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(a) Treat the lake as a heat bath at fixed temperature $T_0=288.15\mathrm{K}$ and the block's heat capacity as constant. Its initial temperature is $T_i=473.15\mathrm{K}$, and $C=mc=355\mathrm{J}{\mathrm{K}}^{-1}$. Although the cooling is irreversible, entropy is a state function, so evaluate the block's change along an imagined reversible thermodynamic process:The lake receives heat $Q=C(T_i-T_0)=65675\mathrm{J}$, and thereforeAdding the two contributions givesThe positive result is consistent with the Second law of thermodynamics for heat transfer across a finite temperature difference. The calculation uses $c=710\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$ exactly as printed in both the original PDF and the transcription; it does not replace the supplied value by a tabulated value for lead.

(b) The block starts and eventually finishes at the lake's temperature, so its net entropy change is approximately zero. Neglecting buoyancy, air resistance and the final immersion depth, the lost gravitational potential energy $mgh$ becomes internal energy in the lake after the motion and waves have dissipated. The lake remains at $T_0$, givingTemporary heating of the block or temporary storage in kinetic energy does not change this final energy and entropy balance once thermal equilibrium is restored.

For a fixed amount of a simple compressible system, write the internal energy as $U(S,V)$. Its exact differential isAssuming the thermodynamic potential has continuous second partial derivatives, equality of its mixed partial derivatives yields the first Maxwell relation:For the second Maxwell relation, use the Legendre transform $H=U+pV$, the enthalpy. Its natural variables are $S,p$ sinceThus $T=(\partial H/\partial S{)}_p$ and $V=(\partial H/\partial p{)}_S$. Equality of the mixed partial derivatives now gives

Use an ideal gas to illustrate the Carnot cycle, with hot and cold heat baths at $T_h=T_2$ and $T_c=T_1$. The four processes are all reversible thermodynamic processes. Lettered states below belong only to this illustration; the engine later in the question uses numbered states.

Starting at $A$, perform the following four stages:

The clockwise enclosed area in the pressure-volume diagram is the positive net work per cycle, $W=\oint pdV=Q_h-Q_c$.

Take inertial axes parallel, with frame $S^{\prime }$ moving at $+v\hat{\mathbf{x}}$ relative to $S$ and origins coincident at zero time. The Lorentz factor is ${\gamma }_v=(1-v^2/c^2{)}^{-1/2}$; $\Delta x,\Delta y,\Delta t$ are separations of the same pair of events in $S$, and primed quantities are their separations in $S^{\prime }$. The transverse coordinates are unchanged. For two events on an inertial clock's worldline,The proper time $\Delta \tau$ is that clock's elapsed time; uniform speed $w$ gives $\Delta t={\gamma }_w\Delta \tau$. Coordinate intervals between spatially separated events must instead be Lorentz-transformed with their spatial separation included.

For the round trip in the station frame, the rendezvous distance is $u{t}_{\mathrm{out}}$, and the return has the same speed and distance. Thus $t_{\mathrm{round}}=2t_{\mathrm{out}}=10a/(\sqrt{7}c)$. Its departure and arrival both occur at the station, so $\Delta x=0$ for these endpoints:This uses the return speed as measured by the station. The elapsed time aboard A, apart from an ideal instantaneous turn, would be $\Delta t_{S,\mathrm{round}}/{\gamma }_u$.

For the crash, A's station-frame velocity components are $(3c/5,\sqrt{7}c/5)$, while B's are $(3c/5,0)$. Here ${\gamma }_u=5/3$ and ${\gamma }_v=5/4$. Four-momentum conservation, with no separately escaping radiation, gives the wreckage totals in the station frame:In B's original inertial frame the same totals areThe invariant wreckage mass is $m\sqrt{14/3}$, exceeding $2m$ because internal energy remains in the wreckage. Its momentum cannot be specified without naming a frame; both natural frames have been given.

The total internal energy and particle number are conserved. The outer container is rigid and isolated, so there is no external work or heat transfer. Its total volume stays $V_0$, while the volume accessible to the gas increases from $V_0/3$ to $V_0$. Neither the gas temperature nor its entropy is generally conserved: this is an irreversible free expansion, with increasing entropy.

