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For the stipulated Strong-shock Rankine-Hugoniot conditions, ${\rho }_d=4{\rho }_u$ and $v_d=v_u/4$. The ideal gas law is $P_d={\rho }_d{k}_B{T}_d/\mu$, with $\mu$ a dimensional mean particle mass. The specific enthalpy of a nonrelativistic monatomic gas is $5k_{B}T/(2\mu )$. Dividing the energy flux equation in d by the common mass flux givesThereforeAs an independent check, conservation of momentum gives $P_d={\rho }_u{v}_u^2-4{\rho }_u(v_u/4{)}^2=3{\rho }_u{v}_u^2/4$, and division by ${\rho }_d=4{\rho }_u$ gives exactly the same temperature. The numerical coefficient relies on $\gamma =5/3$ and the absence of dynamically important radiative or magnetic stresses.

A thin spherical shell has outward pressure force $-4\pi r^2(dP/dr)dr$ and inward gravitational force $GM(r)4\pi r^2\rho (r)dr/r^2$. Hydrostatic equilibrium requiresA uniform-density solar model gives $M(r)=M_{\odot }(r/R_{\odot }{)}^3$, $\bar{\rho }=3M_{\odot }/(4\pi R_{\odot }^3)$, and after integration from $R_{\odot }$ to zero,Thus the appropriate orders of magnitude are $10^{14}$–$10^{16}\mathrm{Pa}$ and $10^7\mathrm{K}$. The uniform-density approximation underestimates the actual central pressure because the Sun is centrally concentrated. Radiation pressure is neglected in applying the ideal gas law.

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.

Write $m(r)$ for enclosed mass. The stellar hydrostatic equation is $dP/dr=-Gm\rho /r^2$. If radiation pressure dominates, compare it directly with the radiative diffusion result for $d{P}_r/dr$:This is the limiting Eddington luminosity scaling, assuming fixed composition and electron-scattering opacity.

If gas pressure dominates, use stellar homology: characteristic density $\rho \sim M/R_{\ast }^3$ and pressure $P\sim G{M}^2/R_{\ast }^4$. The ideal gas law then givesEstimating $|dT/dr|\sim T/R_{\ast }$ in radiative diffusion givesThus at fixed $\mu$ and constant electron-scattering opacity, $\boxed{L\propto M^3}$. These are stellar homology estimates for radiative stars, not exact relations valid for arbitrary composition and internal transport.

A thin spherical shell has outward pressure force $-4\pi r^2(dP/dr)dr$ and inward gravitational force $GM(r)4\pi r^2\rho (r)dr/r^2$. Hydrostatic equilibrium requiresA uniform-density solar model gives $M(r)=M_{\odot }(r/R_{\odot }{)}^3$, $\bar{\rho }=3M_{\odot }/(4\pi R_{\odot }^3)$, and after integration from $R_{\odot }$ to zero,Thus the appropriate orders of magnitude are $10^{14}$–$10^{16}\mathrm{Pa}$ and $10^7\mathrm{K}$. The uniform-density approximation underestimates the actual central pressure because the Sun is centrally concentrated. Radiation pressure is neglected in applying the ideal gas law.

The tropopause is the transition between the lower, often convecting troposphere and the more stably stratified atmosphere above, where the decrease of temperature with altitude weakens or reverses. Its definition concerns the thermal profile, not a fixed universal altitude. An optical depth is a dimensionless path-integrated extinction, ${\tau }_{\nu }=\int {\kappa }_{\nu }\rho ds$ for mass extinction coefficient ${\kappa }_{\nu }$; without emission into the ray its direct intensity is attenuated by $e^{-{\tau }_{\nu }}$.

In hydrostatic equilibrium, the upward pressure force balances weight:The second equality uses the ideal gas law with specific gas constant $R_a$. For constant $T$ and $g$,For variable temperature the corresponding result is $p(z)=p(z_0)\exp [-{\int }_{z_0}^{z}g/(R_{a}T(z^{\prime }))d{z}^{\prime }]$. The fixed-temperature derivative does not require the entire column to be isothermal.

