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An adiabat is a curve representing a quasistatic adiabatic change. In a reversible description it is also a curve of constant entropy. For an ideal gas with constant heat-capacity ratio $\gamma$, it obeys $p{V}^{\gamma }=\mathrm{constant}$. A general irreversible adiabatic process need not follow this equilibrium-state curve and can produce entropy.

Adiabatic compression decreases the volume of a sample without heat transfer across its boundary. For the fixed-composition ideal gas and endpoint assumptions in adiabatic process, compression work increases internal energy; positive heat capacity at constant volume then means a higher final temperature. The effect comes from supplied work, even though no heat enters. The amount of temperature increase depends on the work actually delivered to the gas, not on the adiabatic process condition alone.

The reversible ideal-gas adiabat supplies a particularly useful quantitative benchmark. Assume a fixed amount of substance $n$, a quasistatic process, negligible friction, only pressure-volume work, and constant heat capacity ratio $\gamma =C_p/C_V>1$. Here $C_p$ and $C_V$ are the whole sample's heat capacity at constant pressure and heat capacity at constant volume. With $R$ the molar gas constant, Mayer's relation gives $C_V=nR/(\gamma -1)$. The proof outline is $C_{V}dT=-pdV$ from the first law of thermodynamics, followed by $p=nRT/V$ from the ideal gas law and integration. The result isThese finite power laws belong to this reversible, constant-$\gamma$ ideal gas model. If heat capacity ratio varies, use the integral relation in heat capacity ratio; if the compression is irreversible, use the actual work balance, as in irreversible adiabatic piston compression. A short stroke by itself justifies neither power law. ^[ref]

Writing $r=V_1/V_2>1$ for the compression ratio, the reversible model predictsAll temperatures in these ratios are absolute. The larger pressure rise than in an isothermal compression, where $p_2/p_1=r$, explains the greater local stiffness described by isentropic compressibility.

For an original stipulated example, set $\gamma =7/5$, $p_1{V}_1=nR{T}_1=1\mathrm{kJ}$ and $r=2$. Then $T_2/T_1=2^{0.4}\simeq 1.31951$ and $p_2/p_1=2^{1.4}\simeq 2.63902$. The input work and internal energy increase are both $0.79877\mathrm{kJ}$, while $Q_{\mathrm{in}}=0$. No initial temperature or apparatus performance is implied by the chosen energy scale. Independent integration gives the signed outward workwhich is negative during compression and agrees with $-\Delta U$. This integral uses equilibrium gas pressure only because the path is reversible.

In the sensor experiment of adiabatic process, compare the pressure-volume integral with $C_V\Delta T$ as an energy check only when internal pressure is sufficiently uniform and piston dynamics permit that boundary-work estimate. Account for friction separately before comparing it with hand input work. PASCO's manual supplies a compression procedure and an illustrative logarithmic-slope result $1.39$, compared with the model value $1.40$ for a diatomic gas; that is its sample result, not new measurements here. ^[ref]

Adiabatic expansion increases the accessible volume without heat transfer. In a loaded piston expansion of the fixed-composition ideal gas considered in adiabatic process, positive outward work comes from internal energy, so the final temperature decreases. This provides a cooling mechanism without removing heat during the stroke. In the Carnot cycle, reversible adiabatic expansion connects two different temperatures between its isothermal processes.^[ref]

For the same constant-$\gamma$ reversible ideal-gas adiabat, put $b=V_2/V_1>1$. The scaling and first law of thermodynamics giveWith the figure's stipulated $\gamma =1.4$, $p_1{V}_1=1\mathrm{kJ}$ and $b=2$, the final ratios are $T_2/T_1\simeq 0.75786$ and $p_2/p_1\simeq 0.37893$. Outward work is $0.60535\mathrm{kJ}$ and $\Delta U=-0.60535\mathrm{kJ}$. Reversing this particular path from its cooled endpoint would require the same work input; the figure's separate compression starts at the original state, so its input is different. Endpoint entropy supplies another independent check:The two logarithm terms cancel using $C_V(\gamma -1)=nR$, confirming the reversible path is an isentropic process. ^[ref]

PASCO's expansion demonstration first compresses trapped air and waits for thermal equilibrium before the rapid expansion. Its manual warns that friction-warmed cylinder regions contact the gas during expansion, compromising quantitative comparison. ^[ref] The hold separates cooling caused by expansion from simply undoing the earlier hot compression. As proposed controls, observe the temperature dip and subsequent warming, compare rapid and slow strokes from matched initial states, and assess wall heat transfer and probe lag before fitting a reversible model.