Equating the initial and final internal energies, with the same additive constant,soThis is cooling in free expansion of a van der Waals gas. The gas cools because separating attractive particles raises their potential energy at the expense of kinetic energy. The excluded-volume coefficient $b$ cancels from this caloric model. The result assumes a homogeneous equilibrium gas with $V_0/3>Nb$ and positive $T_f$; a two-phase final state would require a coexistence calculation rather than this homogeneous formula. The ideal-gas limit $a=0$ has no temperature change.

Work in the locally planar shock frame, taking $v_u,v_d>0$ towards the surface. For a stationary normal shock wave with negligible radiative loss inside the discontinuity, the Rankine-Hugoniot conditions for a perfect gas express conservation of mass, conservation of momentum, and conservation of energy per unit area asHere $h$ is specific enthalpy, including the pressure work; using only internal energy would omit that work. The gravitational potential is continuous across an infinitesimally thin front. For cold upstream gas, $P_u$ and $h_u$ are negligible, while a monatomic gas has $\gamma =5/3$ and $h=5P/(2\rho )$.

The hydrodynamic equations apply to flow parallel to the guiding magnetic field when its stress contributions cancel across the front. An oblique magnetohydrodynamic shock requires magnetic stresses and Poynting flux; a radiation-mediated front or appreciable escaping radiation requires the corresponding radiative momentum and energy flux terms. These are additional physics, not terms hidden in $h$ here.

With $\gamma =(1-v^2/c^2{)}^{-1/2}$, four-momentum conservation gives total energy $E=(\gamma m_1+m_2)c^2$ and momentum $\mathbf{p}=\gamma m_1\mathbf{v}$. The compound system therefore hasHere $M^2{c}^4=E^2-c^2{p}^2$. The excess over $m_1+m_2$ is retained internal energy, assuming absorption without escaping radiation.

One mole of original atoms contains $(1-\zeta )$ moles of neutral atoms, $\zeta$ moles of protons and $\zeta$ moles of electrons. All are monatomic translational species, so their combined thermal energy is $\frac{3}{2}RT(1+\zeta )$. Taking the neutral ground state as zero internal energy, ionization stores $\zeta N_A{E}_I$. ThusRadiation, excited states and nonideal interactions have been excluded consistently with the elementary Saha ionization equation.

At fixed volume, logarithmic differentiation of the Saha ionization equation givesDifferentiation the internal energy then yieldsAt low temperature $\zeta \to 0$ and $C_V\to 3R/2$. At high nonrelativistic temperature the gas is fully ionized, $\zeta \to 1$, and $C_V\to 3R$. The large intermediate excess reflects energy spent increasing ionization, as well as heating and creating additional translational particles. “Very high” here remains below regimes where relativistic, radiative or pair-creation contributions invalidate the model.

Let the bed be $z=0$ and the free surface $z=h(x,t)$. In a long, shallow layer, vertical acceleration is smaller than gravity, so the vertical Euler equation gives hydrostatic pressure $p=p_{\mathrm{atm}}+\rho g(h-z)$. Incompressibility makes the vertical speed smaller than the horizontal speed by the aspect ratio; irrotationality then makes the leading horizontal velocity depth-independent. The horizontal Euler equation becomesIntegrating $u_x+w_z=0$ across the depth, using bed impermeability $w(0)=0$ and the surface condition $w(h)=h_t+u{h}_x$, yieldsCombining these gives the conservative momentum equation $(hu{)}_t+(h{u}^2+g{h}^2/2{)}_x=0$. Integral mass and momentum balance across a stationary thin jump therefore requireThese conservation laws still apply across a jump where the smooth hydrostatic/irrotational derivation does not resolve the internal turbulent structure.

Put $A=h_{-}$, $B=h_{+}$ and $q=A{u}_{-}=B{u}_{+}>0$, orienting the flow from minus to plus. For a nontrivial jump the momentum condition gives $q^2=gAB(A+B)/2$. Its change in mechanical-energy flux isA passive jump cannot create mechanical energy, so the change is nonpositive. Since $q>0$, $\boxed{h_{+}>h_{-}}$ for a genuine jump; equality describes no jump. The upstream flow is supercritical and the downstream flow subcritical, since $F{r}_{-}^2=B(A+B)/(2A^2)>1$ and $F{r}_{+}^2=A(A+B)/(2B^2)<1$.

