Thermodynamic Definition of Entropy

Drop an ice cube into warm water and something one-directional happens: heat flows, the cube melts, and the process never reverses itself. Physical chemists needed a number that tracks this one-way arrow of change. It had to be measurable with a thermometer and a calorimeter. It could not just be argued about with words like "disorder." That number is entropy, and its operational definition is remarkably simple. Entropy matters because it turns the vague Second Law statement into something precise. That statement is: heat does not flow spontaneously from cold to hot. Entropy makes this a calculable, testable quantity. Once you can compute $\Delta S$ , you can predict which way a reaction runs. You can also predict how efficient an engine can be. And you can see why melting a solid absorbs heat without changing its temperature. This article builds the thermodynamic definition of entropy from reversible heat and thermodynamic temperature. You will see where the definition comes from. You will see why it must use $T$ . You will also see why reversibility is essential. Then you will apply it to two fully worked numerical examples.

Key takeaways

Entropy is defined by $dS = dq_{\text{rev}}/T$ , using heat exchanged along a reversible path only.
$S$ is a state function: $\Delta S$ depends only on initial and final states, not on the path taken.
Irreversible processes still have a well-defined $\Delta S$ ; compute it via any reversible path between the same two states.
The Clausius inequality, $dS \ge dq/T$ , tells you that real (irreversible) processes generate entropy beyond $dq/T$ .
Units of entropy are J K⁻¹ (or J K⁻¹ mol⁻¹ molar), never J K⁻¹ s⁻¹ or dimensionless.
Key idea: entropy is the state function whose reversible heat-to-temperature ratio, $dq_{\text{rev}}/T$ , is path-independent, making it a measurable thermodynamic property rather than a philosophical abstraction.

The core insight

Entropy is defined thermodynamically as $dS = dq_{\text{rev}}/T$ : dividing a small reversible heat increment by the thermodynamic temperature at which it is transferred converts a path-dependent quantity ( $q$ ) into an exact differential ( $dS$ ), making $S$ a genuine state function you can tabulate, measure, and use to predict spontaneity.

The physical picture

Heat, $q$ , is not a state function. If you take a gas from state A to state B by two different routes — one via a fast, irreversible expansion and one via a slow, reversible expansion — the total heat absorbed differs between the routes, even though the internal energy change is identical. This is exactly why the First Law needs both $q$ and $w$ : neither is path-independent alone. Clausius's insight, refined through the Carnot cycle analysis, was that although $dq$ itself is not an exact differential, the combination $dq_{\text{rev}}/T$ is. Divide the reversible heat by the thermodynamic temperature at which it is exchanged, and path-dependence disappears. Sum (integrate) that ratio around any closed reversible cycle and you get exactly zero — the hallmark of a state function. This ratio was named entropy, symbol $S$ , from the Greek for "transformation." The word "reversible" in the definition is not decorative — it is load-bearing. A reversible process proceeds through a continuous sequence of equilibrium states, with the system's temperature (or pressure) differing infinitesimally from its surroundings at every step. Only along such an idealized path does dividing by a single, well-defined $T$ make sense, because system and surroundings share essentially the same temperature throughout. Along an irreversible path, $T$ may not even be well-defined for the whole system (think of a gas expanding into a vacuum, with no uniform temperature or pressure during the process), so $dq_{\text{irrev}}/T$ is not the entropy change — it is smaller than $dS$ , consistent with the Clausius inequality you met earlier. Because $S$ is a state function, you never need the actual irreversible path to compute $\Delta S$ for an irreversible process. You invent — on paper — any convenient reversible path connecting the same initial and final states, compute $q_{\text{rev}}$ along that fictitious path, and divide by $T$ . The real process might be violent and irreversible; the entropy change is the same either way, because $\Delta S$ only cares about the endpoints.

Figure

Reversible vs irreversible paths between the same two states A and B

Figure

Isothermal melting: reversible heat transfer at constant T

The core relations

dS = \frac{dq_{\text{rev}}}{T}

where: $dS$ is the infinitesimal entropy change, $dq_{\text{rev}}$ is heat transferred reversibly, $T$ is the thermodynamic (Kelvin) temperature at the point of transfer.

\Delta S = \int_{i}^{f} \frac{dq_{\text{rev}}}{T}

where: the integral is evaluated along any reversible path connecting the same initial state $i$ and final state $f$ .

\oint \frac{dq_{\text{rev}}}{T} = 0

where: the cyclic integral vanishing for every reversible cycle is the mathematical signature that $S$ is a state function.

dS \ge \frac{dq}{T} \quad (\text{Clausius inequality, general process})

where: equality holds only for a reversible step; for real (irreversible) processes $dS$ strictly exceeds $dq/T$ , reflecting internally generated entropy.

\Delta S_{\text{isothermal}} = \frac{q_{\text{rev}}}{T}

where: valid whenever the reversible heat transfer occurs at one fixed temperature $T$ , e.g. phase transitions at their transition temperature.

Derivation

Derivation AFrom the Carnot cycle to

dS = dq_{\text{rev}}/T

Goal. Show that $dq_{\text{rev}}/T$ is an exact differential, so its integral over a closed reversible cycle is zero, justifying its identification as $dS$ .

