Class 11 · Chapter 12

Thermodynamics

Overview, notes, short notes, formula sheet, daily practice problems, previous year questions, and videos for this chapter — all in one place.

Thermodynamics Short Notes

6 min read

Condensed revision points for Thermodynamics — for quick recall before exams, not a substitute for the full notes.

PV = μRT. State fixed by ANY 2 of (P, V, T) — third follows automatically.
Homogeneous: uniform throughout. Heterogeneous: not uniform. Isolated: no exchange of matter OR energy.
ΔU = U_f − U_i depends ONLY on initial/final states — NOT on path (state function). Q and W DO depend on path.

dU = dU_k + dU_p. Ideal gas (no intermolecular forces): dU = dU_k = μC_vdT.
Internal energy of an ideal gas depends ONLY on temperature — never on volume or pressure directly.

Heat gained: +. Heat lost: −. Work done BY system: +. Work done ON system: −. ΔU increase: +, decrease: −.
W = ∫PdV = area under the P–V curve. V increasing ⇒ W > 0. V decreasing ⇒ W < 0. V constant ⇒ W = 0.
Cyclic process: net W = enclosed area of the loop. Clockwise loop: W > 0 (+area). Anticlockwise: W < 0 (−area).

dQ = dU + dW (Q = ΔU + W) — conservation of energy applied to heat/work/internal energy.
dQ − dW is path-independent (= dU) even though dQ and dW individually ARE path-dependent.
Applies to EVERY process, all 3 phases of matter. Q, ΔU, W must be in the SAME units.
Melting: Q = ΔU + PΔV (ΔV negligible ⇒ mostly ΔU). Boiling: mL = (U_v−U_l) + P(V_v−V_l) (work term significant).

V = constant. P/T = constant. W = 0 always.
Q = ΔU = μC_vΔT — ALL heat goes to internal energy, none to work.
P–V slope = ∞ (vertical line). Examples: rigid sealed container, pressure cooker heating.

P = constant. V/T = constant. W = PΔV = μRΔT.
Q = ΔU + W = μC_vΔT + μRΔT = μC_PΔT.
ΔU/Q = 1/γ ; W/Q = 1−1/γ. P–V slope = 0 (horizontal line).
Examples: heating water/boiling at atmospheric pressure, melting at atmospheric pressure.

T = constant. PV = constant (Boyle's law) — isotherm is a HYPERBOLA.
ΔU = 0 always (ideal gas, U depends only on T) ⇒ Q = W entirely.
W = μRT ln(V₂/V₁) = 2.303μRT log₁₀(V₂/V₁) = 2.303μRT log₁₀(P₁/P₂).
Isotherms of the same gas at different T NEVER intersect. Slope dP/dV = −P/V.
Examples: melting, boiling, very slow process in a perfectly conducting cylinder.

Q = 0 (insulated or too fast for heat exchange). PV^γ = const; TV^(γ−1) = const; T^γP^(1−γ) = const.
dU = −dW. Expansion ⇒ gas COOLS (T falls). Compression ⇒ gas HEATS UP (T rises).
W = (P₁V₁−P₂V₂)/(γ−1) = μR(T₁−T₂)/(γ−1).
Slope dP/dV = −γP/V — STEEPER than isothermal slope by factor γ (γ>1 always).
Examples: tyre bursting, sound wave propagation, diesel engine ignition (no spark plug).

Insulated gas rushes into an evacuated, insulated container. Q = 0 AND W = 0 (nothing to push against).
⇒ dU = 0 ⇒ T unchanged. This is the key proof that ideal-gas U depends ONLY on T, not on V.
NOT a standard adiabatic process — irreversible, not quasi-static, does NOT follow PV^γ=const.