The gas-dynamic analogue is a compressive normal shock: density and pressure increase and flow changes from supersonic to subsonic. Entropy increases; directed mechanical energy is irreversibly converted into internal energy. Total gas energy, including internal energy, is conserved across an adiabatic shock. This distinguishes that total-energy balance from the reduced shallow-water mechanical-energy loss.

Let $E$ denote the fluctuating microscopic energy and $U=⟨E⟩$ its mean, at fixed $N,V$. In a canonical ensemble with $\beta =1/(k_{B}T)$,Since $dT/d\beta =-k_B{T}^2$, this givesHere $\Delta U$ denotes an energy fluctuation about the mean, as in the question. The relation is not restricted to ideal gases when the Hamiltonian operator has no explicit temperature dependence. Alternatively, the stated equilibrium free-energy curvature gives a Gaussian energy weight proportional to $\exp [-(\Delta U{)}^2/(2k_B{T}^2{C}_V)]$, with the same variance.

For the relative energy fluctuation of a monatomic ideal gas, $U=3N{k}_{B}T/2$ and $C_V=3N{k}_B/2$, henceTaking room conditions as approximately $T=300$ K and $p=10^5$ Pa, with $V=10^{-6}$ m${}^3$, givesThe mean energy is about $0.15$ J and its standard deviation about $2.49\times 10^{-11}$ J. The tiny relative fluctuation explains why macroscopic equilibrium internal energy behaves as a sharply defined thermodynamic quantity. The estimate uses a thermal reservoir, not a particle reservoir; allowing $N$ itself to fluctuate changes the relevant fluctuation problem.

Excited-state saturation. Consider first a uniform three-dimensional noninteracting gas of identical spinless bosons in a large box. Set the one-particle ground energy to zero. The Bose-Einstein distribution is $\bar{n}(ϵ)=[z^{-1}{e}^{\beta ϵ}-1{]}^{-1}$, with quantum gas fugacity $z=e^{\beta \mu }\le 1$. For dispersion $ϵ=p^2/(2m)$, the excited density of states isUsing the thermal de Broglie wavelength ${\lambda }_T=\sqrt{2\pi {\hslash }^2/(m{k}_{B}T)}$, integration givesThe polylogarithm is ${\mathrm{Li}}_s(z)={\sum }_{\ell \ge 1}{z}^{\ell }/{\ell }^s$. Its finite value at $z=1$ implies a maximum excited population $V\zeta (3/2)/{\lambda }_T^3$, with $\zeta (3/2)\simeq 2.612$. Once this is less than $N$, the excess particles occupy the ground state macroscopically. This is Bose-Einstein condensation, rather than occupation of a state by merely one or a few atoms.

Transition and condensate fraction. Setting the saturation number equal to $N$ gives the homogeneous ideal Bose gas condensation temperatureBelow this temperature, $\mu$ approaches zero from below in the thermodynamic limit, and particle conservation givesAbove $T_c$ there is no macroscopic condensate and $\mu <0$ adjusts to keep $N$ fixed. At finite $N$, the chemical potential remains slightly below the ground energy and the transition is rounded.

Energy and the role of geometry. The excited-state internal energy isThe zero-energy condensate contributes no term to this energy in the ideal model. Below $T_c$,Condensation depends on low-energy state counting: for a uniform quadratic-dispersion gas, a finite-temperature bulk condensate requires spatial dimension greater than two. A trap changes the density of states and therefore the temperature law; a three-dimensional harmonic trap gives a cubic, rather than $3/2$, condensate-fraction exponent. Finally, an ideal-gas condensate should not automatically be called a superfluid: its free quadratic spectrum has zero critical speed by the Landau criterion. Interactions matter for the robust collective-flow properties of real dilute condensates.

