(Câ‚‚Oâ‚„)3]^{3- is octahedral with three bidentate oxalate ligands, lacking symmetry, and forms two optical isomers (d and l enantiomers). This follows from NCERT principle where the relation explains the outcome clearly for students in simple steps.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Relevant Chemistry topic covering principles and examples as per NCERT.

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CHEMISTRY

What is the magnetic moment (in BM) of [Ni(NH₃)6]^{2+ ?

Given: What is the magnetic moment (in BM) of [Ni(NH₃)6]^{2+ ? These values define the system as per NCERT data. Formula: Magnetic moment = sqrt2(2+2) = sqrt8 approx 2.83 BM.. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Ni²⁺ ( d⁸ ) with NH₃ (moderate field) in an octahedral field is high spin ( t_{2g⁶ e_g² ), with 2 unpaired electrons. Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Relevant Chemistry topic covering principles and examples as per NCERT.

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