Assume a well-mixed ideal gas with constant $g,c_p$, no latent heat, and sufficiently strong surface heating to maintain dry convection near the surface. An adiabatically displaced unit mass has specific volume $V=1/\rho$, so the stated relation and hydrostatic equilibrium giveThis is the dry neutral lapse rate; infrared absorption alone does not guarantee that a radiative profile is steep enough to trigger convection. Combining the same equations with the ideal gas law yields $T\propto p^{R_a/c_p}$ along that adiabat.

High above the main infrared-emitting region, optical depth to space becomes small. A transparent-to-sunlight atmosphere has no substantial direct solar heating there; a thin grey layer absorbs the upwelling planetary infrared flux $Q=(1-\alpha )S_0/4$ and emits from both faces. In the optically thin radiative-equilibrium limit,The factor four in $Q$ is projected-area versus whole-sphere averaging; the additional factor two counts emission upward and downward by the tenuous layer. For Earth-like $S_0\simeq 1360\mathrm{W}{\mathrm{m}}^{-2}$ and $\alpha \simeq 0.30$, this is about $214\mathrm{K}$. Thus temperature approaches a height-independent skin value as $\tau \to 0$, rather than continuing the dry adiabat to zero. This is an asymptotic grey-atmosphere approximation, not a proof that an entire finite stratosphere is exactly isothermal.

Hydrostatic balance and the ideal gas law give $dp/dz=-pg/(R_{a}T)$, hence approximately $p=p_0{e}^{-z/H}$ with $H=R_{a}T/g$ for weak temperature variation. For well-mixed absorber abundance, ${\rho }_{\mathrm{abs}}\propto p/T$, while pressure broadening gives $k\propto p$. Thus $k{\rho }_{\mathrm{abs}}\propto e^{-2z/H}$ and its integral above height $z$ has the same scale. ThereforeThe factor two depends on both pressure dependence of absorption and constant mixing ratio; it is not the ordinary density scale height.

Integrating $dP/dr=-G{M}_r\rho /r^2$ in order of magnitude, with $\rho \sim M_{\ast }/R_{\ast }^3$, givesThe ideal gas law independently gives $P_c\sim M_{\ast }{k}_B{T}_c/(\mu m_H{R}_{\ast }^3)$. Equating the two estimates yieldsFor similar composition and an approximately fixed hydrogen-burning central temperature, $R_{\ast }\propto M_{\ast }$. Calibrating this rough relation on the Sun and using part a gives a nominal mass range of about $0.05$ to $5M_{\odot }$.

The endpoint luminosity ratio is $10^8$, whereas the inferred mass ratio is about $10^2$, giving $L\propto M^4$ across that range. The exponent is therefore consistent at this schematic level. The solar normalization and individual endpoint masses are much less reliable: using the luminosity relation itself gives approximately $0.1$ to $10M_{\odot }$. Central temperatures are not truly constant, low-mass stellar structure differs, and the uncorrected $V$ luminosities in the diagram are not bolometric. These limitations prevent treating the rough radius-based masses as a contradictory precision measurement.

A thin spherical shell of thickness $dr$ has mass $4\pi r^2\rho dr$. Balancing its inward gravitational force with the pressure difference across its faces givesFor a transparent estimate take constant density $\rho =3M_{\odot }/(4\pi R_{\odot }^3)$ and negligible surface pressure. Then $M_r=M_{\odot }{r}^3/R_{\odot }^3$, so integration givesCombining with the ideal gas law and $\mu \simeq 0.6$ yields $T_c\simeq \mu m_{H}G{M}_{\odot }/(2k_B{R}_{\odot })\simeq 7\times 10^6$ K. The less specific virial scaling gives $T_c\sim \mu m_{H}G{M}_{\odot }/(k_B{R}_{\odot })\sim 10^7$ K.

