Entropy — Clausius definition dS = dQ_rev - T

WHAT is entropy?

The subscript rev is the heart of the definition. To compute $\Delta S$ between two states you must imagine some reversible path connecting them — even if the real process was irreversible. Because $S$ is a state function, any reversible path gives the same answer.

WHY does $dQ/T$ work but $dQ$ doesn't?

HOW we know it's path-independent — the Carnot link:

For a Carnot cycle running reversibly between hot reservoir $T_H$ and cold reservoir $T_C$ : $\frac{Q_H}{T_H} = \frac{Q_C}{T_C}$ This comes directly from the Carnot efficiency $\eta = 1 - \dfrac{Q_C}{Q_H} = 1 - \dfrac{T_C}{T_H}$ , which rearranges to the above. Counting heat in as positive and heat out as negative: $\frac{Q_H}{T_H} - \frac{Q_C}{T_C} = 0 \quad\Rightarrow\quad \oint \frac{dQ_{rev}}{T}=0 \ \ \text{(for a Carnot cycle)}.$

HOW we generalise (Clausius theorem): any reversible cycle can be tiled by infinitely many tiny Carnot cycles. The internal boundaries cancel, so $\boxed{\oint_{rev}\frac{dQ_{rev}}{T}=0}$ A quantity whose integral around every closed loop is zero must be the differential of a state function. Call that function $S$ . That is the existence proof for entropy.

Figure — Entropy — Clausius definition dS = dQ_rev - T

Derivation of $\Delta S$ for an ideal gas (from scratch)

Start from the First Law for a reversible process: $dU = dQ_{rev} - dW_{rev} = dQ_{rev} - P\,dV \ \Rightarrow\ dQ_{rev} = dU + P\,dV.$

Why this step? The First Law is energy conservation; for reversible work $dW = P\,dV$ exactly.

For an ideal gas $dU = nC_V\,dT$ and $P = \dfrac{nRT}{V}$ . Divide everything by $T$ : $dS=\frac{dQ_{rev}}{T}=\frac{nC_V\,dT}{T}+\frac{nR\,dV}{V}.$

Why divide by $T$ ? That is literally the Clausius definition — it turns the inexact $dQ$ into the exact $dS$ .

Integrate from $(T_1,V_1)$ to $(T_2,V_2)$ : $\boxed{\;\Delta S = nC_V\ln\frac{T_2}{T_1} + nR\ln\frac{V_2}{V_1}\;}$

Why are both terms clean logs? Because each term was already an exact differential after dividing by $T$ — confirming $1/T$ did its job as integrating factor.

Worked examples

Common mistakes (steel-manned)

Active recall

Recall Quick self-test (hide answers)

What makes $1/T$ special? ::: It's the integrating factor that makes inexact $dQ_{rev}$ into an exact differential $dS$ .
Why is $S$ a state function? ::: Because $\oint dQ_{rev}/T = 0$ (Clausius theorem from tiling Carnot cycles).
Entropy change of free expansion? ::: $nR\ln(V_2/V_1)>0$ , via a reversible isothermal substitute path.

Recall Feynman: explain to a 12-year-old

Imagine heat is like water poured into a tank. How much it spreads out depends not just on how much you pour, but on how "crowded" the tank already is — and temperature tells you the crowding. Pour the same heat into a cold thing and it makes a big mess (big entropy jump); pour it into something already hot and it barely notices (small jump). Entropy is just the running total of "mess made," and it only counts properly if you pour super gently (reversibly). Once messed up, the universe never tidies back — that's why time runs forward.

What is the Clausius definition of entropy change?

dS = dQ_{rev}/T

;

\Delta S=\int_A^B dQ_{rev}/T

, path-independent.

Why must the heat be reversible in

dS=dQ_{rev}/T

Only the reversible path makes

dQ/T

an exact (path-independent) differential equal to the true state change.

What is the integrating factor for heat?

1/T

— it converts inexact

dQ_{rev}

into exact

dS

State the Clausius theorem (reversible cycle).

\oint dQ_{rev}/T = 0

State the Clausius inequality (any cycle).

\oint dQ/T \le 0

, equality only for reversible cycles.

\Delta S

for ideal gas between two states?

nC_V\ln(T_2/T_1)+nR\ln(V_2/V_1)

\Delta S

for reversible isothermal expansion?

nR\ln(V_2/V_1)

\Delta S

for heating mass

m

from

T_1

T_2

mc\ln(T_2/T_1)

\Delta S

of gas in free expansion into vacuum?

nR\ln(V_2/V_1)>0

(use reversible isothermal substitute;

Q_{actual}=0

doesn't apply).

Units of entropy?

\text{J K}^{-1}

For a reversible adiabatic process

\Delta S = ?

0

(isentropic).

Connections

Carnot Cycle and Efficiency — source of $Q_H/T_H = Q_C/T_C$
First Law of Thermodynamics — gives $dQ_{rev}=dU+PdV$
Second Law of Thermodynamics — $\Delta S_{universe}\ge 0$
Reversible vs Irreversible Processes
Exact and Inexact Differentials — why $1/T$ is an integrating factor
Statistical Entropy — Boltzmann S = k ln W — microscopic meaning
Clausius Inequality

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, entropy ka pura funda yeh hai: sirf heat $dQ$ se kaam nahi chalta kyunki heat "path-dependent" hota hai — same do states ke beech alag-alag raaste se alag-alag heat flow ho sakti hai. Lekin agar tum heat ko us temperature $T$ se divide kar do jis par woh transfer ho rahi hai, aur process ko reversible (bahut dheere, gentle) maan lo, toh $dQ_{rev}/T$ ek jaadui cheez ban jaati hai jo path par depend hi nahi karti. Isi ko hum entropy $S$ kehte hain, aur $dS = dQ_{rev}/T$ . Yeh ek state function hai — matlab sirf starting aur ending state matter karti hai.

Yeh path-independence kaise pata chali? Carnot cycle se. Reversible Carnot mein $Q_H/T_H = Q_C/T_C$ hota hai, jiska matlab cycle ke around $\oint dQ_{rev}/T = 0$ . Aur koi bhi reversible cycle ko chhoti-chhoti Carnot cycles se bhar sakte ho, toh general result bhi $\oint dQ_{rev}/T = 0$ ban jaata hai. Jo cheez har closed loop par zero deti hai, woh zaroor kisi state function ka differential hoti hai — bas wahi hamara $S$ hai.

Sabse important trap: irreversible process mein bhi $\Delta S$ nikaalne ke liye tumhe ek imaginary reversible path sochna padta hai jo same do states ko jode. Jaise vacuum mein free expansion — wahan actual $Q=0$ hai, par $\Delta S = 0$ nahi hota! Reversible isothermal path use karke $\Delta S = nR\ln(V_2/V_1) > 0$ aata hai. Isliye yaad rakho: " $Q=0$ toh entropy zero" sirf tab sahi hai jab process reversible AND adiabatic dono ho. Yeh chhoti si baat exam mein bahut number deti hai.

Entropy — Clausius definition dS = dQ_rev - T

WHAT is entropy?

WHY does $dQ/T$ work but $dQ$ doesn't?

Derivation of $\Delta S$ for an ideal gas (from scratch)

Worked examples

Common mistakes (steel-manned)

Active recall

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Thermodynamics

Connections