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Constant temperature does not generally mean constant internal energy. At fixed amount and composition, within one smooth thermodynamic phase of a simple compressible system, an equilibrium identity isThe bracket vanishes for the ideal gas law $pV=nRT$. It need not vanish for a real gas: changing the volume changes the average interaction potential energy, even when the random-motion contribution stays at the same temperature. The identity follows by writing $dU=TdS-pdV$ and using a Maxwell relation; it does not require the actual process to be reversible. ^[ref]

For a Van der Waals equation model with volume per mole $v=V/n$, $p=RT/(v-b)-a/v^2$ gives $(\partial U/\partial V{)}_{T,n}=a/v^2$. Thus an isothermal expansion from $v_1$ to $v_2>v_1$, at fixed amount of substance $n$, increases internal energy bySeparating attractive particles makes their interaction potential energy less negative. For stipulated $n=1\mathrm{mol}$, $a=0.10\mathrm{Pa}{\mathrm{m}}^6\mathrm{mo}{\mathrm{l}}^{-2}$, $b=10^{-5}{\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$, $T=300\mathrm{K}$, $v_1=0.010$ and $v_2=0.020{\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$, the calculated increase for this one-mole sample is $5\mathrm{J}$, despite $\Delta T=0$. This dilute stable-branch model example is not a measurement of a named gas. ^[ref]

A phase transition gives another reason. During equilibrium vaporization of a pure substance at fixed pressure, the temperature can stay at its phase coexistence value while the phase fraction changes. Molecular separation changes interaction potential energy.^[ref] If the closed system does only pressure-volume work, its latent heat input obeysHere $H=U+pV$ is enthalpy. A vaporization latent heat therefore is not automatically the internal energy increment: some input supplies expansion work. A fixed temperature reading alone cannot tell how much internal energy changed.

The gas-only condition $\Delta U=0$ requires negligible heat exchange with the apparatus. If the insulated isolated system instead includes the walls, sensors and water, the first law of thermodynamics fixes their combined energy. Once everything returns to a common final temperature, a constant apparatus heat capacity $C_a$ givesDefine $A=U_g(T_1,V_2)-U_g(T_1,V_1)$, the gas's internal energy increase for the same expansion evaluated at the initial temperature. For a small temperature change, linearizing at the final volume givesHere $C_{V,g}$ is the whole gas sample's heat capacity at constant volume at the reference state. Compared with an isolated gas sample's $-A/C_{V,g}$, the common final signal is reduced by $C_{V,g}/(C_{V,g}+C_a)$. This energy accounting explains why a massive bath can conceal a real-gas effect. Fitzpatrick gives the corresponding apparatus correction for the Van der Waals equation. ^[ref]

For a toy calculation, stipulate $A=5\mathrm{J}$, $C_{V,g}=10\mathrm{J}{\mathrm{K}}^{-1}$ and $C_a=990\mathrm{J}{\mathrm{K}}^{-1}$, all constant in this example. The predicted cooling magnitude changes from $0.5\mathrm{K}$ for the gas alone to $0.005\mathrm{K}$ for the combined apparatus. An illustrative effective detection threshold of $0.01\mathrm{K}$ would miss the latter. These are chosen numbers, not Joule's measurements. In this approximation a null observation only bounds $|A|\lesssim (C_{V,g}+C_a){ϵ}_T$, where ${ϵ}_T$ is the detectable temperature change after allowing for drift and other experimental uncertainties; a display's smallest digit is not automatically this threshold.

The historical limitation was recognized explicitly: Joule and Thomson later estimated that the common apparatus in Joule's experiment would cool by only about $0.003\mathrm{K}$, explaining why the earlier external thermal effect was undetectable. This was a later calculation about the apparatus, not a measured cooling trace. ^[ref]

The Van der Waals equation is a simple real gas equation of state that can also describe a liquid-gas critical point. With particle count $N$ fixed, write the volume per particle as $v=V/N$:or equivalently $(p+a{N}^2/V^2)(V-Nb)=N{k}_{B}T$. Here $p$ is pressure, $T$ is absolute temperature, and $k_B$ is the Boltzmann constant. In SI, $v$ and $b$ have units ${\mathrm{m}}^3$ per particle, $a$ has units $\mathrm{Pa}{\mathrm{m}}^6=\mathrm{J}{\mathrm{m}}^3$, and $k_B$ has units $\mathrm{J}{\mathrm{K}}^{-1}$; the particle count is dimensionless. The pole at $v=b$ excludes smaller volume values from this model.

The $b$ correction represents excluded volume; the attraction term lowers pressure at fixed temperature and particle number density $1/v$. This is a mean-field theory, replacing detailed intermolecular structure by two parameters. At $v\gg b$ with $a/v^2\ll k_{B}T/v$, it approaches the ideal gas law $pv=k_{B}T$. It supplies a useful qualitative route from a dilute gas to phase coexistence, without a claim of quantitative accuracy for every fluid.^[ref]

Do not mix this particle convention with the molar volume convention elsewhere in this article. With amount of substance $n_{\mathrm{mol}}=N/N_A$, Avogadro constant $N_A$, and molar gas constant $R=N_A{k}_B$, one has $v_m=N_{A}v$, $b_m=N_{A}b$, and $a_m=N_A^{2}a$, so $p=RT/(v_m-b_m)-a_m/v_m^2$. The molar parameters have units ${\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$ and $\mathrm{Pa}{\mathrm{m}}^6\mathrm{mo}{\mathrm{l}}^{-2}$, respectively.

Johannes Diderik van der Waals introduced this equation of state in 1873, connecting molecular attraction with departures from the dilute ideal gas law. Its historical significance is the unified treatment of gases and liquids, rather than a universally exact fit.^[ref]