#gas properties

Specific Heat Capacity and Latent Heat

Why does the specific heat capacity of a gas differ at constant pressure and constant volume?

**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. At constant pressure ( C_p ), heat supplies energy for both internal energy increase and work done due to expansion ( Δ Q = Δ U + P Δ V ). At constant volume ( C_v ), no work is done ( Δ V = 0 ), so heat

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Isobaric and Isothermal Processes Work Calculation

A gas undergoes an adiabatic expansion from 22 L to 66 L , reducing its pressure from 12 atm to 2 atm . What is the valu

**Work calculation** isobaric W = P(V₂ - V₁), e.g., P=1 atm=1.013×10⁵ Pa ΔV=0.01 m³ W=1013 J. Isothermal W = n R T ln(V₂/V₁), for n=1 mol T=300 K V₂/V₁=2 W=1×8.314×300×ln2=1729 J, work done by gas during expansion, area under P-V curve. For adiabatic: P₁ V₁^γ = P₂ V₂^γ . 12 × 22^γ = 2 × 66^γ . 12 / 2 = ((66)/(22))^γ ⇒ 6 = 3^γ . 3^γ = 3¹.63 , γ ≈ 1.63 ≈ 1.67 (standard value from context). Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R T ln(V₂/V₁),

Ref: NCERT > Physics Book > Thermodynamics > Isobaric and Isothermal Processes Work Calculation

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Partial Pressures and Gas Mixtures

The mean free path of a gas is 7 × 10⁻⁷ m with a number density of 1.5 × 10²⁵ m⁻³. What is the molecular diameter?

**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)/(√(2) n π d²), d² = (1)/(√(2) n π l).d² = (1)/(1.414 × 1.5 × 10²⁵) × 3.14 × 7 × 10⁻⁷ = (1)/(4.66 × 10⁻¹⁹) ≈ 2.14 × 10⁻²⁰.d = √(2.14 × 10⁻²⁰) ≈ 1.46 × 10⁻¹⁰ m. Substituting values gives 1.46 × 10⁻¹⁰ m, which matches expected

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

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Molecular Mass Density and Ideal Gas Equation

The mean free path of a gas is 1 × 10⁻⁷ m with a molecular diameter of 2 × 10⁻¹⁰ m. What is the number density of the ga

**Ideal gas equation** P V = n R T = (m/M) R T, density ρ = m/V = P M/(R T), molecular mass M (kg/mol), P pressure (Pa), T temperature (K). At given P,T density proportional to M, heavier gases denser, e.g., at 1.5 atm 300 K V=24 L n= P V/(R T)=1.5×1.013×10⁵×0.024/(8.314×300)≈1.46 mol. l = (1)/(√(2) n π d²), n = (1)/(√(2) π d² l).n = (1)/(1.414 × 3.14 × (2 × 10⁻¹⁰))² × 1 × 10⁻⁷ = (1)/(1.77 × 10⁻²⁶) ≈ 5.65 × 10²⁵ m⁻³. Substituting values gives 5.65 × 10²⁵ m⁻³, which matches expected kinetic theory result, confirming mean free path λ

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

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RMS Speed and Temperature Dependence

The mean free path of a gas is 1.8 × 10⁻⁷ m with a number density of 3.0 × 10²⁵ m⁻³. What is the molecular diameter?

**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. l = (1)/(√(2) n π d²), d² = (1)/(√(2) n π l).d² = (1)/(1.414 × 3.0 × 10²⁵) × 3.14 × 1.8 × 10⁻⁷ = (1)/(2.4 × 10⁻¹⁹) ≈ 4.17 × 10⁻²⁰.d = √(4.17 × 10⁻²⁰) ≈ 2.04 × 10⁻¹⁰ m. Substituting values gives 2.0 × 10⁻¹⁰ m, 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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Internal Energy of Ideal Gases

A gas mixture has equal volumes of helium and argon at the same temperature and pressure. What is the ratio of their rms

**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. v_rms ∝ (1)/(√(m)), v_Hev_Ar = √(m_Ar)m_He = √((39.9)/(4)) ≈ √(10) ≈ 3.16. Substituting values gives 3.16:1, 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 > Internal Energy of Ideal Gases

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Internal Energy of Ideal Gases

The mean free path of a gas molecule is 1.5 × 10⁻⁶ m at 0.2 atm. What will it be at 0.8 atm if temperature remains const

