Latest questions in this category.
Work done: w = -PΔ V = -2 × (-0.3) × 101.325 = 60.795 J . Given q = -800 J , Δ U = q + w = -800 + 60.795 = -739.21 ≈ -739.2 J .
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Thermodynamic Terms - System Surroundings and Types of Systems
For an exothermic process, heat is released (qp < 0), and at constant pressure, Δ H = qp, so Δ H < 0.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Thermodynamic Terms - System Surroundings and Types of Systems
The Δ Hf for CO can be found from the second reaction: Δ Hf (CO) = Δ Hf (CO₂) + 283 = -393.5 + 283 = -110.5 kJ/mol.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Thermodynamic Terms - System Surroundings and Types of Systems
Δ U = q + w. Here, q = -200 J (heat released), w = +150 J (work done on system), so Δ U = -200 + 150 = -50 J.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Thermodynamic Terms - System Surroundings and Types of Systems
For the surroundings, Δ Ssurr = -Δ H / T = -(-890.00 × 10³) / 298 ≈ 2986.6 J/K .
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Thermodynamic Terms - System Surroundings and Types of Systems
For Δ ng = 3 - 0 = 3 , RT = 8.314 × 298 × 10⁻³ = 2.4776 kJ . Using Δ H = Δ U + Δ ng RT , Δ U = 128.10 - 3 × 2.4776 = 128.10 - 7.4328 = 120.67 ≈ 120.7 kJ/mol .
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Thermodynamic Terms - System Surroundings and Types of Systems
In an isolated system (q = 0), spontaneity requires Δ Stotal = Δ Ssys > 0, as Δ Ssurr = 0.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Thermodynamic Terms - System Surroundings and Types of Systems
For Δ ng = 1 - 3 = -2 , RT = 8.314 × 298 × 10⁻³ = 2.4776 kJ . Using Δ H = Δ U + Δ ng RT , Δ U = -890.00 - (-2 × 2.4776) = -890.00 + 4.9552 = -885.04 ≈ -885.0 kJ/mol .
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Thermodynamic Terms - System Surroundings and Types of Systems
For Δ ng = 3 - 2 = 1 , RT = 8.314 × 298 × 10⁻³ = 2.4776 kJ . Using Δ H = Δ U + Δ ng RT , Δ U = 197.80 - 1 × 2.4776 = 195.32 kJ/mol .
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Thermodynamic Terms - System Surroundings and Types of Systems
Intensive properties do not depend on the amount of substance. Temperature depends on the system’s state, not its size, unlike volume or enthalpy.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Thermodynamic Terms - System Surroundings and Types of Systems
At constant volume, Δ U = qv = nCvΔ T. So, 800 = n × 20 × 10, n = 800 / 200 = 4 mol.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Thermodynamic Terms - System Surroundings and Types of Systems
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