U(per mole) = (f/2)RT. C_v = (f/2)R. C_P = C_v+R = [(f+2)/2]R. γ = C_P/C_v = 1+2/f.
Monoatomic: f=3, C_v=3R/2, C_P=5R/2, γ=5/3≈1.67.
Diatomic/linear triatomic: f=5, C_v=5R/2, C_P=7R/2, γ=7/5=1.4.
Non-linear polyatomic: f=6, C_v=3R, C_P=4R, γ=4/3≈1.33.
1 < γ < 2 always. More degrees of freedom ⇒ SMALLER γ. Mixture's γ lies between component γ's.
Process PVˣ=const: C = R/(1−x)+C_v. Isobaric (x=0): C=C_P. Isothermal (x=1): C=∞. Adiabatic (x=γ): C=0.

C_P − C_v = R, derived from first law applied to the isobaric case.
C_P > C_v always — constant-pressure heating must ALSO supply the work of expansion, constant-volume heating doesn't.

η = W/Q₁ = (Q₁−Q₂)/Q₁ = 1−Q₂/Q₁. Three parts: source (T₁), sink (T₂), working substance.
η = 1 (100%) is IMPOSSIBLE for any engine — second law (Kelvin–Planck), not just a practical limitation.

β = Q₂/W = Q₂/(Q₁−Q₂). Carnot refrigerator: β = T₂/(T₁−T₂).
Q₁ = Q₂+W always. Relation to engine efficiency (same reservoirs): β = (1/η)−1.
Unlike η (always ≤1), β has NO upper limit of 1 — can be, and often is, greater than 1.

Kelvin–Planck: can't completely convert heat from ONE reservoir into work as the sole result.
Clausius: can't transfer heat cold→hot as the sole result (no external work). Both statements are EQUIVALENT.
ΔS = ∫dQ/T. Reversible process: ΔS=0. Irreversible: ΔS>0. ΔS<0 is NEVER observed (universe entropy never decreases).
Reversible process needs: quasi-static (mechanical+thermal+chemical equilibrium at every stage) AND no friction/dissipation — an ideal limit, never exactly achieved.
Irreversible = the RULE in nature: friction, Joule heating, unbalanced expansion, diffusion, gas leaks, etc.

4 reversible steps: Isothermal expansion (A→B, absorbs Q₁ at T₁) → Adiabatic expansion (B→C) → Isothermal compression (C→D, rejects Q₂ at T₂) → Adiabatic compression (D→A).
η_Carnot = 1−T₂/T₁ = 1−Q₂/Q₁ — the MAXIMUM possible efficiency between T₁ and T₂.
η=100% only if T₂=0 K — physically unreachable, so even an ideal reversible engine can't be perfectly efficient.
Carnot's theorem: NO engine (reversible or not) between two given temperatures can beat a Carnot engine's efficiency operating between the same two temperatures.
To raise efficiency: lowering T₂ by ΔT helps MORE than raising T₁ by the same ΔT (η more sensitive to the smaller temperature in the T₂/T₁ ratio).

Free expansion has Q=0 AND W=0, so ΔU=0 — but it is NOT an adiabatic process in the textbook sense (irreversible, doesn't follow PV^γ=const). Don't apply adiabatic formulas to it.
Isothermal: ΔU=0 but Q≠0 (Q=W). Adiabatic: Q=0 but ΔU≠0 (dU=−dW). Easy to swap these two under pressure.
C_P − C_v = R is in MOLAR heat capacity terms; don't confuse with specific heat (per unit mass), which needs dividing by molar mass M.
γ = 1+2/f — more degrees of freedom give a SMALLER γ, not larger; a common inversion mistake.
Carnot efficiency η = 1−T₂/T₁ uses ABSOLUTE (Kelvin) temperatures ONLY — plugging in Celsius values directly is a frequent, costly error.
C.O.P. (β) of a refrigerator has NO upper bound of 1, unlike efficiency η which is always ≤ 1 — don't apply engine intuition directly to refrigerator problems.
Work done in a cyclic process = the ENCLOSED area, not the sum of all individual path integrals taken at face value without tracking direction (clockwise vs anticlockwise) and sign.

Message Ajay Sir directly on WhatsApp for doubt support on this chapter.