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This is the ideal Diesel cycle. In the $(p,V)$ plane, $A\to B$ is an adiabatic curve running up and left, $B\to C$ is a horizontal constant-pressure segment running right, $C\to D$ is an adiabatic curve running down and right, and $D\to A$ is a vertical constant-volume segment running down. The traversal is clockwise, so the enclosed area is positive work output.

In the $(T,S)$ plane, the two reversible adiabats are vertical because entropy is constant: $A\to B$ rises at $S_A=S_B$, and $C\to D$ falls at $S_C=S_D$. Along $B\to C$, the heat increment at constant pressure issoAlong the constant-volume curve $D\to A$,which givesThus $B\to C$ runs up and right and $D\to A$ runs down and left between the two vertical adiabats.

An intensive quantity is unchanged when the size of a homogeneous system is scaled; temperature and pressure are examples. An extensive quantity scales in proportion to system size; volume and particle number are examples. Energy and entropy are also extensive, while chemical potential is intensive. These are intensive and extensive thermodynamic quantities.

At a first-order transition the Gibbs free energy is continuous but a first derivative such as entropy or volume jumps; there is latent heat. At a second-order transition the first derivatives are continuous while a second derivative, such as heat capacity or susceptibility, is discontinuous or divergent, and there is no latent heat.

Entropy is nonnegative. Gibbs' inequality also givesDivide by $n$ and pass to the assumed limit to obtain

From $E=\rho V$, $P=\rho /3$, and $dE=T,dS-P,dV$,Using $\rho =4\sigma T^4/c$ givesThereforeup to an irrelevant additive entropy constant.

When $\Gamma \gg H$, reactions maintain thermal and chemical equilibrium, so the expansion is quasistatic and adiabatic. With no change in the equilibrium degrees of freedom, comoving entropy and photon number are conserved. Sinceone gets $T^3{a}^3=\text{constant}$ and hence $T\propto a^{-1}$.

Before electron-positron annihilation, the electromagnetic plasma haswhereas afterward only the two photon polarizations remain, so $g_{\ast s}=2$. Entropy conservation in the decoupled electromagnetic sector givesNeutrinos receive none of this entropy and continue cooling as $T_{\nu }\propto a^{-1}$. Therefore

For $B=J=0$, $m=0$ remains stationary. At large $\beta K$, evaluating the free energy near a positive $m$ makes the negative cubic energy dominate the bounded entropy cost, so some $m>0$ has lower free energy than zero. At high temperature entropy uniquely favors zero. Continuity in temperature therefore forces a transition at an intermediate ${\beta }_{\ast }$.

If otherwise comparable isolated planets of one mass follow $L_{\mathrm{int}}=C{t}^{-\alpha }$, their intrinsic luminosity ratio is $L_{\mathrm{int}}(t_1)/L_{\mathrm{int}}(t_2)=(t_2/t_1{)}^{\alpha }$. The normalization depends on composition and initial entropy. Strong stellar irradiation, deep heating, or accretion can invalidate extrapolating the same cooling law.

A thermodynamically reversible process can be reversed without a net change of the system and surroundings. Its entropy change equals reversible heat transfer divided by temperature. A reversible adiabatic process conserves entropy; adiabaticity alone does not forbid internal entropy production in an irreversible evolution.

After thermal decoupling in cosmology, two homogeneous sectors that evolve adiabatically without entropy injection satisfy $s_{\mathrm{SM}}{a}^3=\mathrm{constant}$ and $s_{\varphi }{a}^3=\mathrm{constant}$, hence their entropy density ratio $\xi$ is fixed. At a common relativistic temperature before decoupling, $\xi =g_{\ast s}^{\mathrm{SM}}/g_{\ast s}^{\varphi }$. Internal annihilations can change each sector's temperature and effective entropy degrees of freedom without changing its total comoving entropy.

Specific entropy is entropy per unit mass. For a perfect gas with constant specific heat capacity $c_v$ at constant volume and specific-heat ratio $\gamma$, differences obeywhere $p$ is pressure and $\rho$ is mass density. Thus a homentropic flow has constant $p/{\rho }^{\gamma }$.