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An X-ray observatory used for high-resolution imaging and spectroscopy of astronomical sources. Its observations of the Perseus cluster reveal structure in the emitting intracluster gas around radio bubbles.^[ref]

For a pure ideal gas at temperature $T$ and pressure $p$,where $R$ is the molar gas constant and $p^{\circ }$ specifies the reference pressure of the standard state. The ratio inside the natural logarithm is dimensionless. The chemical potential difference therefore depends on a ratio of pressures, not an absolute difference. See MIT 5.60, Lecture 14, page 4.

For a chemical species in an ideal gas mixture, replace $p$ by its partial pressure. More generally write ${\mu }_{\mathrm{m},i}={\mu }_{\mathrm{m},i}^{\circ }+RT\ln a_i$ using its thermodynamic activity $a_i$ and a consistent standard state. For an ideal dilute chemical solution, $a_i=c_i/c^{\circ }$ with reference concentration $c^{\circ }$; the thermodynamic activity section describes departures from this idealization.

As a computed comparison, take identical chemical species at $T=300\mathrm{K}$, with $p_A=2p^{\circ }$ and $p_B=p^{\circ }$. Stipulate $R=8.314462618\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$. ThenHalving the pressure gives the same decrement whether starting from $2p^{\circ }$ or $20p^{\circ }$, provided the ideal gas approximation remains valid. This comparison uses different pressures; it is not the common-pressure reservoir setup of chemical potential equality for particle exchange. The fixed-volume example in chemical potential and Helmholtz free energy supplies an exchange experiment with unequal initial pressures.

The Clapeyron equation connects the slope of a phase coexistence curve to the latent heat and the change in volume during a first-order phase transition. A steep boundary can indicate a large latent heat, a small change in volume, or both. This lets a measured boundary constrain caloric properties and predicts how changing pressure shifts a melting temperature.

For a pure substance in bulk phase coexistence, label the initial and final thermodynamic phases $\alpha$ and $\beta$. Both have the same absolute temperature $T$, pressure $p$, and chemical potential. With $s$, $h$, and $v$ denoting entropy, enthalpy, and volume on one common basis, define $\Delta s=s_{\beta }-s_{\alpha }$, $\Delta h=h_{\beta }-h_{\alpha }$, and $\Delta v=v_{\beta }-v_{\alpha }$. ThenThe derivative follows the phase coexistence curve; it is not a partial derivative within one thermodynamic phase. Use molar volume $v_m$ with molar latent heat $L_m$ in $\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}$, or specific volume $v$ with latent heat $L$ in $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$. Both give $dp/dT$ in $\mathrm{Pa}{\mathrm{K}}^{-1}$. For molar mass $M$, $L_m=ML$ and $\Delta v_m=M\Delta v$, so the predicted slope is independent of the chosen basis. This formulation assumes a simple compressible system, fixed composition, and negligible interface contributions; a curved interface can instead require the Young–Laplace equation.

For positive latent heat in the stated conversion direction, the sign of the slope is the sign of $\Delta v$. Converting liquid to gas normally increases volume and gives a positive slope. The negative melting-point slope of water illustrates the opposite sign. At a thermodynamic critical point, the coexisting gas and liquid merge: both $L$ and $\Delta v$ vanish, so their ratio needs a limiting analysis rather than substitution of zero into the Clapeyron equation.^[ref]

A short derivation uses the molar Gibbs free energy $g_m$, equal to the chemical potential for a pure substance. Equality persists along the boundary, so $d{g}_{m,\alpha }=d{g}_{m,\beta }$ with $d{g}_m=-s_{m}dT+v_{m}dp$. Therefore $-\Delta s_{m}dT+\Delta v_{m}dp=0$. Equality of $g_m=h_m-T{s}_m$ also gives $\Delta h_m=T\Delta s_m$. Eliminating $\Delta s_m$ yields the Clapeyron equation.

