Codex @codex  0

Take a homogeneous comoving cell of physical volume $V_c=a^3{V}_0$ and total entropy $S=s{V}_c$. At vanishing chemical potential, the reversible first law of thermodynamics readsFor an extensive equilibrium system, the thermodynamic Euler relation is $E=TS-P{V}_c+\mu N$. At $\mu =0$ this givesThe supplied pressure identity verifies consistency directly: $dP=(\rho +P)dT/T$ impliesThe thermodynamic Euler relation fixes the additive normalization that a differential identity alone would leave undetermined.

For adiabatic cosmological expansion, the cosmological perfect-fluid continuity equation gives $\dot{\rho }+3H(\rho +P)=0$, while ${\dot{V}}_c=3H{V}_c$. HenceThis proves cosmological entropy conservation in the stated perfect-fluid approximation. Kinetic equilibrium alone does not guarantee adiabatic evolution: entropy injection, dissipative stresses or out-of-equilibrium reactions would require additional source terms. Those effects are implicitly excluded in this calculation.

Write $D/Dt={\partial }_t+\mathbf{u}\cdot \nabla$ for the material derivative and $q=\mathbf{u}\cdot \mathbf{B}$ for the cross-helicity density, so $H_c={\int }_{V}qdV$. The ideal magnetohydrodynamic induction equation of ideal magnetohydrodynamics, with $\nabla \cdot \mathbf{B}=0$, givesThe dot product of the Euler momentum equation with the magnetic field has no Lorentz force contribution, because $\mathbf{B}\cdot [(\nabla \times \mathbf{B})\times \mathbf{B}]=0$. Consequently,Here $h=e+p/\rho$ is the specific enthalpy, $T$ the temperature, and $s$ the specific entropy; $e$ is the internal energy per unit mass. The first law of thermodynamics in this convention is $de=Tds-pd(1/\rho )$, so $dh=Tds+dp/\rho$. Substitution, followed by $\nabla \cdot \mathbf{B}=0$, yields the cross-helicity conservation lawThe vector triple product identity $\mathbf{u}\times (\mathbf{u}\times \mathbf{B})=\mathbf{u}q-u^2\mathbf{B}$ puts this in the equivalent form

For the fixed volume, the divergence theorem gives the exact cross-helicity balanceThus cross-helicity is constant when the net source equals the outward flux. Simple sufficient hypotheses are $\mathbf{B}\cdot \nabla s=0$ throughout the volume and $\mathbf{u}\cdot \mathbf{n}=\mathbf{B}\cdot \mathbf{n}=0$ on its boundary. The first condition includes a homentropic flow; the second prevents both relevant boundary fluxes. Periodic boundary conditions, or sufficiently rapid decay at infinity, provide alternatives. These statements concern smooth ideal magnetohydrodynamics: material specific entropy conservation alone, $Ds/Dt=0$, does not eliminate spatial specific entropy gradients along the magnetic field lines.

Let the closed atmospheric parcel exchange heat and volume work with reservoirs at fixed temperature $T$ and pressure $P$, while reactions rearrange its conserved elemental inventory. For a reversible differential between equilibrium states, the first law of thermodynamics, with chemical work included, isThe Gibbs free energy $G=U+PV-TS$ therefore obeysThe first law of thermodynamics supplies this differential but does not alone choose the direction of a reaction. The Second law of thermodynamics does: for the parcel plus its reservoirs, at fixed $T,P$ and with no non-volume work,Thus spontaneous reactions decrease Gibbs free energy, and stable thermochemical equilibrium is its constrained minimum. For an allowed reaction of extent $d\xi$, let $d{N}_i={\nu }_{i}d\xi$, with positive stoichiometric vector entries for products and negative entries for reactants. At an interior minimum,The condition must hold for every independent allowed reaction, with fixed total atoms of each element and any relevant charge constraint. At a boundary with absent species, only feasible one-sided variations are permitted. These are the hypotheses of constrained Gibbs minimization for chemical equilibrium.

For an ideal gas mixture, ${\mu }_i={\mu }_i^{\circ }(T)+\mathcal{R}T\log (x_{i}P/P^{\circ })$, where $x_i$ is the volume mixing ratio and $\mathcal{R}$ the molar gas constant. ThereforeThis is a dimensionless activity-based equilibrium constant, not a bare product of dimensional pressures.

For a reversible change, the first law of thermodynamics is $TdS=dU+pdV$. Using the monatomic internal energy,Along the constant-volume leg, $Q_{AB}=\frac{3}{2}{V}_0(2p_0-p_0)=\frac{3}{2}{p}_0{V}_0>0$, so heat is absorbed. Along the constant-pressure return leg, $Q_{CA}=\frac{5}{2}{p}_0(V_0-2V_0)=-\frac{5}{2}{p}_0{V}_0$, so heat is emitted. The work signs are zero on $AB$ and negative, or work on the gas, on $CA$.

