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Because the working substance returns to its original thermodynamic state, its net entropy change is zero. There is no entropy change on either reversible ideal-gas adiabat; on each isothermal process the entropy transfer is the exchanged heat divided by the bath's absolute temperature. HenceThe first law of thermodynamics over a complete cycle gives $W=Q_h-Q_c$. The thermal efficiency is thereforeThe use of kelvins is essential. This result does not depend on the working substance. For the ideal gas illustration, the two reversible ideal-gas adiabats imply $V_C/V_B=V_D/V_A=(T_h/T_c{)}^{1/(\gamma -1)}$, so $\ln (V_B/V_A)=\ln (V_C/V_D)$; the explicit heat expressions in (a) give the same ratio.

For a diatomic ideal gas in the usual regime with three translational and two rotational quadratic degrees of freedom, but frozen vibrational excitation, the equipartition theorem givesThe Brayton cycle efficiency is thereforeThe constant-pressure expansion has $T_3/T_2=V_3/V_2=2$, while the compression has $T_2/T_1=10^{2/7}=1.93070$. The same relations give $T_4/T_1=2$. Thus, in units of the minimum temperature,The maximum and minimum temperatures are $T_3$ and $T_1$. A Carnot cycle between those extremes hasThe difference is about $25.9$ percentage points. In the Brayton cycle, heat is supplied and rejected over ranges of temperature; the whole input is not supplied at $T_3$, nor is the whole rejection made at $T_1$. This explains the lower efficiency even for mechanically ideal stages. Absolute state temperatures are not supplied, so $\gamma =7/5$ is an explicit standard diatomic-gas approximation rather than a conclusion about all possible rotational and vibrational excitation regimes.

Potential temperature and a parcel. For a dry ideal gas of fixed composition, the first law of thermodynamics per unit mass isA reversible adiabatic process has $\delta q=0$. Taking $c_p,R_d$ constant gives $d\ln T=\kappa d\ln p$, where $\kappa =R_d/c_p$. Integrating to a fixed reference pressure $p_0$ defines the potential temperatureIt is the temperature a parcel would have after reversible dry adiabatic process taking it to $p_0$. Equivalently $ds=c_{p}d\ln \Theta$ under these assumptions, making potential temperature a thermodynamic entropy label useful for diagnosing stratification and air-mass motion. Latent heating, radiation, mixing and variable composition require modifying the conservation statement.

An air parcel is a small material volume followed with the flow. The parcel model treats it as internally approximately uniform, large compared with molecular scales and small compared with environmental variation scales. Rapid acoustic wave adjustment makes its pressure approximately equal to the surrounding pressure, while slower heat and mass exchange can be neglected during an idealized displacement. It is not a rigid container or an isolated molecule.

At the parcel's height, environmental hydrostatic pressure supplies an upward pressure force ${\rho }_{e}g$ per unit volume. Dividing by parcel density gives the upward accelerationThe last steps use equal pressure, equal gas constants and the same reference pressure. Moisture-dependent density requires virtual temperature or its potential-temperature counterpart. Nonhydrostatic pressure perturbations and entrainment are excluded from this elementary buoyancy expression.

The convective available potential energy is the positive buoyancy work per unit mass between the level of free convection and the equilibrium level:One sums positive-buoyancy layers when there is more than one. Negative work below free convection contributes convective inhibition rather than positive CAPE. In a lossless parcel estimate, conversion of CAPE to vertical kinetic energy gives $w_{\max }\simeq \sqrt{2\mathrm{CAPE}}$; pressure work beyond this model, mixing and precipitation can change that speed.

Linear stability. Let $\xi$ denote displacement from a neutrally buoyant reference altitude. Dry motion through an adiabatic process retains ${\Theta }_p={\Theta }_e(0)$, while ${\Theta }_e(\xi )={\Theta }_e(0)+\xi {\Theta }_e^{\prime }(0)+O({\xi }^2)$. The linear vertical momentum equation is thereforeFor $N_0^2>0$, $\xi =C\cos N_{0}t+D\sin N_{0}t$: stable buoyancy oscillations with period $2\pi /N_0$. For $N_0^2=0$, $\xi =C+Dt$: neutral stratification with no restoring force. For $N_0^2<0$, growing and decaying exponentials occur, indicating convective instability. The growing solution is generic unless the initial data select only the decaying branch. These are local small-displacement results with an approximately constant environmental gradient.

