Questions covering heat capacity, calorimetry principles, and methods for measuring enthalpy changes in chemical reactions.
For an irreversible adiabatic expansion (q = 0), Δ U = w = -Pext Δ V. Since Δ V > 0, w < 0, and thus Δ U < 0.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Heat Capacity and Calorimetry and Measurement of Enthalpy
Δ G = Δ H - TΔ S, 30 = 90 - 300 Δ S, Δ S = (90 - 30) / 300 = 60 / 300 = 0.2 kJ/K = 200 J/K.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Heat Capacity and Calorimetry and Measurement of Enthalpy
For Δ ng = -1 , RT = 8.314 × 298 × 10⁻³ = 2.4776 kJ . Using Δ H = Δ U + Δ ng RT , Δ U = -100.00 - (-1 × 2.4776) = -100.00 + 2.4776 = -97.52 kJ/mol .
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Heat Capacity and Calorimetry and Measurement of Enthalpy
For freezing, Δ H = -n × Δ Hfus = -3 × 5.5 = -16.5 kJ, Δ Ssurr = -Δ H / T = -(-16.5 × 10³) / 250 = 66 J/K.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Heat Capacity and Calorimetry and Measurement of Enthalpy
For adiabatic expansion, T₂ = 500 × (5/10)⁰.⁴ ≈ 500 × 0.7579 = 378.9 K . Then, Δ T = 378.9 - 500 = -121.1 K . Work done: w = nCvΔ T = 1 × 20.785 × (-121.1) ≈ -2516 J .
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Heat Capacity and Calorimetry and Measurement of Enthalpy
For Δ ng = 1 , RT = 8.314 × 350 × 10⁻³ = 2.9099 kJ . Using Δ H = Δ U + Δ ng RT , Δ U = 32.00 - 2.9099 = 29.09 kJ/mol .
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Heat Capacity and Calorimetry and Measurement of Enthalpy
Multiply first equation by 2: 2C(s) + 2O₂(g) → 2CO₂(g), Δ H = -787 kJ. Reverse second: 2CO₂(g) → 2CO(g) + O₂(g), Δ H = 566 kJ. Add: Δ H = -787 + 566 = -221 kJ.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Heat Capacity and Calorimetry and Measurement of Enthalpy
Moles = 27 / 27 = 1 mol, Δ T = 20 K, q = nCpΔ T = 1 × 24 × 20 = 480 J = 0.48 kJ.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Heat Capacity and Calorimetry and Measurement of Enthalpy
For Δ ng = 2 - 2 = 0 , Δ H = Δ U + Δ ng RT . Since Δ ng = 0 , Δ U = Δ H = -41.20 kJ/mol .
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Heat Capacity and Calorimetry and Measurement of Enthalpy
For an ideal gas, Cp = Cv + R = 20 + 8.314 = 28.314 J/mol·K.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Heat Capacity and Calorimetry and Measurement of Enthalpy
Δ G = Δ H - TΔ S = 0, -50 × 10³ - T(-100) = 0, T = 50000 / 100 = 500 K.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Heat Capacity and Calorimetry and Measurement of Enthalpy
Δ ng = 2 - 2 = 0, so Δ H = Δ U + Δ ng RT = Δ U. Thus, Δ U = 52.96 kJ/mol.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Heat Capacity and Calorimetry and Measurement of Enthalpy