Step 1. Start from the Carnot cycle efficiency. For a reversible Carnot engine operating between a hot reservoir at $T_h$ and cold reservoir at $T_c$ , absorbing heat $q_h$ from the hot reservoir and releasing $q_c$ to the cold one, the thermodynamic temperature scale is defined so that

\frac{q_c}{q_h} = \frac{T_c}{T_h}.

This is not an empirical accident — it is precisely how the Kelvin scale is constructed from Carnot's theorem (a result you met when the thermodynamic temperature scale itself was introduced).

Step 2. Assign a sign convention and rearrange. Let $q_h$ be heat absorbed by the system (positive) and $q_c$ be heat absorbed by the system from the cold reservoir; since the engine releases heat there, $q_c$ is negative. The relation above, written with signs, becomes

\frac{q_h}{T_h} + \frac{q_c}{T_c} = 0.

This says the sum of (heat in)/(temperature at which it entered) around one full Carnot cycle is exactly zero.

Step 3. Generalize to an arbitrary reversible cycle. Any reversible cyclic process can be approximated arbitrarily closely by a sequence of many infinitesimal Carnot cycles, each exchanging heat $dq_{\text{rev}}$ at the local temperature $T$ of the system at that stage. Step 2 applied to each infinitesimal cycle, summed over the whole path, gives

\oint \frac{dq_{\text{rev}}}{T} = 0.

Step 4. Invoke the mathematical criterion for a state function. A quantity whose cyclic integral vanishes for every possible cycle must be the differential of some function of state — this is a standard result from the theory of exact differentials (the same logic that shows internal energy $U$ is a state function from $\oint dU = 0$ ). Because $\oint dq_{\text{rev}}/T = 0$ holds for any reversible cycle, there must exist a state function $S$ such that

dS \equiv \frac{dq_{\text{rev}}}{T}.

Step 5. Confirm path-independence explicitly. For two different reversible paths, 1 and 2, both connecting state A to state B,

\int_{A,\,\text{path 1}}^{B} \frac{dq_{\text{rev}}}{T} = \int_{A,\,\text{path 2}}^{B} \frac{dq_{\text{rev}}}{T},

because reversing path 2 and appending it to path 1 reconstructs a closed cycle whose integral is zero by Step 3. Both integrals therefore equal the same quantity, $\Delta S = S_B - S_A$ , independent of route.

\boxed{dS = \frac{dq_{\text{rev}}}{T}, \qquad \Delta S = \int_i^f \frac{dq_{\text{rev}}}{T} \text{ is path-independent}}

Worked examples

Example 1 — Isothermal reversible expansion of an ideal gas (foundational). One mole of an ideal gas expands reversibly and isothermally at $T = 298\ \text{K}$ from $V_1 = 10.0\ \text{L}$ to $V_2 = 20.0\ \text{L}$ . Find $\Delta S$ . Since $\Delta U = 0$ for an isothermal ideal-gas process, the First Law gives $q_{\text{rev}} = -w_{\text{rev}}$ . For reversible isothermal expansion,

w_{\text{rev}} = -nRT\ln\frac{V_2}{V_1}, \qquad q_{\text{rev}} = nRT\ln\frac{V_2}{V_1}.

Plugging in numbers:

q_{\text{rev}} = (1.00\ \text{mol})(8.314\ \text{J K}^{-1}\text{mol}^{-1})(298\ \text{K})\ln\left(\frac{20.0}{10.0}\right)

q_{\text{rev}} = (2478\ \text{J})(0.6931) = 1717\ \text{J}.

Since $T$ is constant throughout this reversible path,

\Delta S = \frac{q_{\text{rev}}}{T} = \frac{1717\ \text{J}}{298\ \text{K}} = 5.76\ \text{J K}^{-1}.

Sanity check: entropy increased, matching intuition — the gas occupies more volume/more accessible states, and this is a spontaneous-direction (expansion) process at constant $T$ . The magnitude, about $R\ln 2 = 5.76\ \text{J K}^{-1}$ , is a textbook-standard benchmark value worth memorizing.

Example 2 — Entropy change for an irreversible free expansion, computed via a reversible substitute path (applied/harder). One mole of ideal gas at $298\ \text{K}$ undergoes a free (Joule) expansion into vacuum from $V_1 = 10.0\ \text{L}$ into a total volume $V_2 = 20.0\ \text{L}$ inside an insulated, rigid container. No work is done ( $w=0$ , expansion against zero external pressure) and no heat is exchanged with surroundings ( $q = 0$ , insulated). Find $\Delta S_{\text{system}}$ , $\Delta S_{\text{surroundings}}$ , and $\Delta S_{\text{universe}}$ .

Step 1 — resist the trap. You might be tempted to write $\Delta S = q/T = 0/298 = 0$ . This is wrong, because $q = 0$ here is the actual irreversible heat, not $q_{\text{rev}}$ . The defining formula demands reversible heat.