Cooling is not a universal consequence of adiabatic expansion: free expansion of an ideal gas has no outward work and returns to the same equilibrium temperature. Its entropy increases, so the reversible power laws do not describe it. Reuse free expansion, Joule's free-expansion experiment and temperature sensitivity of a free-expansion experiment for that apparatus, its energy budget and the real-gas qualifications. The useful distinction is the load: expansion cools this ideal gas when it must spend internal energy doing work.

For fixed-composition monatomic ideal gas plus equilibrium radiation pressure, specific internal energy is $u=3\mathcal{R}T/2+a_{\mathrm{r}}{T}^4/\rho$. An isentropic process satisfies $du=Pd\rho /{\rho }^2$, giving ${\Gamma }_3-1=(8-6\beta )/(24-21\beta )$ andThe stellar gas-pressure fraction must be allowed to change along the adiabat. The pure-radiation and pure-gas limits give ${\Gamma }_1={\Gamma }_2=4/3$ and $5/3$, respectively. Partial ionization or different gas heat capacities change these formulas.

A dry parcel in rapid pressure balance with an ambient ideal gas has $d{T}_g/dP=T_g\mathcal{R}/(P{C}_p)$. Ambient hydrostatic equilibrium gives $dP/dz=-Pg/(\mathcal{R}{T}_{\mathrm{env}})$, hence $d{T}_g/dz=-(g/C_p)T_g/T_{\mathrm{env}}$. At a temperature-matched launch point this equals the dry adiabatic lapse rate, supplying the local linear stability test. A substantially hotter parcel does not exactly follow the ambient density's hydrostatic adiabat.

For a fixed-composition ideal gas, the reversible adiabatic process relation $c_{p}dT/T=Rdp/p$ gives $A\ln (T/T_s)+B(T-T_s)=R\ln (p/p_s)$. Under hydrostatic equilibrium, $gz=A(T_s-T)+B(T_s^2-T^2)/2$. Increasing positive $B$ at fixed surface conditions increases temperature at a given smaller pressure and raises the model's formal zero-temperature height. That finite-height mathematical boundary is not a physical absolute-zero prediction.

The translational Helmholtz free energy cost of restricting a dilute ideal gas from a volume $V$ to an area $A$ at the same temperature. For identical particle number and mass,Here $\Lambda$ is the thermal de Broglie wavelength. This contribution is positive for $V/A\gg \Lambda$; a favourable binding energy must be added separately to predict net adsorption.

Consider an original simplified model containing only nitrogen and argon, with argon mole fraction $x$. Let their molar masses be $M_N$ and $M_A$. Adding the component masses gives the mixture's mean molar massFor an ideal gas mixture at the same pressure and temperature as a pure-nitrogen reference, the gas density ratio isThe inversion requires known identities, the binary-mixture assumption and $M_A\ne M_N$. A density ratio cannot uniquely recover an arbitrary multicomponent mixture. Nor should $x$ be interpreted as the argon fraction of untreated air: the model sample has only the two specified components.

Choose round model values $M_N=28\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$ and $M_A=40\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$. For $x=0.010$, the calculated relative density excess isThus a one-percent mole fraction does not mean a one-percent gas density change. In a chosen $2.00\mathrm{L}$ globe at $100\mathrm{kPa}$ and $300\mathrm{K}$, the ideal gas law gives approximately $2.24508\mathrm{g}$ for pure nitrogen and $2.25471\mathrm{g}$ for this mixture: only $9.62\mathrm{mg}$ separates the weighings. These are model calculations with rounded molar masses, not historical measurements.

