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Combining $\Phi =-pV$ with part (i) and the expression for $⟨N⟩$ gives the ideal-gas equation of state

The one-particle translational partition function isFor a classical ideal gas,and hence

The zero-temperature internal energy isUsing part (i),Consequently the adiabatic equation of state isThe pressure is degeneracy pressure: it remains nonzero at zero temperature because of the Pauli exclusion principle. By comparison, a classical ultrarelativistic ideal gas at fixed temperature obeys $pV=N{k}_{B}T$, so its corresponding volume exponent is one and its pressure vanishes as $T\to 0$.

For $N$ particles, the ideal gas equation of state isThe internal energy of an ideal gas depends only on temperature and obeysFor constant heat capacity and a convenient choice of energy zero,

The one-dimensional mass flux and momentum flux are respectivelyIn a steady flow, the conservation laws make both the mass flux and the given energy fluxconstant. Wherever $J_m\ne 0$, their ratio is therefore constant:where $W=p/(\gamma -1)$ is the internal energy per unit volume of the ideal gas. This is the required steady-flow energy integral.

For a family in stellar homology, dimensionless radial profiles are fixed. Mass conservation and hydrostatic equilibrium then give the central scalingsThe ideal gas equation of state, with fixed mean molecular weight, consequently giveswhere fixed dimensional constants such as $G$ are suppressed in homology relations.

The proton–proton chain law $ϵ\propto \rho T^{7/2}$ gives the nuclear luminosityOn the other hand, radiative diffusion in a star givesUsing the Kramers' opacity law ${\kappa }_c\propto {\rho }_c{T}_c^{-7/2}$,Thermal equilibrium requires $L_{\mathrm{nuc}}=L_{\mathrm{rad}}$. Their common mass factor cancels, leaving $R^{-13/2}\propto R^{-1/2}$, and hence

This model captures the gas-pressure support, pp-chain burning, and strongly mass-dependent luminosity of the lower main sequence, including the Sun approximately. Its fully radiative assumption is an idealization: the Sun has a convective envelope, and sufficiently low-mass red dwarfs become largely or fully convective, so real radii are not exactly constant.

Write the ideal gas law as $P=\mathcal{R}\rho T/\mu$. Since $P=g\Sigma$,The prescribed opacity and burning laws becomeThe radiative equation is consequentlyWith $y=T^7$ and $x={\Sigma }^2/2$,The energy equation similarly givesDifferentiating the first relation therefore produces the nonlinear ordinary differential equation

At the idealized zero-temperature surface, $x=0$ and $y=0$. The outward flux there is $F_s=L/(4\pi R^2)$, soAt the base, $x_0={\Sigma }_0^2/2$ and $y=T_0^7$. The core supplies no luminosity in this model, so all flux has been generated in the overlying hydrogen envelope and $F(x_0)=0$. Hence

Let the mean particle mass be $\mu m_p$. For a monatomic ideal gas with $\gamma =5/3$,A uniform sphere of mass $M=(4\pi /3){\rho }_0{r}^3$ has Newtonian gravitational energyAt the virial threshold, the pressure term $3\int PdV=3M{k}_{B}T/(\mu m_p)$ balances $|W|$. ThereforeEliminating $r$ gives the Jeans massThe numerical coefficient depends on the convention used to identify a finite cloud with a Jeans mode. For example, assigning the mass inside a sphere of radius half the standard Jeans instability wavelength givesBoth conventions have the physically invariant scaling

For a homologous star, hydrostatic equilibrium, mass conservation, and the ideal gas equation give the central scalingswhere $\mu$ is the mean molecular weight. Integrating the nuclear energy-generation law over a fixed homologous profile givesRadiative stellar structure gives independentlyEquating the two luminosities yieldsAt fixed zero-age composition, the stars are therefore homologous with

The effective temperature satisfies $L=4\pi R^2\sigma T_e^4$, so $T_e\propto M^{3/8}$. The zero-age main sequence consequently hasIt is a steep line rising toward high luminosity and high temperature on a Hertzsprung-Russell diagram.

