= Solution
Along a change in <volume> at fixed <temperature>, the <first law of thermodynamics> gives
$$
\left(\frac{\partial U}{\partial V}\right)_T
=T\left(\frac{\partial S}{\partial V}\right)_T-P.
$$
Use the Helmholtz <Maxwell relation> derived in (a) to replace the <entropy> <derivative>. This proves
$$
\boxed{\left(\frac{\partial U}{\partial V}\right)_T
=T\left(\frac{\partial P}{\partial T}\right)_V-P}.
$$
For the stated <Van der Waals equation>, temperature-independent $a,b$ give $P_T|_V=R/(V-b)$. Hence $U_V|_T=a/V^2$. Integrating with respect to $V$ yields
$$
U(V,T)=-\frac aV+h(T).
$$
The dilute-limit condition holds at every fixed $T$ and forces $h(T)=cT$, so
$$
\boxed{U(V,T)=cT-\frac aV}.
$$
The result is the <internal energy of a van der Waals gas> in the fixed-amount convention of the question. Its <volume> <derivative> is positive even though the attractive energy correction itself is negative.
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