A direct control calculation explains why the filling conditions matter. At fixed pressure, pure nitrogen athas the same gas density as that mixture at $300\mathrm{K}$. A roughly $1.28\mathrm{K}$ mismatch can therefore imitate the entire stipulated composition signal. For small changes, the ideal gas law givesAn independently measured pressure and temperature, matched weighing procedure, and checks for residual vapour distinguish composition from these alternative causes. The model explains the need for controls; it does not supply the complete corrections for a real gas-weighing instrument.

Equal volumes of ideal gases at the same temperature and pressure contain equal amounts of substance. Named after Amedeo Avogadro. From $pV=nRT$, $V/n=RT/p$ is common to them.

A steady outward radial flow with negligible pressure force conserves $v^2/2-GM/r$. If its speed at a reference radius equals the mechanical escape velocity, this constant is zero and $v\propto r^{-1/2}$. Conservation of mass gives $\rho \propto r^{-3/2}$. An adiabatically evolving ideal gas carried along that leading flow has $p\propto r^{-3\gamma /2}$ and $T\propto r^{-3(\gamma -1)/2}$.

Finite specific enthalpy changes the dynamical problem: the full conserved Bernoulli equation is $v^2/2+h-GM/r=B$. At the same boundary mechanical escape speed, $B=h_e$ rather than zero. A warm wind therefore needs its enthalpy and equation of state specified before the explicit ballistic velocity can be replaced by a consistent pressure-driven solution.

Consider $N$ noninteracting bosons in one internal quantum state, at thermal equilibrium in the three-dimensional harmonic oscillator potentialHere $m$ is the mass, the positive ${\omega }_i$ are angular frequencies, and $\bar{\omega }$ is their geometric mean. Measure the one-particle energy $ϵ$ and chemical potential $\mu$ from the ground-state energy. The quantum harmonic oscillator levels then have $ϵ=\hslash ({\omega }_x{n}_x+{\omega }_y{n}_y+{\omega }_z{n}_z)$ with nonnegative integer $n_i$ and reduced Planck constant $\hslash =h/(2\pi )$.

When $ϵ\gg \hslash {\omega }_i$, counting lattice points can be approximated by the volume of a tetrahedron in the positive octant of $(n_x,n_y,n_z)$:This is a density of states, with units of inverse energy. The factor $1/6$ is the unit tetrahedron volume; differentiating produces the factor $1/2$. Treat the single ground state separately: its macroscopic population would be lost by replacing every level with a smooth density of states.

For $k_{B}T\gg \hslash {\omega }_i$, integrate the Bose-Einstein distribution over excited energy eigenstates. As $\mu$ approaches zero from below, their maximum population isHere $k_B$ is the Boltzmann constant and $\zeta$ the Riemann zeta function. A short proof uses $1/(e^x-1)={\sum }_{\ell \ge 1}{e}^{-\ell x}$ and ${\int }_0^{\infty }{x}^2{e}^{-\ell x}dx=2/{\ell }^3$. The nonnegative terms permit exchanging summation and integration by Tonelli's theorem, giving ${\int }_0^{\infty }{x}^2/(e^x-1)dx=2\zeta (3)$. Near zero the integrand is proportional to $x$, so its integral converges. That finite excited-state capacity is the mechanism of Bose-Einstein condensation.

Set $N_{\mathrm{ex}}^{\mathrm{sat}}(T_c)=N$. Particle conservation gives, to this continuum accuracy,Above $T_c$, the macroscopic condensate fraction is zero and the chemical potential adjusts below zero. This cubic power belongs to a three-dimensional harmonic trap; it is not the homogeneous three-dimensional ideal gas exponent. For $N=10^5$, $\bar{\omega }/(2\pi )=100\mathrm{Hz}$ and $\zeta (3)\simeq 1.202056903$, the model gives $T_c\simeq 209.5\mathrm{nK}$. At $T=0.8T_c\simeq 167.6\mathrm{nK}$, $N_0/N=0.488$, or approximately $48800$ condensed bosons. These are chosen parameters and calculated values.

