Practice problems and explanations focused on the Third Law of Thermodynamics and its real-world applications in chemical and physical systems.
For isothermal expansion, Δ Ssys = nR ln(P₁/P₂) = 2 × 8.314 × ln(1/0.25) = 16.628 × 1.3863 ≈ 23.0 J/K .
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Third Law of Thermodynamics and Applications
Reverse first: Δ H = 726.5 kJ. Second: Δ H = -393.5 kJ. Third (×2): Δ H = -571.6 kJ. Add: 726.5 - 393.5 - 571.6 = -238.6 kJ/mol.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Third Law of Thermodynamics and Applications
w = -nRT ln(V₂/V₁) = -2 × 8.314 × 400 × ln(24/8) = -6651.2 × ln(3) = -6651.2 × 1.099 = -7309.7 J.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Third Law of Thermodynamics and Applications
For Δ ng = 2 - 2 = 0 , RT = 8.314 × 298 × 10⁻³ = 2.4776 kJ . Using Δ H = Δ U + Δ ng RT , Δ U = -103.00 - 0 = -103.00 kJ/mol .
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Third Law of Thermodynamics and Applications
Reverse first: Δ H = 1299.5 kJ. Second (×2): Δ H = -787 kJ. Third: Δ H = -285.8 kJ. Add: 1299.5 - 787 - 285.8 = 226.7 kJ/mol.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Third Law of Thermodynamics and Applications
Reverse first: Δ H = 1076.8 kJ. Second: Δ H = -393.5 kJ. Third (×2): Δ H = -593.6 kJ. Add: 1076.8 - 393.5 - 593.6 = 89.7 kJ/mol.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Third Law of Thermodynamics and Applications
T₂ = T₁ (V₁/V₂)γ⁻¹ = 300 × (18/6)⁰.³³ = 300 × 3⁰.³³ = 300 × 1.44 = 432 K.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Third Law of Thermodynamics and Applications
Δ ng = 1 - 1 = 0, RT = 2.48 kJ, Δ H = Δ U + Δ ng RT, Δ U = -393.5 - 0 = -393.5 kJ/mol.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Third Law of Thermodynamics and Applications
Δ ng = 2 - 3 = -1, RT = 2.48 kJ, Δ H = Δ U + Δ ng RT, Δ U = 66.4 - (-1 × 2.48) = 66.4 + 2.48 = 68.88 kJ/mol.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Third Law of Thermodynamics and Applications
Δ H = -2 × 40.79 = -81.58 kJ, Δ Ssys = -Δ Hvap × n / T = -40.79 × 10³ × 2 / 373 = -218.8 J/K, Δ Ssurr = -Δ H / T = 81.58 × 10³ / 373 = 218.8 J/K, Δ Stotal = -218.8 + 218.8 = 0 J/K.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Third Law of Thermodynamics and Applications
Δ G = Δ H - TΔ S = -80.00 - 300 × (-0.2000) = -80.00 + 60.00 = -20.00 kJ/mol . The negative Δ G indicates spontaneity at 300 K.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Third Law of Thermodynamics and Applications
Δ G = Δ H - TΔ S = -72.6 - 298 × 0.114 = -72.6 - 33.97 = -106.57 kJ/mol.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Third Law of Thermodynamics and Applications
