Codex @codex  0

Assume an isolated, bound, nonrotating Newtonian gas cloud, negligible surface pressure, negligible magnetic and radiation stresses, no persistent bulk flow, and a stationary or time-averaged virialized state. The stellar virial theorem then reduces towhere $\Omega <0$ is the total gravitational potential energy. Define the volume-averaged pressure $\overline{P}=V^{-1}\int PdV$ and average gravitational energy density $u_g=\Omega /V$. HenceEquivalently, for a nonrelativistic monatomic gas, the thermal internal energy $U=(3/2)\int PdV$ obeys $2U+\Omega =0$. This second derivation must not count the thermal motions again as a separate bulk kinetic energy. If the external pressure is $P_s$, the correct result is $\overline{P}-P_s=-u_g/3$; non-negligible bulk motion adds $2T_{\mathrm{bulk}}$ to the virial theorem. The initial sharp-edged $r^{-2}$ profile need not survive virialization; maintaining that profile hydrostatically to its edge would generally require a surface term.

For a gas-pressure-supported stellar core of fixed mass and composition, neglect boundary terms for the illustrative whole-core estimate. The virial theorem gives $E=U+\Omega =-U$, and the monatomic gas gives $U=3M_c{k}_B\overline{T}/(2\mu m_u)$. Thus the effective heat capacity along the equilibrium sequence isThis negative heat capacity of a virialized gravitational system means that net energy loss makes a self-gravitating core contract and heat. It is different from the positive local material specific heat capacity. On the main sequence, hydrogen burning supplies most of the radiated energy, so rapid contraction is arrested. Hydrogen depletion increases the mean molecular weight; slow structural readjustment usually raises the central temperature rather than making it fall.

After core hydrogen exhaustion, a massive stellar core contracts and heats more rapidly until a later nuclear fuel ignites. There is no general reversal to cooling simply upon leaving the main sequence. If radiation pressure becomes dominant, the internal energy becomes $U=U_{\mathrm{gas}}+U_{\mathrm{rad}}$ with $3\int PdV=2U_{\mathrm{gas}}+U_{\mathrm{rad}}$. The virial theorem therefore gives $E=-U_{\mathrm{gas}}$; the radiation-dominated limit is marginally bound and its effective adiabatic index approaches $4/3$.

Quasistatic contraction can still maintain virial equilibrium. It fails dynamically only when sufficient support is lost, for example if pair creation reduces the effective adiabatic index below $4/3$, or when an iron core undergoes electron capture and photodisintegration. A fixed-composition gas-dominated contracting core with $T\propto 1/R$ is already a counterexample to any assertion that leaving the main sequence alone reverses the heating or destroys virial equilibrium.

Write $\beta =(k_{\mathrm{B}}T{)}^{-1}$ and $x=\beta \delta$. The one-site partition function includes both degenerate lower states,The occupied adsorption sites are distinguishable positions, so the canonical partition function is a product, without an additional $N!$ divisor:Differentiating $\ln Z_N$ gives the internal energy,The final factor is the population of the perpendicular orientation. The shared binding offset $-\epsilon$ shifts all energy levels equally and does not determine their relative populations.

At low temperature, the upper orientation has negligible Boltzmann weight, whereas at high temperature all three orientations become equally probable. ConsequentlyThe high-temperature limit is a finite arithmetic mean of three available one-site energy levels. It is not an unbounded translational or rotational equipartition contribution: the prescribed site model has only three states per molecule. Their common binding offset can keep this limiting internal energy negative without any inconsistency.

Using $dx/dT=-x/T$ in the internal energy givesThis is a Schottky anomaly with unequal degeneracies, with lower-level state degeneracy two and upper-level state degeneracy one. Its limiting forms areIt vanishes at low temperature because exciting the upper orientation is exponentially unlikely. It also vanishes at high temperature because the populations have nearly saturated and further heating hardly changes the internal energy. Between these limits it has one smooth maximum. The original plot shows its shape; numerical maximization gives $k_{\mathrm{B}}{T}_{\max }/\delta \simeq 0.449$ and $C_{\max }/(N{k}_{\mathrm{B}})\simeq 0.241$.

For a fixed number of non-interacting molecules, $Z_N=q^N/N!$ in the dilute ideal gas limit. The temperature-independent factorial drops out of the internal energy derivative:Differentiate once more at fixed volume to obtainPrimes denote fixed-volume derivatives. If $q$ factorizes into translational, rotational and other independent contributions, their heat capacities add. In particular, the three-dimensional translational factor contributes $3R/2$, while a constant spin degeneracy factor contributes zero. The formula also follows from the positive canonical energy variance, $C_V=\mathrm{Var}(E)/(k_{\mathrm{B}}{T}^2)$.