Lapse-rate estimate. Define $\Gamma =-dT/dz$. Hydrostatic balance and the ideal gas law imply $p^{\prime }/p=-g/(R_{d}T)$, soIn the free-parcel-oscillator interpretation, identifying the ridge encounter interval with the buoyancy period gives $P=10^4/20=500\mathrm{s}$ and $N_0=2\pi /P=0.01257{\mathrm{s}}^{-1}$. With $T=293.15\mathrm{K}$, $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$ and $c_p\simeq 1004\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$,The frequency identification is an extra modeling assumption. A stationary mountain gravity wave generally obeys ${\hat{\omega }}^2=N_0^2{k}_h^2/(k_h^2+m_z^2)$ without rotation, with $\hat{\omega }=U{k}_h$. Ridge spacing fixes $\hat{\omega }$, not the vertical wavenumber $m_z$. For example, $m_z=k_h$ allows the same spacing and wind with $N_0=\sqrt{2}\hat{\omega }$, and a lapse rate about $0.3\mathrm{K}\mathrm{k}{\mathrm{m}}^{-1}$. Thus $5.1$ is the requested parcel-period estimate, not a uniquely identifiable lapse rate for every possible mountain-wave field; a resolved wave field needs its vertical structure as well.

Take a weak line in the radiative sense: the integrated slant optical depth is small wherever absorption materially contributes, not merely a small greenhouse-gas mass mixing ratio. Normalize every local spectral line profile so that $\int {\kappa }_{\nu }(p)d\nu =S$, with the line strength independent of pressure and, as stipulated, temperature. Keep a finite Doppler or natural core, or a Voigt profile, so that a pressure-only zero-width core at the top does not invalidate a uniform optical-depth expansion. The line must also be narrow enough to use $B_{{\nu }_0}$ across it.

To first order,Thus the detailed normalized line shape drops out, even if its width varies with pressure. Substituting in the deficit in c givesThe neglected term is second order when the relevant squared optical depths are integrable and uniformly small. A very narrow saturated line need not be weak simply because $q\ll 1$. More generally the displayed first-order integrated limit follows from the finite integrated opacity even if an idealized line centre makes a pointwise expansion nonuniform; a finite physical core is the straightforward sufficient assumption.

For an explicitly perturbative temperature estimate, evaluate the bracket at $T_{\mathrm{bare}}$:At fixed $p_s,g,q,S$ and radiative conditions, increasing $\kappa =R_b/c_{p,b}$ makes every level $0<r<1$ cooler relative to the surface, enlarging the bracket and the warming. Increasing $c_{p,b}$ at fixed $R_b$ has the opposite effect. The limit $\kappa \to 0$ is isothermal and gives zero line-induced deficit. A change of background molecular mass alone need not change this leading result: for ideal gases with the same active degrees of freedom, $R_b$ and $c_{p,b}$ scale together and their ratio is unchanged. Comparing only the height-coordinate lapse rate $g/c_{p,b}$ would miss that the absorber column is naturally a pressure-coordinate quantity.

For a unit mass of ideal gas, the first law of thermodynamics in enthalpy form is $\delta Q=c_{p}dT-vdp$. A reversible adiabatic displacement has $\delta Q=0$, and the ideal gas law gives $v=RT/p$. ThusThe atmospheric profile approaches this dry adiabat when convection efficiently mixes potential temperature and there is no condensation or other strong diabatic source. The first law by itself determines a parcel path, not every possible environmental temperature profile; the convective mixing assumption is needed. Hydrostatic balance gives the corresponding height lapse rate $dT/dz=-g/c_p$.

For rigid linear molecules with translation and rotation active but vibration frozen, the molecular heat capacities are $C_V=(5/2)R_u$ and $C_P=(7/2)R_u$, hence $R/c_p=2/7$. Earth is dominated by the diatomic gases nitrogen and oxygen. Mars is dominated by carbon dioxide, which is triatomic but linear and has the same two rotational degrees of freedom. In the rigid-molecule approximation it therefore has the same ratio. The value is not exact for carbon dioxide when its bending vibrations contribute appreciably to heat capacity, and variable heat capacities give a nonconstant dry-adiabatic exponent. Here $\gamma$ denotes $R/c_p$, not the distinct adiabatic index $c_p/c_v$.

For an ideal gas of fixed molecular composition, translational energy is supplemented by rotational, vibrational and eventually electronic excitations as their level spacings become thermally accessible. Activating more internal degrees of freedom generally increases the heat capacity. In a harmonic vibrational mode the contribution rises from nearly zero when $k_{B}T\ll \hslash \omega$ to its classical limit when $k_{B}T\gg \hslash \omega$; summing modes with different frequencies gives a smooth increase over an intermediate temperature range.