The real Sun is strongly centrally concentrated: a uniform-density model substantially underestimates central pressure and is not a detailed solar model. Its value is a justified scale estimate, while more realistic structure produces central temperatures of order $1.5\times 10^7$ K and central pressures of order $10^{16}$ Pa. Hydrostatic equilibrium alone does not uniquely determine both profiles without an equation of state and a density/thermal model.

The material derivative $\boxed{D/Dt={\partial }_t+\mathbf{u}\cdot \nabla }$ measures change while following a moving gas parcel. The four equations respectively express mass conservation, momentum conservation, internal-energy balance, and the ideal gas law. The first accounts for density change under expansion or compression. The second is Newton's law with pressure as the only force. In the third the specific internal energy is $c_{v}T$ and pressure work $-p\nabla \cdot \mathbf{u}$ heats compression and cools expansion. The fourth relates thermodynamic pressure, density and temperature; $R$ is the gas constant per unit mass.

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.

The pressure coefficient compares a local static pressure with a specified reference static pressure, using a fixed reference dynamic pressure:The reference mass density, speed, pressure and location must accompany a reported pressure coefficient. Keep the same $q_{\infty }$ throughout a surface distribution; replacing it by local $\rho U^2/2$ gives a different quantity and fails at a stagnation point. The definition itself requires neither Bernoulli's principle nor inviscid flow, so measured pressure coefficients remain useful when those idealizations fail.

Positive $C_p$ means $p>p_{\infty }$; zero means equality; negative $C_p$ means $p<p_{\infty }$. This sign does not specify the sign of absolute pressure. For stipulated $p_{\infty }=101325\mathrm{Pa}$ absolute and $q_{\infty }=240\mathrm{Pa}$, $C_p=-3$ gives $p=100605\mathrm{Pa}$ absolute: a $720\mathrm{Pa}$ deficit, not a negative absolute pressure. If $p_{\infty }$ equals atmospheric pressure, the numerator is gauge pressure; a wind tunnel's reference static pressure need not equal room atmospheric pressure. These are reference choices, not different physical pressures.

Under the same-height, constant-mass density $\rho ={\rho }_{\infty }$, steady flow and inviscid flow hypotheses of the Bernoulli equation, and with the same Bernoulli invariant as the reference flow,The first equation immediately gives the second after subtracting $p_{\infty }$ and dividing by $q_{\infty }$. Hence a lossless stagnation point has $C_p=1$, $U=U_{\infty }$ gives $C_p=0$, and $U=2U_{\infty }$ gives $C_p=-3$. For a real stationary wall satisfying the no-slip boundary condition, the velocity is zero right at the wall; inserting that wall velocity in an inviscid flow formula would wrongly predict $C_p=1$ everywhere. A thin boundary layer approximation instead relates the wall pressure to the outer-flow pressure and uses the outer velocity where the Bernoulli equation applies.

For potential flow around a circular cylinder of radius $a$ with zero circulation, the surface tangential velocity is $u_{\theta }=-2U_{\infty }\sin \theta$. Thus the ideal pressure coefficient isWith $\theta$ measured from downstream, the front is at $\pi$ and the rear at zero. Both ideal stagnation points have $C_p=1$, while the sides at $\pi /2$ and $3\pi /2$ have $C_p=-3$. This exact model is a useful baseline for an experiment; its complete rearward pressure recovery is not a prediction for a separated real fluid wake.

To obtain a force, integrate the local normal contributions with their directions. Let $\mathbf{n}$ be the unit normal vector from the body into the fluid and $\Sigma$ its closed wetted surface:The second equality removes a spatially constant reference pressure over the closed surface; it does not remove a nonuniform hydrostatic pressure or justify dropping the reference term on an isolated open patch. A negative $C_p$ contributes outwards in this reference-subtracted calculation, even though a positive absolute pressure still pushes inwards locally. Project ${\mathbf{F}}_p$ onto the chosen lift and drag directions. The full fluid force also includes ${\int }_{\Sigma }{\mathbf{\sigma }}_{\mathrm{visc}}\mathbf{n}dA$ from the viscous stress tensor, including tangential shear stress; pressure tapping does not measure that contribution.