Finite $N$ gives a smooth crossover rather than a sharp phase transition. The exact ideal gas sum over excited quantum harmonic oscillator levels differs from the continuum integral, especially when $k_{B}T$ is not large relative to every $\hslash {\omega }_i$. A sharp trapped-gas limit takes $N\to \infty$, $\bar{\omega }\to 0$ with $N{\bar{\omega }}^3$ fixed and fixed frequency ratios. Interactions also alter the critical temperature, condensate fraction and release profile; a dilute gas can have appreciable interaction energy compared with the trap level spacing. The cubic law is a useful baseline, not a fit guaranteed for either 1995 experiment.

Compare uniform three-dimensional ideal gases at the same number density n, with fermion mass $m_F$, spin degeneracy $g_F$, boson mass $m_B$ and a single boson spin component. The Fermi temperature has $k_B{T}_F={\hslash }^2(6{\pi }^{2}n/g_F{)}^{2/3}/(2m_F)$, while Bose saturation gives $k_B{T}_B=2\pi {\hslash }^2[n/\zeta (3/2){]}^{2/3}/m_B$. Eliminating density yieldsThe degeneracy is essential; using a spinless Fermi sea changes the numerical estimate. This is an ideal-gas comparison and does not remove interaction effects in a dense liquid.

Boyle's law states that a fixed amount of substance of an ideal gas at constant absolute thermodynamic temperature has constant product of absolute pressure and total volume:Here $p$ is measured relative to a vacuum, not relative to the surrounding atmospheric pressure. Halving the volume doubles the absolute pressure; doubling the volume halves it. A real gas approximately follows Boyle's law when it is sufficiently dilute and far from a phase transition.^[ref]

For an original calculated example, suppose a sealed gas sample has $p_1=100\mathrm{kPa}$ and $V_1=100\mathrm{c}{\mathrm{m}}^3$. After isothermal compression to $V_2=50\mathrm{c}{\mathrm{m}}^3$,Both states have $C=10000\mathrm{kPa}\mathrm{c}{\mathrm{m}}^3=10\mathrm{J}$, since $1\mathrm{kPa}\mathrm{c}{\mathrm{m}}^3=10^{-3}\mathrm{J}$. The volume ratio is dimensionless, so using the same volume units in numerator and denominator leaves the pressure in kilopascals. Although $pV$ has units of energy, its value is not the compression work: isothermal ideal-gas work involves an integral along the path.

The blue curves show calculated ideal gas states, not measurements. The left panel is a branch of a hyperbola; the right panel plots exactly the same absolute pressures against $1/V$. The dashed line uses gauge pressure with a stipulated constant atmospheric pressure of $100\mathrm{kPa}$. Its displaced intercept explains why absolute pressure in Boyle's law matters.

A short derivation starts from the ideal gas law, $pV=nRT$. Fixing the amount of substance $n$ and absolute thermodynamic temperature $T$ fixes $C=nRT$, where $R$ is the molar gas constant. The microscopic picture in ideal gas pressure from wall collisions is equally appealing: at fixed temperature, the typical particle kinetic energy stays fixed, while compression increases the number density and wall momentum transfer per unit surface area and time. A larger absolute pressure follows without requiring hotter particles.

The constraints are essential. A leak changes the amount of substance. Rapid compression can raise the temperature before heat transfer restores thermal equilibrium, as in the isothermal gas syringe experiment. Molecular size and interactions in a dense real gas invalidate the ideal gas approximation; the Van der Waals equation describes one such correction. Equal endpoint temperatures allow an endpoint test of Boyle's law, but do not establish that every intervening state lies on an isothermal process.

For fixed mass $m$, mass density satisfies $\rho =m/V=mp/C$, so it is proportional to absolute pressure. In a dilute bubbly flow, Boyle's law allows bubble volume and its contribution to buoyancy to increase as hydrostatic pressure decreases, provided gas mass is retained, dissolution is negligible, and the temperature stays approximately fixed. Its usefulness is that a simple ratio predicts a substantial size change without tracking individual molecules.

The ideal Brayton cycle consists of adiabatic compression, constant-pressure heat addition, adiabatic expansion, and constant-pressure heat rejection. Its ideal working substance is an ideal gas with constant heat capacities, and the insulated stages are reversible thermodynamic processes.

