🔥 Thermodynamics – Stage 3

Page 2 : Hard Numerical Problems & Strategy

This page trains you to think like a JEE Advanced examiner.

Problem 1️⃣ : Variable Pressure Process

A gas expands such that pressure varies with volume as
P = kV from volume V₁ to V₂.
Strategy:
  • Recognize non-constant pressure
  • Work = ∫P dV
W = ∫ kV dV = (k/2)(V₂² − V₁²)
❌ Using W = PΔV here gives wrong answer.

Problem 2️⃣ : Zero Net Work but Heat Exchange

A gas undergoes a cyclic process. Net work = 0.
Question: Is heat exchange zero?
Key Idea:
For cycle, ΔU = 0
So Q = W
Since W = 0 ⇒ Net heat = 0
But heat absorbed ≠ heat rejected

Problem 3️⃣ : Free Expansion (Classic Trap)

Gas expands freely into vacuum.
Instant Results:
  • W = 0
  • Q = 0
From first law:
ΔU = 0
For ideal gas → Temperature remains constant
❌ This is NOT isothermal process (no heat exchange)

Problem 4️⃣ : Adiabatic Compression

Gas compressed adiabatically.
Use Relations:
  • PVγ = constant
  • TVγ−1 = constant
❌ Do not use ideal gas equation alone.

Problem 5️⃣ : Entropy Based Feasibility

A system absorbs heat Q at temperature T from reservoir.
ΔS(system) = Q / T
ΔS(reservoir) = −Q / T
ΔS(universe) = 0 → reversible
Any deviation ⇒ irreversible

6️⃣ Two-Process Comparison

Same initial & final states, different paths.
ΔU is same for both.
❌ Work & heat depend on path.

7️⃣ Examiner’s Favorite Trick

If pressure, volume & temperature all change:
Examiner wants you to:
  • Identify dominant law
  • Ignore useless data

8️⃣ Rank Booster Tip

When stuck:
✔ Write symbolic equations first
✔ Substitute limits later
✔ Units check saves marks

✅ Stage 3 – Page 2 Completed
➡️ Next: Stage 3 – Page 3 (Extreme JEE Advanced PYQs & Multi-Concept)

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