Gibbs free energy and its minimum. For fixed atom number $N$, there are $N+n$ sites andconfigurations with $n$ indistinguishable vacancies. Their configurational entropy is $k_B\ln W$. Relative to the vacancy-free Gibbs free energy $G_0$, creating $n$ vacancies costs energy $nϵ$ and pressure-volume work $npv$. Put $A=ϵ+pv>0$. ThenAt leading thermodynamic order, Stirling's approximation givesThe minimum obtained from the stationary point below isIf the perfect crystal's internal energy and other entropy are chosen as zero, its pressure term is $G_0=pNv$. More generally the information given fixes the vacancy contribution rather than the absolute bulk free energy. The pressure-volume term must be retained because the number of sites, and hence the volume, changes.

Equilibrium vacancy population. Differentiating the extensive free energy at fixed $N,T,p$ givesThe stationary point is consequently the unique interior minimum. Write $q=e^{-A/(k_{B}T)}$; its condition is $n/(N+n)=q$, giving the requested thermodynamic result for equilibrium vacancies with variable lattice size,Substitution back into $G$ cancels the terms containing $n_{\ast }$ and yields the boxed minimum. In the dilute-defect limit $A\gg k_{B}T$, $n_{\ast }\simeq N{e}^{-A/(k_{B}T)}$. The Bose-like denominator is a consequence of the changing number of lattice sites and combinatorial entropy; it does not assert that vacancies here are an ideal gas of bosonic particles.

There is a finite-size correction to a variable-site vacancy model if “average” is interpreted exactly. The microscopic count written above gives the relative isothermal-isobaric sumsoIts relative equilibrium free energy is $-k_{B}T\ln \mathcal{Y}=(N+1)k_{B}T\ln (1-q)$. Thus the source's $N$ numerator is correct to leading extensive order, but is not the exact finite-$N$ mean for that stated configuration count. A free-energy minimum is likewise not generally identical to the mean of a finite distribution. These distinctions vanish relatively as $N\to \infty$.

Thermal volume. Each atom or vacancy occupies volume $v$, so at the same thermodynamic accuracyFor the exact counting model, replace $n_{\ast }$ by $(N+1)q/(1-q)$ in $v(N+⟨n⟩)$. The vacancy-free limit is $Nv$ as $T\to 0$. This model isolates expansion from vacancies, taking $v$ and $ϵ$ constant; it omits ordinary phonon thermal expansion. Its growth at high temperature should not be extrapolated to arbitrarily large vacancy fractions in a stable crystal, and $A>0$ is essential for a normalizable variable-site ensemble.

Availability and total entropy. The reservoir changes by $d{U}_R=-dU$ and $d{V}_R=-dV$. Its particle number and the system's particle number are fixed. The reservoir's first law of thermodynamics therefore givesAdd the system's entropy change. Because $T_R,p_R$ are constant,Thus define the availabilityAn additive constant in $A$ is immaterial; its value above the equilibrium reference is the exergy. The Second law of thermodynamics gives $\Delta S_{\mathrm{tot}}\ge 0$, hence $\Delta A\le 0$. A stable equilibrium maximizes total entropy over the allowed variations and therefore minimizes $A$. The environmental $T_R,p_R$ enter this definition even when the system has not yet reached their values. Because both thermal and mechanical exchanges are allowed, equilibrium here includes mechanical balance; equality of temperatures alone does not ensure the minimum.

Probability and curvature. For a fixed value of a macroscopic coordinate $x$, let $S(x)$ count the compatible system microstates. The product of the system and reservoir multiplicities isAll factors independent of $x$ enter a normalization constant, givingHere $A(x)$ is the constrained or effective availability for this coordinate, including the multiplicity of any eliminated variables. A nonuniform coordinate measure must likewise be retained or incorporated in that effective availability.

At a stable interior minimum, $A^{\prime }(x_0)=0$ and $A^{\prime \prime }(x_0)>0$. Its Taylor expansion givesKeeping the quadratic term yields a normal distribution,This is availability curvature and equilibrium fluctuations. The quoted fluctuation formula holds at quadratic order in the Gaussian fluctuation approximation; it is not an exact identity for an arbitrary nonlinear $A(x)$. Higher terms can shift the mean away from its most probable value $x_0$.

