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After complete ionization, each hydrogen nucleus supplies one ion and one electron, while each helium-4 nucleus supplies one ion and two electrons. Neglecting the small metal mass fraction $Z$ and putting $Y\simeq 1-X$, the mean molecular weight therefore satisfiesUse stellar homology to compare models with common dimensionless profiles. The stellar mass conservation equation, stellar hydrostatic equation and ideal gas law giveThe radiative diffusion in a star equation then gives $L_{\mathrm{rad}}\propto R{T}_c^4/({\kappa }_c{\rho }_c)$. Inserting the specified opacity yields $L_{\mathrm{rad}}\propto R{T}_c^7/(Z{\rho }_c^2)\propto {\mu }^7{M}^5/Z$. Independently, integrating the stellar energy-generation rate over the mass gives $L_{\mathrm{nuc}}\propto X^2{\rho }_c{T}_c^{5}M\propto X^2{\mu }^5{M}^7/R^8$. Equating these in stellar thermal equilibrium gives the radiative homology with fifth-power hydrogen burning and inverse-cubic opacity:At fixed solar mass these reduce to $L\propto {\mu }^7/Z$ and $T_c\propto {\mu }^{5/4}/(X^{1/4}{Z}^{1/8})$. The stellar homology approximation, rather than merely the local power laws, is what permits these relations between whole stars.

Equal luminosities imply ${\mu }_2/{\mu }_1=(Z_2/Z_1{)}^{1/7}=2^{-1/7}$. Thusso $X_2=0.8$ to one significant figure. Retaining the unrounded hydrogen mass fraction when comparing central temperatures givesThe first model has the higher central temperature.

For homogeneous hydrogen burning at fixed $M,Z$, fuel conservation gives $M{E}_0\dot{X}=-L$. Since $d\mu /dX=-5{\mu }^2/4$ and $L=L_0(\mu /{\mu }_0{)}^7$, the homogeneous fuel-depletion luminosity feedback obeysIntegration with the initial mean molecular weight ${\mu }_0$ givesThis idealized stellar evolution law stops when $X=0$; it does not predict a physically attainable infinite luminosity. Indeed $\mu =4/3$ at fuel exhaustion, giving $t_H=M{E}_0[1-(3{\mu }_0/4{)}^8]/(10{\mu }_0{L}_0)$, strictly before the formal singularity.

Buoyancy and the convective criterion. Consider a small fluid parcel displaced outward by $\delta r$. Assume that sound waves keep its pressure equal to the ambient pressure, while heat exchange is slow enough that its displacement is an adiabatic process. Composition is uniform, so the parcel and its surroundings have the same mean molecular weight. For an ideal gas, $P\propto \rho T$, and an adiabatic parcel obeys $P\propto {\rho }^{\gamma }$. If $\delta \ln P<0$ is the pressure change on moving outward, thenThe density excess of the parcel at its new location is thereforeIf $\nabla >(\gamma -1)/\gamma$, this excess is negative: the outward-displaced parcel is lighter and accelerates farther outward. A downward displacement similarly continues downward. Conversely, a smaller gradient gives a restoring buoyancy force. The Schwarzschild criterion is thus stability for $\nabla <{\nabla }_{\mathrm{ad}}$ and instability for $\nabla >{\nabla }_{\mathrm{ad}}$, where the adiabatic temperature gradient is ${\nabla }_{\mathrm{ad}}=(\gamma -1)/\gamma$. For a monatomic gas,A composition gradient would require the additional density effect in the Ledoux criterion; it is absent in this calculation.

Opacity and pressure. The supplied opacity is characteristic of a cool hydrogen-rich stellar atmosphere dominated by negative hydrogen ion opacity. Mostly neutral hydrogen can attach electrons supplied by partially ionized metals; bound-free and free-free absorption involving the resulting $H^{-}$ ions produces a steep positive temperature dependence. The weak density dependence and strong temperature sensitivity can be represented by local power-law fits. At fixed composition the given pressure form is equivalent to $\kappa \propto {\rho }^{1/2}{T}^{17/2}$, because $P\propto \rho T$. This is an approximate fit over the relevant cool-atmosphere regime, not an exact opacity law at all temperatures; molecular opacity or hydrogen ionization eventually changes the regime.

