Skip to content
Login Register

#moles

9 public questions tagged with this topic.

Specific Heat Capacity and Latent Heat

An ideal gas absorbs 1000 J of heat in an isochoric process, increasing its temperature by 20 K . What is the number of

**Heat capacity** at constant pressure C_p and volume C_v, C_p = C_v + R per mole, for solids Dulong-Petit C_v≈3R≈25 J/mol·K. Specific heat and latent heat govern temperature changes and phase transitions, Q = m c ΔT for heating, Q = m L for melting/boiling at constant T. For isochoric process: Δ Q = μ C_v Δ T . Δ Q = 1000 , C_v = 20 , Δ T = 20 . 1000 = μ × 20 × 20 ⇒ μ = (1000)/(400) = 2.5 moles . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P

Ref: NCERT > Physics Book > Thermodynamics > Specific Heat Capacity and Latent Heat

View Details
Partial Pressures and Gas Mixtures

What is the volume of 0.2 moles of an ideal gas at 1.5 atm and 227°C? (R = 8.31 J mol⁻¹ K⁻¹)

**Partial pressure concept** for mixture of non-reacting gases, total pressure sum of partials, each gas behaves independently, kinetic theory still holds with effective n = Σ n_i, mean free path depends on total n and cross-sections, mixture properties weighted by mole tions. PV = μ R T, V = (μ R T)/(P).T = 227 + 273 = 500 K, P = 1.5 × 1.01 × 10⁵ = 1.515 × 10⁵ Pa.V = (0.2 × 8.31 × 500)/(1.515 × 10⁵) = 5.485 × 10⁻³ m³ ≈ 5.49 litres. Substituting values gives 5.49 litres, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Partial Pressures and Gas Mixtures

View Details
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

View Details
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

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Molecular Mass Density and Ideal Gas Equation

View Details
Molecular Mass Density and Ideal Gas Equation

What is the volume of 0.5 moles of an ideal gas at 2 atm and 627°C? (R = 8.31 J mol⁻¹ 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. PV = μ R T, V = (μ R T)/(P).T = 627 + 273 = 900 K, P = 2 × 1.01 × 10⁵ = 2.02 × 10⁵ Pa.V = (0.5 × 8.31 × 900)/(2.02 × 10⁵) = 1.854 × 10⁻² m³ ≈ 18.54 litres. Substituting values gives 18.5 litres, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Molecular Mass Density and Ideal Gas Equation

View Details
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

View Details
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 mea

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

View Details
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

View Details