Bubble equilibrium. Keep the number of air molecules fixed and include the surface energy in the system's total internal energy. For constant surface tension $\Gamma$, this contribution is $\Gamma \Omega =\Gamma \alpha V^{2/3}$, with no surface entropy under the stated approximation. Its differential isHenceIndependent thermal and volume variations give $T=T_R$ andThe air pressure is evaluated at $V_0$; since the bubble is closed it generally depends on volume, so this equation is implicit until an air equation of state is supplied. For a sphere, $\alpha =(36\pi {)}^{1/3}$, $\Omega =4\pi R^2$, and the pressure jump becomes $2\Gamma /R$, the Young–Laplace equation for a single gas–liquid interface. There is no second surface as in a thin soap film.

Isothermal volume fluctuations. With the air maintained at $T_R$, eliminate its entropy in favor of its Helmholtz free energy:The identity $(\partial F_{\mathrm{air}}/\partial V{)}_{T,N}=-p_{\mathrm{air}}$ givesUsing the curvature result,These are the volume fluctuations of a closed gas bubble. Stability requires the denominator to be positive. The negative surface contribution reflects the concavity of $V^{2/3}$; gas compression supplies the positive restoring curvature. For ideal air, ${\kappa }_T^{-1}=p_{\mathrm{air}}$, and mechanical equilibrium reduces the denominator toThe units are consistent: $D$ is pressure, so $k_B{T}_R{V}_0/D$ has units ${\mathrm{m}}^6$, the units of a volume variance. Holding $p_{\mathrm{air}}$ externally fixed while varying $V$ would remove the gas-compression term and change this stability problem; that is not the fixed-particle-number situation here.

Numerical size of the fluctuations. The volume conversion is important:Use the specified $\alpha =4.84$, $\Gamma =0.072\mathrm{N}{\mathrm{m}}^{-1}$ and $p_R=10^5\mathrm{Pa}$. ThenWith the Boltzmann constant $k_B=1.380649\times 10^{-23}\mathrm{J}{\mathrm{K}}^{-1}$,Thus the fractional root mean square fluctuation and fractional variance are respectivelyThe absolute root mean square fluctuation is $2.01\times 10^{-19}{\mathrm{m}}^3$. The very small fractional value supports the local Gaussian fluctuation approximation. This estimate neglects exchange of dissolved air, variations of surface tension, shape fluctuations and temperature fluctuations, consistently with the prescribed fixed-number, isothermal bubble model.

With a thermally insulating boundary, $\delta Q=0$. If the slow piston motion is frictionless and reversible, it is an isentropic process:The first law of thermodynamics gives $dU=-pdV$, so expansion lowers the internal energy and temperature. For constant heat capacity ratio $\gamma$, $p{V}^{\gamma }$ and $T{V}^{\gamma -1}$ are constant combinations of state functions. Insulation alone does not guarantee constant entropy: an irreversible adiabatic process can produce entropy. The conclusion uses the intended reversible, quasistatic piston idealization.

A state function is determined by the equilibrium state, rather than the path used to reach it. Its differential is exact and its change around a closed thermodynamic cycle is zero. Internal energy, entropy and enthalpy are examples; transferred heat and work are not state functions.

For slow expansion in contact with a large thermal reservoir, the fixed temperature of that reservoir maintains an isothermal process. For a fixed amount of ideal gas, internal energy and enthalpy depend only on temperature, soThe ideal gas law also gives $pV=\mathrm{constant}$. For a reversible expansion the absorbed heat equals the work done by the gas, while its entropy increases by $n_{\mathrm{mol}}R\ln (V_2/V_1)$; constant internal energy does not mean zero heat transfer.

In the one-mole Van der Waals equation, $b>0$ represents excluded volume: the volume available to molecular motion is $V-b$, rather than $V$. Subtracting $b$ in the denominator therefore increases the repulsive contribution to pressure. The $a>0$ term represents intermolecular attraction, which reduces the outward momentum flux at a wall and hence appears as the negative correction $-a/V^2$. The two minus signs do not mean that both physical effects lower pressure.