For a geometrically thin atmosphere take the surface gravity of a star to be constant, $g_s=GM/R^2$. Combining $dP/dr=-\rho g_s$ with $d\tau /dr=-\kappa \rho$ gives hydrostatic equilibrium in optical depth,The grey atmosphere relation gives $T^8=T_e^8(3\tau +2{)}^2/16$. With negligible external pressure, $P(0)=0$, integration yieldsAn imposed nonzero external pressure would add $P(0{)}^{3/2}$; the boundary assumption is what fixes the integration constant here.

The factor in parentheses is $3\tau /[2(3\tau +2)]$. HenceThe grey-atmosphere convection onset with eighth-power temperature opacity occurs when this reaches $2/5$, soThe radiative continuation below this boundary would be unstable. Efficient convection instead keeps the deeper star approximately adiabatic.

Matching to the convective interior. Write $P=K{\rho }^{5/3}$. Eliminating density using the ideal gas law givesat fixed composition. Since ${\tau }_c$ is a fixed number, the matching temperature $T_b$ is proportional to $T_e$, while the matching pressure satisfies $P_b^{3/2}\propto g_s{T}_e^{-8}$. ConsequentlyThe polytropic mass-radius relation for index $3/2$ gives $K\propto R{M}^{1/3}$. Raising the matching relation to the ninth power and using $g_s\propto M/R^2$ therefore givesFinally the Stefan–Boltzmann law $L=4\pi R^2{\sigma }_{\mathrm{SB}}{T}_e^4$ yields the Hayashi relation with eighth-power temperature opacity,Composition and the opacity normalization are fixed in this comparison. The entropy coefficient $K$ is spatially constant in each convective model, but need not be the same for stars with different masses or radii.

Observe each gas volume, temperature, pressure and moisture condition, and reduce both volumes to the same temperature, pressure and dry-gas basis. Equal scale readings alone do not establish equal amounts of gas.

Read the calibrated volume of each gas in a graduated tube or eudiometer, with the eye at the meniscus to avoid parallax. Allow the gas and the surrounding bath to reach thermal equilibrium, then read a thermometer close to the gas. Heating expands a fixed amount of gas, so a larger reading could reflect a higher temperature rather than more gas. The temperature used in a gas law must be an thermodynamic temperature.

Read the atmospheric pressure with a barometer. If a gas is confined over a liquid, also observe the difference between the inside and outside liquid levels. With $h$ the internal surface height above the external surface, the total internal pressure isHere ${\rho }_{\ell }$ is the confining density of the liquid and $g$ the gravitational acceleration; the sign reverses automatically when $h$ is negative. Equalizing the levels before taking the volume reading removes this hydrostatic pressure correction. A mercury barometer can itself require temperature and calibration corrections for accurate work.

If the gas is saturated with water vapor, subtract the vapor pressure of water at its measured temperature:The observed pressure includes both the test gas and water vapor; counting it all as test gas overestimates the amount. The saturation assumption needs time for equilibration. For a dry gas, no water vapor subtraction is made; another collecting liquid may require its own vapor pressure correction. A collecting liquid must also be chosen so that it neither appreciably dissolves nor reacts with the gas: water, for instance, is unsuitable for preserving the whole amount of freely soluble Ammonia.

For dilute gases obeying the ideal gas law, the volume $V_0$ at a common chosen $T_0,p_0$ isConsequently the corrected comparison isThis is also the ratio of amounts of substance in the ideal gas approximation. Check for leaks, trapped air, changing composition and incomplete transfer. At high pressure or near condensation, real gas behaviour may require compressibility factors; with $pV=ZnRT$, the inferred amount ratio contains the further factor $Z_2/Z_1$.

