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#chemistry problem

253 public questions tagged with this topic.

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,

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

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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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RMS Speed and Temperature Dependence

A mixture of 1 mole of neon and 2 moles of argon is at 500 K in a 30-litre container. What is the total pressure? (R = 8

**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).Total moles = 1 + 2 = 3, V = 30 × 10⁻³ m³.P = (3 × 8.31 × 500)/(30 × 10⁻³) = 4.155 × 10⁵ Pa ≈ 4.16 atm. Substituting values gives 4.16 atm, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π

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

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RMS Speed and Temperature Dependence

What is the volume of 0.6 moles of an ideal gas at 3 atm and 527°C? (R = 8.31 J mol⁻¹ K⁻¹)

**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. PV = μ R T, V = (μ R T)/(P).T = 527 + 273 = 800 K, P = 3 × 1.01 × 10⁵ = 3.03 × 10⁵ Pa.V = (0.6 × 8.31 × 800)/(3.03 × 10⁵) = 1.317 × 10⁻² m³ ≈ 13.17 litres. Substituting values gives 13.2 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

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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CHEMISTRY

How many Faradays of electricity are required to deposit 0.635 g of copper from CuSO₄ solution? (Molar mass of Cu = 63.5

Given: How many Faradays of electricity are required to deposit 0.635 g of copper from CuSO₄ solution? (Molar mass of Cu = 63.5 g/mol) These values define the system as per NCERT data. Formula: Moles of Cu = 0.635/63.5 = 0.01 mol. This is standard NCERT relation. Substitution & Calculation: Reaction: Cu²⁺ + 2e⁻ → Cu(s), 2F deposits 63.5 g. . Faradays = 0.01 × 2 = 0.02 F . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

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

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