For fully ionized hydrogen and helium with $Y=1-X$,while $\kappa \propto 1+X$. At fixed mass,and the corresponding radius relation is

At the pure-hydrogen zero-age point $X=1$,whereasHenceandAs hydrogen is consumed, $X$ falls, so both $L$ and $T_e$ rise. In the usual diagram with temperature increasing leftward, the evolutionary track initially moves upward and leftward from the zero-age main sequence.

If $Pdx+Qdy=dF$ for a twice continuously differentiable potential $F$, then $P=F_x$ and $Q=F_y$. Equality of mixed partial derivatives gives the necessary condition for an exact differential:This condition alone does not ensure global exactness on a domain with holes.

For the Helmholtz free energy $A=U-TS$, use the stated identity for internal energy and the product rule:Thus $A_T{|}_V=-S$ and $A_V{|}_T=-p$. Equality of mixed partial derivatives yields the Maxwell relationThe temperature and volume are the independent variables here, with composition and amount of gas fixed.

Next express $S=S(T,V)$ and substitute its total differential into $dU=TdS-pdV$:Matching the coefficient of $dV$ givesThis is the thermodynamic internal pressure identity: an equation of state determines how internal energy changes under an isothermal change of volume.

For the Dieterici equation, set $c=an/(VR)$ while differentiating at fixed $V$. ThenSubstitute into the thermodynamic internal pressure identity. The term proportional to $T$ cancels $p$, leavingThe physical gas domain has $T>0$ and $V>nb$.

For the ideal gas instead, $p=nRT/V$ gives $T(\partial p/\partial T{)}_V-p=0$. Therefore $U$ is independent of volume at fixed amount of gas. Integrating the remaining condition $(\partial U/\partial T{)}_V=C_V$, with constant heat capacity at constant volume, giveswhere $U_0$ is independent of both $T$ and $V$ and fixes the reference for internal energy.

Differentiate $H=U+pV$ at fixed pressure and subtract the heat capacity at constant volume:Now use $U=U(T,V)$ and the multivariable chain rule along a constant-pressure path:Substitution cancels the fixed-volume derivative and gives the volume derivative form of the heat-capacity difference,For the ideal gas, the first term in brackets vanishes and $V=NkT/p$, so $(\partial V/\partial T{)}_p=Nk/p$. Consequently $\boxed{C_p-C_V=Nk}$ for the entire gas, with fixed particle number $N$. The extra heat capacity accounts for expansion work at constant pressure.

Holding $y$ fixed, the chain rule giveswhere the local change of variable from $x$ to $u$ is valid.

For the internal energy $U(S,V)$, its exact differential identifies $T=U_S$ and $p=-U_V$. The Maxwell relation from equality of mixed partial derivatives is $T_V{|}_S=-p_S{|}_V$. At fixed volume, the constant-volume heat capacity satisfiesso $T_S{|}_V=T/C_V$, assuming nonzero $C_V$ and a locally invertible relation. Applying the chain rule to $p(T,V)$ gives the isentropic temperature-volume derivativeFor the stated one-mole ideal gas, the equation of state gives $p_T{|}_V=R/V$ and $C_V=3R/2$, so the right side is $-2T/(3V)$. Along an isentropic process, $p\propto V^{-5/3}$ and $T=pV/R\propto V^{-2/3}$, giving the same left side:

The molecular formula is $\boxed{{\mathrm{C}}_4{\mathrm{H}}_8{\mathrm{O}}_2}$, interpreting the vapour density as relative to air. The organic elemental analysis first determines a simpler empirical formula; the vapor-density molar mass measurement then fixes its multiple.

Take 100 mass units of the chemical substance and divide each elemental contribution by its rounded relative atomic mass:The empirical formula is therefore ${\mathrm{C}}_2{\mathrm{H}}_4\mathrm{O}$, with empirical formula mass approximately 44. Its calculated mass percentages are 54.55% carbon, 9.09% hydrogen and 36.36% oxygen. The small discrepancies from the given organic elemental analysis are consistent with rounded experimental data; they should not be converted into large, meaningless integer subscripts.

