JEE Main · 2019 · Shift-IeasyTHERMO-015

Which one of the following equations does not correctly represent the first law of thermodynamics for the given…

Thermodynamics · Class 11 · JEE Main Previous Year Question

Question

Which one of the following equations does not correctly represent the first law of thermodynamics for the given processes involving an ideal gas? (Assume non-expansion work is zero)

Options
  1. a

    Isochoric process: ΔU=q\Delta U = q

  2. b

    Isochoric process: q=wq = -w

  3. c

    Cyclic process: q=wq = -w

  4. d

    Adiabatic process: ΔU=w\Delta U = -w

Correct Answerb

Isochoric process: q=wq = -w

Detailed Solution

🧠 Which first-law equation doesn't belong to the isochoric process? For an isochoric (constant volume) process: w=0w = 0, so ΔU=q\Delta U = q. The equation q=wq = -w describes a cyclic process, not an isochoric one.

🗺️ Checking each option (a) Isochoric: ΔU=q\Delta U = q — Correct. w=0w = 0, so ΔU=q\Delta U = q. ✓

(b) Isochoric: q=wq = -wWrong. For isochoric, w=0w = 0, so this becomes q=0q = 0. That would mean no heat exchange at all — only true if the process is also adiabatic. ✗

(c) Cyclic: q=wq = -w — Correct. ΔU=0\Delta U = 0 for cyclic, so q=wq = -w. ✓

(d) Adiabatic: ΔU=w\Delta U = -w — Correct. q=0q = 0, and using the convention where ww = work done by system, ΔU=w\Delta U = -w. ✓

⚠️ Trap: Option (b) copies the cyclic-process equation and labels it as isochoric. The equations look similar but apply to completely different processes.

Answer: (b)\boxed{\text{Answer: (b)}}

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