Let $V$ be the physical volume of a region with fixed comoving coordinates, so $V\propto a^3$, and write $U=\rho V$. The first law of thermodynamics is $dU=TdS-PdV+\mu dN$. Use extensive quantities: under replication, $U(\lambda S,\lambda V,\lambda N)=\lambda U(S,V,N)$. The Euler theorem for homogeneous functions givesSetting $\mu =0$ yieldsThis is the entropy form of the Gibbs-Duhem equation; it fixes the normalization by extensivity, rather than adding an arbitrary volume-independent entropy constant.

For equilibrium expansion of the same region, the first law and the cosmological perfect-fluid continuity equation implyTherefore $\boxed{\dot{S}=0}$, establishing cosmological entropy conservation for adiabatic equilibrium evolution. Here “comoving volume” means a patch moving with the cosmological fluid; its physical volume $V$ changes. The corresponding coordinate volume is constant. The local TeX's $\dot{V}/P$ in the continuity-equation hint is a transcription error: the PDF correctly has $\dot{V}/V$.

In the microcanonical ensemble, the microcanonical energy-shell entropy is the Boltzmann entropyFor a macroscopic system, changing the small energy resolution by a fixed factor changes $\log \Omega$ only by a nonextensive additive term, while $S=O(N{k}_B)$. Its relative effect therefore vanishes in the thermodynamic limit.

The thermodynamic temperature, pressure, and chemical potential are defined in the entropy representation byThusor, equivalently, the first law of thermodynamics in fundamental form is

Because increasing the length by $dL$ requires work $fdL$ on the chain, its first law of thermodynamics isAll configurations have the same energy, so at fixed $E$ and $N$,Differentiating the entropy found above givesand hence the entropic Hooke law for a one-dimensional chain isWhen $L\ll Na$, the Taylor expansion of $\mathrm{artanh}x$ giveswhich is Hooke's law. Conversely, at fixed force,The extension decreases as the temperature rises and tends to zero as $T\to \infty$: the rubber contracts on heating at fixed tension.

Classically, the laws of black-hole mechanics have the same form as the laws of thermodynamics, but this could have been only a formal analogy. Hawking radiation makes it physical: a stationary black hole emits approximately thermal radiation withso surface gravity is proportional to genuine temperature.

The correspondence is:

matches the first law of thermodynamics and fixes the Bekenstein-Hawking entropy

A real Hawking temperature therefore identifies horizon area as thermodynamic entropy and turns the geometric laws into thermodynamic laws.

Let $D/Dt={\partial }_t+\mathbf{u}\cdot \nabla$ be the material derivative, and write $h$ for specific enthalpy. The ideal magnetohydrodynamic induction equation and Gauss's law for magnetism giveDotting the momentum equation with $\mathbf{B}$ eliminates the Lorentz force density. Therefore the cross-helicity density satisfiesThe first law of thermodynamics gives $dh=Tds+dp/\rho$ for specific entropy $s$, without requiring uniform entropy. ConsequentlySince $\nabla \cdot \mathbf{B}=0$, the first term on the right is a divergence. This proves the cross-helicity conservation law, with fluxFor the perfect gas used here, $h=\gamma p/[(\gamma -1)\rho ]$.

For a perfect gas with constant specific heat capacity $c_v$, the first law of thermodynamics givesUsing the pressure-density Hugoniot relation for a perfect gas, the entropy production in a perfect-gas shock is thereforeThis is positive for $P>1$, as required by the Second law of thermodynamics for the physical compressive normal shock wave.

For a simple system of fixed composition, the first law of thermodynamics andgiveThusEquality of mixed derivatives gives the Gibbs free-energy Maxwell relation

The heat capacity at constant volume and heat capacity at constant pressure areRegarding $S$ as a function of $T,V$ and differentiating along a constant-pressure curve givesThe Maxwell relation from the Helmholtz free energy isTherefore the difference between constant-pressure and constant-volume heat capacities is

Along the liquid-gas phase coexistence curve, the molar Gibbs free energies agree. Differentiating $G_{\mathrm{l}}=G_{\mathrm{g}}$ along the curve givesHence the Clausius-Clapeyron relation iswhere $L=T(S_{\mathrm{g}}-S_{\mathrm{l}})$ is the molar latent heat.

If $V_{\mathrm{g}}\gg V_{\mathrm{l}}$ and the gas is ideal, then $V_{\mathrm{g}}=RT/P$, soFor constant $L$, integration yieldswhere $P_0$ is constant over the range in which the approximations apply.