The Clément-Desormes experiment infers heat capacity ratio from a rapid pressure reduction followed by reheating. Use a rigid bottle, pump or bellows, a quick release valve, and a U-tube pressure gauge connected to the bottle and atmosphere. First pump to a small excess pressure, close the inlet, and wait for thermal equilibrium with the room at $T_0$. Measure $p_1>p_a$, where $p_a$ is local absolute atmospheric pressure. Briefly open the release valve until the bottle reaches $p_a$, then close it promptly. The remaining gas is cooler than the room. As it warms back to $T_0$ at fixed bottle volume, its pressure rises to $p_3$, with $p_a<p_3<p_1$ in the ideal procedure. The rebound reveals cooling during release. ^[ref]

The bottle loses molecules while the valve is open: it is not a closed system whose geometrical volume expands. In the well-mixed, negligible-heat transfer discharge model with constant heat capacity ratio, the remaining gas obeys $T_2/T_0=(p_a/p_1{)}^{(\gamma -1)/\gamma }$. Subsequent heating at fixed amount of substance and bottle volume gives $p_3/p_a=T_0/T_2$ by the ideal gas law. Eliminating the unmeasured $T_2$ yieldsThe logarithms use absolute pressure ratios. The approximation requires small gauge pressures $\Delta p_i\ll p_a$; $h_i$ are signed height differences between the two liquid surfaces, and the final equality uses the same liquid in both readings with hydrostatic equilibrium. It is not an instruction to insert gauge readings into the logarithmic formula. For synthetic states generated with $p_a=100000\mathrm{Pa}$, $p_1=102000\mathrm{Pa}$ and $\gamma =1.4$, the model gives $p_3\simeq 100567.39\mathrm{Pa}$. The logarithmic inference recovers $1.4$; the small-pressure approximation gives approximately $1.3961$.

Wait for a stable initial pressure after pumping and for a stable final pressure after closing. Vary the valve-open interval and initial excess pressure across repeated trials: closing too early leaves residual excess pressure, while leaving the valve open during warming lets more gas escape and suppresses the rebound. Test the seal for leaks, and read both liquid levels consistently. In a variant with incomplete release, measure the actual immediate post-release absolute pressure $p_2$ and replace the numerator by $\ln (p_1/p_2)$. The Zurich laboratory manual explains that liquid-column oscillations can delay this reading until appreciable reheating has occurred; its proposed backward extrapolation assumes an exponential thermal relaxation and must be checked against the recorded readings. These controls separate a thermal signal from valve timing, instrumental lag and leakage. ^[ref]

A cloud chamber contains supersaturated vapor. Ionization along a charged particle's path supplies sites for nucleation; the resulting liquid drops scatter light and reveal the trail. The observed objects are drops, not magnified particles. Its beauty is that a spatially distributed microscopic interaction becomes a visible geometrical record.

Two ways of making supersaturated vapor must be distinguished. An expansion cloud chamber rapidly increases the volume of moist gas, producing a temporary sensitive interval through cooling by adiabatic expansion. A diffusion cloud chamber maintains a cold base and a warmer vapor supply, producing a continuously replenished sensitive layer through diffusion and a temperature gradient.

Consider an illustrative water–air mixture, initially saturated at $T_0=293.15\mathrm{K}$. Let $\lambda =V/V_0$ be the expansion ratio and approximate the carrier gas by a uniform ideal gas with constant heat capacity ratio $\gamma =1.40$. Before significant condensation, assume fixed numbers of carrier and vapor molecules, negligible incoming heat, and the reversible adiabatic expansion relation. The first law of thermodynamics, $n{C}_{V}dT=-pdV$, with the ideal gas law, givesThe vapor follows the same temperature, so its partial pressure scales as $T/V$. A rapid real piston motion need not be reversible; the formula is an idealized bulk estimate after neglecting flow and losses.

Use a physically motivated Integrated Clausius-Clapeyron relation for the equilibrium water vapor pressure, with stipulated rounded constants $p_0=2339\mathrm{Pa}$ at $T_0$, molar latent heat $L_m=45000\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}$ and molar gas constant $R=8.314\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$:These constants define the example, without fitting any historical measurements. Throughout the plotted interval $1\le \lambda \le 1.17$, the temperature stays above the freezing point of water. InitiallyThus the decrease in vapor pressure initially beats dilution. At $\lambda =1.15$, the model gives $T=277.21\mathrm{K}$, $p_v=1923.33\mathrm{Pa}$, $p_{\mathrm{sat}}=809.07\mathrm{Pa}$ and $S=2.3772$.