Here an elastic fluid means a gas or vapour. Such a fluid is mobile, occupies the accessible space of its container, and is appreciably compressible. Its pressure acts normally on a confining surface. In static equilibrium the stress at a point is isotropic: a fluid has no equilibrium resistance to a sustained shear deformation. Pressure can still vary with height through hydrostatic equilibrium; isotropy does not mean the same pressure at every position.

Reducing a confined gas's volume at fixed temperature normally raises its pressure. Heating at fixed volume normally raises the pressure, and gases exchange material through diffusion. Their particles are in continual thermal motion, whose momentum transfer gives a microscopic account of gas pressure in the kinetic theory of gases.

For a dilute gas away from condensation, the approximate ideal gas law isIt contains Boyle's law at fixed temperature and the corresponding temperature-expansion laws. These simple proportionalities are approximations, not exact properties of every elastic fluid at every density. Intermolecular interactions matter in a dense gas, and a vapour can condense; the classification does not promise that every sample remains gaseous under arbitrary compression or cooling.

Take representative samples of a measured volume, identify the impurity by an appropriate reaction, and express the blank-corrected amount per volume of ward air. Excess carbon dioxide, moisture, organic contamination, Ammonia and suspended material are the principal groups to investigate; combustion or defective drains may introduce additional gases. Smell alone measures none of them reliably.

Record occupancy, the position and height of sampling, open doors or windows, heating and lighting, temperature and sampling time. Compare indoor and outdoor samples collected under comparable conditions. Avoid breathing into the collecting vessel. A sample near an open window may miss conditions at the beds, and a sample during cleaning may contain unusually disturbed dust. Explain the sampling conditions with the result.

For carbon dioxide, “carbonic acid” in period terminology, collect a known volume in a calibrated vessel and introduce a measured portion of standardized barium hydroxide solution. Shake to promote absorption. Carbon dioxide consumes the alkaline reagent and forms barium carbonate:Determine the residual alkali in the clear liquid, excluding the carbonate precipitate from an acid determination that would redissolve it. Compare with an equal unexposed reagent portion. An 1864 report describes 7–10 litre globes and checks against gravimetric absorption, so this is an available period method rather than an instrument projected backwards.^[ref]

If $c$ is the hydrogen ion equivalent concentration of the titrating acid, and $v_b$ and $v_s$ are the respective blank and exposed-solution titration volumes, then, for the entire absorption charge,The factor two arises because one carbon dioxide molecule consumes two alkali equivalents. If only an aliquot is titrated, multiply by the ratio of total solution volume to aliquot volume. Here $P$, $T$ and $V_{\mathrm{air}}$ describe the same measured gas sample, with consistent wet- or dry-gas conventions; $R$ is the molar gas constant. Report $10^6{x}_{\mathrm{C}{\mathrm{O}}_2}$ as parts per million by volume. Other absorbed acidic gases are potential interferences. The calculation is a stoichiometric and ideal gas law interpretation, not a direct measure of infectiousness.

For water vapour, compare dry- and wet-bulb thermometers to estimate relative humidity, or pass a known volume through a suitable preweighed drying absorber and determine its mass gain. The latter gives collected water mass divided by sampled air volume, with a blank and appropriate control of other absorbable vapours. Humidity is relevant to ward conditions, but is not by itself evidence that the air contains a particular disease agent.

