A Carnot engine operates between a heat source at 600 K and a heat sink at 300 K. If the engine absorbs 1200 J of heat from the source, calculate (i) the efficiency of the engine and (ii) the work done by the engine.
📋 Given
T_H = 600 K (hot source)
T_C = 300 K (cold sink)
Q₁ = 1200 J (heat absorbed)
🎯 What Examiner Tests
Application of η = 1 − T_C/T_H
Using η = W/Q₁ to find work
Temperature in Kelvin (not Celsius)
✏️ Solution
η = 1 − T_C/T_H = 1 − 300/600
η = 1 − 0.5 = 0.5 = 50%
W = η × Q₁ = 0.5 × 1200
W = 600 J
⚡ Shortcut Insight
When T_H = 2T_C, efficiency is always 50%. No calculation needed. Memorize: ratio determines efficiency. If T_C = 0, η = 100% — but Third Law says T = 0K is unreachable.
Internal Energy in Isothermal Expansion
An ideal gas undergoes isothermal expansion at temperature T. Determine the change in internal energy, heat absorbed, and work done by the gas in terms of T and relevant constants.
🧠 Concept Selection
Isothermal → T = constant → for ideal gas, ΔU = f(T) only
Since ΔT = 0 → ΔU = 0
First Law: Q = ΔU + W = 0 + W = W
✏️ Solution
ΔU = 0 (isothermal, ideal gas)
W = nRT ln(V₂/V₁)
Q = W = nRT ln(V₂/V₁)
Common Mistake Alert
Many students say "isothermal means no heat exchange." That is ADIABATIC, not isothermal. Isothermal means constant temperature — heat exchange DOES happen to keep T constant.
Diatomic Gas in Two Processes
2 moles of a diatomic ideal gas undergo two processes: (A) Isobaric heating at P = 2 × 10⁵ Pa from V₁ = 10 L to V₂ = 20 L. (B) Isochoric cooling from state 2 back to original temperature. Find: (i) Work done in process A, (ii) ΔU in process A, (iii) Heat in process A, (iv) ΔU in process B.
📋 Given & Setup
n = 2 mol, diatomic (f = 5, Cv = 5R/2, Cp = 7R/2, γ = 7/5)
Process A: P = 2×10⁵ Pa, V₁ = 10 L = 0.01 m³, V₂ = 20 L = 0.02 m³
From PV = nRT: T₁ = PV₁/nR = (2×10⁵ × 0.01)/(2 × 8.314) = 120.3 K
T₂ = PV₂/nR = (2×10⁵ × 0.02)/(2 × 8.314) = 240.6 K → ΔT = 120.3 K
✏️ Process A (Isobaric)
W_A = PΔV = 2×10⁵ × (0.02 − 0.01) = 2000 J
ΔU_A = nCvΔT = 2 × (5/2 × 8.314) × 120.3 = 5003 J ≈ 5000 J
Q_A = ΔU + W = 5000 + 2000 = 7000 J
Check: Q_A = nCpΔT = 2×(7R/2)×120.3 ≈ 7000 J ✓
✏️ Process B (Isochoric)
V = constant → W_B = 0
Cools back to T₁ → ΔT_B = −120.3 K
ΔU_B = nCvΔT = 2×(5R/2)×(−120.3) ≈ −5000 J
⚡ Shortcut Insight
Notice: ΔU_A + ΔU_B = 5000 + (−5000) = 0. Makes sense — the gas returns to T₁ in process B. State functions (U, T) have zero net change in a complete cycle-like path.
PV Diagram Work Calculation
In a PV diagram, a gas undergoes a rectangular cycle: A(P₀, V₀) → B(2P₀, V₀) → C(2P₀, 2V₀) → D(P₀, 2V₀) → A. Calculate the net work done by the gas in one complete cycle.
2P₀ │ B----C
│ │ │
P₀ │ A----D
└────────── V
V₀ 2V₀
Clockwise rectangular cycle on PV diagram
✏️ Solution
Net work = Area enclosed in PV diagram
Width = ΔV = 2V₀ − V₀ = V₀
Height = ΔP = 2P₀ − P₀ = P₀
Area = ΔP × ΔV = P₀ × V₀ = P₀V₀
Clockwise → W_net = +P₀V₀ (engine)
Exam Insight — Graph Problems
"This is where most students lose marks." For rectangular PV cycles, W = ΔP × ΔV. For triangular cycles, W = ½ × base × height. For irregular curves, you integrate. JEE Advanced 2020 had a problem with overlapping isothermal and adiabatic — always compare slopes to identify the process.
Efficiency and Temperature
Assertion (A): The efficiency of a Carnot engine increases if the temperature of the cold reservoir is decreased, while the temperature of the hot reservoir remains unchanged.
Reason (R): The efficiency of Carnot engine is given by η = 1 − T_C/T_H.
✏️ Analysis
If T_C decreases (T_H fixed), T_C/T_H decreases → (1 − T_C/T_H) increases → η increases. ✓ TRUE
η = 1 − T_C/T_H is the correct Carnot formula. ✓ TRUE
R correctly explains A. Both are true and R is the correct explanation.
⚡ Shortcut for Assertion-Reason
Always verify: (1) Is A true? (2) Is R true? (3) Does R explain A? If both true but R doesn't explain A → answer is "both true, R not correct explanation." This 3-step check prevents silly errors.
Adiabatic vs Isothermal Expansion
An ideal monoatomic gas starts at state (P₀, V₀, T₀). It undergoes two separate expansions to volume 2V₀: Path 1 (Isothermal), Path 2 (Adiabatic). Answer the following: (a) Which path results in lower final pressure? (b) Which path does more work? (c) Which path results in lower final temperature?
✏️ Analysis
Isothermal: P₁_final = P₀V₀/2V₀ = P₀/2
Adiabatic (γ=5/3): P₂_final = P₀(V₀/2V₀)^γ = P₀ × (1/2)^(5/3) = P₀/3.17
P₂_final < P₁_final → Adiabatic has lower pressure
Area under isothermal > adiabatic (isothermal is less steep)
Isothermal does more work
Isothermal: T remains T₀. Adiabatic: T falls below T₀.
Adiabatic has lower final temperature
Thinking Step
In isothermal, heat from surroundings helps the gas maintain temperature and do more work. In adiabatic, the gas uses its own internal energy for work — temperature drops. This is the physical reasoning behind all 3 answers.