A monatomic gas has no molecular rotations or vibrations, so its mass-specific $c_p$ is nearly constant until electronic excitation or ionization becomes important: $B$ is then small. A simple diatomic gas can also have a small slope when its rotations are already active but its vibrational modes remain frozen. Large flexible polyatomic molecules have many internal modes, often with relatively low excitation energies, and can give a larger slope over their activation range.

The linear law is a local approximation, not an asymptotic result. A finite set of modes approaches finite classical heat-capacity contributions, whereas $A+BT$ grows without bound. At still higher temperatures dissociation and ionization change the composition and gas constant, invalidating the fixed-molecule ideal-gas model as well. The constants have units $[A]=\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$ and $[B]=\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-2}$.

Adding the two flux solutions givesThe supplied pressure–optical-depth law gives $D-\tau =Dp/p_0$. With the exponential pressure model $p=p_0{e}^{-z/H}$,It approaches the finite atmospheric top temperature $T_{\infty }$ and has the lowermost-air value $T(0)=T_{\infty }(1+D{)}^{1/4}$. Its lapse rate isThe stated approximate sign is appropriate physically: a nonisothermal ideal gas does not have an exactly constant hydrostatic scale height $RT/g$. Treating $H$ as a representative constant is an additional vertical-coordinate approximation. Algebraically the displayed profile is exact for the prescribed exponential pressure and two-stream equations.

Let $R_A,R_B$ be the specific specific gas constants and $c_{pA},c_{pB}$ the specific heat capacities of the noncondensible gas and vapour. Write $e_s(T)$ for the saturation vapor pressure, $r$ for vapour mass per unit dry-gas mass, and $ϵ=R_A/R_B$. For a dilute ideal gas mixture,The second equation is the Clausius-Clapeyron equation, neglecting condensed-phase volume and taking constant latent heat $L$. The first law of thermodynamics for saturated ascent, with condensate removed and negligible condensate heat capacity, is to leading orderSince $dr=r[LdT/(R_B{T}^2)-d\ln p]$, the resulting moist adiabat has logarithmic slopeConsequently a sufficient dry adiabat criterion isThe first two conditions justify neglecting the vapour's effects on the gas law and sensible heat capacity; the last two suppress latent heating. With $x=R_A/c_{pA}$ and $\beta =L/(R_{B}T)$, the latent conditions are $\beta e_s/p\ll 1$ and $x{\beta }^2{e}_s/p\ll 1$. Usually $x\beta \gg 1$, so the squared-latent term gives the stronger condition. At sufficiently low surface temperature, the exponential smallness of $e_s(T_s)$ makes these conditions hold andThese are local conditions along the column. They can be checked at the surface when $x{\beta }_s>1+2x$: on this dry profile the strongest latent parameter is proportional to $\exp [(1+2x)\zeta -{\beta }_s(e^{x\zeta }-1)]$ and decreases upwards. Without that restriction a surface-only test need not bound every altitude. If suspended condensate is retained, its own heat capacity and loading must also be negligible; saturation alone does not specify those quantities.

Assume an ideal gas dominated by carbon dioxide, constant mean molar heat capacity $C_p$, and a dry convective adiabat connecting the stated emitting layer to the ground. Conservation of entropy gives $C_{p}d\ln T=Rd\ln p$, where $R$ is the universal molar gas constant. ThusThe estimate extrapolates across a pressure ratio of $450$. It neglects radiatively stable layers, variable heat capacity, departures from an ideal gas and changes in composition; these approximations limit its accuracy. A brightness temperature observed from space cannot generally be equated to a unique material level, but this calculation uses the layer interpretation explicitly supplied by the question.

Notation. The source uses $p_s$ for both the hypothetical dry surface pressure and the actual surface pressure after evaporation. Distinguish them as $P_d$ and $P$, respectively, and interpret the specified pressure thickness $\Delta p$ as fixed. This conserves the original hydrogen inventory and permits $P$ to vary. Reading both occurrences as one fixed number would be inconsistent with adding water vapor to a closed atmospheric column mass budget.