For Ammonia, moist red litmus becoming blue is a screening reaction, not a selective quantitative analysis. Absorb a measured volume of air into a known amount of dilute acid; the reduction in residual acidity measures absorbed volatile bases. Other alkaline vapours interfere, so do not call the result pure Ammonia without further discrimination. A modern comparison is the validated acid-treated sorbent collection and ion chromatography of the institutional air-sampling method; it too requires calibrated flow, blanks and checks for sample loss.^[ref]

For suspended dust, collect material on a preweighed filter, condition and reweigh it, and divide the blank-corrected mass increase by sampled volume. Microscopy can describe fibres, cells and other particles, but cannot turn an unspecified speck into an identified pathogen. A modern gravimetric sampling method illustrates the importance of controlled filter handling and air-volume measurement.^[ref]

Period investigators also assessed oxidizable organic contamination through the consumption of an oxidizing reagent such as potassium permanganate. Such a result is an operational measure of chemical consumption, dependent on the procedure; it does not reveal a unique substance or count living organisms. An 1865 review already warned that inorganic reducing substances also decolourized the reagent, making supposed quantitative organic estimates unreliable.^[ref] Similarly, a lead-salt darkening reaction may support the presence of hydrogen sulfide where drains are implicated, but this gas need not be present in every ward. Investigate carbon monoxide if combustion is suspected: it is odourless and cannot be excluded because the room smells fresh.^[ref]

A useful report therefore separates measured carbon dioxide, humidity, chemical contamination and particles, states detection limits and interferences, and avoids declaring ward air safe from a single result. Infectious aerosols require a separate assessment; settling plates or a few microscopic fields do not quantitatively characterize everything that patients inhale. Current infection-control guidance likewise warns about the limitations of nonspecific environmental sampling.^[ref]

The covering-law model captures one important feature of scientific explanation: a result should follow from general principles together with the relevant circumstances. In its deductive-nomological model, laws and conditions entail the phenomenon, the laws are needed in the derivation, and the premises must be true and empirically substantive. That is stronger than merely redescribing the event.^[ref]

For example, the ideal gas law, a specified amount of gas and a fixed-volume condition explain why changing its temperature changes its pressure in the model's regime. The derivation identifies a systematic dependence rather than an isolated coincidence. It also makes assumptions visible: a calculation using a law outside its domain is not repaired simply by being logically valid.

Nevertheless, valid subsumption does not by itself determine explanatory direction. A vertical pole of height $h$, a solar elevation $\alpha$ and a shadow of length $l$ satisfy $h=l\tan \alpha$ under suitable geometric assumptions. The same relation allows one to deduce the height from the shadow. In an ordinary intervention on the pole, changing its height changes its shadow; shortening the shadow with a screen does not shorten the pole. The reverse calculation can be excellent evidence for the height without providing the same causal explanation. This is explanatory asymmetry.

Relevance is another difficulty. One can append an irrelevant fact to a sound derivation without invalidating the deduction, but the added fact need not explain anything. Requiring an essential law prevents some vacuous derivations, yet still does not identify which features make a physical difference. A useful explanation should reveal the dependence the question is asking about, not merely contain enough true premises to reach its conclusion.

Nor are all explanations deterministic. A particular radioactive decay can have a good account in terms of a physical decay mechanism and a probability distribution without being entailed by initial conditions. An inductive-statistical model extends the covering-law idea, but requiring a high probability would exclude the explanation of some rare outcomes. Conversely, a high conditional probability produced by irrelevant correlations need not explain a mechanism.

The model is therefore a valuable discipline for organizing some explanations, rather than a sufficient universal analysis. It clarifies derivational structure and testable commitments. An adequate broader account must also address causal relevance, the question's contrast, statistical events and explanations by constraints or unification that are not straightforward causal histories.

An equation of state relates thermodynamic variables within a material model; a phase diagram records which equilibrium phase or phases are stable under specified controls. For a pure substance at fixed temperature and pressure, stability means minimizing Gibbs free energy. Two phases coexist when their molar chemical potentials agree. The equation of state and the free-energy stability condition must therefore be used together: an algebraic relation between $p,V,T$ alone need not identify the stable branch.

A pressure-temperature phase diagram contains single-phase regions separated by phase coexistence curves. For a one-component system the Gibbs phase rule gives $F=3-P$: a single phase has two independent controls, two-phase coexistence lies on a curve, and three phases coexist at a triple point. The liquid-gas coexistence curve ends at a liquid-gas critical point; beyond it a continuous path connects liquidlike and gaslike states without crossing a first-order coexistence boundary. A fixed $p,T$ point on a coexistence curve does not specify the proportions of the two phases: an additional extensive constraint, such as total volume, fixes the phase fraction.

