= Internal energy at constant temperature
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 is
$$
dU=C_V\,dT+\left[T\left(\frac{\partial p}{\partial T}\right)_V-p\right]dV.
$$
The bracket vanishes for the <ideal gas law> $pV=nRT$. It need not vanish for a <real gas>: changing the <volume (thermodynamics)> changes the average interaction <potential energy>, even when the random-motion contribution stays at the same <temperature>. The identity follows by writing $dU=T\,dS-p\,dV$ and using a <Maxwell relation>; it does not require the actual process to be reversible. <Thermodynamics and Statistical Mechanics by Richard Fitzpatrick/Free Expansion of Gas>{ref}
For a <Van der Waals equation> model with <volume (thermodynamics)> 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> by
$$
\Delta U=na\left(\frac1{v_1}-\frac1{v_2}\right)>0.
$$
Separating attractive particles makes their interaction <potential energy> less negative. For stipulated $n=1\,\mathrm{mol}$, $a=0.10\,\mathrm{Pa\,m^6\,mol^{-2}}$, $b=10^{-5}\,\mathrm{m^3\,mol^{-1}}$, $T=300\,\mathrm K$, $v_1=0.010$ and $v_2=0.020\,\mathrm{m^3\,mol^{-1}}$, 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>. <Thermodynamics and Statistical Mechanics by Richard Fitzpatrick/Van der Waals 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>.<University Physics Volume 2 by Samuel J. Ling and others/1.5 Phase Changes>[§1.5, figure 1.18 and equations 1.7–1.8: molecular attractions and constant-temperature phase conversion]{ref} If the <closed system> does only <pressure>-<volume (thermodynamics)> <work>, its <latent heat> input obeys
$$
Q_{\rm in}=\Delta H,\qquad
\Delta U=\Delta H-p\Delta V.
$$
Here $H=U+pV$ 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.
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