For the two-dimensional cylinder, force per unit span follows from $dA=ad\theta$ per unit span and $\mathbf{n}=(\cos \theta ,\sin \theta )$:Substituting $C_p=1-4{\sin }^2\theta$ gives $(0,0)$: opposite points have identical pressure coefficients and opposite normal vectors. Zero drag despite a strongly nonuniform pressure is the instructive limitation associated with D'Alembert's paradox. Real viscosity, boundary layers and flow separation can prevent rearward pressure recovery and produce pressure drag. Thus the measured curve can test the model and estimate a pressure contribution without proving that total drag vanishes.

A motivating measurement is Virginia Tech's Experiment 3 - Flow Past a Circular Cylinder. Its open-jet wind tunnel holds a rotatable $140\mathrm{mm}$ cylinder with end plates and 36 one-millimetre pressure tappings at ten-degree intervals. A reference Pitot-static tube supplies $p_{0\infty }$ and $p_{\infty }$ to one digital manometer; a second connects a surface tap to its positive port and reference static pressure to its negative port. Record the simultaneous readings:The last equality uses the low-speed stopping relation, not an additional speed measurement. The manual requires sealed tubing and zero checks with the tunnel off. It also provides a water-column manometer: read each column's initial level, and subtract the independently measured reference-to-atmosphere pressure from the tap-to-atmosphere readings. Record tap positions after rotation. Compare the two instruments and check spanwise uniformity; end plates do not guarantee two-dimensional flow. ^[ref]

The inferential distinction matters: the manometers directly measure pressure differences, the ratio estimates pressure coefficients, and a surface integral estimates the pressure part of the force. Measure room absolute pressure and temperature if converting the reference difference to speed through the ideal gas law $\rho =p_{\mathrm{abs}}/(RT)$; use thermodynamic temperature in that equation. For force reconstruction, cover the complete contour and use numerical integration with the tap geometry. Finite tap spacing, unknown end effects, probe disturbance and fluctuations limit that reconstruction; tubes and instruments also average or delay changing pressures. Retain the signed raw readings, repeated readings and reference readings so that these limits can be assessed. A small reference difference magnifies any zero error in the ratio, rather than producing a reliably large $C_p$.

An atmospheric vertical coordinate using pressure instead of geometric height. Under hydrostatic balance, pressure decreases upward and geopotential obeys ${\partial }_p\Phi =-1/\rho$. With the ideal gas law, ${\partial }_{\ln p}\Phi =-RT$. Horizontal gradients taken at fixed $p$ lie on an isobaric surface. A log-pressure coordinate is a useful related coordinate with the opposite upward sign.

A quasistatic process is an ideal limit in which a changing thermodynamic system passes through a sequence of states arbitrarily close to thermodynamic equilibrium. Its intermediate pressure, temperature and volume are therefore well defined and obey its equation of state. Driving must be slow compared with the system's own equilibration, rather than merely taking a long time on a clock. A reversible thermodynamic process additionally allows both system and surroundings to be restored without net change; friction can produce entropy even in a quasistatic process.^[ref]

For a computed example, take a fixed amount of substance $n$ of ideal gas in a sealed cylinder with a frictionless piston, in contact with a heat bath at absolute temperature $T$. Reduce the external load in infinitesimal steps so that the piston expands from $V_0$ to $2V_0$. In the reversible limit, the gas pressure exceeds the external pressure by an infinitesimal amount during expansion, and the heat transfer occurs across an infinitesimal temperature difference. Exact mechanical equilibrium would give no driving force; the limiting imbalance supplies motion while preserving equilibrium intermediate states.^[ref]