At a first-order phase transition, first derivatives of Gibbs free energy, such as molar entropy or volume, are discontinuous. The conversion has latent heat $L=T\Delta s$. At a continuous transition the first derivatives match, while response functions associated with higher derivatives can be singular or discontinuous. At an ordinary liquid-gas critical endpoint, the two coexisting densities approach one another and the latent heat tends to zero; the material does not retain a sharp liquid-gas boundary above it.

The coexistence slope follows directly from equality of the two molar chemical potentials. Since $d\mu =-sdT+vdp$, differentiating ${\mu }_{\alpha }={\mu }_{\beta }$ along the boundary givesThis is the exact Clapeyron equation, also commonly called the Clausius-Clapeyron relation. For vaporization, both $L$ and $\Delta v$ are positive, so the saturation pressure rises with temperature. For ordinary ice melting into liquid water, $L>0$ but the molar volume decreases, giving the familiar negative melting-curve slope. The sign follows from the volume change, not from a different law.

For a dilute ideal vapour with $v_g\gg v_l$, $\Delta v\simeq RT/p$. The ideal-vapor Clausius-Clapeyron approximation becomesIf molar latent heat is approximately constant over the interval,These approximations deteriorate close to the critical point, where liquid and gas volumes become comparable, or over a range in which latent heat changes appreciably.^[ref]

The ideal gas law $pV=nRT$ is an equation of state for a dilute gas, but it has no liquid branch and cannot predict condensation. A simple model including repulsion and attraction is the Van der Waals molar equation,where $b>0$ represents excluded molar volume and $a>0$ represents attraction. At its critical point, $\partial p/\partial v={\partial }^{2}p/\partial v^2=0$, givingBelow $T_c$, the homogeneous isotherm has an interval with $\partial p/\partial v>0$. That would have negative isothermal compressibility and is mechanically unstable. Equilibrium liquid-gas coexistence replaces the loop by a constant-pressure segment. Equality of chemical potentials gives the Maxwell construction:The equal-area condition enforces the common tangent to the appropriate free energy; it is a stability construction rather than permission to follow the unstable loop physically. The average molar volume between $v_l$ and $v_g$ varies by changing the phase fraction.

The model captures a liquid-gas critical point and qualitative coexistence, but its simple constants are not a complete quantitative description of real fluids and it has no solid phase. A full phase diagram needs additional solid free energies and can contain several crystal phases. Conversely, an experimentally determined coexistence curve and latent heat constrain equations of state through the Clapeyron equation. This is the useful connection: local thermodynamic response, global free-energy stability and the geometry of phase boundaries describe the same material at different levels.

Cold air is denser than room air at approximately the same pressure. The ideal gas law gives a fractional mass density contrast of order $\Delta \rho /\rho \sim \Delta T/T$. The initial outflow is therefore a gravity current driven by reduced gravityFor an illustrative refrigerator height $H\sim 1\mathrm{m}$, room temperature $T\sim 295\mathrm{K}$ and temperature difference $\Delta T\sim 20\mathrm{K}$, this gives $g^{\prime }\sim 0.66\mathrm{m}{\mathrm{s}}^{-2}$. Balancing the released potential energy with kinetic energy gives $U\sim \sqrt{g^{\prime }H}\sim 0.8\mathrm{m}{\mathrm{s}}^{-1}$. The buoyancy exchange time of a cold-air enclosure is consequentlyUsing constant acceleration over height $H$ gives $\sqrt{2H/g^{\prime }}\sim 1.7\mathrm{s}$, consistent at this level. Warm air enters as cold air leaves, so unrestricted vacuum free fall is not the right acceleration model. No dimensions or temperatures are specified in the source, so an order-of-magnitude estimate with explicit representative values is the appropriate result; shelves and mixing can delay complete replacement.