For pressure ratio $r_p=p_h/p_l$ and heat capacity ratio $\gamma =C_p/C_V$, a reversible ideal-gas adiabat has $T\propto p^{(\gamma -1)/\gamma }$. Thus $T_2/T_1=T_3/T_4=k=r_p^{(\gamma -1)/\gamma }$, and the constant-pressure heat transfers have $Q_{\mathrm{out}}/Q_{\mathrm{in}}=(T_4-T_1)/(T_3-T_2)=1/k$. The thermal efficiency isThis formula requires constant heat capacities and internally reversible insulated stages. An arbitrary dissipative adiabatic process does not satisfy the same pressure-temperature relation.

For uniform composition, upward coordinate $z$ and downward acceleration $g>0$, a pressure-equilibrated adiabatic parcel in a hydrostatic ideal gas has $\ddot{\zeta }=-N^2\zeta$ with $N^2=g[(d\ln p/dz)/\gamma -d\ln \rho /dz]$. A positive value gives restoring buoyancy; for a calorically perfect gas it is also $(g/c_p)ds/dz$. Heat or composition exchange requires a modified stability test.

A Carnot cycle realizes the reversible bound in Carnot's theorem. Its four steps for an ideal gas working substance are:

The temperature versus entropy plot in the Carnot's theorem figure is therefore a rectangle traversed clockwise. Because $\delta Q_{\mathrm{rev}}=TdS$, its width is $\Delta S=Q_h/T_h$. For the illustrated engine, $\Delta S=5/3\mathrm{J}{\mathrm{K}}^{-1}$: the upper step absorbs $600\times 5/3=1000\mathrm{J}$, the lower step rejects $300\times 5/3=500\mathrm{J}$, and the enclosed area is the $500\mathrm{J}$ net work. Geometry displays the thermal efficiency without solving the gas motion. The two vertical steps are isentropic processes, not constant-temperature steps. ^{[ref] [ref]}

Reversible heat transfer is a limit of vanishing temperature differences; ordinary finite-rate transfer introduces entropy production. The Carnot cycle is a performance benchmark rather than a guarantee of useful power. Running it backward gives the ideal operation of a refrigerator or heat pump, with work supplied to move heat toward the hotter thermal reservoir. ^[ref]

For spherical hydrostatic equilibrium with negligible surface pressure, ideal gas support and constant mean molecular weight $\mu$, the stellar virial theorem gives $3k_B\int Tdm/(\mu m_p)=\int Gmdm/r$. Since $r<R$ throughout the interior, the right side exceeds $G{M}^2/(2R)$. Division by $M$ gives the stated mass-weighted temperature bound. The central bound additionally assumes that the central temperature is maximal. A central pressure lower bound alone is insufficient without controlling central mass density.

The chemical potential measures how a thermodynamic potential changes when matter is added under specified constraints. It plays for matter exchange the role that temperature plays for heat exchange: a difference supplies a direction, and equality characterizes thermodynamic equilibrium when the corresponding exchange is allowed. This connects diffusion, phase coexistence and chemical equilibrium through one quantity.

For a single component, the particle convention iswhere $U$ is internal energy, $N$ counts particles, $S$ is entropy and $V$ is volume. This partial derivative holds entropy and volume fixed; it measures an incremental change rather than the average energy per particle. See David Tong, Statistical Physics, section 1.4.

In chemistry the molar convention instead uses amount of substance $n$, measured in moles:Here $G$ is Gibbs free energy, $T$ is temperature, $p$ is pressure and $N_{\mathrm{A}}$ is the Avogadro constant. The units are $\mathrm{J}$ per particle versus $\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}$. For several chemical species, define each chemical potential by varying only its own particle count or amount of substance, with all other components' counts or amounts fixed. Many texts use $\mu$ for either convention, so the units and the variable being differentiated must identify it.

For a simple compressible system with variable component inventories, the fundamental differentials areThey describe changes between states in thermodynamic equilibrium with only pressure-volume work. If every component inventory is fixed, the sums vanish. The Gibbs free energy form follows by differentiating $G=U-TS+pV$ and using $d{N}_i=N_{\mathrm{A}}d{n}_i$; this is why the two definitions describe the same chemical potential in different units.

