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A liquid-gas thermodynamic critical point is the endpoint of the coexistence curve where distinct liquid and gas densities merge. In the van der Waals illustration, write the original constants as $a_v,b_v$ to distinguish them from the later polynomial coefficients. The critical isotherm has a horizontal inflection in $p(V)$; setting its first two volume derivatives to zero givesBelow this temperature the homogeneous van der Waals loop contains an unstable region; physical equilibrium instead separates into liquid and gas.^[ref]

For the critical point of a cubic density equation of state, assume $b,c>0$. Its density derivatives areBecause $n=N/V$ is a smooth monotone function of $V$ at finite positive density, a stationary inflection obeys $p_n=p_{nn}=0$ as well. Solving these equations and substituting back givesThe symbols $b,c$ here have different units from the van der Waals excluded-volume parameter. With $x=n/n_c$, $t=T/T_c$, the isotherms becomeThe spinodal densities of a cubic gas delimit the two-extrema regime. The curves are monotone for $t>1$, have a stationary inflection at $(x,p/p_c)=(1,1)$ for $t=1$, and have two extrema $x=1\pm \sqrt{1-t}$ for $0<t<1$.

For the critical-density compressibility of a cubic gas, since ${\kappa }_T=(1/n)(\partial n/\partial p{)}_T$, evaluation at the critical density givesIt diverges as $T\to T_c^{+}$ in this mean-field model. For $T<T_c$ it is negative on the homogeneous branch at $n_c$, violating mechanical stability: a pressure increase would expand rather than compress that branch.

The actual equilibrium below criticality is coexistence of two phases, with equal pressure and chemical potential and different densities. A sample's mean density can change by changing phase proportions. The unstable homogeneous branch is not a realizable equilibrium isotherm. The Maxwell construction in density coordinates requires care: the equal-area construction is performed in $p$ versus volume, not by using unweighted area in $p$ versus density. The truncated polynomial is an approximate mean-field equation, so its predictions need not give quantitative critical behaviour for a real gas.

The Boltzmann distribution assigns probability $P_i=e^{-E_i/(k_{B}T)}/Z$ to each microstate. For the equilibrium height distribution in a uniform gravitational field, take a dilute gas above a floor in a constant gravitational field. Uniform cross-section and no relevant upper boundary give the normalized height densityThusThe gravitational potential is linear in height, so its mean is not a quadratic-mode contribution of $k_{B}T/2$.

The equipartition states that each independent quadratic term in a classical equilibrium Hamiltonian contributes mean energy $k_{B}T/2$, provided the relevant degrees of freedom are thermally accessible. Equivalently, $⟨x_i\partial H/\partial x_i⟩=k_{B}T$ when the phase-space integration by parts has vanishing boundary terms. Three translational velocity components therefore contribute $3k_{B}T/2$.

For rotational equipartition in a rough-ball gas, collisions exchange both translational and rotational energy. In the intended three-dimensional classical equilibrium model there are three translational and three rotational quadratic terms, giving $3k_{B}T$ in total kinetic energy per ball. Adding $⟨U_g⟩=k_{B}T$, and conserving the initial energy $mgh$ per ball, givesRoughness is relevant because it allows rotation to thermalize. The effective mechanical temperature of released rough balls describes the balls' mechanical degrees of freedom, not their internal material temperature; finite ball size and excluded-volume effects are neglected.^[ref]

The height tail isThis can be nonzero because collisions redistribute energy between balls; an individual ball is not restricted to its original energy. The exponential distribution is the large-number statistical approximation under the stated effective-equilibrium assumption.

Use the number-density virial expansion, following the density versus molar second virial convention, with $n=N/V$:Thus $B_2$ is the coefficient of the first density correction and has units of volume. With molar density instead, the molar coefficient is $N_A{B}_2$. For a classical pair potential $u(r)$,Attractive separations contribute negatively and short-range repulsion contributes positively. For inert-gas interactions the attractive contribution dominates at sufficiently low temperature, $B_2$ crosses zero at the Boyle temperature, and repulsion dominates above it.

The soft-core high-temperature second virial coefficient illustrates why the far-high-temperature tail depends on the model. A hard-core estimate $B_2\simeq b-a/(k_{B}T)$ tends to positive excluded volume $b$. A soft repulsive potential, such as the Lennard-Jones potential in the original diagram, allows a positive maximum followed by decay toward zero. The sign change is the main attraction-versus-repulsion feature; neither sketch should be interpreted as measured data for a specified gas. This classical, dilute-gas sketch omits low-temperature quantum-exchange corrections.^[ref]