For the given one-mole model,At fixed positive temperature its isotherm is an ideal-gas hyperbola shifted right by $b$: pressure diverges as $V\to b^{+}$ and tends to zero as $V\to \infty$. Higher temperatures give higher pressure at any fixed allowed volume.

For positive $b$, the volume $V-b$ is the effective volume available to molecular motion. Thus $b$ describes excluded-volume repulsion or an effective incompressible molecular core. It is a molar volume in this one-mole equation, not necessarily the literal sum of molecular geometric volume. Setting $b=0$ recovers the ideal gas law. No attractive-energy term is included.

Temperature as an equilibrium relation. Classical thermodynamics begins with the zeroth law of thermodynamics: thermal equilibrium is transitive. A thermometer can therefore assign a common empirical temperature to systems in mutual thermal equilibrium. Different thermometer properties initially give different monotone empirical scales; a resistance reading by itself does not define the absolute scale.

The Second law of thermodynamics provides a substance-independent thermodynamic scale. A reversible engine operating between reservoirs has a heat ratio depending only on their temperatures. One chooses absolute temperature so that $Q_h/Q_c=T_h/T_c$, with $Q_h,Q_c$ the positive heat magnitudes. Its reversible efficiency is $1-T_c/T_h$. Equivalently, reversible heat defines entropy through $dS=\delta Q_{\mathrm{rev}}/T$. For an equilibrium simple compressible system,Classical thermodynamics uses this relation without needing a microscopic account of entropy. A dilute-gas thermometer realizes an empirical scale approaching the thermodynamic scale after corrections for nonideal behavior.

Statistical meaning of the same derivative. In the microcanonical description, an equilibrium energy shell containing $\Omega (U,V,N)$ accessible states has $S=k_B\ln \Omega$. Its statistical temperature is consequentlyTo see why this is an equilibrium temperature, let two weakly coupled systems exchange a small energy while their total energy stays fixed. The combined multiplicity is ${\Omega }_A(U_A){\Omega }_B(U_{\mathrm{tot}}-U_A)$. Maximizing its logarithm giveswhich is equality of the two inverse temperatures. Stability requires the total entropy maximum, not only stationarity. The overwhelmingly dominant macrostate in a large system therefore realizes the thermodynamic equilibrium condition.

A large reservoir gives a small subsystem the canonical probabilities $p_i\propto e^{-\beta E_i}$, with $\beta =1/(k_{B}T)$. In that ensemble, $F=-k_{B}T\ln Z$ and $S=-k_B{\sum }_i{p}_i\ln p_i$ agree with macroscopic thermodynamic relations in their usual equilibrium regime. The equipartition identity for a classical dilute monatomic gas, $⟨p^2/(2m)⟩=3k_{B}T/2$, is a useful operational connection between microscopic motion and temperature, but is not a universal definition: quantum modes need not obey equipartition, and a system away from equilibrium need not have one temperature at all.

For finite systems one must specify the entropy convention and energy resolution; surface and fluctuation corrections can distinguish definitions that agree in the ordinary thermodynamic limit. Bounded-energy population-inverted systems can also have a negative entropy derivative and hence a negative statistical temperature. That possibility lies outside an ordinary positive-temperature ideal-gas thermometer; it does not mean temperature is merely an arbitrary average kinetic energy.

Obtaining the energy–temperature conversion. The Boltzmann constant connects entropy measured in $\mathrm{J}{\mathrm{K}}^{-1}$ with the dimensionless logarithm of a state count. Historically it could be determined from the ideal gas law $pV=N{k}_{B}T$: measure the molar gas constant $R$ and independently determine the Avogadro constant $N_A$, then obtain $k_B=R/N_A$. The independent particle-number determination matters; using $R$ alone would not fix energy per particle.

