45 public questions tagged with this topic.
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
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
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
For an ideal gas, Cp = Cv + R = 12.5 + 8.314 = 20.814 J/mol·K.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Third Law of Thermodynamics and Applications
For adiabatic process, T₂ = T₁ (V₁/V₂)γ⁻¹ = 400 × (2/4)⁰.⁶⁷ ≈ 400 × 0.62996 = 252.0 K .
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Third Law of Thermodynamics and Applications
In an isochoric process (Δ V = 0), work done w = -PΔ V = 0, regardless of other conditions.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Thermodynamic Terms - System Surroundings and Types of Systems
In a reversible adiabatic process (q = 0), Δ U = w. For an ideal gas, Δ U = nCvΔ T, and since expansion cools the gas, Δ T < 0, making Δ U < 0.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Thermodynamic Terms - System Surroundings and Types of Systems
T₂ = 600 × (10/20)⁰.⁴ = 600 × 0.7579 = 454.74 K, Cv = R/(γ-1) = 8.314 / 0.4 = 20.785 J/mol·K, w = nCvΔ T = 2 × 20.785 × (454.74 - 600) = -6042 J.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Thermodynamic Terms - System Surroundings and Types of Systems
For adiabatic reversible expansion, T₂ = T₁ (V₁/V₂)γ⁻¹ = 300 × (10/20)⁰.⁴ ≈ 300 × 0.7579 = 227.4 K . Then, Cv = R/(γ-1) = 8.314 / 0.4 = 20.785 J/mol·K . Work done: w = nCvΔ T = 1 × 20.785 × (227.4 - 300) ≈ -1509.2 J .
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Thermodynamic Terms - System Surroundings and Types of Systems
w = -nRT ln(P₁/P₂) = -1 × 8.314 × 300 × ln(1/0.5) = -2494.2 × 0.693 = -1728 J.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Thermodynamic Terms - System Surroundings and Types of Systems
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
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