Along this reversible isothermal expansion, the ideal gas law givesHere $R$ is the molar gas constant; specifying $nRT$ fixes an energy scale, without choosing $n$ and $T$ separately. The initial and final equilibrium states are $(V_0,p_0,T)$ and $(2V_0,p_0/2,T)$. Count work delivered to the surroundings as positive. The reversible work is the shaded area beneath the pressure-volume curve:The integral of $1/V$ supplies the natural logarithm. Since internal energy of an ideal gas depends only on temperature, $\Delta U=0$; the first law of thermodynamics, $\Delta U=Q_{\mathrm{in}}-W_{\mathrm{out}}$, gives $Q_{\mathrm{in},\mathrm{rev}}=0.693147\mathrm{kJ}$ to the displayed precision. The heat bath replaces the energy delivered as work.^[ref]

Now start from the same initial state, but abruptly lower the external pressure to the constant value $p_{\mathrm{ext}}=p_0/2=0.5\mathrm{MPa}$. There is a finite initial pressure imbalance, so this is an irreversible expansion rather than a quasistatic process. Assume the piston begins and ends at rest, has no net change in stored energy, and has no external friction; relaxation occurs within the gas and its heat bath. After motion has settled and the heat bath has restored temperature $T$, mechanical equilibrium and the ideal gas law give $V_f=nRT/p_{\mathrm{ext}}=2V_0$. The work delivered to the constant load isThis work follows directly from load force times piston displacement, using $dV=Adx$ for piston area $A$. Intermediate gas states need not have a single equilibrium pressure or temperature. In particular, neither $p(V)=nRT/V$ at the bath temperature nor $p(V)=p_{\mathrm{ext}}$ describes the actual gas during this irreversible motion. The rectangle below plots the known external load; its upper edge is not an equilibrium gas path.

The entropy accounting makes the difference precise. Because entropy is a state function, both expansions have the same gas entropy change, computed along the reversible isothermal expansion:The thermal reservoir stays at $T$ and loses heat $Q_{\mathrm{in}}=W_{\mathrm{out}}$, so $\Delta S_{\mathrm{bath}}=-W_{\mathrm{out}}/T$. Assume the ideal load stores work without an entropy change. The combined entropy production is thereforeNo numerical entropy unit is implied by $nRT=1\mathrm{kJ}$ alone: $nR$ or $T$ must also be specified. The lost work opportunity is $T\Delta S_{\mathrm{total}}=0.193147\mathrm{kJ}$ for this comparison. This calculation uses equilibrium endpoints and the reservoir's temperature, without assigning equilibrium intermediate states to the irreversible gas. ^[ref]

For another boundary choice, let the same ideal gas expand into an initially evacuated compartment of volume $V_0$ inside a rigid, insulated vessel of total volume $2V_0$. Removing the partition causes free expansion: no load is moved and no heat crosses the outer boundary, so $W_{\mathrm{out}}=Q_{\mathrm{in}}=0$ and $\Delta U=0$. After equilibration the temperature is again $T$, the pressure is $p_0/2$, and the gas entropy has increased by $nR\ln 2$. The outer vessel has fixed volume; the gas's accessible volume doubles. Restoring the initial gas state requires compression and removal of heat, changing the surroundings. Equal endpoints do not make free expansion reversible. ^[ref]

A slowly moving piston with dry friction supplies a different caution: its gas can remain arbitrarily close to thermodynamic equilibrium, yet rubbing still converts work into heat and produces entropy. If the frictional force has a nonzero slow-motion limit, its dissipated energy over a fixed travel does not vanish with speed. Reversing the motion reverses friction, so infinitesimally changing the external load cannot retrace the same operation and restore the surroundings. The heat capacity calculations additionally exclude friction and other dissipative work.^[ref]

The usefulness of a quasistatic process is that it replaces detailed motion by a readable sequence of equilibrium states. An equation of state and an integral then turn a pressure-volume drawing into an energy budget. The beautiful distinction is that internal energy and entropy depend on the endpoints, while heat and work record how the change was performed. Reversible benchmarks expose how much work an irreversible device sacrifices, while entropy production explains why simply making it slower need not recover that opportunity.

