🔥 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.
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
For cycle, ΔU = 0
So Q = W
Since W = 0 ⇒ Net heat = 0
But heat absorbed ≠ heat rejected
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
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(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:
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
✔ 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)
➡️ Next: Stage 3 – Page 3 (Extreme JEE Advanced PYQs & Multi-Concept)
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🔥 Thermodynamics – Objective Final Revision
Thermodynamics – Set 1
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