Let $E(T)$ denote the saturation pressure of water, let $x$ be the mole fraction of water vapor in the mixed layer, and let $M_d,M_v$ be the molar masses of hydrogen and water. Surface phase equilibrium and the ideal gas law giveBecause the layer has constant temperature, its relative humidity isThus every level strictly above the surface is subsaturated: hydrostatic balance makes $p<P$, whereas the saturation pressure stays constant. The result does not require a dilute water vapor approximation.

The Clausius-Clapeyron relation, neglecting the specific volume of liquid water and treating the water vapor as an ideal gas, givesHere $T_t,E_t$ are the triple point values, $L_v$ is the specific latent heat, and $R$ is the molar gas constant. All temperatures in these expressions are absolute. As throughout the question, this is a constant-latent heat, ideal gas model, not an extrapolation valid through the real critical point of water.

For a well-mixed ideal gas mixture, its mean molar mass and water vapor mass fraction areBy hydrostatic balance, the mixed layer has total mass per unit area $\Delta p/g$, of which $w_v\Delta p/g$ is water vapor. The remaining atmosphere contains no water vapor. Consequently conservation of hydrogen mass givesEquivalently the requested quadratic equation isFor each $T_s$, calculate $E$, take the positive root of this quadratic equation, and retain it only if $0\le x<1$ and $P>\Delta p$. The last condition leaves a nonempty dry layer. In particular the dry constituent's surface partial pressure is not generally $P_d$: $P_d$ measures its column weight, and the hydrogen and water vapor have different molar masses.

A convenient parameter is the water vapor mole fraction itself. The mass conservation calculation yieldsBoth functions increase for $0\le x\le 1$. At the endpoint,Therefore the physically admissible hydrogen–water vapor mixture requiresBeyond this bound, the formal positive quadratic equation root requires $x>1$, equivalently a negative hydrogen partial pressure. It is the physical constraints, rather than loss of real roots, that end the solution branch. For a concrete example, put $P_d=\Delta p=1$ in consistent pressure units, $ϵ=9$, and $E=3$. The positive root of $P^2+23P-51=0$ is about $2.037$, giving $x\simeq 1.473$, which is impossible.

The limiting temperature is obtained from $E(T_{\ast })=P_d+\Delta p$:This finite limit exists above $T_t$ if $E_t<P_d+\Delta p<E_t\exp [L_v/(R_v{T}_t)]$. The upper qualification matters mathematically: the stipulated constant-latent heat formula has a finite limiting saturation pressure as $T\to \infty$, so for an arbitrarily large unspecified $P_d+\Delta p$ the printed assertion of a finite breakdown temperature is not unconditional. For a concrete formal counterexample, choose $P_d+\Delta p=10^{15}\mathrm{Pa}$: with the usual triple-point constants the model has $E(\infty )\simeq 2.5\times 10^{11}\mathrm{Pa}$, so it never reaches the proposed inventory limit. This is a counterexample inside the stipulated equations, not a physically credible high-pressure ideal gas ocean. Actual liquid water also ceases to exist above its critical point, before such an extrapolation would make sense.

Physically, the imposed finite-pressure-thickness layer can hold no more than $\Delta p/g$ of water vapor per unit area. As $x\to 1$, it expels all its hydrogen into the dry upper layer. A still larger surface saturation pressure cannot coexist with this geometry. The layer must deepen, the temperature profile must change, or the assumed phase equilibrium must fail. If $P_d<\Delta p$, the additional restriction $P>\Delta p$ can also exclude the low-temperature end of the branch.

The constants $R_a,R_b,R_d$ are effective reservoir capacities: stored carbon dioxide mass anomaly per unit effective concentration anomaly, in GtCO${}_2$ per ppmv. Only $C_a$ is a literal atmospheric mole fraction; $C_b,C_d$ are scaled reservoir variables chosen to share its equilibrium value. The exchange coefficients $r,s$ are conductances, in GtCO${}_2$ per ppmv per year: $r(C_a-C_b)$ is transfer from the atmosphere to the fast reservoir and $s(C_b-C_d)$ transfer from that reservoir to the deep ocean. They are not by themselves inverse times; division by a reservoir capacity gives an inverse-time rate.

Hydrostatic balance estimates total atmospheric mass as $M_{\mathrm{air}}=4\pi R_E^2\bar{p}/g$. One ppmv of a dilute ideal gas is a mole fraction $10^{-6}$, so, with dry-air molar mass $M_{\mathrm{mol},\mathrm{air}}\simeq 28.97$ g/mol and carbon dioxide molar mass $44.01$ g/mol,The final factor converts kilograms to gigatonnes. This is carbon dioxide mass, not the smaller carbon mass.