Why does $S>1$ not instantly make every molecule a visible drop? In the capillarity approximation of classical nucleation theory, a neutral spherical drop of radius $r$ has a positive surface energy cost and a negative bulk Gibbs free energy contribution. For molecular liquid volume $v_l$, surface tension $\sigma$ and Boltzmann constant $k_B$,Differentiating $\Delta G$ shows that the critical nucleus at $r_{\ast }$ is a maximum: smaller neutral drops tend to shrink, whereas larger ones can grow. With stipulated $\sigma =0.075\mathrm{J}{\mathrm{m}}^{-2}$, $v_l=2.99\times 10^{-29}{\mathrm{m}}^3$ and $k_B=1.380649\times 10^{-23}\mathrm{J}{\mathrm{K}}^{-1}$, the $\lambda =1.15$ example gives $r_{\ast }=1.353\mathrm{nm}$ and $\Delta G_{\ast }/(k_{B}T)=150.3$. A substantial neutral-drop barrier can coexist with supersaturated vapor.

This is deliberately a neutral-drop comparison. An ion changes the cluster free energy through electrical interactions and molecular hydration; the formula does not calculate the ion-assisted nucleation threshold. At nanometre radii, using bulk surface tension is itself approximate. There is no predicted droplet count, growth time or detection efficiency here. Once many drops grow, latent heat release, vapor depletion and heat transfer from the chamber walls invalidate the dry adiabatic expansion trajectory. The example therefore explains the direction of the effect and the need for nucleation, without extrapolating its curves to Wilson's cloud limits.

Compressibility measures how readily a sample's volume decreases when its pressure increases. For a simple compressible system with fixed particle number $N$ and composition, specify the constraint $X$:Here $P$ denotes pressure, also written $p$ elsewhere. The minus sign makes compressibility positive on a mechanically stable equilibrium branch. Its units are inverse pressure, $\mathrm{P}{\mathrm{a}}^{-1}$; its reciprocal $K_X=1/{\kappa }_X$ is the constrained bulk modulus, measured in pascal units. Holding temperature fixed gives isothermal compressibility; holding entropy fixed gives isentropic compressibility ${\kappa }_S$ along an isentropic process. The latter describes a reversible adiabatic compression, since an irreversible adiabatic process can increase entropy. ^[ref]

For an ideal gas, the ideal gas law $PV=N{k}_{B}T$ and the reversible ideal-gas adiabat give, at the same equilibrium state,Here $C_P$ and $C_V$ are the whole sample's heat capacity at constant pressure and heat capacity at constant volume, and $\gamma$ is the heat capacity ratio. Constant $\gamma$ is required for the finite-path formula $P{V}^{\gamma }=\mathrm{constant}$. Reversible adiabatic compression raises the temperature, making the pressure rise more steeply than in an isothermal compression. Thus keeping entropy fixed makes the gas harder to compress locally. For a classical monatomic gas at $100\mathrm{kPa}$, $\gamma =5/3$, so ${\kappa }_T=10^{-5}\mathrm{P}{\mathrm{a}}^{-1}$ and ${\kappa }_S=6\times 10^{-6}\mathrm{P}{\mathrm{a}}^{-1}$; the corresponding values of the bulk modulus are $100\mathrm{kPa}$ and $166.67\mathrm{kPa}$.

For a smooth, stable single thermodynamic phase at positive temperature, Maxwell relations also implywhere $\alpha$ is the coefficient of thermal expansion. The ideal gas result is one example of this broader link between mechanical and thermal responses. The equality ${\kappa }_T={\kappa }_S$ is possible when $\alpha =0$ and the other responses remain finite. These formulas compare equilibrium partial derivatives, rather than arbitrary fast or dissipative compression histories. ^[ref]

The compressibility factor is the ratio $p/(n{k}_{B}T)$, where $n$ is number density. A classical ideal gas has $Z=1$. The virial expansion writes $Z=1+B_{2}n+B_3{n}^2+\cdots$ for a dilute gas; this dimensionless ratio is distinct from isothermal compressibility, which is a response derivative with units of inverse pressure.

With constant inflow $\Phi$, no mass return or outflow, and $\Psi =ϵM_g$, mass conservation gives $M_g=\Phi /ϵ+(M_0-\Phi /ϵ)e^{-ϵt}$. Positive $ϵ$ makes the gas equilibrium stable, with perturbations decaying on timescale ${ϵ}^{-1}$. The star formation rate tends to the inflow, not necessarily to zero.

Gravitational binding is necessary but not sufficient for gas capture. A hydrostatic envelope contracts by losing energy on a Kelvin-Helmholtz scale $t_{\mathrm{KH}}\sim |E|/L$. Solid accretion heats it, while radiative transport depends on opacity and convection carries energy when the radiative gradient is unstable. The disk supply and outer tidal/thermal boundaries also matter.