A chemical potential can be negative. Holding entropy fixed while adding matter need not increase internal energy; Tong explains this explicitly for a classical ideal gas in Statistical Physics, section 2.2.4. Its absolute sign also depends on the reference energy: replacing $U$ by $U+{ϵ}_{0}N$ shifts ${\mu }_{\mathrm{particle}}$ by ${ϵ}_0$. A common reference shift cancels from differences for the same chemical species, which are what determine exchange. Thus “energy cost” means a signed increment, not an always positive amount that must be supplied.

Rigid compartments impose a different constraint: fixed volume rather than fixed pressure. With Helmholtz free energy $F=U-TS$,This follows immediately by differentiating $F=U-TS$ in the chemical potential differential above. It expresses the same chemical potential through the thermodynamic potential appropriate to fixed temperature and volume.

Consider a closed system consisting of two rigid compartments of equal volume $V$, maintained at a common fixed temperature $T$. A valve permits exchange of the same ideal gas; the partition stays in place. Initially $n_A=9\mathrm{mol}$ and $n_B=3\mathrm{mol}$, with conserved total amount of substance $n_{\mathrm{tot}}=12\mathrm{mol}$. These are macroscopic moles, modeled continuously, not twelve individual particles. By the ideal gas law, $p_B/p_A=n_B/n_A=1/3$, so initiallyAt $300\mathrm{K}$ with the stipulated $R$, this is approximately $-2740.311\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}$, favoring transfer from A to B.

At fixed $T$ and equal $V$, integrating the ideal-gas chemical potential gives the total Helmholtz free energy, up to a term independent of the allocation,Here $n_{\mathrm{ref}}$ is a positive reference amount of substance; any omitted common term linear in $n_A+n_B$ is constant along this exchange. Let $x=n_A/n_{\mathrm{tot}}$ be the dimensionless fraction in A. The relative Helmholtz free energy then takes the particularly simple formIts derivative along the allowed transfer and its second derivative areThus the Helmholtz free energy is a strictly convex function of the allocation for positive amounts. Its unique minimum has $x=1/2$, or $n_A=n_B=6\mathrm{mol}$, and equal chemical potentials. Initially its excess above this minimum is $n_{\mathrm{tot}}RT[\frac{3}{4}\ln (3/2)+\frac{1}{4}\ln (1/2)]\simeq 3915.474\mathrm{J}$ at $300\mathrm{K}$. Relaxation decreases the total Helmholtz free energy by this amount in the model; the final pressures are also equal because both compartments have the same volume.

The beautiful connection is that the direction of matter exchange and the slope of a free energy curve are the same calculation. The entropy gained by redistributing the ideal gas makes equal allocation favorable here. As with chemical equilibrium, a free energy minimum identifies the favored state, while transport properties determine how quickly it is reached. The fixed-temperature, fixed-pressure Gibbs free energy criterion is summarized in 16.4 Free Energy; the rigid experiment instead uses Helmholtz free energy.

Consider two internally equilibrated reservoirs A and B at the same imposed temperature and pressure. They may have different compositions or represent different phases. Transfer a small positive amount of substance $\delta n$ of the same chemical species from A to B, holding the other inventories fixed. To first order, their total Gibbs free energy changes byHence exchange toward lower chemical potential lowers Gibbs free energy. If transfer is allowed in both directions and the two reservoirs contain the exchangeable component, thermodynamic equilibrium requires equal chemical potentials for that component. See MIT 5.60, Lecture 14, pages 1–2.

An impermeable partition can maintain unequal chemical potentials; equality is a condition on an allowed exchange. Two pure samples of the same ideal gas already have equal chemical potentials at the same temperature and pressure. Different compositions at a common pressure can instead give different chemical potentials for one component through its thermodynamic activity.

Equality concerns the same exchangeable component in different places, not all different chemical species. For a permitted chemical reaction, the corresponding interior chemical equilibrium criterion is ${\sum }_i{\nu }_i{\mu }_{\mathrm{m},i}=0$, with signed stoichiometric vector entries ${\nu }_i$. The existing chemical equilibrium and constrained Gibbs minimization for chemical equilibrium sections develop those composition constraints.