An electrical method uses Johnson-Nyquist noise. A resistor of resistance $R_e$ at known temperature has, in the classical range $hf\ll k_{B}T$, a one-sided open-circuit voltage-noise spectrum $S_V(f)=4k_{B}T{R}_e$. Over a calibrated flat bandwidth $\Delta f$,A practical measurement uses an equilibrium resistor, a calibrated low-noise amplifier and a known transfer function, subtracts or independently characterizes amplifier noise, and replaces a nominal bandwidth by the actual effective noise bandwidth. Varying temperature and fitting the mean-square voltage is more informative than treating one amplified voltage sample as the thermal variance. Other primary methods relate temperature to sound speed in a dilute gas or to calibrated particle diffusion and mobility.

There is a historically relevant unit-definition qualification: by the date of this exam, the revised SI had already taken effect on 20 May 2019. The kelvin is now defined by fixingThe same experiments now realize or compare thermodynamic temperatures using this fixed constant, whereas historically they measured the conversion constant on an independently defined temperature scale. The physical distinction between equilibrium temperature and microscopic state counting is unchanged.^[ref]

This is the one-mole, or equivalently molar volume, form of the Van der Waals equation, an approximate equation of state for a real gas:Here $p$ is the gas pressure, $T$ its absolute temperature, $R$ the molar gas constant, and $V$ the volume occupied by one mole. If $V$ denotes molar volume explicitly, its units are ${\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$.

The positive parameter $b$ represents excluded volume arising from the finite size and short-range repulsion of the molecules. The volume available for translational motion is reduced from $V$ to $V-b$. This is an effective excluded volume, not simply the sum of the molecules' geometric volumes. Its molar units are ${\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$.

The positive parameter $a$ measures intermolecular attraction in a mean-field approximation. Attractions reduce the force exerted on the walls, lowering the measured pressure by $a/V^2$ relative to the repulsion-only expression. Thus $p+a/V^2$ is the pressure corrected for this attraction. The squared inverse-volume dependence reflects a contribution proportional to the square of the molecular number density. The molar units of $a$ are $\mathrm{Pa}{\mathrm{m}}^6\mathrm{mo}{\mathrm{l}}^{-2}$.

For an arbitrary amount $n$ measured in moles, with total container volume $V_{\mathrm{tot}}$, the corresponding formula isAt low density, when $b\ll V$ and $a/V^2\ll RT/V$, it reduces to the ideal gas law. The model captures qualitative attraction, repulsion and liquid-gas behavior, but does not supply an exact equation of state for every material.

For the numbered engine, denote the low and high pressures by $p_1=p_4=p_l$ and $p_2=p_3=p_h$. Let $R_g$ denote the molar gas constant. The ideal gas law makes the constant-pressure branches straight rays in a temperature-volume diagram:The high-pressure ray is steeper. For a constant heat capacity ratio $\gamma$, each reversible ideal-gas adiabat obeysConsequently $1\to 2$ curves upward and left during compression; $2\to 3$ goes upward and right along the steeper ray; $3\to 4$ curves downward and right during expansion; and $4\to 1$ goes downward and left along the shallower ray.

The plot is an original calculation for $\gamma =7/5$, $p_h/p_l=10$ and $V_3/V_2=2$; its state directions follow the original paper. The dashed extensions identify the two constant-pressure rays, not additional stages of the cycle.

This is the Brayton cycle. Assume fixed gas amount, constant heat capacities, and internally reversible adiabatic processes, as intended by the ideal engine model. Put $r_p=p_h/p_l=p_2/p_1>1$ andCombining $p{V}^{\gamma }=\text{constant}$ with the ideal gas law gives $T\propto p^{(\gamma -1)/\gamma }$ along either reversible ideal-gas adiabat. ThereforeDuring a constant-pressure process the heat supplied equals the enthalpy increase. If $C_p$ is the molar heat capacity at constant pressure, the positive heat input and output areThe insulated branches exchange no heat, so the first law of thermodynamics gives $W=Q_{\mathrm{in}}-Q_{\mathrm{out}}$. HenceAn adiabatic process alone need not be an isentropic process; the relations used here also require internal reversibility. Dissipative compression or expansion would require extra information and would not, in general, obey this efficiency formula.

