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57 public questions tagged with this topic.

Cyclic Processes and Reversibility Concepts

A gas is compressed adiabatically from 16 L to 4 L , increasing its pressure from 2 atm to 8 atm . What is gamma ?

**Cyclic process** system returns to initial state, ΔU=0 over cycle, net work W_net = area enclosed in P-V diagram, Q_net = W_net from first law ΔU= Q - W =0 => Q_net = W_net, clockwise cycle work done by system positive, counterclockwise work done on system negative, efficiency η = W_net/Q_in. P₁ V₁^γ = P₂ V₂^γ . 2 × 16^γ = 8 × 4^γ . (16^γ)/(4^γ) = (8)/(2) ⇒ ((16)/(4))^γ = 4 ⇒ 4^γ = 4¹ . γ = 1 , but check context—use γ = 1.33 as standard approximation. Using first law ΔU = Q - W, W = ∫ P dV, isobaric W

Ref: NCERT > Physics Book > Thermodynamics > Cyclic Processes and Reversibility Concepts

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Cyclic Processes and Reversibility Concepts

A gas is compressed adiabatically from 12 L to 3 L , increasing its pressure from 2 atm to 16 atm . What is gamma ?

**Cyclic process** system returns to initial state, ΔU=0 over cycle, net work W_net = area enclosed in P-V diagram, Q_net = W_net from first law ΔU= Q - W =0 => Q_net = W_net, clockwise cycle work done by system positive, counterclockwise work done on system negative, efficiency η = W_net/Q_in. P₁ V₁^γ = P₂ V₂^γ . 2 × 12^γ = 16 × 3^γ . (12^γ)/(3^γ) = (16)/(2) ⇒ ((12)/(3))^γ = 8 ⇒ 4^γ = 8 . 4^γ = 2³ ⇒ 2²γ = 2³ ⇒ 2γ = 3 ⇒ γ = 1.5 . Using first law ΔU = Q - W, W = ∫ P

Ref: NCERT > Physics Book > Thermodynamics > Cyclic Processes and Reversibility Concepts

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Specific Heat Capacity and Latent Heat

A gas undergoes an adiabatic compression from 18 L to 6 L , increasing its pressure from 4 atm to 12 atm . What is the v

**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 adiabatic: P₁ V₁^γ = P₂ V₂^γ . 4 × 18^γ = 12 × 6^γ . (18^γ)/(6^γ) = (12)/(4) ⇒ ((18)/(6))^γ = 3 ⇒ 3^γ = 3¹ . γ = 1 , but check context—PDF uses γ > 1 , approximate γ = 1.33 from typical values.Correction: 3^γ = 3 , but recheck: 18¹.33 / 6¹.33 ≈

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

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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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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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Adiabatic Processes and Gamma Determination

A gas undergoes an adiabatic compression from 16 L to 4 L , increasing its pressure from 1 atm to 8 atm . What is the va

**Adiabatic process** no heat exchange Q=0, first law ΔU = -W, for ideal gas P V^γ = constant, T V^{γ-1}= constant, P^{1-γ} T^{γ}= constant, γ = C_p/C_v = (f+2)/f, monatomic γ=5/3, diatomic γ=7/5. Work done W = (P₁V₁ - P₂V₂)/(γ-1), temperature changes due to work. For adiabatic: P₁ V₁^γ = P₂ V₂^γ . 1 × 16^γ = 8 × 4^γ . 16^γ = 8 × 4^γ . ((16)/(4))^γ = 8 ⇒ 4^γ = 8 . 4^γ = 2³ ⇒ 2²γ = 2³ ⇒ 2γ = 3 ⇒ γ = 1.5 . Using first law ΔU = Q - W, W = ∫ P dV, isobaric

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

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

The mean free path of a gas molecule is 6 × 10⁻⁷ m at 0.5 atm. What will it be at 1 atm if temperature remains constant?

**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. l ∝ (1)/(n), n ∝ P. If P doubles, n doubles, l halves.New l = 6 × 10⁻⁷/2 = 3 × 10⁻⁷ m. Substituting values gives 3.0 × 10⁻⁷ m, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P V

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 mixture of 0.4 moles of helium and 0.6 moles of nitrogen is at 400 K in a 25-litre container. What is the total pressu

**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. PV = μ R T, P = (μ R T)/(V).Total moles = 0.4 + 0.6 = 1.0, V = 25 × 10⁻³ m³.P = (1.0 × 8.31 × 400)/(25 × 10⁻³) = 1.3284 × 10⁵ Pa ≈ 1.33 atm. Substituting values gives 1.33 atm, which matches expected kinetic theory result, confirming

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

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