Thermodynamics

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Zeroth, First, and Second Laws — The Tripod:

Zeroth Law of Thermodynamics: If two systems A and B are each in thermal equilibrium with a third system C, then A and B are in thermal equilibrium with each other. This defines temperature and is the basis for using thermometers.

First Law of Thermodynamics: Energy conservation applied to thermodynamic systems. ΔU = Q - W, where ΔU is the change in internal energy of the system, Q is heat added to the system (positive when heat enters), and W is work done by the system on the surroundings (positive when system expands against external pressure). This is the energy accounting equation.

Second Law of Thermodynamics: Two equivalent formulations — (1) Kelvin-Planck statement: no heat engine can convert all absorbed heat into work (no 100% efficient engine), (2) Clausius statement: no refrigerator can transfer heat from a cold body to a hot body without external work input. Both statements express the same truth: natural processes have a preferred direction — toward disorder (higher entropy).

Thermodynamic Processes — Quick Reference:

Process	Condition	Key Relation	Work Done
Isothermal	ΔT = 0, ΔU = 0	P₁V₁ = P₂V₂	W = nRT ln(V₂/V₁)
Adiabatic	ΔQ = 0	PV^γ = constant	W = (P₁V₁ - P₂V₂)/(γ - 1)
Isochoric	ΔV = 0	P/T = constant	W = 0
Isobaric	ΔP = 0	V/T = constant	W = PΔV = nRΔT

⚠️ Note: Always use absolute temperature (Kelvin). Conversion: T(K) = T(°C) + 273.

⚡ JEE Tip: For cyclic processes, ΔU = 0 (since the system returns to its initial state). Therefore Q_net = W_net = area enclosed by the cycle on a P-V diagram. For a clockwise cycle on a P-V diagram, the system does net work on surroundings (positive W); for counter-clockwise, net work is done on the system (negative W, or equivalently, heat flows out).

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🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Carnot Cycle — The Ideal Engine:

The Carnot cycle consists of four reversible processes:

1. Isothermal expansion at temperature T_hot: absorbs heat Q_hot from the hot reservoir, does work W₁, internal energy unchanged
2. Adiabatic expansion: no heat exchange, temperature falls from T_hot to T_cold, does work W₂
3. Isothermal compression at T_cold: rejects heat Q_cold to the cold reservoir, work W₃ done on system
4. Adiabatic compression: no heat exchange, temperature rises from T_cold to T_hot, work W₄ done on system

Carnot efficiency: η = 1 - T_cold/T_hot (all temperatures in Kelvin). This is the MAXIMUM possible efficiency for any heat engine operating between two temperatures — no real engine can exceed this. Carnot efficiency depends ONLY on the temperatures, not on the working substance.

Entropy — The Second Law Quantified:

Entropy S is a state function (path-independent). For a reversible process: ΔS = ∫(dQ_rev/T). For an irreversible process, ΔS_universe = ΔS_system + ΔS_surroundings > 0 (total entropy always increases).

Physical meaning: entropy measures the number of microscopic configurations (Ω) corresponding to a macroscopic state: S = k_B ln Ω (Boltzmann equation). Higher entropy means more disorder, more possible microstates. This connects thermodynamics to statistical mechanics.

⚡ JEE Tip: For mixing two ideal gases (same temperature and pressure), the entropy increase is ΔS = n₁R ln(V_final/V₁) + n₂R ln(V_final/V₂). This is positive because V_final > V₁ and V_final > V₂, showing irreversibility. Even though the mixing is spontaneous and natural, it increases total entropy — consistent with the second law.

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Adiabatic Process — Derivation and Characteristics:

For an adiabatic process (Q = 0), using the first law: ΔU = -W. For an ideal gas: ΔU = nC_VΔT and W = ∫P dV. Using the ideal gas equation PV = nRT, we can derive:

For a monatomic gas (C_V = 3R/2, γ = C_P/C_V = 5/3): PV^γ = constant. For a diatomic gas (C_V = 5R/2, γ = 7/5 at room temperature): PV^γ = constant.

Derivation: From first law, nC_V dT = -P dV. Using PV = nRT, P = nRT/V. So nC_V dT = -(nRT/V) dV. Dividing by nRT: (C_V/R)(dT/T) = -dV/V. Since C_V/R = 1/(γ - 1), we get dT/T = -(γ - 1) dV/V. Integrating: ln T = -(γ - 1) ln V + constant → T V^(γ-1) = constant. Since PV = RT, P V^γ = constant. QED.

Van der Waals Equation — Real Gas Corrections:

The ideal gas equation PV = nRT assumes point particles with no volume and no intermolecular forces. The Van der Waals equation corrects for both: (P + an²/V²)(V - nb) = nRT, where:

• a corrects for intermolecular attraction (reduces pressure)
• b corrects for finite molecular volume (reduces available volume)

At low pressure and high temperature, Van der Waals → ideal gas. For CO₂, a = 3.59 L²bar/mol², b = 0.0427 L/mol. At the critical point, these corrections determine the liquefaction behaviour — above the critical temperature T_c, no amount of pressure can liquefy the gas.

Clausius Inequality:

∮(dQ/T) ≤ 0, where the equality holds for a completely reversible cycle and the strict inequality for any irreversible cycle. This is the most general statement of the second law. It implies that for a reversible process, ΔS = ∮(dQ_rev/T) = 0, confirming entropy is a state function.

Heat Pump — Coefficient of Performance:

A refrigerator or heat pump is essentially a heat engine run in reverse. Using external work, it transfers heat from a cold reservoir to a hot reservoir. Coefficient of Performance (COP) for a refrigerator: COP = Q_cold/W_input = T_cold/(T_hot - T_cold). For an ideal (Carnot) refrigerator. Higher COP means more efficient cooling per unit work input. Residential air conditioners typically have COP ≈ 2.5–3.5.

⚡ JEE Advanced Pattern: JEE Advanced frequently combines thermodynamics with kinetic theory (PV diagrams for Carnot cycles), with mechanics (work done in expansion against variable external pressure), and with calorimetry. The 2021 Paper 2 Qn 44 involved a cyclic process with two isochoric and two isobaric segments — students needed to calculate net work done (area inside the cycle), net heat added (using ΔU = 0 for the cycle), and the efficiency compared to a Carnot engine operating between the same temperature extremes. The key insight: for any cyclic process, the efficiency η = W_net/Q_hot < η_Carnot (when both operate between the same temperatures).