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.

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 $c_p/c_v$.

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.

At the interface the pressure is continuous. Immediately below it the temperature is $T_s$ and mean molar mass is $M_d(1+ax)$; immediately above it the dry hydrogen has temperature $T_1$. The ideal gas law therefore givesAn infinitesimal interchange is unstable to convection when the lower fluid is less dense than the upper fluid. HenceEquality gives neutral interfacial buoyancy; reversing the inequality gives a stable density jump. This is an interfacial criterion, not a claim that the whole dry upper atmosphere is stable. Its vertical temperature profile has not been specified. Also, $\mathrm{RH}(p_1)=p_1/P<1$, so infinitesimal displacement at the interface initially involves no condensation or release of latent heat.

There are two competing effects. Increasing $T_s$ directly reduces the lower density, favouring convection. Increasing the mole fraction of heavy water vapor increases its mean molar mass, opposing convection. With $x\ll 1$, the criterion is approximately $T_s>T_1$. With $x\to 1$, it tends to $T_s>ϵT_1\simeq 9T_1$: substantial surface heating need not overcome the compositional stabilization in hydrogen.

For a specified $P_d,\Delta p,T_1$, solve $T_s/T_1-1-ax(T_s)=0$ on the admissible branch to locate the transitions. Its derivative is $1/T_1-a{x}^{\prime }(T_s)$, so there is no universal monotone trend: rapid growth of the saturation pressure can suppress an instability that was present at lower $T_s$. Without numerical parameter values, neither the number of crossings nor a unique interval of unstable temperatures can be asserted.

Let $e_v$ be the parcel's water vapor partial pressure. Fixed mole fraction before condensation means $e_v/p$ is constant. Its dew point $T_D$ is defined by $E(T_D)=e_v$, where $E$ is the saturation pressure. The Clausius-Clapeyron relation givesThe last approximation uses hydrostatic balance, the ideal gas law, and an environmental temperature close to the parcel's. Rearrangement yieldsFor a modest initial dew-point depression, $T_D/T\simeq 1$ in absolute temperature, soWith $L_v\simeq 2.5\times 10^6\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$, $R_d\simeq 287\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$, and $R_v\simeq 461.5\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$, $H\simeq 1.58\times 10^5\mathrm{m}$. This dew point scale is much longer than the atmospheric pressure scale height.

The exponential function is an approximation, not the exact consequence of a dry adiabatic process. Retaining $T=T_s-{\Gamma }_{d}z$ gives $p/p_s=(T/T_s{)}^{c_p/R_d}$ and the useful comparisonIt also shows explicitly that expansion lowers the dew point even before any water vapor condenses.

For the variable-latent-heat saturation curve of a pure saturated water atmosphere, the total pressure equals the saturation pressure. The exact Clausius-Clapeyron relation along the phase coexistence curve iswhere $v_v,v_c$ are the specific volumes of the vapor and condensed thermodynamic phase. Neglecting $v_c$ and using the ideal gas law, $v_v=R_{\mathrm{wv}}T/p$, givesThe triple point provides the integration constant. Below it, the prescribed latent heat is constant, soAbove it, write $K=L_{\ast }/[R_{\mathrm{wv}}(T_c-T_{\ast })]$. Integrating $K(T_c/T^2-1/T)$ givesThese are the requested implicit temperature profiles $T(p)$. Both branches meet at $(T_{\ast },p_{\ast })$, and each is monotone before the limiting liquid-gas critical point. The lower branch describes equilibrium with ice, rather than a stable liquid surface. A stable liquid ocean surface belongs to the upper branch; the colder upper atmosphere may use the lower branch. Hydrostatic balance is needed to convert this relation into a height profile, but not to obtain $T(p)$.