Practice questions exploring the first law of thermodynamics, enthalpy, and internal energy. Helps students understand energy conservation in chemical systems.
For adiabatic compression, T₂ = 400 × (20/5)⁰.⁴ ≈ 400 × 1.741 = 696.4 K . Then, Δ T = 696.4 - 400 = 296.4 K . Internal energy change: Δ U = nCvΔ T = 1 × 20.785 × 296.4 ≈ 6160 J .
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: First Law of Thermodynamics and Enthalpy and Internal Energy
Δ U = q + w. Here, q = -500 J (heat released), w = -300 J (work done by system), so Δ U = -500 - 300 = -800 J.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: First Law of Thermodynamics and Enthalpy and Internal Energy
Work done: w = -PΔ V = -2 × 0.5 × 101.325 = -101.325 J . Given q = 1200 J , Δ U = q + w = 1200 - 101.325 = 1098.68 ≈ 1098.7 J .
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: First Law of Thermodynamics and Enthalpy and Internal Energy
A state function depends only on the initial and final states, not the path taken. Internal energy (U) is a state function, while work (w) and heat (q) are path-dependent.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: First Law of Thermodynamics and Enthalpy and Internal Energy
Reverse the third equation and add the first two: Δ H = -393.5 + (-2 × 285.8) + 890.3 = -74.8 kJ/mol.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: First Law of Thermodynamics and Enthalpy and Internal Energy
Moles = 39 / 78 = 0.5 mol, heat released = 0.5 × 3267.6 = 1633.8 kJ.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: First Law of Thermodynamics and Enthalpy and Internal Energy
For reversible isothermal compression, w = -nRT ln(V₂/V₁) = -1 × 8.314 × 350 × ln(3/15) = -2909.9 × (-1.6094) ≈ 4685 J .
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: First Law of Thermodynamics and Enthalpy and Internal Energy
Δ H = Δ Hf° (C₂H₆) - [Δ Hf° (C₂H₂) + 2 × Δ Hf° (H₂)] = -84.7 - [226.7 + 0] = -84.7 - 226.7 = -311.4 kJ/mol.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: First Law of Thermodynamics and Enthalpy and Internal Energy
At constant volume, w = -pΔ V = 0, so from the first law, Δ U = qv, the heat absorbed at constant volume.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: First Law of Thermodynamics and Enthalpy and Internal Energy
At constant volume, w = -PΔ V = 0. First Law: Δ U = q + w = 600 + 0 = 600 J, so w = 0 J.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: First Law of Thermodynamics and Enthalpy and Internal Energy
For Δ ng = 1 - 2 = -1 , RT = 8.314 × 298 × 10⁻³ = 2.4776 kJ . Using Δ H = Δ U + Δ ng RT , Δ U = -137.00 - (-1 × 2.4776) = -137.00 + 2.4776 = -134.52 kJ/mol .
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: First Law of Thermodynamics and Enthalpy and Internal Energy
For reactions at constant temperature (ideal gas), Δ H = Δ U + Δ ng RT . With Δ ng = 0 , Δ H = Δ U = 250 J .
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: First Law of Thermodynamics and Enthalpy and Internal Energy
