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#ideal gas law

53 public questions tagged with this topic.

Work Heat Distinction and Energy Transfer Modes

A gas at 10 atm and 80^circ C in a 8 L container is heated isochorically to 140^circ C . What is the final pressure?

**Modes of energy transfer** work is force times displacement, e.g., gas expansion W = P ΔV, heat is due to temperature gradient, internal energy change same for different combinations of Q and W, e.g., same ΔU can be achieved by adding heat at constant volume or doing work adiabatically, illustrating equivalence but distinction in mechanism. For isochoric: (P₁)/(T₁) = (P₂)/(T₂) . P₁ = 10 atm , T₁ = 80 + 273 = 353 K , T₂ = 140 + 273 = 413 K . (10)/(353) = (P₂)/(413) ⇒ P₂ = (10 × 413)/(353) ≈ 11.7 atm . Using first law ΔU = Q - W,

Ref: NCERT > Physics Book > Thermodynamics > Work Heat Distinction and Energy Transfer Modes

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Work Heat Distinction and Energy Transfer Modes

In an isobaric process, 1.1 moles of an ideal gas expand from 7 L to 14 L at 390 K . What is the work done by the gas? (

**Energy transfer** first law ΔU = Q - W, W includes P-V work, shaft work, electrical work, Q includes conduction Fourier law, convection, radiation Stefan-Boltzmann, distinction important because work is controllable, heat spontaneous from hot to cold, entropy associated with heat not work, explaining why heat engine efficiency

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Cyclic Processes and Reversibility Concepts

A gas at 8 atm and 60^circ C in a 7 L container is heated isochorically to 120^circ C . What is the final pressure?

**Reversibility** reversible process can be reversed by infinitesimal change, no entropy production, quasi-static without friction, e.g., Carnot cycle reversible, irreversible processes involve friction, free expansion, heat transfer across finite temperature difference, entropy increases, most real processes irreversible. For isochoric: (P₁)/(T₁) = (P₂)/(T₂) . P₁ = 8 atm , T₁ = 60 + 273 = 333 K , T₂ = 120 + 273 = 393 K . (8)/(333) = (P₂)/(393) ⇒ P₂ = (8 × 393)/(333) ≈ 9.44 atm . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R T ln(V₂/V₁), adiabatic

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

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Isochoric Processes and Pressure-Temperature

What is the primary reason a real gas deviates from the ideal gas equation?

**Pressure-temperature relation** at constant volume Gay-Lussac law P ∝ T, for V constant, P₁/T₁ = P₂/T₂, if T doubles from 300 K to 600 K P doubles, e.g., P₁=1 atm at 300 K P₂=2 atm at 600 K, no work done, ΔU = n C_v ΔT = Q. Real gases deviate from the ideal gas equation ( P V = μ R T ) due to intermolecular forces, which are negligible in ideal gases but significant in real gases, especially at high pressures or low temperatures. Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal

Ref: NCERT > Physics Book > Thermodynamics > Isochoric Processes and Pressure-Temperature

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Specific Heat Capacity and Latent Heat

A gas at 9 atm and 70^circ C in a 6 L container is cooled isochorically to 10^circ C . 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. For isochoric: (P₁)/(T₁) = (P₂)/(T₂) . P₁ = 9 atm , T₁ = 70 + 273 = 343 K , T₂ = 10 + 273 = 283 K . (9)/(343) = (P₂)/(283) ⇒ P₂ = (9 × 283)/(343) ≈ 7.42 atm . Using first law ΔU = Q -

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

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Zeroth Law Thermal Equilibrium and Quasi-static

An ideal gas expands isothermally at 540 K from 10 L to 30 L with 0.3 moles . What is the work done by the gas? ( R = 8.