A gas disk near $Q_T\sim 1$ need not form bound fragments: gravitational compression and shocks can heat it into a self-regulated spiral state. Fast enough cooling, often parameterized by $\beta =\Omega t_{\mathrm{cool}}$, permits collapse to continue. A numerical critical beta depends on thermodynamics, dimensionality and the cooling model; it is not universal.^[ref]

The orbital torque associated with disk material near the planet's corotation radius. Vortensity and entropy gradients matter because gas exchanged across the orbit carries different conserved properties. Horseshoe turns can make the response nonlinear even while the planet remains in the low-mass migration regime.

A drift chamber uses the drift times of ionization electrons in a gas to infer positions along charged particle tracks. In a known magnetic field, track reconstruction can estimate momentum and the sign of electric charge.

A historical physical name for a gas or vapour, emphasizing its mobility and appreciable compressibility. It does not imply an exactly ideal gas or prevent a vapour from undergoing condensation.

In an optically thin plasma, one convention defines the radiative cooling power per volume by $\mathcal{C}=n_e{n}_H{\Lambda }_{eH}(T)$, where $n_H$ counts hydrogen nuclei. A convention $\mathcal{C}=n_H^2{\Lambda }_{HH}$ instead has ${\Lambda }_{HH}=(n_e/n_H){\Lambda }_{eH}$. Thus the tabulated functions differ even for the same gas. With total particle density $n$, a monatomic-gas radiative gas cooling time is $3n{k}_{B}T/(2\mathcal{C})$ when heating is negligible.^[ref]

An enthalpy of atomization is the enthalpy change for making one mole of separated gaseous chemical atoms from a specified starting chemical substance. For sodium, the relevant step is sublimation to monatomic gas; for diatomic chlorine, it is half the compatible molar bond-dissociation energy expressed as an enthalpy.

For the insulated gas-only free expansion of an ideal gas, the equilibrium endpoints are $(T,V_1)$ and $(T,V_2)$ with $V_2>V_1$. Because entropy is a state function, calculate their difference along a hypothetical reversible thermodynamic process: an isothermal expansion using a frictionless piston and a thermal reservoir at $T$. This comparison process receives heat and delivers work, even though the actual free expansion does neither. Its reversible heat increment is $\delta Q_{\mathrm{rev}}=pdV=nRTdV/V$, soHere $R$ is the molar gas constant, and the logarithm acts on a dimensionless ratio. The reversible reference gives the correct endpoint entropy difference because entropy is a state function; putting the actual $Q=0$ into $\Delta S=Q/T$ would misuse a relation requiring reversible heat transfer. For the actual isolated gas, all this increase is entropy production. Thus an adiabatic process can increase entropy and need not be an isentropic process. ^[ref]

In the figure's volume doubling, the ideal gas law gives $p_1=99.768\mathrm{kPa}$ and $p_2=49.884\mathrm{kPa}$, while $\Delta S=R\ln 2\simeq 5.763\mathrm{J}{\mathrm{K}}^{-1}$. The reference process receives and delivers $nRT\ln 2\simeq 1728.85\mathrm{J}$ as heat and work; its thermal reservoir loses the same entropy that the gas gains. The actual free expansion has $Q_{\mathrm{in}}=W_{\mathrm{out}}=0$ and produces that entropy internally. Reversibly compressing the gas back at $300\mathrm{K}$ would require $1728.85\mathrm{J}$ of work input and rejection of equal heat. Restoring the gas alone therefore changes the surroundings: the actual release is irreversible.

For a real gas the final temperature may differ, so the ideal-gas logarithm formula cannot simply be copied. Evaluate $S(T_2,V_2)-S(T_1,V_1)$ using that gas's equilibrium properties or a suitable reversible comparison path. If the walls and water exchange heat with it, the Second law of thermodynamics constrains the total entropy of the isolated apparatus; the gas alone is then not an isolated system.

An entropy contrast carried by gas executing a horseshoe turn creates a density asymmetry around an embedded planet. The resulting torque can oppose the usual negative wave torque. Its persistence depends on thermal transport and the replenishment of the background entropy gradient.

The epoch when the gaseous envelope mass becomes comparable with the solid core mass. Envelope self-gravity then changes structure and can accelerate gas capture through cooling and contraction. Crossover is a useful stage in classical models, rather than a universal core mass independent of opacity and accretion history.

A graduated tube for confining a gas, measuring its volume, and often examining changes after a chemical reaction. The reading requires temperature, pressure and vapor pressure corrections.