For an original idealized mathematical model, let $n_{{\mathrm{O}}_2}(t)$ be the amount of substance of molecular oxygen consumed since the initial reading and $n_{{\mathrm{CO}}_2}(t)$ the carbon dioxide produced. Suppose a fraction $f$ of that carbon dioxide is removed from the gas, all other gas inventories remain fixed, and temperature $T$ and gas pressure $P$ stay constant. The ideal gas law gives the positive gas contractionwhere $R$ is the molar gas constant and $V$ the gas volume. With complete absorption, $f=1$, the contraction reflects molecular oxygen uptake regardless of the amount of carbon dioxide made. For a uniform pipette bore of area $A$, an inward boundary displacement $x$ gives $D=Ax$. A graduated pipette can instead supply the displaced volume directly.

Let $D_s$ denote sample contraction and $D_b$ the matched blank contraction. Define $D_{\mathrm{corr}}=D_s-D_b$ with both taken relative to their own initial readings. Under the preceding assumptions,The slope of corrected volume against time estimates uptake over the chosen interval. The approximate sign acknowledges that a liquid boundary also responds to hydrostatic and capillary pressure, while the gas can exchange water vapour. A fixed-volume apparatus would instead show a pressure decrease; its ideal conversion is $n_{{\mathrm{O}}_2}\simeq V(P(0)-P(t))/(RT)$ after complete absorption and appropriate blank correction.

The ideal gas law also exposes drift without any biology: for fixed gas inventory, $V\propto T/P$. A one-percent rise in absolute temperature at fixed pressure produces a one-percent expansion. Matching the initial gas spaces makes such common changes easier to subtract. Subtracting unmatched blanks can leave a physical signal that looks biological.

The second virial coefficient is the leading interaction correction to the ideal gas law: at fixed temperature, $B_2(T)={\lim }_{n\to 0}(Z-1)/n$, where $Z=p/(n{k}_{B}T)$ and $n$ is number density. For identical classical particles in a homogeneous three-dimensional gas, with an isotropic pair potential $u(r)$ and $r$ the centre-to-centre separation,The function $f$ is the Mayer f-function. The factor $4\pi r^{2}dr$ counts the volume of a thin spherical shell; the factor $1/2$ counts each unordered pair once. The result has units of volume. Repulsive regions, $u>0$, contribute positively, while attractive regions, $u<0$, contribute negatively. ^[ref]

This is a striking bridge from a microscopic pair potential to a macroscopic pressure reading. At sufficiently small number density, $B_2>0$ raises pressure relative to an ideal gas at the same temperature and number density, while $B_2<0$ lowers it. A single value of $B_2$, however, cannot determine the entire pair potential: many different functions have the same weighted integral. Measurements over several temperatures supply more information.

The integral must exist. For example, a bounded short-range attraction outside a hard core has no difficulty. If instead $u(r)\sim A/r^s$ as $r\to \infty$, then $1-e^{-u/(k_{B}T)}\sim u/(k_{B}T)$ and the radial tail behaves like $\int r^{2-s}dr$: it converges only for $s>3$. A bare unscreened Coulomb potential therefore falls outside this formula's ordinary short-range setting. This condition on the tail does not by itself establish the stability or convergence of the full virial expansion.

For an exactly integrable toy attraction, choose the following square-well pair potential, with diameter $\sigma >0$, attraction depth $ϵ>0$, and range multiplier $a>1$:Set $T^{\ast }=k_{B}T/ϵ$ and $b_{\mathrm{HS}}=2\pi {\sigma }^3/3$, the second virial coefficient of a hard-sphere gas. Integrating the core and attractive shell separately gives the exact toy resultThe core contributes $b_{\mathrm{HS}}$; the shell contributes $-b_{\mathrm{HS}}(a^3-1)(e^{1/T^{\ast }}-1)$. Heating weakens the relative Boltzmann factor enhancement of attractive configurations, so $B_2$ approaches the positive hard-core value as $T^{\ast }\to \infty$. For the illustrative choice $a=3/2$, the zero occurs atAt this temperature, $Z=1+B_3{n}^2+\cdots$: cancellation removes the linear correction, not every correction. At low $T^{\ast }$, the negative shell contribution becomes large, requiring a correspondingly smaller number density for a useful truncated virial expansion. The square well is a deliberately chosen model, not a fit to the carbon dioxide experiment.