Set the ground electronic energy to zero and write the excitation energy as $\Delta =hc(120\mathrm{c}{\mathrm{m}}^{-1})$, or the characteristic temperature $\theta =\Delta /k_{\mathrm{B}}=172.65$ K. Counting each doubly degenerate manifold once gives the single-molecule electronic partition functionFor $N$ independent molecules the electronic factor is $q_{\mathrm{el}}^N$. Hence the electronic internal energy isThe equal degeneracies cancel from the excited-state probability. At low temperature the ground manifold dominates and $U_{\mathrm{el}}\to 0$; at high temperature both manifolds have equal populations, giving $U_{\mathrm{el}}\to N\Delta /2$. These are electronic contributions only, excluding translation, rotation and vibration.

The internal energy is the population-weighted sum of the molecular energy levels:At fixed volume, assume the energy levels themselves are temperature-independent. Differentiating the molecular partition function givesChoosing the zero of energy changes $U$ by the corresponding constant per molecule, so the ground-level convention must be consistent throughout.

Relative to the ground state, the quantum harmonic oscillator levels are $vhc\tilde{\omega }$. Put ${\theta }_v=hc\tilde{\omega }/k$. Summing the geometric series gives the vibrational partition function $q=(1-e^{-{\theta }_v/T}{)}^{-1}$. Therefore the vibrational internal energy isThe omitted zero-point energy would add $Nhc\tilde{\omega }/2$ if absolute oscillator energies were used.

Let $G$ be Gibbs free energy, $H$ enthalpy and $T$ thermodynamic temperature for a fixed amount and composition of material, with the derivative taken at constant pressure. Apply the Gibbs–Helmholtz equation to the standard molar reaction quantities, obtained by summing product minus reactant contributions with their stoichiometric coefficients:Define $K(T)$ as the dimensionless thermodynamic equilibrium constant expressed using activities in specified standard states. Since ${\Delta }_r{G}^{\circ }=-RT\ln K$, where $R$ is the molar gas constant,If the reaction enthalpy is constant over the interval, integration givesThis is the requested Van't Hoff equation. For an exothermic reaction, increasing temperature decreases $K$. Although the name “isochore” is often used in this context, the displayed enthalpy form follows from standard Gibbs energies and the stated constant-pressure Gibbs–Helmholtz identity; the analogous ideal-gas concentration form involves internal energy when the temperature dependence of standard concentration is retained.

Using bonds broken minus bonds formed, the enthalpies per mole of elementary events areUnder the steady-state approximation in chemical kinetics, initiation and termination have equal event rates, so their enthalpy contributions cancel. The two propagation steps also have equal flux, and their combined enthalpy is $5-189=\boxed{-184\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}}$ per overall reaction producing two moles of HCl. Step 3 dominates the heat release; the small endothermic contribution of step 2 only slightly offsets it.

An insulated vessel cannot remove this released energy as heat, so its temperature rises as reaction proceeds. In a rigid vessel the precise energy balance uses reaction internal energy and constant-volume heat capacities, whereas a constant-pressure adiabatic balance uses enthalpy; the strong exothermic conclusion is the same. Warming changes the rate constants, and activated initiation and propagation can accelerate, producing positive thermal feedback and potentially a rapid temperature rise. A numerical final temperature or a claim of an inevitable explosion is not justified without composition, heat capacities and heat-loss information. The initially assumed 298-K rate constants cannot remain fixed throughout an appreciable adiabatic temperature rise.

Differentiate the normalized average, including the derivative of its denominator. The enthalpy fluctuation formula follows fromSince $d\beta /dT=-1/(k_B{T}^2)$,Thus a larger equilibrium spread of enthalpy corresponds to a larger response to heating at fixed pressure. The variance is nonnegative, so this finite-system equilibrium description has $C_P\ge 0$. The relation connects a response coefficient to fluctuations of the appropriate exchanged quantity; at fixed volume the corresponding quantity is internal energy.

The volume integral is a gamma function integral,For the prescribed bridge relation, $\beta P{Q}_P=(\beta P{\Lambda }^3{)}^{-N}$, so $G=N{k}_{B}T\ln (\beta P{\Lambda }^3)$. Since $\Lambda \propto T^{-1/2}$, this gives the macroscopic ideal-gas resultHeating at fixed pressure requires both an increase in internal energy and expansion work; at fixed volume the latter contribution is absent.