This is the Joule-Thomson coefficient, ${\mu }_{JT}=(\partial T/\partial P{)}_H$, describing a temperature change during an isenthalpic pressure change. Here $H$ is enthalpy, $C_P=(\partial H/\partial T{)}_P$ is the heat capacity at constant pressure, and $V$ is volume, all for the same chosen amount of substance. Starting with $dH=TdS+VdP$ and using the Gibbs free-energy Maxwell relation $(\partial S/\partial P{)}_T=-(\partial V/\partial T{)}_P$ givesSetting $dH=0$ provesIn a steady insulated throttle or porous plug, negligible kinetic- and potential-energy changes make the upstream and downstream enthalpies equal. A positive coefficient gives cooling on a pressure decrease; a negative one gives warming. For an ideal gas, $V\propto T$ at fixed $P$, so the coefficient is zero. Real-gas interactions permit cooling and an inversion curve, making the effect useful in gas refrigeration and liquefaction.

The measurement of the Joule-Thomson effect with a porous plug compares pressures and thermometers on the two sides of an insulated flow restriction. Suppressing external heat exchange and separating the readings from high-speed jets are part of making the enthalpy argument applicable; an arbitrary adiabatic expansion need not be isenthalpic.

Entropy is a thermodynamic state function. For a reversible heat transfer, $dS=\delta Q_{\mathrm{rev}}/T$; the Clausius inequality generalizes this to irreversible processes. The combined entropy of a system and its surroundings does not decrease. A subsystem can lose entropy by exporting heat or matter, so the second-law statement must specify the whole isolated combination. In a simple compressible system,which identifies temperature, pressure and chemical potential as derivatives of the energy representation of the state function. Thermodynamic equilibrium maximizes entropy under fixed total energy, volume and particle number. A heat reservoir instead leads to minimizing the Helmholtz free energy at fixed $T,V,N$, or the Gibbs free energy at fixed $T,P,N$.

Statistical mechanics explains the state count. In a microcanonical ensemble, equally probable compatible microstates give Boltzmann's entropy formula, $S=k_B\ln \Omega$. With general discrete probabilities, the Gibbs entropy isThe microcanonical formula follows by setting every nonzero probability to $1/\Omega$. For a canonical ensemble, $p_j=e^{-\beta E_j}/Z$, soThis turns microscopic energy levels and their populations into a measurable thermodynamic quantity. Classical phase-space counting needs a consistent microscopic normalization and indistinguishability convention; quantum state counting supplies that normalization. Independent systems have multiplicative state counts and therefore additive entropies.

For an ideal gas freely expanding from volume $V$ to $2V$ in an insulated enclosure, energy remains fixed while each particle gains twice the accessible position volume. The state count increases by $2^N$, giving $\boxed{\Delta S=N{k}_B\ln 2}$. The actual irreversible process transfers no heat. A reversible isothermal comparison path transfers heat $N{k}_{B}T\ln 2$ and computes the same entropy change because the initial and final thermodynamic equilibrium states are the same.

Calorimetry supplies another connection to experiment: at fixed pressure and within one phase, $S(T)-S(0)={\int }_0^T{C}_P(T^{\prime })d{T}^{\prime }/T^{\prime }$, with latent-heat contributions $\mathcal{L}/T$ at first-order transitions. The third law of thermodynamics fixes $S(0)=0$ for an ideal perfect crystal with a unique ground state. Degeneracy or frozen disorder can produce residual entropy. Entropy thus links heat-capacity measurements, phase thermodynamic equilibrium and the number of accessible microscopic configurations.

For spherical hydrostatic equilibrium, the stellar structure equations includeEliminate $\rho$ to obtain $dP/dm=-Gm/(4\pi r^4)$. Since $r(m)<R_{\odot }$ through the interior, integration givesWith negligible surface pressure, this provesThis is a conservative lower bound; concentrating mass near the centre increases the required pressure.

For the temperature inference assume predominantly ideal gas support, a nearly fully ionized hydrogen–helium mixture with mean molecular weight $\mu \simeq 0.61$, and the hottest region at the centre. A rigorous scale estimate follows from the stellar virial theorem, rather than substituting the mean mass density for the unknown central mass density. With negligible surface pressure,The ideal gas law gives $\int PdV=k_B\int Tdm/(\mu m_p)$. The mass-weighted mean temperature consequently obeysThis establishes several million kelvin under the stated physical assumptions. The central pressure inequality alone does not give a central temperature bound without an equation of state and information about mass density or an integrated argument.