For a reversible isothermal process along a homogeneous gas branch with $V_2>V_1>b$, the work done by the gas isTo find the internal energy change without guessing a caloric law, combine $dU=TdS-pdV$ with the Helmholtz free energy Maxwell relation $(\partial S/\partial V{)}_T=(\partial p/\partial T{)}_V$. This givesThereforeThe attractive contribution $-a/V$ becomes less negative as the gas expands: separating attracting particles requires energy. This differs from the ideal gas result $\Delta U=0$ at fixed temperature. The required heat is $Q=\Delta U+W_{\mathrm{by}}=RT\ln [(V_2-b)/(V_1-b)]$. Below the critical temperature, an equilibrium path crossing a liquid–gas coexistence region needs the equilibrium coexistence construction rather than the unstable homogeneous Van der Waals isotherm.

For a single-particle state of energy $ϵ$, put $z=e^{\mu /(k_{B}T)}$. The equilibrium mean occupancies areThe Maxwell-Boltzmann distribution is the dilute classical limit. The Fermi-Dirac distribution applies to identical fermions with at most one particle in each complete single-particle state by the Pauli exclusion principle. The Bose-Einstein distribution applies to identical bosons with no such upper occupation bound. The chemical potential fixes total particle number; these are mean state occupancies, not three separately normalized single-particle probability densities. For a number-conserving ideal Bose gas, $\mu$ cannot exceed the lowest single-particle energy.

Take $\mathbf{k}$ to be a wavevector and count periodic-box states with cell volume $(2\pi {)}^3/V$ in momentum space, parameterized by $\mathbf{k}=\mathbf{p}/\hslash$. Write $g_s$ for internal degeneracy. The number of states in a spherical shell is $g_{s}V4\pi k^{2}dk/(2\pi {)}^3$. Since $ϵ=\alpha k$,Thus the one-particle canonical partition function isHere $\beta =1/(k_{B}T)$, $g_s=1$ for one nondegenerate branch, and $g_s=2$ for the spin-half case below. The integral uses ${\int }_0^{\infty }{u}^2{e}^{-u}du=2$. For classical indistinguishable particles, $Z_N=Z_1^N/N!$, givingThe factor three in the energy arises from the linear dispersion relation and three-dimensional state counting; the usual nonrelativistic quadratic-dispersion value $3N{k}_{B}T/2$ does not apply.

For spin-half fermions at $T=0$, all states up to the Fermi energy are occupied. With $g_s=2$,Integrating the occupied energy states givesThese are the corresponding ultrarelativistic ideal Fermi gas formulae with general slope $\alpha$. The nonzero ground-state internal energy and degeneracy pressure reflect the Pauli exclusion principle: particles cannot all occupy the lowest-momentum state. Formally extrapolating the classical equations gives $U,p\to 0$, but Maxwell–Boltzmann statistics is not valid at fixed positive $n$ as $T\to 0$.

At large $T$, the number equation gives $z\simeq N/Z_1\ll 1$, so expanding the Fermi-Dirac distribution recovers the Maxwell-Boltzmann distribution. A convenient crossover scale is the Fermi temperature,For $T\gg T_F$, both gases have $p\simeq n{k}_{B}T$. At fixed $n$, the Fermi curve begins at $p_0=n{ϵ}_F/4$, has zero initial slope, lies above the classical line, and approaches it at high temperature. For example its low-temperature expansion is $p/(n{ϵ}_F)=1/4+({\pi }^2/6)(T/T_F{)}^2+\cdots$. The plot evaluates the finite-temperature occupation integrals at fixed number density.

If the source's symbol $\mathbf{k}$ is interpreted literally as momentum rather than a wavevector, the state-space cell is $h^3/V$. In that convention replace $\alpha$ in the wavevector-based coefficients and energy scales above by $\hslash \alpha$. The dimensionless thermodynamics and the relation $p=U/(3V)$ are unchanged.

A photon rocket converts stored mass-energy into a collimated beam of photons. In its instantaneous rest frame, energy conservation gives $c^{2}dM=-d{E}_{\gamma }$, and momentum conservation gives $Mcd\eta =d{E}_{\gamma }/c$. Therefore a monotone change in rapidity costs $M_f/M_i=e^{-|\Delta \eta |}$. The rocket's rest mass includes its fuel and stored internal energy; inefficiency or an angular spread of the exhaust increases the fuel requirement.