Step 2 — construct a reversible replacement path. The gas's initial and final states (same $T = 298\ \text{K}$ for an ideal gas, since $\Delta U = 0$ in a free expansion with no work and no heat; and $V_1 \to V_2$ ) are identical to those of Example 1. So we use the same reversible isothermal path to connect them and get the same answer, because $S$ is a state function and does not know or care how the real process happened:

\Delta S_{\text{system}} = nR\ln\frac{V_2}{V_1} = (1.00)(8.314)\ln(2.00) = 5.76\ \text{J K}^{-1}.

Step 3 — surroundings. In the actual irreversible free expansion, the container is insulated and rigid, so the surroundings exchange no heat and undergo no change at all:

\Delta S_{\text{surroundings}} = \frac{q_{\text{actual, surroundings}}}{T} = \frac{0}{298} = 0\ \text{J K}^{-1}.

Step 4 — total.

\Delta S_{\text{universe}} = \Delta S_{\text{system}} + \Delta S_{\text{surroundings}} = 5.76 + 0 = 5.76\ \text{J K}^{-1} > 0.

Sanity check: $\Delta S_{\text{universe}} > 0$ confirms this spontaneous, irreversible process is consistent with the Second Law. Contrast with Example 1's reversible path at the same $T$ , $V_1$ , $V_2$ : there, heat flowed in from surroundings, so $\Delta S_{\text{surroundings}} = -5.76\ \text{J K}^{-1}$ exactly cancels $\Delta S_{\text{system}}$ , giving $\Delta S_{\text{universe}} = 0$ — the reversible-process benchmark. The irreversible free expansion generates net entropy; the reversible expansion generates none.

Limiting cases & common traps

Situation	What to do
Process is irreversible	Never plug actual $q$ into $dS = dq/T$ . Invent a reversible path between the same endpoints instead.
Process is reversible and isothermal	Use $\Delta S = q_{\text{rev}}/T$ directly (phase changes, isothermal ideal-gas steps).
Adiabatic AND reversible	$dq_{\text{rev}} = 0$ at every step, so $\Delta S = 0$ — a reversible adiabat is a constant-entropy (isentropic) path.
Cyclic process, any working substance	$\oint dS = 0$ always, since $S$ is a state function, regardless of reversibility of the cycle's individual steps.
Adiabatic but irreversible	$q = 0$ does not imply $\Delta S = 0$ ; entropy can still increase (e.g. free expansion above).

The chemist behind it

Rudolf Clausius coined the term "entropy" in 1865, deliberately choosing a Greek root related to "transformation" to parallel "energy." He had already shown, through his analysis of Carnot's reversible heat engine, that $dq_{\text{rev}}/T$ integrates to zero around any closed reversible cycle — the mathematical fingerprint of a state function. Clausius then generalized this into the inequality $dS \ge dq/T$ for real processes, giving the Second Law its first fully quantitative, calculable form and setting the stage for every later statistical interpretation of entropy.

Origins · 1 minRudolf Clausius and how this was discovered

Rudolf Clausius

1822-1888 · German physicist who defined entropy from Carnot-cycle reasoning and gave the Second Law its modern mathematical form

Ask the tutor: How does this reversible-heat definition connect to entropy as a state function with exact-differential properties like $dU$ and $dH$ ?

Why must the heat in dS = dq_rev/T be reversible, not just any heat?

Because only along a reversible path is the system's temperature well-defined and essentially equal to that of the surroundings at every instant, letting you divide by a single, meaningful $T$ . Along an irreversible path the system may pass through non-equilibrium states with no single temperature, and dividing actual heat by a boundary temperature systematically undercounts entropy production, per the Clausius inequality $dS \ge dq/T$ .

If entropy is defined using reversible heat, how can we ever compute ΔS for a real, irreversible process?

Entropy is a state function, so $\Delta S$ depends only on initial and final states, never on the path actually taken. To find $\Delta S$ for any irreversible process, mentally replace it with a convenient reversible path connecting the same two endpoints — same starting temperature, pressure, volume, and same final ones — compute $q_{\text{rev}}$ along that imagined path, and divide by $T$ . The real process can be as violent as you like; the answer is identical.

What are the units of entropy, and what is a typical order of magnitude?

Entropy carries units of joules per kelvin, J K⁻¹, or per mole, J K⁻¹ mol⁻¹, since it is heat divided by temperature. Typical molar entropy changes for phase transitions or gas expansions run from a few J K⁻¹ mol⁻¹ up to several tens of J K⁻¹ mol⁻¹; for reference, $R\ln 2 = 5.76\ \text{J K}^{-1}\text{mol}^{-1}$ and water's entropy of fusion is about $22.0\ \text{J K}^{-1}\text{mol}^{-1}$ .

Does dS = dq_rev/T mean entropy only changes when heat is transferred?

For a closed system exchanging only heat and expansion work, yes — every entropy change traces back to reversible heat exchange, even if that heat is a bookkeeping device along an imagined path. However, entropy can also be generated internally, with no heat crossing the boundary at all, whenever a process is irreversible; the Clausius inequality $dS > dq/T$ captures exactly this internally generated contribution.