6 public questions tagged with this topic.
**Second law Kelvin-Planck statement** no process possible whose sole result is absorption of heat from reservoir and complete conversion to work, heat engine must have at least two reservoirs hot and cold, efficiency η = W/Q_h =1 - Q_c/Q_h
Ref: NCERT > Physics Book > Thermodynamics > Second Law Heat Engines and Kelvin-Planck
**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
**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. For monatomic gas, U = (3)/(2) μ R T.U = (3)/(2) × 2 × 8.31 × 400 = 9972 J ≈ 9.97 kJ . Substituting values gives 9.97 kJ, 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
**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. C_p - C_v = R, C_v = C_p - R.C_v = 35.4 - 8.31 = 27.09 J mol⁻¹ K⁻¹ ≈ 27.1 J mol⁻¹ K⁻¹. Substituting values gives 27.1 J mol⁻¹ K⁻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 > Molecular Mass Density and Ideal Gas Equation
**RMS speed** v_rms = √(3 R T/M) = √(3 k_B T/m) where M molar mass (kg/mol), m molecular mass (kg), k_B=1.38×10⁻²/³ J/K, R=8.314 J/mol·K, T absolute temperature (K). Proportional to √T and 1/√M, lighter gases faster at same T, e.g., H₂ faster than O₂, temperature increase raises v_rms as √T. For a diatomic gas (rigid rotator), 5 degrees of freedom (3 translational + 2 rotational).C_v = (5)/(2) R = (5)/(2) × 8.31 = 20.775 J mol⁻¹ K⁻¹ ≈ 20.8 J mol⁻¹ K⁻¹ . Substituting values gives 20.8 J mol⁻¹ K⁻¹, 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 > RMS Speed and Temperature Dependence
**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. C_p = C_v + R.C_p = 20.4 + 8.31 = 28.71 J mol⁻¹ K⁻¹ ≈ 28.7 J mol⁻¹ K⁻¹. Substituting values gives 28.7 J mol⁻¹ K⁻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