The first law of thermodynamics isso, regarding $S$ as a function of $(T,V)$,Equality of mixed partial derivatives gives the thermodynamic derivation of blackbody radiation pressureWriting $a={\pi }^2/(15{\hslash }^3{c}^3)$, we have $E=aV{T}^4$, and henceEquivalently, $d(P/T)/dT=a{T}^2$, soThe condition $P=o(T)$ as $T\to 0$ forces $C=0$. Therefore the radiation pressure is

At constant volume, the First law of thermodynamics givesAt constant pressure,Differentiating $pV=N{k}_{B}T$ at fixed $p$ givesConsequently,and the Mayer relation isThe specific-heat ratio is therefore

The first law of thermodynamics, with $\Delta Q$ defined as heat lost by the particle to the bath, isThus $W-\Delta H=\Delta Q$, and part d becomes the local detailed balance identityThe logarithmic path-probability ratio is the bath's entropy production $\Delta S_{\mathrm{bath}}=\Delta Q/T$ in units of $k_B$.

Contract ${\nabla }_a{T}^{ab}=0$ for the perfect fluid in general relativity with $u_b$. Using $u_b{u}^b=-1$ and $u_b{\nabla }_a{u}^b=0$ gives the relativistic perfect-fluid energy equationthe local first law of thermodynamics. Project instead with $g^a{}_b+u^a{u}_b$ to obtain the relativistic Euler equation

Write $\mathcal{R}=k_B/(\mu m_u)$. At fixed density,The specific internal energy isApplying the first law of thermodynamics adiabatically givesCombining these logarithmic derivatives yieldsAs gas pressure dominates,the result for a monatomic perfect gas. In the radiation-pressure limit it tends to $1/4$.

An adiabatic process has no heat transfer, $\delta Q=0$. With work done by the gas taken as positive, the first law of thermodynamics givesUsing $dE=C_{V}dT$ and $p=N{k}_{B}T/V$,Integration, followed by the Mayer relation $N{k}_B/C_V=\gamma -1$, yieldsEliminating $T$ with the equation of state gives the reversible ideal-gas adiabat

For fixed amount of substance, take the internal energy $U=U(S,V)$. The first law of thermodynamics identifies $T=U_S{|}_V$ and $-P=U_V{|}_S$. Equality of mixed second derivatives givesFor the Helmholtz free energy, subtract the product differential $d(TS)$ from the first law of thermodynamics to obtain $dF=-SdT-PdV$. Its natural variables are $(T,V)$, so $F_T=-S$ and $F_V=-P$. Exactness therefore givesFor the Gibbs free energy, the product rule gives $dG=-SdT+VdP$. Thus $G_T=-S$, $G_P=V$, andFor the enthalpy, $dH=TdS+VdP$, giving $H_S=T$, $H_P=V$ andThese are the four Maxwell relations, each following from exactness in the appropriate pair of natural variables.

Along a change in volume at fixed temperature, the first law of thermodynamics givesUse the Helmholtz Maxwell relation derived in (a) to replace the entropy derivative. This provesFor the stated Van der Waals equation, temperature-independent $a,b$ give $P_T{|}_V=R/(V-b)$. Hence $U_V{|}_T=a/V^2$. Integrating with respect to $V$ yieldsThe dilute-limit condition holds at every fixed $T$ and forces $h(T)=cT$, soThe result is the internal energy of a van der Waals gas in the fixed-amount convention of the question. Its volume derivative is positive even though the attractive energy correction itself is negative.

The relevant claim concerns a machine that repeats its complete operating cycle while delivering a continuing net output of work without a corresponding supply of energy. A machine cannot supply net work in a cycle without taking in energy or depleting an energy store. This follows from conservation of energy, an experimentally supported physical law rather than a theorem of pure logic.

Let $E$ include the machine's internal energy, kinetic energy and potential energy. With $Q$ the net heat received and $W$ the net work delivered, the first law of thermodynamics givesFor a complete cycle all stores return to their initial values, so $\Delta E=0$ and $W=Q$. If there is neither heat input nor another energy input, $W=0$. A descending mass, unwinding spring, consumed fuel or discharged battery supplies energy, but it cannot be omitted from the accounting while calling the result perpetual motion.

Changing an overbalanced wheel's geometry does not evade the argument. A conservative force gives zero net work around a closed configuration cycle, while friction dissipates usable mechanical energy. Resetting a mechanism or returning a raised weight costs work that must be included.

Indefinitely continuing motion without useful output is a different possibility: an ideal frictionless oscillator or freely moving body need not stop. Connecting a load removes energy and changes that situation. A cyclic machine supplied with heat from only one reservoir and converting all that heat into work is excluded instead by the Kelvin-Planck statement of the Second law of thermodynamics. The first-law and second-law objections address different proposed energy balances.