For equal volumes of ideal gases at equal temperature and pressure, the density ratio equals the molar mass ratio. On the air convention,Doubling ${\mathrm{C}}_2{\mathrm{H}}_4\mathrm{O}$ gives ${\mathrm{C}}_4{\mathrm{H}}_8{\mathrm{O}}_2$; its rounded relative molecular mass of 88 predicts an air-relative vapor density of approximately $88/28.97=3.04$, close to 3.05. If the stated vapor density were instead relative to hydrogen, it would imply a relative molecular mass near 6.1, smaller than the empirical formula mass and incompatible with the analysis. That interpretation cannot describe a pure, undecomposed molecular vapor with the stated composition.

Under the older equivalent weight convention $\mathrm{C}=6$, $\mathrm{O}=8$, $\mathrm{H}=1$, the same composition and formula weight could be written ${\mathrm{C}}_8{\mathrm{H}}_8{\mathrm{O}}_4$. This is a change of notation, not eight modern carbon chemical atoms in the molecule. The result specifies composition and molar mass; it does not distinguish among structural isomers, such as ethyl acetate and butyric acid.

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 $Z_2/Z_1$.

At the same temperature and pressure, one volume of HCl corresponds to $\frac{1}{2}$ volume of hydrogen and $\frac{1}{2}$ volume of chlorine; one of steam to one of hydrogen and $\frac{1}{2}$ of oxygen; one of ${\mathrm{N}}_2\mathrm{O}$ to one of nitrogen and $\frac{1}{2}$ of oxygen; one of NO to $\frac{1}{2}$ of each element; and one of ammonia to $\frac{1}{2}$ of nitrogen and $\frac{3}{2}$ of hydrogen. All elemental volumes here refer to their ordinary diatomic gases.

The simplest way to avoid an erroneous factor of two is to balance each chemical reaction using ${\mathrm{H}}_2$, $\mathrm{C}{\mathrm{l}}_2$, ${\mathrm{N}}_2$ and ${\mathrm{O}}_2$. Under Avogadro's law, their stoichiometric coefficients are also gas volume ratios, provided all gases are compared at the same temperature and pressure and ideal gas behaviour is a good approximation.

If the common chosen volume of each compound is $V$, the corresponding ordered elemental volumes are therefore $(V/2,V/2)$, $(V,V/2)$, $(V,V/2)$, $(V/2,V/2)$ and $(V/2,3V/2)$. The ratios concern molecular counts and hypothetical free elemental gases, not separate compartments of elemental gas still present inside a compound.

“Hydrochloric acid” must here mean gaseous hydrogen chloride, rather than its aqueous chemical solution; water must remain as steam. If water condenses or Ammonia dissolves in the collecting liquid, measured volumes no longer show these simple ratios. For an experimental comparison, dry-gas and hydrostatic pressure corrections are those discussed in 3.

Take dry air as unity, and compare each gas at the same temperature and pressure. In the ideal gas approximation, $\rho =pM/(RT)$, so the relative density isHere $M$ is molar mass. The following values use modern approximate molar masses; rounding and historical atomic weight conventions can shift the last figures.

“Olefiant gas” is confirmed as a historical synonym of Ethylene in the NIST Chemistry WebBook, SRD 69. In this gas list, “carbonic acid” means carbon dioxide, not an isolated vapour of aqueous ${\mathrm{H}}_2{\mathrm{CO}}_3$. “Alcohol” and “ether” refer to ethanol and diethyl ether, rather than their entire modern chemical classes.^[ref]

The order is therefore Ammonia, then carbon monoxide and Ethylene nearly tied, then air, then carbon dioxide and nitrous oxide nearly tied, then ethanol vapour, then diethyl ether vapour. Nearly equal molar masses do not imply equal chemical behaviour.

If instead molecular hydrogen is chosen as unity, divide each molar mass by about $2.016\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$ and state that convention. In the older nominal ${\mathrm{H}}_2=2$ approximation, the corresponding values are about $22,14,22,14.5,8.5,14,23,37$ in the source order. These are a different reference scale, not different substances.

