 Internal energy of a van der Waals gas

For fixed amount of substance, thermodynamic consistency givesIndeed, the first law of thermodynamics gives $U_V{|}_T=T{S}_V{|}_T-P$, and the Helmholtz free energy Maxwell relation gives $S_V{|}_T=P_T{|}_V$. For the fixed-amount convention $P=RT/(V-b)-a/V^2$ with temperature-independent $a,b$, substitution gives $U_V{|}_T=a/V^2$. Integration yields $U=h(T)-a/V$. If the dilute-limit reference is $U\to cT$ as $V\to \infty$ for every $T$, then $h(T)=cT$.

The same formula may be interpreted molarly if $V,U,a,b$ all use that convention. Mixing molar and total quantities changes the amount factors. The attraction controls the volume correction to internal energy, while excluded volume alone does not contribute to this derivative when $a,b$ are independent of temperature.