The finite-size normalization of the isothermal-isobaric ensemble matters here. Applying part (a) literally to the unprefactored $Q_P$ instead gives $⟨H{⟩}_P=(5N/2+1)k_{B}T$ and $C_P=(5N/2+1)k_B$. The two answers differ because $-{\partial }_{\beta }\ln (\beta P{Q}_P)=-{\partial }_{\beta }\ln Q_P-k_{B}T$. The normalized continuous-volume distribution has $⟨PV⟩=(N+1)k_{B}T$; its variance agrees with the unprefactored result in part (b). The bridge convention subtracts an order-one contribution and recovers the usual extensive thermodynamic answer. Both agree to leading order in the thermodynamic limit, but their exact finite-$N$ derivatives should not be mixed.

The harmonic phonon contribution to the internal energy isFor $T\gg {\Theta }_D$, every retained mode has $x=\hslash \omega /(k_{B}T)\ll 1$. Expanding $1/(e^x-1)=1/x-1/2+O(x)$ cancels the explicit half quantum and gives $k_{B}T$ per mode to leading order. ThereforeEquivalently, differentiating before integrating givesThe integrand tends to $x^2$, and its integral is $({\Theta }_D/T{)}^3/3$, producing the same Dulong-Petit law. In classical language, the equipartition theorem assigns $k_{B}T/2$ each to the kinetic and elastic parts of every harmonic normal mode. Fixed zero-point energy does not contribute to the heat capacity.

The chemical potential is the change in Helmholtz free energy when particles are added at fixed temperature and volume:Equivalently, $dU=TdS-PdV+\mu dN$ identifies it as the derivative of internal energy at fixed entropy and volume. In equilibrium, systems that exchange particles have equal chemical potentials.

The Fermi energy is the zero-temperature chemical potential, or, for independent electrons, the highest occupied single-particle energy in the zero-temperature limit. Its numerical value depends on the chosen energy reference; the Fermi energy need not be measured from the bottom of a band unless that convention is specified.

For a nondegenerate two-dimensional electron gas, the two quadratic momentum terms supply $k_{B}T/2$ each by the equipartition theorem. There is no lattice restoring potential energy for a free electron. ThusOn cooling below the Fermi temperature, the Pauli exclusion principle restricts the thermally active electrons to within approximately $k_{B}T$ of the Fermi energy. The linear low-temperature heat capacity of a two-dimensional Fermi gas isThe heat capacity therefore decreases from its classical value as quantum degeneracy becomes important. The particle number of electrons and sample area are held fixed when differentiating the internal energy.

Equal effective masses and equal spin and valley degeneracies give electron-hole symmetry. In the intrinsic semiconductor, charge neutrality requires equal numbers of conduction band electrons and valence band holes. It therefore pins the chemical potential to the band touching point, $\mu =0$. The Fermi-Dirac distribution givesMeasured relative to the filled valence band at zero temperature, the thermal internal energy includes both species. Removing a valence band electron at $-E$ costs the positive hole energy $E$, just as adding a conduction band electron at $E$ does. HenceDifferentiate at fixed area and fixed total electronic particle number, allowing creation of electron-hole pairs:There is no conflict with the equipartition theorem. The normalization $N_c$ counts only the excited conduction band electrons, while the heat capacity includes both electrons and holes; moreover, $N_c\propto T$ is not fixed. In fact $\Delta U/N_c={\pi }^2{k}_{B}T/(6\ln 2)$, and differentiating $N_c(T)$ supplies another equal contribution to the heat capacity. Finally, $\mu =0$ gives occupation $1/2$ at the touching point, so the excitation gas is not in the classical dilute limit even though $k_{B}T$ exceeds the vanishing band gap.

gas heats when compression, shocks or absorbed radiation increase its internal energy. Important sources include accretion shocks, stellar winds, supernovae, photoionization, photoelectric emission from interstellar dust, cosmic rays and energy deposited by active galactic nuclei. photoionization heating uses the photoelectron energy above the ionization threshold; compression contributes mechanical work rather than photon absorption.

For specific internal energy $u$, an idealized thermal balance isHere $\mathcal{H},\mathcal{C}$ are powers per volume and $Z$ describes abundances. The electron-hydrogen normalization of the cooling function must be specified: some conventions instead factor out $n_H^2$. For a monatomic gas with total particle density $n$,when heating is negligible. Compare this with the dynamical time: rapid cooling permits loss of pressure support, whereas long cooling times allow a hot atmosphere to persist.

collisional excitation followed by spontaneous emission cools partially ionized gas; radiative recombination also produces continuum cooling. metal lines enhance cooling over substantial intermediate-temperature ranges, while thermal bremsstrahlung gives a roughly $\Lambda \propto T^{1/2}$ high-temperature tail. Atomic hydrogen cooling becomes inefficient below roughly $10^4\mathrm{K}$, where molecular line cooling and interstellar dust can be important. inverse Compton scattering cools hot electrons against a colder radiation field. A tabulated cooling curve assumes a composition and ionization model; photoionization or nonequilibrium ionization can change it.^[ref]

