Free expansion
= Free expansion
Free expansion occurs when a <gas> spreads into an initially evacuated region without driving an external load. The standard idealization uses a rigid, insulated vessel: initially a partition confines the <gas> to <volume (thermodynamics)> $V_1$, while the remaining compartment contains <vacuum>. Open a passage and wait for <thermodynamic equilibrium>; the same fixed <amount of substance> $n$ now occupies the larger accessible <volume (thermodynamics)> $V_2$. The outer vessel never changes size. This simple release separates three questions: what <energy> is conserved, whether <temperature> changes, and why <entropy> increases. <Thermodynamics and Statistical Mechanics by Richard Fitzpatrick/Free Expansion of Gas>{ref}
\Image[/free-expansion-vessel-and-thermometry.png]
{title=<Free expansion> changes the accessible <volume (thermodynamics)>; the apparatus affects the observed <temperature>}
{description=Original schematic, with equal compartments chosen for illustration. The top row assumes an insulated <gas> with negligible wall <heat capacity>. The bottom left sketches the water-bath arrangement of <Joule's free-expansion experiment>, rather than a reconstruction to scale. The bottom right uses stipulated <heat capacities> and an illustrative <energy> demand to show signal suppression; it contains no measured data.}
Choose the <gas> as the <closed system>, neglect <heat> exchange with the walls, and assume negligible changes in bulk <kinetic energy> and <potential energy> between the equilibrium endpoints. Insulation gives $Q_{\rm in}=0$; no moving outer boundary or external load receives <work>, so $W_{\rm out}=0$. The <first law of thermodynamics> therefore gives
$$
\Delta U=Q_{\rm in}-W_{\rm out}=0.
$$
The gas can develop internal motion during the release; this does not supply external <work>. The intermediate state need not have one equilibrium <pressure> or <temperature>, so there is no actual <quasistatic process> curve $p(V)$ to integrate. In particular, substituting the equilibrium <ideal gas law> into $\int p\,dV$ would describe a different process. https://openstax.org/books/university-physics-volume-2/pages/4-1-reversible-and-irreversible-processes[OpenStax, section 4.1: free expansion and the absence of equilibrium intermediate states]{ref}
For an <ideal gas>, <internal energy of an ideal gas> depends only on <temperature>. With positive <heat capacity at constant volume>, $U(T_2)=U(T_1)$ implies $T_2=T_1$, and the <ideal gas law> gives $p_2/p_1=V_1/V_2$. Doubling the accessible <volume (thermodynamics)> thus halves the final <pressure> without changing the final <temperature>. This is an endpoint statement; it does not assert a spatially uniform, constant <temperature> during the rush. Insulation alone does not imply constant <temperature>: a reversible <adiabatic expansion> cools an <ideal gas> because it supplies <work>. <Statistical Physics by David Tong/4 Classical Thermodynamics>{ref}
For a <real gas>, $U(T,V)$ can depend on <volume (thermodynamics)> as well as <temperature>, so unchanged <internal energy> need not mean unchanged <temperature>. For a fixed-composition <simple compressible system> in a single <thermodynamic phase>, equilibrium identities using a <Maxwell relation> give
$$
dU=C_V\,dT+\left[T\left(\frac{\partial p}{\partial T}\right)_V-p\right]dV,
\qquad
\left(\frac{\partial T}{\partial V}\right)_U=
\frac{p-T(\partial p/\partial T)_V}{C_V}.
$$
Here $C_V$ is the whole sample's <heat capacity at constant volume>. The sign depends on the <equation of state>; for an <ideal gas> the numerator vanishes. For a finite <free expansion>, solve $U(T_2,V_2)=U(T_1,V_1)$, or integrate this slope along equilibrium states of equal <internal energy>; that mathematical curve does not describe the actual release. <Thermodynamics and Statistical Mechanics by Richard Fitzpatrick/Free Expansion of Gas>{ref}
Free expansion is useful because it tests how <internal energy> depends on <number density> while eliminating external expansion <work>. Its beautiful surprise is that an <ideal gas> can finish with the same <temperature> and <internal energy> but greater <entropy>: a <thermometer> alone cannot diagnose reversibility. See <entropy change in free expansion> for the calculation and <temperature sensitivity of a free-expansion experiment> for the experimental qualification. The continuous-flow throttling associated with the <Joule-Thomson coefficient> has a different balance: under its usual assumptions it preserves <enthalpy>, whereas the gas-only idealization here preserves <internal energy>. https://farside.ph.utexas.edu/teaching/sm1/Thermalhtml/node68.html[Fitzpatrick, Joule-Thompson Throttling: steady-flow enthalpy balance]{ref}