Thermodynamics

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Three Laws and a State Function:

Zeroth Law: Establishes the concept of temperature. If A is in thermal equilibrium with B, and B is in thermal equilibrium with C, then A is in thermal equilibrium with C. This allows us to define thermometers and temperature scales.

First Law — Energy Conservation: ΔU = Q - W. The change in internal energy of a system equals heat added minus work done by the system. Heat and work are path-dependent (process-dependent), but internal energy is a state function — it depends only on the initial and final states.

Second Law — Direction of Natural Processes: Heat flows spontaneously from hot to cold, not the reverse. No heat engine can be 100% efficient (Kelvin-Planck statement). No refrigerator can operate without external work (Clausius statement). Both are equivalent expressions of the same fundamental truth.

⚡ CUET Tip: The efficiency of a heat engine η = W/Q_hot = (Q_hot - Q_cold)/Q_hot = 1 - Q_cold/Q_hot. Always convert temperatures to Kelvin: T(K) = T(°C) + 273. Never substitute °C temperatures directly into Carnot efficiency formula.

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🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Thermodynamic Processes Summary:

Isothermal (constant temperature, ΔT = 0): For an ideal gas, ΔU = 0, so Q = W = nRT ln(V₂/V₁). Since T is constant, P₁V₁ = P₂V₂ (Boyle’s law). Work done equals heat absorbed — the system does work by expanding using internal energy, but gains exactly the same amount of heat from the reservoir to keep temperature constant.

Adiabatic (no heat exchange, ΔQ = 0): Work done equals change in internal energy: W = -ΔU = nC_V(T₁ - T₂). For an ideal gas: PV^γ = constant, TV^(γ-1) = constant, TP^(1-γ) = constant. Adiabatic curves on P-V are steeper than isotherms because pressure drops faster as volume increases (no heat input to maintain pressure).

Isochoric (constant volume, ΔV = 0): W = 0, so ΔU = Q. All heat added goes into increasing internal energy. Pressure increases directly with temperature: P/T = constant.

Isobaric (constant pressure): W = PΔV = nRΔT. More heat is required for a given temperature change than in an isochoric process because some heat does work of expansion while the rest increases internal energy.

Carnot Cycle and Efficiency:

An ideal (theoretical) Carnot engine between temperatures T_hot and T_cold (in Kelvin) has efficiency η_Carnot = 1 - T_cold/T_hot. This is the maximum possible efficiency. No real engine operating between the same temperatures can exceed this. The Carnot cycle consists of two isotherms and two adiabats — all reversible processes, hence unattainable in practice with finite time.

⚡ CUET Tip: For reversible processes, entropy change of the universe = 0. For irreversible processes, ΔS_universe > 0. The Second Law can be expressed as: natural processes always increase the total entropy of the universe.

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Entropy — Quantitative Definition:

For a reversible process: ΔS = ∫(dQ_rev/T). This is path-independent for reversible paths — the entropy change between two equilibrium states is the same regardless of the reversible path taken.

For an irreversible process: calculate ΔS by finding a reversible path between the same initial and final states, then computing ΔS along that reversible path.

Physical interpretation: S = k_B ln Ω, where Ω is the number of microstates consistent with the macrostate. Boltzmann’s entropy formula connects thermodynamics to statistical mechanics. For mixing two ideal gases at the same temperature and pressure, ΔS_mixing = n₁R ln(V₁_final/V₁_initial) + n₂R ln(V₂_final/V₂_initial) — this is positive because volume increases for each gas, demonstrating that mixing is irreversible and entropy increases.

Adiabatic Derivation for Ideal Gases:

Starting from the first law with dQ = 0: dU + PdV = 0. For n moles of ideal gas: dU = nC_V dT and PV = nRT. So nC_V dT + (nRT/V) dV = 0. Dividing by nRT: (C_V/R)(dT/T) + dV/V = 0. Since C_V/R = 1/(γ - 1), we get dT/T + (γ - 1) dV/V = 0. Integrating: ln T + (γ - 1) ln V = constant → T V^(γ-1) = constant. Substituting T = PV/R: PV^γ = constant. QED.

Heat Pumps and Refrigerators:

A refrigerator uses work input to transfer heat from a cold reservoir to a hot reservoir. COP (Coefficient of Performance) = Q_cold/W_input = T_cold/(T_hot - T_cold) for a Carnot refrigerator. Residential refrigerators have COP ≈ 2.5–5. An air conditioner works on the same principle but is sized for room-scale cooling.

⚡ CUET Pattern: The most frequently tested CUET thermodynamics concepts are: (1) identifying isothermal vs adiabatic curves on P-V diagrams (adiabatic is steeper), (2) calculating work done from P-V graph areas, (3) Carnot efficiency comparisons, and (4) entropy change in reversible processes. CUET 2022 Section II Qn 87 asked students to identify which process had maximum work done for a given initial state — the answer was the isothermal expansion (because it does work while absorbing heat, whereas adiabatic does work without heat input, so the P-V curve is lower).