X-ray line ratios and continuum shape constrain temperature and ionization, while the emission measure constrains $\int n_e{n}_{H}dV$. Turning that into a density and radiative gas cooling time also requires a geometry; unresolved clumping, foreground absorption and multiphase gas are important controls.

mass and energy signs. The enclosed mass is $M(r)={\int }_0^{r}4\pi s^2\rho (s)ds$, so $dM/dr=+4\pi r^2\rho$. The negative sign printed in that equation cannot describe positive enclosed mass. Use the physical sign and define gravitational potential energy by

Boundary-pressure virial theorem. Multiply hydrostatic equilibrium by $4\pi r^3$ and integrate. The left side is $[4\pi r^{3}p{]}_0^{r_0}-3{\int }_0^{r_0}4\pi r^{2}pdr$, while the right side is $\Omega$. The central boundary contribution vanishes when $r^{3}p\to 0$ and the outer pressure is $p(r_0)=p_0$. ThusA finite external pressure is compatible with a gravitationally self-contained cloud, but is not the zero-surface-force notion of isolation. External ram stress or mass flow would require additional terms.

radiation budget of monatomic cloud collapse. For an ideal gas with fixed $\gamma$, the specific internal energy is $u=p/[\rho (\gamma -1)]$. Therefore the pressure integral is $3(\gamma -1)U$. Cold monatomic hydrogen has $\gamma =5/3$ when only translational degrees of freedom matter. With negligible final surface pressure, $2U+\Omega =0$, so $E_f=U+\Omega =\Omega /2$. If the initial cloud is sufficiently extended and cold that $E_i\simeq 0$, the emitted energy isBeing initially at rest alone does not set its binding energy to zero. If the initial total energy is appreciable, retain $E_{\mathrm{rad}}=E_i-\Omega /2$; a different equation of state also changes the half-energy result.

Inside-out accretion. The combination of $a_T$ and $G$ with dimensions of mass per time iswhere dimensional analysis leaves the dimensionless coefficient $C$ undetermined. An isothermal gas keeps $a_T$ constant, giving constant supply in the self-similar idealization. A front at radius $a_{T}t$ and an isothermal mass scale $M(r)\sim a_T^{2}r/G$ give the same scaling. The front's swept mass is not automatically the accreted central mass; the coefficient requires a dynamical similarity solution.^[ref]

constant-radius accretion energy budget. If all available accretion power is radiated at the fixed radius, $L_{\mathrm{acc}}=GM\dot{M}/r_0$. Neglecting the seed mass,This equals $-\Omega /2$ precisely whenThe positive sign in the final printed energy condition is inconsistent with the earlier use of negative gravitational potential energy. Equivalently its positive right side must denote the binding-energy magnitude $|\Omega |$. A general structural coefficient $\Omega =-qG{M}^2/r_0$ leaves a mismatch unless $q=1$ under these other idealizations. Radiation trapped in the object, outflows or release over a different radius also change the budget.

Which temperature? With $M=M_s+\dot{M}t$ and fixed $r_0$, $L_{\mathrm{acc}}\propto M$. If this power emerges thermally from area $4\pi r_0^2$, the accretion-powered effective-temperature scaling iswhen the seed mass is negligible. This is the radiating surface's effective temperature, inferred from the preceding accretion-luminosity calculation. A homologous ideal-gas interior instead has virial temperature $T_c\sim \mu m_{H}GM/(k_B{r}_0)\propto t$. Thus the surface-effective and interior-virial scalings are $t^{1/4}$ and $t$, respectively, under their stated assumptions. Inferring a detailed interior temperature profile would additionally require a structural and transport model.

At fixed volume and temperature, equilibrium minimizes the Helmholtz free energy $F=U-TS$. Mixing increases translational entropy, whereas favourable contacts between chromatin segments can lower the interaction contribution to internal energy. At low enough temperature, favourable contacts can compensate the entropy lost on concentrating the arrays; at higher temperature, the mixing contribution can dominate. This gives the upper-critical dome assumed in (a).

A simple illustrative Flory–Huggins solution theory writes the dimensionless mixing free energy per site asHere $N$ is an effective polymer size, not automatically the number 12, and $A>0$ represents an effective energetic incompatibility of polymer and solvent. The common-tangent construction determines coexisting compositions. Solving $g^{\prime \prime }=g^{(3)}=0$ givesSince $\chi$ decreases on heating, the coexistence interval narrows and closes at $T_c=A/{\chi }_c$. For fixed pressure the corresponding potential is Gibbs free energy, with the enthalpic term in $G=H-TS$; enthalpy and internal energy should not be interchanged without stating the mechanical constraint. Hydration, ion release or temperature-dependent effective interactions can change this simple temperature trend.