**Thermal equilibrium** achieved when temperatures equal, zeroth law allows definition of temperature, quasi-static process approximates equilibrium at each step, enabling calculation of work as area under P-V curve, reversible processes are quasi-static without dissipative effects. For isothermal: W = μ R T ln((V₂)/(V₁)) . μ = 0.3 , R = 8.3 , T = 540 , V₂ = 30 , V₁ = 10 . W = 0.3 × 8.3 × 540 × ln((30)/(10)) = 1344.6 × ln(3) . ln(3) ≈ 1.0986 , W ≈ 1344.6 × 1.0986 ≈ 1477 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric

Ref: NCERT > Physics Book > Thermodynamics > Zeroth Law Thermal Equilibrium and Quasi-static

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Zeroth Law Thermal Equilibrium and Quasi-static

A gas at 4 atm and 300 K in a 5 L container is compressed isothermally to 2 L. What is the work done on the gas? ( R = 8

**Zeroth law of thermodynamics** if two systems A and B each in thermal equilibrium with third C, then A and B in equilibrium with each other, defines temperature as property that is same for systems in thermal equilibrium, basis for thermometer, temperature scale, thermal equilibrium means no net heat flow, same temperature. Isothermal: W = μ R T ln((V₂)/(V₁)) . P₁ V₁ = μ R T ⇒ 4 × 5 = μ × 8.3 × 300 ⇒ μ = (20)/(2490) ≈ 0.008 mol . W = 0.008 × 8.3 × 300 × ln((2)/(5)) = 19.92 × (-0.916) ≈ -18.25 J (work by gas negative). Work

Ref: NCERT > Physics Book > Thermodynamics > Zeroth Law Thermal Equilibrium and Quasi-static

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Zeroth Law Thermal Equilibrium and Quasi-static

A gas at 5 atm in a 6 L container is heated from 20°C to 60°C at constant volume. What is the final pressure?

**Thermal equilibrium** achieved when temperatures equal, zeroth law allows definition of temperature, quasi-static process approximates equilibrium at each step, enabling calculation of work as area under P-V curve, reversible processes are quasi-static without dissipative effects. Isochoric: (P₁)/(T₁) = (P₂)/(T₂) . P₁ = 5 atm , T₁ = 20 + 273 = 293 K , T₂ = 60 + 273 = 333 K . (5)/(293) = (P₂)/(333) ⇒ P₂ = (5 × 333)/(293) ≈ 5.68 atm . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R T ln(V₂/V₁), adiabatic P V^γ =

Ref: NCERT > Physics Book > Thermodynamics > Zeroth Law Thermal Equilibrium and Quasi-static

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Isobaric and Isothermal Processes Work Calculation

An ideal gas expands isothermally at 420 K from 7 L to 21 L with 0.4 moles . What is the work done by the gas? ( R = 8.3

**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 isothermal: W = μ R T ln((V₂)/(V₁)) . μ = 0.4 , R = 8.3 , T = 420 , V₂ = 21 , V₁ = 7 . W = 0.4 × 8.3 × 420 × ln((21)/(7)) = 1394.4 × ln(3) . ln(3) ≈ 1.0986 , W ≈ 1394.4 × 1.0986 ≈ 1532 J . Using first law ΔU = Q - W,

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

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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

A gas mixture has 2 g of helium and 16 g of oxygen. What is the ratio of their partial pressures?

**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 = (μ RT)/(V), P_HeP_O₂ = μ_Heμ_O₂.μ_He = (2)/(4) = 0.5 mol, μ_O₂ = (16)/(32) = 0.5 mol.Ratio = (0.5)/(0.5) = 1:1. Substituting values gives 1:1, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P V = n R T

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 pressure exerted by 0.1 mole of an ideal gas in a 2-litre container at 127°C? (R = 8.31 J mol⁻¹ K⁻¹)

**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. PV = μ R T, P = (μ R T)/(V).T = 127 + 273 = 400 K, V = 2 × 10⁻³ m³.P = (0.1 × 8.31 × 400)/(2 × 10⁻³) = 1.663 × 10⁵ Pa ≈ 1.66 atm (1 atm ≈ 10⁵ Pa). Substituting values gives 1.66 atm, which matches expected kinetic theory result, confirming mean free path λ =

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

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