4 public questions tagged with this topic.
w = -PΔ V = -2.5 × 0.4 × 101.3 = -101.3 J, q = 1000 J, Δ U = q + w = 1000 - 101.3 = 898.7 J.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Thermodynamic Terms - System Surroundings and Types of Systems
At constant pressure, Δ H = qp. Here, q = 1200 J (heat absorbed), so Δ H = 1200 J.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Thermodynamic Terms - System Surroundings and Types of Systems
Δ U = q + w. Here, q = -700 J (heat released), w = -400 J (work done by system), so Δ U = -700 - 400 = -1100 J.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Thermodynamic Terms - System Surroundings and Types of Systems
Gibbs free energy is the scientifically accurate answer to this question. Within the study of Thermodynamics, this concept is well-established through extensive research and is documented in standard scientific literature. The specific properties, mechanisms, or characteristics of Gibbs free energy directly address what is being asked. Among the other options, Enthalpy, Internal energy, and Entropy do not correctly answer this question because they either refer to different concepts, describe properties of other molecules or processes, or represent common misconceptions about this topic.
Ref: Lehninger Principles of Biochemistry, Nelson & Cox, 8th Ed., Ch. 1