The vapour comparisons are not comparisons of the liquid densities. Pure ethanol and diethyl ether must be considered at conditions where they remain gaseous, or by a low-pressure gas approximation. Their liquids can be lighter than water while their vapours are heavier than air.

Sound compressions in air are approximately adiabatic, giving $c_s^2=(\partial p/\partial \rho {)}_s=\gamma p/\rho$. With $p=\rho RT/M$,Here dry air is approximated as a diatomic ideal gas with $\gamma =7/5$ and molar mass $M\simeq 0.029\mathrm{kg}\mathrm{mo}{\mathrm{l}}^{-1}$. The ideal-gas law alone is insufficient without this adiabatic response and mean molecular mass; humidity and composition affect a more precise inference.

For a small longitudinal disturbance, let the equilibrium pressure and mass density be $p_0,{\rho }_0$, their perturbations be $p^{\prime },{\rho }^{\prime }$, and the fluid velocity be $v$. Linearizing conservation of mass and the inviscid momentum equation in one dimension givesWhen compression is adiabatic, the equation of state gives $p^{\prime }=c_s^2{\rho }^{\prime }$ with $c_s^2=(\partial p/\partial \rho {)}_s$. Differentiating and eliminating $v$ yieldsfor an ideal gas with heat-capacity ratio $\gamma =C_p/C_v$ and molar mass $M_{\mathrm{mol}}$. Compression raises temperature as well as density, explaining the factor $\gamma$ compared with the isothermal value $\sqrt{p_0/{\rho }_0}$.

A rightgoing plane sound wave has pressure and particle velocity in phase, $p^{\prime }={\rho }_0{c}_{s}v$, and mean intensity $⟨p^{\prime }v⟩={\hat{p}}^2/(2{\rho }_0{c}_s)={\rho }_0{c}_s{\hat{v}}^2/2$ for peak amplitudes. Particle displacement is a quarter-cycle out of phase with particle velocity: particles oscillate locally while the compression pattern transports energy. This continuum linear description requires wavelength much larger than the mean free path, small pressure perturbations and negligible viscous attenuation. Adiabatic propagation requires heat exchange across the compressed region to be slow on the oscillation time; strong thermal coupling would instead favour an isothermal response.

A wave propagates an organized disturbance through space. What oscillates depends on the medium: displacement in a string or solid, pressure and density in a fluid, electric fields and magnetic fields in an electromagnetic wave, or a quantum wavefunction. The propagation law follows from the restoring mechanism, inertia and relevant boundary conditions. A common linear equation is $u_{tt}=c^2{\nabla }^{2}u$, whose plane waves obey $\omega =c|\mathbf{k}|$. The resulting phase velocity and group velocity coincide, so a narrow wave packet can travel without dispersive spreading. More generally the material supplies a dispersion relation $\omega (\mathbf{k})$; the envelope velocity is ${\nabla }_{\mathbf{k}}\omega$, while constant phase surfaces move at the corresponding phase velocity.

Mechanical waves make the balance of restoring force and inertia particularly visible. A flexible stretched string has $c=\sqrt{T/\mu }$: greater tension stiffens the string, whereas greater linear mass density slows its transverse motion. Small-amplitude acoustic waves in a uniform fluid are longitudinal, with $c_s=\sqrt{K_s/\rho }$, where $K_s$ is the adiabatic bulk modulus. For an ideal gas, $c_s=\sqrt{\gamma p/\rho }$. Pressure changes in rapid sound oscillations are approximately adiabatic because little heat crosses a wavelength in one period. Fluids do not support ordinary bulk shear waves, whereas an elastic solid supports both longitudinal and transverse waves. In an isotropic solid their speeds are $c_L=\sqrt{(K+4G/3)/\rho }$ and $c_T=\sqrt{G/\rho }$, where $G$ is the shear modulus.

