🔥 Thermodynamics – Stage 3
Page 4 : Tough Numerical PYQs with Full Logical Solutions
This page contains calculation-heavy + concept-integrated questions exactly at JEE Advanced level.
Problem 1️⃣ : Adiabatic Expansion – Temperature Drop
An ideal gas expands adiabatically from volume V to 2V.
Initial temperature = T.
Key Relation:
T·Vγ−1 = constant
T·Vγ−1 = constant
Final Temperature:
Tf = T / 2(γ−1)
Tf = T / 2(γ−1)
❌ Do not use isothermal logic (T constant)
Problem 2️⃣ : Work Done in Adiabatic Process
Calculate work done during adiabatic expansion.
Formula:
W = (P₁V₁ − P₂V₂)/(γ − 1)
W = (P₁V₁ − P₂V₂)/(γ − 1)
❌ Using W = nRT ln(V₂/V₁) is strictly wrong here
Problem 3️⃣ : Heat Supplied but Temperature Constant
Gas absorbs heat but temperature does not change.
ΔU = 0 ⇒ Q = W
❌ Many students think ΔT ≠ 0 whenever Q ≠ 0
Problem 4️⃣ : Cyclic Process – Net Work
Gas undergoes a closed PV cycle.
Net Work = Area enclosed in PV diagram
❌ Work is NOT zero just because initial = final state
Problem 5️⃣ : Mixing of Two Gases (Entropy)
Two gases at same T and P are mixed.
Entropy Change:
ΔS = nR ln(Vfinal/Vinitial)
ΔS = nR ln(Vfinal/Vinitial)
❌ Even identical gases increase entropy
Problem 6️⃣ : Refrigerator Numerical
Refrigerator operates between T₁ and T₂.
COP = T₂ / (T₁ − T₂)
❌ Students confuse COP with efficiency
Problem 7️⃣ : Free Expansion (Classic Trap)
Gas expands freely into vacuum.
Q = 0, W = 0 ⇒ ΔU = 0 ⇒ ΔT = 0
❌ Volume change does NOT imply temperature change
Problem 8️⃣ : Temperature vs Internal Energy
Same temperature, different gases.
Internal energy depends on:
✔ Temperature
✔ Degrees of freedom
✔ Temperature
✔ Degrees of freedom
Problem 9️⃣ : Concept Linking Test
Examiner checks if you can link:
- First Law
- PV relations
- Entropy condition
✅ Stage 3 – Page 4 Completed
➡️ Next: Stage 3 – Page 5 (Extreme Traps & One-Line Kill Techniques)
➡️ Next: Stage 3 – Page 5 (Extreme Traps & One-Line Kill Techniques)
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