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61 public questions tagged with this topic.

Specific Heat Capacity and Latent Heat

A gas at 3 atm in a 4 L container is heated from 25°C to 75°C at constant volume. What is the final pressure?

**Specific heat capacity** c = Q/(m ΔT) (J/kg·K), molar C = Q/(n ΔT), heat required to raise temperature, Q = m c ΔT, for water c=4186 J/kg·K, latent heat L = Q/m for phase change at constant temperature, fusion L_f and vaporization L_v, Q = m L, e.g., ice melting L_f=3.34×10⁵ J/kg, water vaporization 2.26×10⁶ J/kg. Isochoric: (P₁)/(T₁) = (P₂)/(T₂) . P₁ = 3 atm , T₁ = 25 + 273 = 298 K , T₂ = 75 + 273 = 348 K . (3)/(298) = (P₂)/(348) ⇒ P₂ = (3 × 348)/(298) ≈ 3.5 atm . Using first law ΔU = Q - W,

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

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First Law of Thermodynamics Applications

Which property of an ideal gas simplifies its internal energy calculation?

**First law applications** for isobaric W = P ΔV, Q = n C_p ΔT, ΔU = n C_v ΔT, for isothermal ideal gas ΔU=0 Q=W=n R T ln(V₂/V₁), for adiabatic Q=0 W= -ΔU = (P₁V₁ - P₂V₂)/(γ-1), for isochoric W=0 ΔU=Q=n C_v ΔT, enabling calculation of Q,W,ΔU for any process. For an ideal gas, internal energy ( U ) depends solely on temperature because intermolecular forces are negligible, reducing U to the sum of molecular kinetic energies, independent of pressure or volume interactions. Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n

Ref: NCERT > Physics Book > Thermodynamics > First Law of Thermodynamics Applications

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Adiabatic Processes and Gamma Determination

A gas at 7 atm and 40^circ C in a 5 L container is cooled isochorically to -20^circ C . What is the final pressure?

**Gamma determination** γ = C_p/C_v, C_p - C_v = R, for monatomic f=3 C_v=3/2 R C_p=5/2 R γ=1.67, diatomic f=5 C_v=5/2 R C_p=7/2 R γ=1.4, adiabatic relation P V^γ = const allows γ determination from P-V measurements, slope of log P vs log V = -γ. For isochoric: (P₁)/(T₁) = (P₂)/(T₂) . P₁ = 7 atm , T₁ = 40 + 273 = 313 K , T₂ = -20 + 273 = 253 K . (7)/(313) = (P₂)/(253) ⇒ P₂ = (7 × 253)/(313) ≈ 5.66 atm . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P

Ref: NCERT > Physics Book > Thermodynamics > Adiabatic Processes and Gamma Determination

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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

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Partial Pressures and Gas Mixtures

The rms speed of oxygen molecules is 482 m/s at 300 K. What is the rms speed of helium molecules at the same temperature

**Gas mixtures** ideal gas law applies to each component, P_total = Σ n_i R T/V, partial pressure P_i = n_i R T/V, mole tion X_i = n_i/n_total, P_i = X_i P_total, enabling calculation of individual pressures from composition, important for kinetic theory and chemistry. v_rms ∝ (1)/(√(m)), v_Hev_O₂ = √(m_O)₂m_He.v_He482 = √((32)/(4)) = √(8) ≈ 2.828.v_He = 482 × 2.828 ≈ 1363 m/s. Substituting values gives 1363 m/s, 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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Partial Pressures and Gas Mixtures

A gas at 3 atm and 600 K has a density of 1.44 kg m⁻³. What is its molecular mass? (R = 8.31 J mol⁻¹ K⁻¹, 1 atm = 1.01 ×

**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. P = (ρ R T)/(M), M = (ρ R T)/(P).P = 3 × 1.01 × 10⁵ = 3.03 × 10⁵ Pa.M = (1.44 × 8.31 × 600)/(3.03 × 10⁵) = 0.0237 kg/mol ≈ 23.7 g/mol ≈ 24 g/mol. Substituting values gives 24 g/mol, which matches expected kinetic theory result, confirming mean

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

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Partial Pressures and Gas Mixtures

What is the temperature at which the rms speed of helium atoms is 1000 m/s? (Atomic mass of He = 4 u, k_B = 1.38 × 10⁻²³

**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. v_rms = √((3k_B T)/(m)), m = 4 × 10⁻³⁶.02 × 10²³ = 6.64 × 10⁻²⁷ kg.1000² = 3 × 1.38 × 10⁻²/³ × T6.64 × 10⁻²⁷, T = 10⁶ × 6.64 × 10⁻²⁷/⁴.14 × 10⁻²/³ ≈ 1604 K . Substituting values gives 1604 K, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law

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 at 3 atm and 300 K has a volume of 10 litres. If the temperature rises to 900 K at constant pressure, what is the

**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. Charles’ law: (V₁)/(T₁) = (V₂)/(T₂).V₁ = 10 litres, T₁ = 300 K, T₂ = 900 K.V₂ = V₁ × (T₂)/(T₁) = 10 × (900)/(300) = 30 litres. Substituting values gives 30 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 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

The rms speed of a gas is 350 m/s at 175 K. At what temperature will the rms speed be 700 m/s?

**Collision frequency** Z = √2 n π d² v_avg, n number density, d molecular diameter, v_avg average speed, proportional to n and v_avg, mean free path λ = v_avg/Z =1/(√2 n π d²), inversely proportional to n, so λ ∝1/P at constant T because n ∝ P, collision frequency increases with pressure, λ decreases. v_rms ∝ √(T), (v₂)/(v₁) = √((T₂)/(T₁)).(700)/(350) = √((T₂)/(175)), 2 = √((T₂)/(175)).Square both sides: 4 = (T₂)/(175), T₂ = 700 K. Substituting values gives 700 K, 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 > 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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