A material can be dispersive even when its equilibrium properties are uniform. In a liquid with a free surface, gravity and surface tension act as restoring mechanisms. A simple inviscid capillary-gravity wave haswhere $h$ is depth and ${\gamma }_s$ is surface tension. Long shallow gravity waves have approximately constant speed $\sqrt{gh}$; deep-water gravity waves have $\omega =\sqrt{gk}$ and group velocity half their phase velocity. Short capillary waves have a different dispersion dominated by surface tension. Bending stiffness similarly changes short-wavelength propagation in a string or plate, giving a contribution proportional to $k^4$ in ${\omega }^2$. Different restoring mechanisms can therefore control different wavelength ranges of the same medium.

Electromagnetic waves propagate in vacuum without material displacement, because Maxwell equations couple changing electric and magnetic fields. They are transverse and travel at $c=1/\sqrt{{ϵ}_0{\mu }_0}$. In a simple transparent isotropic material, the phase speed is $1/\sqrt{ϵ\mu }=c/n$, where $n$ is the refractive index. Real material polarization depends on frequency. A Lorentz oscillator model for bound charges explains why permittivity and refractive index become strongly frequency dependent near material resonances, accompanied by absorption. In a birefringent medium, different field polarizations have different propagation constants, permitting wave plates and mode-dependent delays. A electrical conductor additionally supports dissipative currents; its complex wavenumber gives attenuation and, for a good conductor, a skin depth approximately $\sqrt{2/(\mu \sigma \omega )}$. Energy removed from the wave becomes heat rather than disappearing.

At an interface, frequency is conserved for stationary media, and tangential wave phase must match. These conditions give Snell's law for refraction and constrain reflected and transmitted directions. Amplitudes follow from continuity of the appropriate fields or mechanical forces. On strings the relevant mechanical impedance is $\sqrt{T\mu }$; for sound it is related to $\rho c_s$, and for electromagnetic waves to $\sqrt{\mu /ϵ}$. A mismatch gives wave reflection, and a lossless boundary conserves the total incident, reflected and transmitted power. Multiple reflections can reinforce or cancel, as in a Fabry–Pérot etalon. Total internal reflection and other boundary conditions confine waves in waveguides. Transverse modes then become discrete, and even an otherwise nondispersive material develops longitudinal cutoffs and wave dispersion. A membrane strip, for example, obeys ${\omega }_n^2=c_m^2[k^2+(n\pi /a{)}^2]$.

Propagation also involves transport and loss. The Poynting vector gives electromagnetic energy flux; mechanical energy transport is force or pressure times velocity, averaged over a cycle. Wave absorption, viscosity, electrical resistance and imperfect boundaries dissipate energy. Scattering redistributes energy into other directions and modes, while geometric spreading reduces local intensity without necessarily dissipating anything. These mechanisms should be distinguished when interpreting a decreasing measured amplitude. In strongly absorbing or rapidly varying dispersive regimes, the derivative called group velocity is not automatically the speed of a sharp signal front; pulse reshaping and the medium's causal response must be considered.

Finally, quantum matter waves illustrate how the same propagation mathematics can describe a different physical quantity. For a free nonrelativistic particle, $\omega =\hslash k^2/(2m)$, so the group velocity $\hslash k/m$ equals its particle velocity and differs from the phase speed $\hslash k/(2m)$. In a periodic crystal, Bloch's theorem replaces the free dispersion by energy bands. Their slope controls wave-packet velocity, $v_g={\hslash }^{-1}dE/dk$, and their curvature gives band effective mass. Thus mechanical confinement, material response and microscopic periodicity all reshape propagation through their allowed modes and dispersion, while the physical interpretation of the wave and its transported quantities remains essential.

At fixed positive temperature, $-R\ln (p/p_0)$ tends to $+\infty$ as $p\to 0^{+}$. This is a physically meaningful unbounded-volume limit for a fixed mole number: $V_m=RT/p\to \infty$, so the number of accessible translational configurations grows without bound. For an ideal gas the entropy difference is $R\ln (V_{m,2}/V_{m,1})$ at fixed temperature. No finite vessel reaches an infinite entropy, and $p=0$ at finite temperature is not a finite-volume one-mole state. The divergence is approached by dilution; it does not conflict with the third law, which concerns a different temperature limit.