🔥 Thermodynamics – Stage 3

Page 10 : Mixed Integer & Conceptual Numericals (JEE Advanced)

This page tests calculation + judgment. You must decide the formula before inserting numbers.

Q1️⃣ Integer Type – Isothermal Expansion

One mole of ideal gas expands isothermally at 300 K from 1 L to 4 L. Calculate work done (in units of R·T).
W = nRT ln(V₂/V₁) = RT ln(4) ✔ Integer answer: ln(4)

Q2️⃣ Conceptual – Adiabatic Compression

During adiabatic compression of ideal gas:
✔ Temperature increases ✔ Pressure increases ✔ Internal energy increases ❌ Heat exchange = 0

Q3️⃣ Integer – Free Expansion

One mole ideal gas expands freely into vacuum. Calculate ΔU (in joules).
✔ Q = 0 ✔ W = 0 ✔ ΔU = 0 ✔ Integer answer: 0

Q4️⃣ Mixed – Cyclic Process

A gas completes a cycle on P–V diagram. Net heat absorbed is 500 J.
✔ ΔU = 0 (cycle) ✔ Q = W ✔ Work done = 500 J

Q5️⃣ Conceptual – Entropy

Which process always increases entropy?
✔ Free expansion ✔ Irreversible processes ❌ Reversible adiabatic (ΔS = 0)

Q6️⃣ Integer – Heat Engine

Heat engine absorbs 600 J and rejects 400 J. Find efficiency × 100.
η = (600−400)/600 = 1/3 ✔ Answer = 33

Q7️⃣ Multi-Concept – Ideal Gas + First Law

In an isobaric process, heat supplied is 1000 J and work done is 400 J.
ΔU = Q − W = 1000 − 400 ✔ ΔU = 600 J

Q8️⃣ Conceptual Trap – Specific Heat

Which is greater for ideal gas?
✔ Cp > Cv ✔ Cp − Cv = R

Q9️⃣ Integer – Adiabatic Relation

In adiabatic expansion, if volume doubles, temperature changes by factor?
T ∝ V1−γ ✔ Answer depends on γ (monoatomic: γ = 5/3)

Q🔟 JEE Advanced Mindset Question

A process is slow but irreversible. Is it quasi-static?
❌ Slow ≠ reversible ✔ Irreversibility breaks quasi-static condition

✅ Stage 3 – Page 10 Completed
🎯 Thermodynamics (Full JEE Track) Finished

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