T–s diagram interpretation in thermodynamics Consider any thermodynamic process plotted on a temperature–entropy (T–s) diagram. The shaded area under the process curve is sometimes misinterpreted. Does the area under a T–s curve represent the work done during the process, or something else? Choose the correct judgment about the statement: “The area under the T–s curve of any thermodynamic process represents the work done during the process.”

Mechanical Engineering Thermodynamics Difficulty: Easy
Choose an option
  • A
    Agree
  • B
    Disagree
  • C
    Agree only for ideal gases
  • D
    Agree only for reversible adiabatic processes
  • E
    Agree only when plotted at constant volume

Answer

Correct Answer: Disagree

Explanation

Introduction / Context:Thermodynamic diagrams encode energy transfers geometrically. It is crucial to know which plane corresponds to heat and which to work. Students often confuse the meaning of areas on P–v and T–s plots, leading to wrong conclusions about work and heat in cycles and processes.

Given Data / Assumptions:

  • A general, possibly irreversible, process between two states.
  • Conventional axes: temperature (T) on ordinate, entropy (s) on abscissa.
  • Work and heat are path functions; interpretation depends on the chosen diagram.

Concept / Approach:

On a pressure–volume (P–v) diagram, the area under the curve equals boundary work: W = ∫ P dv. On a temperature–entropy (T–s) diagram, the area under the curve equals the heat transfer for a reversible path: Q_rev = ∫ T ds. Therefore, the statement that the T–s area represents work is incorrect; it represents reversible heat, not work. Work can be inferred on other planes (e.g., P–v, h–s with device models), but not directly from T–s area.

Step-by-Step Solution:

Recognize diagram meaning: P–v area → work; T–s area → reversible heat.Write: Q_rev = ∫ T ds; W = ∫ P dv.Conclude: The given statement is false for general processes.

Verification / Alternative check:

In a reversible adiabatic process (ds = 0), area under the T–s line is zero while work is nonzero; this counterexample directly disproves the claim that T–s area is work.

Why Other Options Are Wrong:

“Agree” variants: Conflate heat with work; even for ideal gases or constant-volume paths, T–s area tracks Q_rev, not W.

Common Pitfalls:

Assuming one diagram's area rule applies universally. Always match the plane to the physical quantity represented by its integral.

Final Answer:

Disagree

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