Assume a dilute, nondegenerate, nonrelativistic ideal gas in local thermodynamic equilibrium, with ground-state atomic hydrogen, protons and electrons, and neglect excited atomic levels and interactions. Let the conserved number of hydrogen nuclei per volume be $n_0$, so $n_e=n_p=\zeta n_0$ and $n_H=(1-\zeta )n_0$. Equality of chemical potentials for $H\rightleftharpoons p+e$ gives the Saha ionization equationThe factor two is electron spin; $g_p$ and $g_H$ must be defined consistently with nuclear spin. The small difference between the atomic and proton translational partition functions has been neglected. ThereforeFor the ground-state convention $g_p=1$, $g_H=2$, the statistical prefactor $2g_p/g_H$ is one. At fixed volume $n_0$ is constant, whereas an isobaric heating process would not have a constant $A$.

For a unit mass of ideal gas, $h=u+P/\rho =u+R_{a}T$. Since $u$ depends only on temperature, $c_p=dh/dT=c_v+R_a$, giving $\boxed{R_a=c_p-c_v}$. For a dry parcel in adiabatic, quasistatic ascent, the first law of thermodynamics is $c_{p}dT={\rho }^{-1}dP$. Environmental hydrostatic equilibrium gives $dP/dz=-\rho g$, so a convectively well-mixed layer has the dry adiabatic lapse rateThis assumes negligible condensation, approximately constant $g$, well-mixed composition and efficient convection; a radiative or stably stratified layer need not follow it. For mole fractions $x_N,x_C$ and molar masses $M_N,M_C$, $R_a=R/(x_N{M}_N+x_C{M}_C)$ and $c_p=[x_N{C}_{p,N}+x_C{C}_{p,C}]/(x_N{M}_N+x_C{M}_C)$. Molecular vibrational excitation can make $c_p$ depend on temperature; then ${\Gamma }_d=g/c_p(T)$. Nitrogen dominance near terrestrial temperatures gives the familiar order $10\mathrm{K}\mathrm{k}{\mathrm{m}}^{-1}$.

Neglect disk self-gravity and take $|z|\ll r$ around a central point mass $M_{\star }$. Its vertical gravitational acceleration isVertical hydrostatic equilibrium is $dP/dz=\rho g_z$. For a vertically isothermal ideal gas, $P=\rho c_s^2$ and $c_s$ is independent of $z$, givingHere $v_K=r\Omega =\sqrt{G{M}_{\star }/r}$, $c_s^2=k_{B}T/(\mu m_p)$ is the isothermal sound speed squared, and $h$ is the Gaussian scale height. A different convention writing $e^{-z^2/h^2}$ changes the quoted height by $\sqrt{2}$ but not the physics.

Hydrostatic equilibrium balances the vertical pressure gradient against weight, $dP/dz=-\rho g$. For a dry ideal gas parcel displaced slowly enough to maintain pressure balance but rapidly enough to exchange negligible heat, the first law givesCombining the two equations yields the dry adiabatic lapse rateusing $c_p\simeq 1004\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$. Here temperature in the thermodynamic differential is absolute, $g$ and $c_p$ are treated as constant, composition is fixed, and condensation is excluded. Hydrostatic balance alone does not force the atmosphere to be adiabatic; the parcel assumption supplies the extra relation.

Water condensation releases latent heat, reducing the cooling of a rising saturated parcel and giving a smaller moist adiabatic lapse rate. Radiation, mixing, large-scale motion and incomplete convective adjustment also shape the environmental lapse rate. Thus an observed $6$--$7\mathrm{K}\mathrm{k}{\mathrm{m}}^{-1}$ need not equal the dry parcel value and is broadly consistent with a moist, radiatively influenced atmosphere.

For a spherical star in hydrostatic equilibrium, $dP/dr=-Gm(r)\rho /r^2$. Multiply by $4\pi r^{3}dr$ and integrate to its radius $R$:Integration by parts gives $4\pi R^3{P}_s-3\int PdV=\Omega$. Neglecting the external surface pressure $P_s$, an ideal gas with constant heat capacity ratio $\gamma$ has thermal energy density $u=P/(\gamma -1)$. ThereforeThis is the hydrostatic virial theorem. A nonzero confining pressure retains the right-hand side $4\pi R^3{P}_s$; varying composition, radiation contributions or a spatially varying effective exponent require the integral relation, rather than pulling one $\gamma$ outside it.