A strong, nonradiative shock in a monatomic ideal gas gives $p_{\mathrm{d}}=3{\rho }_{\mathrm{u}}{v}_{\mathrm{u}}^2/4$ and ${\rho }_{\mathrm{d}}=4{\rho }_{\mathrm{u}}$. The ideal gas law therefore gives the temperature in the title, with $\mu$ the mean particle mass. If the upstream speed is a Newtonian free-fall speed, $k_{\mathrm{B}}{T}_{\mathrm{d}}=3\mu GM/(8R)$. Mass alone does not determine this temperature without a radius and composition; electron-ion disequilibrium and radiation can also make the emergent spectrum differ from the ideal gas temperature.

Rojas Cardenas, Graur, Perrier and Méolans studied thermal transpiration in helium using a glass microtube of diameter $490\pm 1\mu \mathrm{m}$ and length $3.05\pm 0.01\mathrm{cm}$ in their 2012 experiment. They heated one reservoir, leaving the other near room temperature, and measured tube-end temperatures of $300$ and $351\mathrm{K}$. Three contact thermometers and an infrared camera checked the thermal conditions; the hot reservoir was warmer than the adjacent tube end, so heater temperature was not an adequate substitute for tube-end temperature. Two capacitance diaphragm gauges recorded the reservoir pressures; a third gauge supported calibration. ^[ref]

Their earlier 2011 paper explains the measurement sequence. A wide bypass links the reservoirs in parallel with the microtube. After isolating the gas supply and pump, keep the bypass open to equalize pressures with the temperature difference already established. Closing the bypass defines $t=0$. The hot-side pressure rises and the cold-side pressure falls; ultimately their slopes approach zero at unequal pressures. Repeating at different initial pressures changes the degree of rarefaction without changing the microtube. The 2011 study also compared helium and nitrogen and imposed temperature differences of $50$, $60$ and $70\mathrm{K}$. ^[ref]

The quantities to record are both $p_c(t)$ and $p_h(t)$, the reservoir volumes, and the temperatures relevant to their particle inventories. In an idealized reservoir of fixed volume $V_i$ and uniform fixed temperature $T_i$, the ideal gas law gives ${\dot{N}}_i=[V_i/(k_B{T}_i)]{\dot{p}}_i$, so opposite inventory changes test particle conservation. If temperature changes, differentiation instead gives ${\dot{N}}_i=(V_i/k_B)[{\dot{p}}_i/T_i-p_i{\dot{T}}_i/T_i^2]$: a thermal drift can imitate particle transfer. The authors' 2011 presentation explicitly checks this thermal correction when extracting an initial mass flow rate from the pressure slope. ^[ref]

The 2012 paper compares the final pressure ratio across near-free-molecular through slip conditions with a kinetic model and empirical corrections. It does not establish the thin-hole square-root law at every measured pressure. ^[ref] For a replication, useful additional controls follow from the interpretation: remove the temperature difference to test the null signal; reverse it to test the direction; check gauge zero offsets and leaks; and wait for stable temperatures before closing the bypass. These are suggested controls, not claims that the cited paper performed each one. A reproducible pressure split at zero net transfer makes the failure of a single continuum wall model observable, while the Knudsen number organizes which molecular and apparatus scales are competing.

An equilibrium thermodynamic state specifies sufficient independent macroscopic properties, together with material identity and constraints, to determine the other equilibrium properties. For a uniform fixed-composition ideal gas, its amount of substance, temperature and volume determine its pressure through the ideal gas law. These properties specify a state, without specifying a thermodynamic process that prepared it.^[ref]