**Internal energy of ideal gas** U = f/2 n R T depends only on temperature, f degrees of freedom, n moles, R=8.314 J/mol·K, for monatomic f=3 U=3/2 n R T, diatomic f=5 at moderate T U=5/2 n R T, independent of pressure or volume, only T matters for ideal gas. l ∝ (1)/(n), n ∝ P. If P increases by 4 times (0.2 to 0.8), n increases 4 times, l reduces to (1)/(4).New l = 1.5 × 10⁻⁶/4 = 3.75 × 10⁻⁷ m. Substituting values gives 3.75 × 10⁻⁷ m, 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 > Internal Energy of Ideal Gases

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Internal Energy of Ideal Gases

The mean free path of a gas molecule is 1.0 × 10⁻⁶ m at 0.5 atm. What will it be at 2 atm if temperature remains constan

**Internal energy of ideal gas** U = f/2 n R T depends only on temperature, f degrees of freedom, n moles, R=8.314 J/mol·K, for monatomic f=3 U=3/2 n R T, diatomic f=5 at moderate T U=5/2 n R T, independent of pressure or volume, only T matters for ideal gas. l ∝ (1)/(n), n ∝ P. If P increases by 4 times (0.5 to 2), n increases 4 times, l reduces to (1)/(4).New l = 1.0 × 10⁻⁶/4 = 2.5 × 10⁻⁷ m. Substituting values gives 2.5 × 10⁻⁷ m, 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 > Internal Energy of Ideal Gases

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Degrees of Freedom and Molar Specific Heat

The mean free path of a gas is 5 × 10⁻⁷ m with a number density of 2 × 10²⁵ m⁻³. What is the molecular diameter?

**Degrees of freedom** f counts independent motions, monatomic 3 translational, diatomic 3 translational +2 rotational =5 at room T, vibrational adds at high T, molar specific heat at constant volume C_v = f/2 R, at constant pressure C_p = C_v + R, ratio γ = C_p/C_v = (f+2)/f, monatomic γ=5/3≈1.67, diatomic γ=7/5=1.4. l = (1)/(√(2) n π d²), d² = (1)/(√(2) n π l).d² = (1)/(1.414 × 2 × 10²⁵) × 3.14 × 5 × 10⁻⁷ = (1)/(4.44 × 10⁻¹⁹) ≈ 2.25 × 10⁻²⁰.d = √(2.25 × 10⁻²⁰) ≈ 1.5 × 10⁻¹⁰ m. Substituting values gives 1.5 × 10⁻¹⁰ m, which matches expected kinetic theory

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Degrees of Freedom and Molar Specific Heat

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Degrees of Freedom and Molar Specific Heat

A gas mixture has equal masses of hydrogen and argon at 300 K. What is the ratio of their rms speeds? (Molecular mass: H

**Molar specific heat** from equipartition C_v = f/2 R, C_p = f/2 R + R, γ = C_p/C_v =1+2/f, for f=3 γ=1.67, f=5 γ=1.4, f=6 γ=1.33, explaining specific heat variation with molecular structure, degrees of freedom determine heat capacity. v_rms ∝ (1)/(√(m)), v_H₂v_Ar = √(m_Ar)m_H₂.v_H₂v_Ar = √((39.9)/(2)) ≈ √(19.95) ≈ 4.47. Substituting values gives 4.47:1, 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 > Degrees of Freedom and Molar Specific Heat

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Gas Laws and Volume-Temperature Relations

The mean free path of a gas is 1.2 × 10⁻⁷ m with a number density of 4.0 × 10²⁵ m⁻³. What is the molecular diameter?

**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. l = (1)/(√(2) n π d²), d² = (1)/(√(2) n π l).d² = (1)/(1.414 × 4.0 × 10²⁵) × 3.14 × 1.2 × 10⁻⁷ = (1)/(2.13 × 10⁻¹⁹) ≈ 4.69 × 10⁻²⁰.d = √(4.69 × 10⁻²⁰) ≈ 2.17 × 10⁻¹⁰ m. Substituting values gives 2.17

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

The rms speed of a gas is 550 m/s at 275 K. At what temperature will the rms speed be 1100 m/s?

**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. v_rms ∝ √(T), (v₂)/(v₁) = √((T₂)/(T₁)).(1100)/(550) = √((T₂)/(275)), 2 = √((T₂)/(275)).Square both sides: 4 = (T₂)/(275), T₂ = 1100 K. Substituting values gives 1100 K, 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 > Gas Laws and Volume-Temperature Relations

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