The Second Law and Entropy

Two Statements, One Law

The second law has two classical formulations that appear to concern different machines.

Kelvin-Planck: no cyclic process can take heat from a single reservoir and convert it entirely into work. Clausius: no cyclic process can transfer heat from a colder body to a hotter one with no other effect.

They are logically equivalent: assume a violation of either and you can construct a violation of the other. Couple a Kelvin-Planck violator to an ordinary refrigerator and the pair moves heat from cold to hot with no net work, breaking Clausius.

Analysing arbitrary cycles gives the Clausius inequality:

$$\oint \frac{\delta Q}{T} \leq 0$$

with equality exactly for reversible cycles. The equality case is the important one: if $\oint \delta Q_{rev}/T = 0$ round every reversible loop, then $\delta Q_{rev}/T$ is the differential of a state function. That function is entropy:

$$dS = \frac{\delta Q_{rev}}{T}$$

The subscript is essential. $S$ is a state function, so to compute $\Delta S$ for an irreversible process you invent any convenient reversible path between the same endpoints and integrate along that instead.

For an isolated system the inequality becomes the entropy increase principle, $\Delta S \geq 0$, with equality only for reversible change. Combining the first and second laws for a simple system gives the thermodynamic identity:

$$dU = T\,dS - P\,dV$$

which contains no $\delta$ at all, every term is a state function differential, which is why it holds for any process whatever, reversible or not.

Mixing two different ideal gases at the same $T$ and $P$ raises the entropy by $\Delta S = -nR\sum x_i \ln x_i$, which is strictly positive: mixing is spontaneous and unmixing costs work.

Common pitfall: using $\Delta S = Q/T$ with the actual heat of an irreversible process. That underestimates the change every time. Entropy is defined through the reversible heat, so you must construct a reversible path between the same two states and use its heat instead.