IITJEE: Thermodynamics P-V Indicator Diagrams & Carnot Engine

Published on

in

Apex Class • iitJEE Thermal Physics Lab

Thermodynamics P-V Indicator Diagrams & Carnot Engine Lab

Compare Isobaric, Isothermal, Adiabatic & Isochoric curves and master Carnot Efficiency (η)!

🔬 Expand from State A (P₀, V₀ = 150, 60) to Volume 2V₀. Select Process:
Pressure (P) P₀ P₀/2 P₀/2^γ V₀ 2V₀ Volume (V) → MAX AREA W = P₀ ΔV = nRΔT 1. Isobaric (Slope dP/dV = 0) Isobaric Energy Split (ΔQ = ΔU + W) Ratio ΔQ : ΔU : W = C_p : C_v : R ΔU = (1/γ) ΔQ (60%) W (40%) • Work Fraction: W / ΔQ = 1 – (1/γ) • Monoatomic (γ=5/3): W = ⅖ ΔQ, ΔU = ⅗ ΔQ 🏆 JEE Shortcut: In Isobaric expansion, W / ΔQ = R / C_p = (γ – 1) / γ (For Diatomic γ=7/5 → W = ²⁄₇ ΔQ)! W = nRT ln(V₂/V₁) 2. Isothermal (Slope = -P/V) Isothermal Process (ΔT = 0) ΔU = n C_v ΔT = 0 J! ΔQ = W = 2.303 nRT log₁₀(V₂/V₁) • Bulk Modulus: B_iso = P • Molar Heat Capacity: C = ΔQ/(nΔT) = ∞ 🌡️ Since Temperature T is constant, Internal Energy ΔU = 0, so 100% of added heat becomes Work Done (ΔQ = W)! Isothermal (Slope = -P/V) 3. Adiabatic (γ× Steeper!) Adiabatic Process (ΔQ = 0) Slope = -γ (P / V) = γ × Slope_iso! W = -ΔU = (P₁V₁ – P₂V₂) / (γ – 1) • Equations: PV^γ = const | TV^(γ-1) = const • Bulk Modulus: B_adia = γ P | C_adia = 0 ⚡ #1 JEE Graph Question: Where Isothermal & Adiabatic curves intersect, the Adiabatic curve is always γ times steeper! 4. Isochoric (Vertical Line: ΔV = 0) Zero Area Under Line → W = 0 J! Isochoric Process (ΔV = 0) Work Done W = ∫ P dV = 0 J! ΔQ = ΔU = n C_v ΔT = (f/2) nRΔT • Gay-Lussac’s Law: P ∝ T • Slope on P-V Graph: dP/dV = ∞ (Vertical) 🔒 Rigid Container Rule: Whenever a gas is heated in a closed rigid vessel, ΔV = 0, so W = 0 and ΔQ = nC_vΔT!
Governing Equation
V / T = Constant P V = Constant P V^γ = Constant P / T = Constant
Ideal Gas Law Form
Work Done (W = ∫ P dV)
P ΔV = n R ΔT (MAX) nRT ln(V₂ / V₁) (P₁V₁ – P₂V₂) / (γ – 1) 0 Joules (ΔV = 0)
W_isobaric > W_iso > W_adia > 0
P-V Curve Slope (dP/dV)
0 (Horizontal Flat) – (P / V) – γ (P / V) [Steeper!] ∞ (Vertical Line)
Slope_adia = γ × Slope_iso
Molar Heat Capacity (C)
C_p = (f/2 + 1) R C_iso = ∞ (Infinite) C_adia = 0 (Zero) C_v = (f/2) R
Polytropic PVⁿ: C = C_v + R/(1-n)
🔥 Carnot Efficiency η = 1 – (T_C / T_H) = W / Q_H. Select Carnot Scenario:
P V → 1. Isothermal (T_H: +Q_H ↘) 2. Adiabatic (Q=0) 3. Isothermal (T_C: -Q_C ↙) 4. Adiabatic A(V₁) B(V₂) C(V₃) D(V₄) Loop Area = W_net Carnot Cycle Laws (ΔU_cycle = 0) Q_C / Q_H = T_C / T_H Volume Ratio: V₂ / V₁ = V₃ / V₄ • Clockwise Loop → W_net > 0 (Engine) • Anti-Clockwise → W_net < 0 (Heat Pump) 🔄 Carnot Theorem: No heat engine operating between T_H and T_C can ever exceed Carnot efficiency η = 1 – T_C/T_H! HOT SOURCE (T_H) 600 K (327°C) Supplies Q_H = 1000 J Q_H = 1000 J CARNOT η = 50% Useful Work W = 500 J! Q_C = 500 J COLD SINK (T_C) 300 K (27°C) Absorbs Q_C = 500 J ⚠️ Mandatory JEE Rule: ALWAYS convert Celsius (°C) to Kelvin (K = °C + 273) before using η = 1 – T_C/T_H! Starting at T_H = 600 K, T_C = 300 K (η₀ = 50%): Which gives higher efficiency—Heating Source or Cooling Sink by 100 K? Option A: Increase T_H by +100 K T_H = 700 K, T_C = 300 K η_A = 1 – 300/700 = 57.1% (Modest +7.1% Gain) 🏆 Option B: Decrease T_C by -100 K T_H = 600 K, T_C = 200 K η_B = 1 – 200/600 = 66.7%! (Massive +16.7% Gain — ALWAYS WINS!) 🎯 Classic JEE Statement Question: Lowering Sink T_C by ΔT is ALWAYS more effective than raising Source T_H by the same ΔT! HOT ROOM (T_H) 300 K (27°C) Receives Q_H = Q_C + W ← Rejects Q_H FRIDGE COP = 9.0 Compressor Work Input W ↑ ← Extracts Q_C FREEZER (T_C) 270 K (-3°C) COP = 270 / (300 – 270) = 9 ❄️ Refrigerator Formula: COP (β) = Q_C / W = T_C / (T_H – T_C) | Relation with η: β = (1 – η) / η!
Carnot Efficiency (η)
η = 1 – (T_C / T_H)
Also η = W / Q_H = 1 – Q_C/Q_H
Heat Ratio Law
Q_C / Q_H = T_C / T_H
Valid ONLY for Reversible Cycles
Refrigerator COP (β)
β = Q_C / W = T_C / ΔT
Where W = Q_H – Q_C
Degrees of Freedom (γ)
γ = 1 + (2 / f)
Mono: 5/3 | Dia: 7/5 | Poly: 4/3
🔥 3 Fast JEE Thermodynamics Shortcuts:
1. Sign Convention (Physics): Expansion (ΔV > 0) → W is Positive (+); Compression (ΔV < 0) → W is Negative (-).
2. Cyclic Process Shortcut: In any closed loop, ΔU_cycle = 0, so Total Heat Q_net = W_net = Area of Loop (Positive if Clockwise, Negative if Anti-Clockwise).
3. Gas Mixture γ_mix: Use (n₁ + n₂)/(γ_mix – 1) = n₁/(γ₁ – 1) + n₂/(γ₂ – 1) — never take a simple average of γ!

Leave a Reply


Apex Class App

Play revision games, read notes, and test your skills anywhere!


🚀

Fresh Updates

Latest notes, lectures & announcements!

Loading latest posts…
View All Posts ➔

Discover more from Apex Class

Subscribe now to keep reading and get access to the full archive.

Continue reading

Enable Notifications OK No thanks